\section{Introduction}

Recognizing and interpreting integrality of sequences defined by recursions is at the same time an extensively studied and a hardly understood topic in number theory. Even for the case of sequences $A(n)$ defined by linear recurrences with polynomial coefficients, the so-called \emph{P-recursive} sequences,
\[
p_r(n) A(n+r) = p_{r-1}(n) A(n+r-1) + \cdots + p_0(n) A(n), \quad p_i(n) \in \mathbb{Z}[n],
\]
neither a criterion nor even an algorithm is known for classifying/deciding integrality. An attempt for such a classification is the famous and widely open conjecture by Christol~\cite[Conjecture 4, p.~55]{christol-glob}. Roughly speaking, it states that a P-recursive sequence $(A(n))_{n\geq0}$ with (at most) geometric growth is
integral
if and only if $(A(n))_{n\geq0}$ is the coefficient sequence of the diagonal of a rational function $R(\boldsymbol{x}) \in \mathbb{Z}(x_1,\dots,x_d) \cap \mathbb{Z}\ps{x_1,\dots,x_d}$ for some $d\geq 1$. Recall that the diagonal of a multivariate power series
\begin{equation} \label{eq:diag}
  R(\boldsymbol{x}) = \sum_{n_1, n_2, \dots, n_d \geq 0} c (n_1, n_2, \dots, n_d)
   x_1^{n_1}  x_2^{n_2}\cdots x_d^{n_d}
\end{equation}
is the univariate power series $\diag(R)$ whose coefficient sequence is given by $A (n) = c (n, n, \ldots, n)$. For a precise statement of Christol's conjecture see \cref{conj:christol} below.

Often integrality of sequences can be explained by the underlying combinatorial nature. For example, the Catalan numbers $C(n)$ satisfying
\[
(n+2) C(n+1) = 2(2n+1) C(n), \quad C(0)=1,
\]
are clearly integers because they count
triangulations of convex polygons with $n+2$ vertices.
On the other hand, for many other integral and P-recursive sequences, combinatorial interpretations are not \textit{a priori} known;
this is the case, for instance, for the Apéry numbers $A(n)$ (associated with the irrationality proof of $\zeta(3)$) defined by
\begin{align*}
(n+1)^3 A(n+1) = (2n + 1)(17n^2 + 17n + 5) A(n) & - n^3 A(n-1), \\ & A(0) = 1, A(1) = 5.
\end{align*}
In both examples above, integrality can be seen from the explicit formulas
\[
C(n) = \binom{2n}{n} - \binom{2n}{n+1} \quad \text{and} \quad A(n) = \sum_{k=0}^n \binom{n}{k}^2\binom{n+k}{k}^2.
\]
Putting Christol's conjecture in practice gives a different justification for the integrality of these two examples. It namely holds that
\begin{align*}
\sum_{n \geq 0} C(n) t^n &= \diag \left(\frac{1-y}{1- x(y+1)^2} \right)
\quad \text{and}\\
\sum_{n \geq 0} A(n) t^n &= \diag \left( \frac{1}{1-(xy+x+y)(zw+z+w)} \right) ,
\end{align*}
and the integrality of $C(n)$ and $A(n)$ follows from that of the coefficients in the Taylor expansions of the corresponding multivariate rational functions.

In the context of the current text, however, we would like to emphasize a slightly different viewpoint, which does not only justify integrality of the two examples, but also implies some interesting arithmetic properties.  Writing $\operatorname{ct}[P(\boldsymbol{x})]$ for the constant term of a Laurent polynomial $P(\boldsymbol{x}) \in \mathbb{Q}[x_1^{\pm1},\dots,x_d^{\pm1}]$, one can prove that \cite[Rem.~1.4]{Straub14}
\begin{align*}
    C(n) & = \operatorname{ct}\left[ (x^{-1} + 2 + x)^n(1-x) \right] \quad \text{and}\\
    A(n) & = \operatorname{ct}\left[ \left(\frac{(x + y)(z + 1)(x + y + z)(y + x + 1)}{xyz}\right)^n \right].
\end{align*}

Similar identities as in the examples of the Catalan and Apéry numbers can be deduced for many other integral P-recursive sequences. This motivates the following definition and the subsequent natural question.
\begin{definition}
A sequence $A(n)$ is a \emph{constant term} if it can be represented as
\begin{equation}
  A (n) = \operatorname{ct} [P (\boldsymbol{x})^n Q (\boldsymbol{x})], \label{eq:ct:pq}
\end{equation}
where $P, Q \in \mathbb{Q} [\boldsymbol{x}^{\pm 1}]$ are Laurent polynomials in
$\boldsymbol{x}= (x_1, \ldots, x_d)$.
\end{definition}
Using the geometric series it is easy to see that generating functions of constant term sequences can be expressed as diagonals of rational functions. The converse is, however, not true in general. This leads to the following question which was raised by Zagier \cite[p.~769, Question~2]{zagier-de} and Gorodetsky \cite{gorodetsky-ct} in the case $Q = 1$ (see \cref{lem:minton:ct} below for an indication why this case is of
particular arithmetic significance).

\begin{question}	 \label{Q1}
    \textit{Which P-recursive sequences are constant terms?}
\end{question}

To our knowledge, \cref{Q1} is widely open. In fact, the initial motivation for the present text was the goal of answering the following very particular sub-question asked by the second author in~\cite[Question 5.1]{s-schemes}:

\begin{question}	 \label{Q2}
    \textit{Is the Fibonacci sequence $(F(n))_{n\geq0}$ a constant term sequence?}
\end{question}
Recall that the Fibonacci sequence $(F(n))_{n\geq0}$ is the coefficient sequence in the Taylor expansion of the univariate rational function $x/(1-x-x^2)$, or equivalently the P-recursive sequence $(F(n))_{n\geq0}$ defined by $F(n+2) = F(n+1) + F(n)$ and $F(0)=0, F(1)=1$.

Already in~\cite{s-schemes} the second author noted that a representation of the Fibonacci numbers as constant terms with $Q = 1$ is impossible since $(F(n))_{n\geq0}$ does not satisfy the so-called \emph{Gauss congruences} (see~\eqref{eq:gauss}). Exploiting the fact that for any prime $p$, the value $F(p) \pmod{p}$ depends on $p \pmod{5}$, we can show (see~\cref{ex:fibonacci}) that the answer to \cref{Q2} is negative. The reason for this is that, as we will prove, for any constant term sequence $A(n)$, the sequence $A(p) \pmod{p}$ must be constant for large enough primes~$p$. Note that this is not a sufficient criterion, since already the Lucas numbers $L(n)$ (defined by the same recursion as the Fibonacci numbers, but with different initial terms $L(0)=2, L(1)=1$, see~\eqref{eq:lucas:L}) do satisfy the Gauss congruences but {are not} constant terms (see \cref{ex:lucas}).

In the present text, we are able to answer \cref{Q1} in the case of diagonals of rational functions
$F (x) \in \mathbb{Q} (x)$ in a single variable. Such sequences are precisely the (rational) \emph{C-finite sequences} (also known as \emph{C-recursive} sequences), and are characterized by the fact that they satisfy a linear recursion with constant rational coefficients. More explicitly, we define a sequence $A(n)$ of rational numbers to be C-finite if there exists a polynomial $P(x) \in \mathbb{Q}[x]$ such that for every $n\geq0$ we have
\begin{equation}\label{def:cfinite0}
P(N) A(n) = 0,
\end{equation}
where $N$ denotes the shift operator $N^\ell(A(n)) \coloneqq A(n+\ell)$ for all $\ell \geq 0$. Equivalently, there exist integers $r>0$ and $n_0\geq0$, and complex numbers $c_0,\dots,c_{r-1}$ with $c_0 \neq 0$ such that
\begin{equation}\label{def:cfinite}
A(n+r) = c_{r-1} A(n+r-1) + \cdots + c_0 A(n) \quad \text{for all $n\geq n_0$}.
\end{equation}
We recall that associated to the recursion~\eqref{def:cfinite}, the \emph{characteristic roots} are usually defined as the roots of
\[
\chi(\lambda) \coloneqq \lambda^r - c_{r-1} \lambda^{r-1} - \cdots - c_0.
\]
For our purpose, however, it is useful to define the characteristic roots of a C-finite sequence $A(n)$ as the roots of $P(x)$, where $P(x)$ is chosen with minimal degree such that \eqref{def:cfinite0} holds. Note that the only difference between considering roots of $\chi$ and $P$ is that $0$ can be a root of the latter. Equivalently, 0 is defined to be a characteristic root of $A(n)$ of multiplicity $m_0$ if the minimal $n_0$ in \eqref{def:cfinite} (chosen so that $r$ is minimal) equals $m_0$. With these definitions we obtain the following:

\begin{proposition}
  \label{thm:C:ct:1:pf}Let $A (n)$ be a C-finite sequence. $A (n)$ is a constant term if and only if it has a single characteristic root
  $\lambda$ and $\lambda \in \mathbb{Q}$.
\end{proposition}

This proposition immediately answers \cref{Q2} but also shows that, for example, the sequence $A(n) = 2^n+1$ is not a constant term sequence either (in both of these cases, there are two different characteristic roots). Evidently, however, it is the sum of two constant terms: we see that the class of constant term sequences is not a ring. Therefore, to fix this issue, it is natural to consider the class of sequences given as $\mathbb{Q}$-linear combinations of constant terms:

\begin{question}	 \label{Q3}
	\textit{Which P-recursive sequences are finite $\mathbb{Q}$-linear combinations of constant terms?}
\end{question}

Again in the case of C-finite sequences, we can answer this question completely with the main result of the present work:
\begin{theorem}
  \label{thm:C:ct} Let $A (n)$ be a C-finite sequence. Then $A (n)$ is an
  $r$-term $\mathbb{Q}$-linear combination of constant terms if and only if it
  has at most $r$ distinct characteristic roots, all of which are rational.
\end{theorem}

Having completed the classification of C-finite sequences
that can be written as (sums of) constant terms,
there are two most natural directions for further work. On the one hand, it is reasonable to go from diagonals in one variable to diagonals in two variables. By the combination of results due to Pólya~\cite{Polya22} and Furstenberg \cite{furstenberg-diag} this is
known to be
exactly the class of algebraic generating functions. One is then lead to the following question which we leave for future work:

\begin{question}	 \label{Q4}
    \textit{Which sequences $A(n)$ with algebraic generating function are constant terms?}
\end{question}

Another reasonable direction is to try to classify those hypergeometric sequences which are constant terms. Recall that a P-recursive sequence $A(n)$ is called \emph{hypergeometric} if it satisfies a recursion of order one, i.e. $\alpha(n) A(n+1) = \beta(n)A(n)$ for some polynomials $\alpha(n), \beta(n) \in \mathbb{Q}[n]$. In this sense, this class of sequences is arguably the simplest (and best understood) among P-recursive ones. Still, Christol's conjecture remains open even in this very special case. In fact, it is still an open question whether the generating function of the sequence
\[
A(n) = \frac{\left(\tfrac{1}{9} \right)_n \left(\tfrac{4}{9} \right)_n
  \left(\tfrac{5}{9} \right)_n}{n!^2 \left(\tfrac{1}{3} \right)_n}
\]
can be represented as the diagonal of a rational function. Here and later, $(x)_n \coloneqq x(x+1)\cdots(x+n-1)$ denotes the rising factorial. We can use the same methods as in the C-finite case to prove that $A(n)$ is not a constant term sequence (see \cref{lem:christol_hypergeom}). By classifying when the family \eqref{eq:hyp:2f1:ct} of hypergeometric sequences is a constant term, we are further able to conclude that not all hypergeometric diagonals are constant terms. The following question, however, remains open in general:

\begin{question}	 \label{Q5}
    \textit{Which hypergeometric sequences are constant terms?}
\end{question}

\medskip

The organization of the paper is as follows: In \cref{sec:trace}, we review properties of C-finite sequences that will be important for our purposes. In particular, we state \cref{thm:minton} which is due to Minton~\cite{minton-cong} and which is a crucial ingredient of our approach. In \cref{sec:cong}, we derive certain congruences that are satisfied by any constant term sequence; these are already enough to answer \cref{Q2}.
By combining these congruences with Minton's Theorem, we prove in \cref{sec:C:ct} our main \cref{thm:C:ct}, thus answering \cref{Q1} and \cref{Q3} in the case of C-finite sequences.
In the short \cref{sec:minton:ct} we prove a statement which is pleasingly similar to Minton's theorem and which allows to classify the constant terms with $Q=1$ among all constant terms. Finally, in \cref{sec:hyp}, we turn our attention to hypergeometric sequences and discuss \cref{Q5}.

\medskip
Throughout the article, $p$ denotes a prime number, $\mathbb{F}_p$ the finite field with $p$ elements
and $\mathbb{Z}_p$ the ring of $p$-adic integers.

\section{Trace sequences}\label{sec:trace}

Let $A (n)$ be a C-finite sequence. Denote by $\lambda_1, \lambda_2, \ldots, \lambda_d \in \overline{\mathbb{Q}}$ the characteristic roots, and let $m_j$ be the multiplicity of the root $\lambda_j$. Recall that $\lambda_0 = 0$ is defined to be a characteristic root of $A(n)$ of multiplicity $m_0$ if the minimal $n_0$ in \eqref{def:cfinite} equals $m_0$. $A (n)$ can be written
as a linear combination
\begin{equation}
  A (n) = A_0(n) + \sum_{j = 1}^d \sum_{r = 0}^{m_j - 1} c_{j, r} n^r \lambda_j^{n}
  \label{eq:C:lincomb}
\end{equation}
for certain coefficients $c_{j, r} \in \overline{\mathbb{Q}}$ (more precisely, $c_{j, r} \in \mathbb{Q} (\lambda_1, \ldots, \lambda_d)$) and $A_0(n)$ a sequence of finite support $\{0,1,\ldots,m_0-1\}$. We refer to \cite{epsw-rec} or \cite[Chapter~4]{kauers-paule-ct}
for introductions to
C-finite sequences. Note that allowing $0$ as a characteristic root is equivalent to not restricting the numerator of the rational generating function of $A(n)$ to have degree less than the degree of its denominator. In the following, we will refer to
\begin{equation}
  A^{\operatorname{sep}} (n) = A_0(0) + \sum_{j = 1}^d c_{j, 0} \lambda_j^n
  \label{eq:C:sep:lincomb}
\end{equation}
as the \emph{separable part} of $A (n)$. We note that, if $A (n) \in
\mathbb{Q}$, then $A^{\operatorname{sep}} (n) \in \mathbb{Q}$.

A sequence $A (n)$ is said to be a \emph{trace sequence} if it is a
$\mathbb{Q}$-linear combination of traces $\operatorname{Tr} (\theta^n) = \theta_1^n +
\cdots + \theta_r^n$ of algebraic numbers $\theta$ with Galois conjugates
$\theta_1 = \theta, \theta_2, \ldots, \theta_r$ (with the understanding that $\operatorname{Tr}(0^n)$ is $1$ for $n=0$ and $0$ otherwise). Equivalently, a trace
sequence is a C-finite sequence for which the multiplicity of each characteristic root is $m_j = 1$ and for which $c_{i, 0} = c_{j, 0}$ in \eqref{eq:C:lincomb} whenever $\lambda_i$ and $\lambda_j$ have the same
minimal polynomial. We further note as in \cite{bhs-gauss} that the
condition to be a trace sequence is equivalent to the property that the
generating function $F (x)$ is $F (0)$ plus a $\mathbb{Q}$-linear combination
of functions of the form $x u' (x) / u (x)$, where $u \in \mathbb{Q} [x]$ is
irreducible and $u (0) = 1$.

\begin{example} \label{ex:fib}
  \label{eg:F:L:trace}For the Fibonacci numbers $F (n)$, the representation
  \eqref{eq:C:lincomb} takes the form
  \begin{equation} \label{eq:fib}
    F (n) = \frac{\varphi_+^n - \varphi_-^n}{\sqrt{5}}, \quad \varphi_{\pm} =
     \frac{1 \pm \sqrt{5}}{2} .
  \end{equation}
  Because the coefficients of $\varphi_+^n$ and $\varphi_-^n$ differ in sign,
  the Fibonacci numbers $F (n)$ are not a trace sequence. On the other hand,
  the Lucas numbers
  \begin{equation}
    L (n) = \varphi_+^n + \varphi_-^n = \operatorname{tr} [M^n], \quad M =
    \begin{bmatrix}
      0 & 1\\
      1 & 1
    \end{bmatrix}, \label{eq:lucas:L}
  \end{equation}
  which satisfy the same recurrence as the Fibonacci numbers, are a trace
  sequence. In particular, it follows from \cref{thm:minton} that the
  Lucas numbers $L (n)$ satisfy the Gauss congruences \eqref{eq:gauss}.
\end{example}

Minton~\cite{minton-cong} classified those C-finite sequences that satisfy
the Gauss congruences \eqref{eq:gauss} (see \cite{bhs-gauss} for another
proof of Minton's result).

\begin{theorem}[Minton, 2014] \label{thm:minton}
Let $A (n)$ be C-finite. Then the following are
  equivalent:
  \begin{enumerate}
    \item For all large enough primes $p$ and for all $r\geq 1$, $A (n)$ satisfies the Gauss
    congruences
    \begin{equation}
      A (p^r n) \equiv A (p^{r - 1} n) \pmod{p^r} .
      \label{eq:gauss}
    \end{equation}
    \item For all large enough primes $p$, $A (n)$ satisfies the congruences
    \begin{equation}
      A (p n) \equiv A (n) \pmod{p} . \label{eq:gauss:r1}
    \end{equation}
    \item $A (n)$ is a trace sequence.
  \end{enumerate}
\end{theorem}

We conclude from Minton's \cref{thm:minton} the following result, which we employ in the proof of our main result (\cref{thm:C:ct}). To see the importance of \cref{lem:C:minton:sep}, we note that, as we will show later (in \cref{cor:ct:pr:large}), the sequences $A(n)$ which are linear combinations of constant terms satisfy the congruences $A (p^r n) \equiv A (p n) \pmod{p}$ for all $r\geq1$ and large enough primes~$p$.

\begin{lemma}
  \label{lem:C:minton:sep}Let $A (n)$ be C-finite. If $A (n)$ satisfies the
  congruences
  \begin{equation}
    A (p^r n) \equiv A (p n) \pmod{p} \label{eq:A:pr:p}
  \end{equation}
  for all $r \geq 1$ and for all large enough primes $p$, then the separable
  part $A^{\operatorname{sep}} (n)$ is a trace sequence.
\end{lemma}

\begin{proof}
  It follows from comparing \eqref{eq:C:lincomb} with
  \eqref{eq:C:sep:lincomb} that for $n$ large enough
  \begin{equation*}
    A (n) = A^{\operatorname{sep}} (n) + n \tilde{A} (n),
  \end{equation*}
  where $A^{\operatorname{sep}} (n)$ and $\tilde{A} (n)$ are rational and satisfy the
  minimal recurrence for $A (n)$. In particular, each of these sequences is in
  $\mathbb{Z}_p$ for large enough $p$, since denominators can only arise from the coefficients of the recurrence and the initial conditions. It follows that
  \begin{equation*}
    A^{\operatorname{sep}} (p n) \equiv A (p n) \pmod{p}
  \end{equation*}
  for all large enough $p$. Consequently, the congruences \eqref{eq:A:pr:p}
  are also satisfied by the C-finite sequence $A^{\operatorname{sep}} (n)$. That is,
  for all $r \geq 1$ and large enough $p$
  \begin{equation}
    A^{\operatorname{sep}} (p^r n) \equiv A^{\operatorname{sep}} (p n) \pmod{p}.
    \label{eq:A:sep:pr:p}
  \end{equation}
  On the other hand, let us consider the C-finite
  sequence $A^{\operatorname{sep}} (n)$ in $\mathbb{F}_p$. To avoid confusion, we denote this reduced sequence by $a_p^{\operatorname{sep}} (n)$. Since the
  characteristic polynomial of $A^{\operatorname{sep}} (n)$ over $\mathbb{Q}$ is
  separable, it is also separable for all large enough primes $p$ (this can be
  seen by looking at the discriminant which, if nonzero over $\mathbb{Q}$, can
  only vanish modulo finitely many primes). Consequently, we have a version of
  \eqref{eq:C:sep:lincomb} with coefficients and roots in
  $\overline{\mathbb{F}}_p$. Namely,
  \begin{equation*}
    a_p^{\operatorname{sep}} (n) = \sum_{j = 1}^d d_j \mu_j^n, \quad d_j, \mu_j \in
     \overline{\mathbb{F}}_p .
  \end{equation*}
  Denoting with $\varphi_p : \overline{\mathbb{F}}_p \rightarrow
  \overline{\mathbb{F}}_p$ the Frobenius automorphism defined by $\varphi_p (z) =
  z^p$, we therefore have
  \begin{equation*}
    a_p^{\operatorname{sep}} (p^s n) = \sum_{j = 1}^d d_j \mu_j^{p^s n} = \sum_{j =
     1}^d d_j (\varphi_p^s (\mu_j))^n
  \end{equation*}
  for each $s \in \mathbb{Z}_{> 0}$. Note that $\varphi_p$ acts as a
  permutation on the roots $\mu_j$. Writing $m$ for the order of this
  permutation, we have $\varphi_p^m (\mu_j) = \mu_j$ and thus
  \begin{equation*}
    a_p^{\operatorname{sep}} (p^m n) = a_p^{\operatorname{sep}} (n) .
  \end{equation*}
  Consequently, the corresponding sequence $A^{\operatorname{sep}} (n)$ satisfies
  \begin{equation*}
    A^{\operatorname{sep}} (p^m n) \equiv A^{\operatorname{sep}} (n) \pmod{p} .
  \end{equation*}
  Combined with the congruences \eqref{eq:A:sep:pr:p}, this implies that
  \begin{equation*}
    A^{\operatorname{sep}} (p n) \equiv A^{\operatorname{sep}} (n) \pmod{p}
  \end{equation*}
  for all large enough $p$. \cref{thm:minton} therefore implies that
  $A^{\operatorname{sep}} (n)$ is a trace sequence.
\end{proof}

\section{Congruences for constant terms}\label{sec:cong}

In this section we will show that if $A(n)$ is a constant term sequence then it must satisfy certain congruences for large enough primes $p$. As a consequence, this allows us to conclude that the Fibonacci numbers are not a constant term sequence, thus answering \cref{Q2} from the introduction.

For a Laurent polynomial $P \in \mathbb{Q} [\boldsymbol{x}^{\pm 1}]$, let $\deg
(P)$ denote the maximal degree with which any variable or its inverse appears
in $P$.

\begin{lemma}
  \label{lem:ct:pr:large}Let $A (n) = \operatorname{ct} [P (\boldsymbol{x})^n Q
  (\boldsymbol{x})]$ with $P, Q \in \mathbb{Z}_p [\boldsymbol{x}^{\pm 1}]$. Then
  \begin{equation*}
    A (p^r n + k) \equiv A (k) \operatorname{ct} [P (\boldsymbol{x})^{p^{r - 1} n}]
     \pmod{p^r}
  \end{equation*}
  for all integers $n, k \geq 0$ and $r \geq 1$, provided that $p >
  \deg (P^k Q)$.
\end{lemma}

\begin{proof}
  Recall that (see, for instance, \cite[Proposition~1.9]{ry-diag13}), for
  any Laurent polynomial $F \in \mathbb{Z}_p [\boldsymbol{x}^{\pm 1}]$,
  \begin{equation}
    F (\boldsymbol{x})^{p^r} \equiv F (\boldsymbol{x}^p)^{p^{r - 1}} \pmod{p^r} . \label{eq:power:pr}
  \end{equation}
  As in \cite{s-schemes}, it follows from \eqref{eq:power:pr} that
  \begin{eqnarray*}
    A (p^r n + k) & = & \operatorname{ct} [P (\boldsymbol{x})^{p^r n} P (\boldsymbol{x})^k
    Q (\boldsymbol{x})]\\
    & \equiv & \operatorname{ct} [P (\boldsymbol{x}^p)^{p^{r - 1} n} P (\boldsymbol{x})^k
    Q (\boldsymbol{x})] \pmod{p^r}\\
    & = & \operatorname{ct} [P (\boldsymbol{x})^{p^{r - 1} n} \Lambda_p [P
    (\boldsymbol{x})^k Q (\boldsymbol{x})]],
  \end{eqnarray*}
  where $\Lambda_p$ denotes the Cartier operator
  \begin{equation*}
    \Lambda_p \left[ \sum_{\boldsymbol{k} \in \mathbb{Z}^d} a_{\boldsymbol{k}}
     \boldsymbol{x}^{\boldsymbol{k}} \right] = \sum_{\boldsymbol{k} \in
     \mathbb{Z}^d} a_{p\boldsymbol{k}} \boldsymbol{x}^{\boldsymbol{k}} .
  \end{equation*}
  If $p > \deg (P^k Q)$, then
  \begin{equation*}
    \Lambda_p [P (\boldsymbol{x})^k Q (\boldsymbol{x})] = \operatorname{ct} [P
     (\boldsymbol{x})^k Q (\boldsymbol{x})] = A (k)
  \end{equation*}
  and the claim follows.
\end{proof}

\begin{example} \label{ex:fibonacci}
  For the Fibonacci numbers $F(n)$, it is a well-known consequence of \eqref{eq:fib} that, modulo any prime $p$, we have the congruences
  \begin{align*}
    F(p) \equiv \begin{cases}
      1, &\text{if $p \equiv 1,4 \bmod{5}$}, \\
      -1, &\text{if $p \equiv 2,3 \bmod{5}$},
    \end{cases}
    \pmod{p}.
  \end{align*}
  Since this is incompatible with \cref{lem:ct:pr:large} (setting $r=n=1$ and $k=0$ implies that $A(p) \equiv A(0)\cdot c \pmod{p}$ for some $c \in \mathbb{Q}$ that is independent of $p$), we see that $F(n)$ is not a constant term sequence.

  On the other hand, by \cref{thm:minton}, the Lucas numbers $L(n)$ (from~\eqref{eq:lucas:L}) satisfy the congruences $L(p^r n) \equiv L(p n) \pmod{p}$ for $r\geq1$ and $p$ large enough.
  As such, \cref{lem:ct:pr:large} is not sufficient to conclude that $L(n)$ is not a constant term sequence.
  However, we will be able to conclude in \cref{ex:lucas} the stronger result that both the Fibonacci numbers and the Lucas numbers cannot be expressed as a $\mathbb{Q}$-linear combination of constant terms.
\end{example}

\begin{corollary}
  Let $A (n) = \operatorname{ct} [P (\boldsymbol{x})^n Q (\boldsymbol{x})]$ with $P, Q \in
  \mathbb{Z}_p [\boldsymbol{x}^{\pm 1}]$. Then
  \begin{equation*}
    A (p^s n + k) \equiv A (p^r n + k) \pmod{p^r}
  \end{equation*}
  for all integers $n, k \geq 0$ and $s \geq r \geq 1$,
  provided that $p > \deg (P^k Q)$.
\end{corollary}

\begin{proof}
  It follows from \cref{lem:ct:pr:large} and \eqref{eq:power:pr} that
  \begin{eqnarray*}
    A (p^s n + k) & \equiv & A (k) \operatorname{ct} [P (\boldsymbol{x})^{p^{s - 1} n}]
    \pmod{p^s}\\
    & \equiv & A (k) \operatorname{ct} [P (\boldsymbol{x}^{p^{s - r}})^{p^{r - 1} n}]
    \pmod{p^r}\\
    & = & A (k) \operatorname{ct} [P (\boldsymbol{x})^{p^{r - 1} n}],
  \end{eqnarray*}
  as claimed.
\end{proof}

The simple but useful special case $r = 1$ and $k = 0$ of the corollary above takes the following
form. Here, $p$ is large enough if $p > \deg (Q)$ and $P, Q \in \mathbb{Z}_p
[\boldsymbol{x}^{\pm 1}]$.

\begin{corollary}
  \label{cor:ct:pr:large}Let $A (n) = \operatorname{ct} [P (\boldsymbol{x})^n Q
  (\boldsymbol{x})]$ with $P, Q \in \mathbb{Q} [\boldsymbol{x}^{\pm 1}]$. If $p$
  is large enough, then, for all integers $n \geq 0$ and $r \geq 1$,
  \begin{equation*}
    A (p^r n) \equiv A (p n) \pmod{p} .
  \end{equation*}
\end{corollary}

\section{\texorpdfstring{$C$}{C}-finite sequences that are constant terms}\label{sec:C:ct}

In this section, we prove our main result, \cref{thm:C:ct} stated in the introduction, thus classify those C-finite sequences that are constant terms or linear combinations of such. We start by proving the following weaker version, since it illustrates well our approach and the usefulness of the congruences proved in \cref{sec:cong}. We then extend the argument to prove \cref{thm:C:ct} in full generality.

\begin{proposition} \label{prop:charroots:rational}
  Let $A (n)$ be a C-finite sequence. $A (n)$ is a $\mathbb{Q}$-linear
  combination of constant terms if and only if all characteristic roots are
  rational.
\end{proposition}

\begin{proof}
  For one direction, note that
  \begin{equation}
    \operatorname{ct} [(x + \lambda)^{n} (\lambda/x)^{r}] = \binom{n}{r} \lambda^{n} =
    \frac{n (n - 1) \cdots (n - r + 1)}{r!} \lambda^{n}. \label{eq:ct:exp}
  \end{equation}
  Varying $r$, the right-hand side forms a basis for the span of the sequences $(n^r \lambda^{n})_{n \geq 0}$. A sequence $A_0(n)$ of finite support can be represented as \[
    A_0(n) = \operatorname{ct}[x^n (A(0)+A(1)x^{-1}+\cdots+A(N)x^{-N})],
  \]
  where $N$ is the largest integer for which $A_0(N)$ is non-zero. It therefore follows with \eqref{eq:C:lincomb} that, if all characteristic roots $\lambda$ are rational, then $A (n)$ can
  be represented as a linear combination of constant terms.

  On the other hand, suppose that $A (n)$ is a linear combination of constant
  terms. Note that this implies that any shift $A (n + k)$, where $k \in
  \mathbb{Z}_{\geq 0}$, is a linear combination of constant terms as
  well. These shifts generate the space $V_A$ of rational solutions of the
  minimal constant-coefficient recursion satisfied by $A (n)$. Thus, any
  sequence in $V_A$ is a linear combination of constant terms. Assume, for
  contradiction, that there is a characteristic root $\lambda$ that is not
  rational. Then among the sequences in $V_A$ there is always a sequence $B (n)$ of
  the form \eqref{eq:C:sep:lincomb} (that is, $B (n)$ equals its separable
  part) which is not a trace sequence.

  For instance, if $\lambda_1, \ldots, \lambda_d$ are the roots of the minimal
  polynomial of~$\lambda$, then the space $V_{\lambda}$ of rational sequences
  of the form $b (n) = c_1 \lambda_1^n + \cdots + c_d \lambda_d^n$, with $c_1,
  \ldots, c_d \in \overline{\mathbb{Q}}$, is a $d$-dimensional subspace of $V_A$.
  Clearly, each sequence in $V_{\lambda}$ is of the form
  \eqref{eq:C:sep:lincomb}. Note that $\lambda_1^n + \cdots + \lambda_d^n$ and
  its multiples are the only trace sequences in $V_{\lambda}$. Since $d \geq 2$,
  we can therefore choose a sequence $B (n)$ in $V_{\lambda}$ that is not a
  trace sequence.

  It follows from \cref{cor:ct:pr:large} that $B (n)$ satisfies the
  congruences
  \begin{equation*}
    B (p^r n) \equiv B (p n) \pmod{p}
  \end{equation*}
  for all $r \geq 1$ and all large enough primes $p$.
  \cref{lem:C:minton:sep} therefore implies that $B^{\operatorname{sep}} (n) = B
  (n)$ is a trace sequence. This is a contradiction, and we conclude that all
  characteristic roots must be rational.
\end{proof}

\begin{example}\label{ex:lucas}
  Recall from \eqref{eq:fib} that the Fibonacci numbers $F(n)$ are C-finite with characteristic roots $(1\pm\sqrt{5})/2$.
  Since these are not rational, it follows from \cref{prop:charroots:rational} that $F(n)$ cannot be expressed as a linear combination of constant terms.

  The same argument applied to \eqref{eq:lucas:L} shows that the Lucas numbers $L(n)$ cannot be expressed as a linear combination of constant terms as well.
  Alternatively, this can also be concluded from the relationship
    \[
    2L(n+1) - L(n) = 5F(n)
    \]
  combined with the fact that Fibonacci numbers are not a sum of constant terms.
\end{example}

We next prove the case $r = 1$ of \cref{thm:C:ct}, that is \cref{thm:C:ct:1:pf}, stating that a C-finite sequence $A (n)$ is a single constant term if and only if it has a single characteristic root $\lambda$ and $\lambda \in \mathbb{Q}$.

\begin{proof}[Proof of \cref{thm:C:ct:1:pf}]
  It follows from \eqref{eq:C:lincomb} and \eqref{eq:ct:exp} that if $A (n)$
  is a C-finite sequence with the single characteristic root $\lambda \in
  \mathbb{Q}$ (possibly repeated or possibly 0), then $A (n)$ is a constant term, namely $A
  (n) = \operatorname{ct} [(x + \lambda)^n Q (x^{- 1})]$ for a suitable polynomial $Q
  (x)$.

  On the other hand, suppose that $A (n) = \operatorname{ct} [P (\boldsymbol{x})^n Q
  (\boldsymbol{x})]$ is a single constant term. Since $A (n)$ is a C-finite
  sequence, it has a representation of the form \eqref{eq:C:lincomb} or,
  equivalently,
  \begin{equation}
    A (n) = A_0 (n) + \sum_{j = 1}^d \lambda_j^n p_j (n)
    \label{eq:C:lincomb:poly}
  \end{equation}
  for pairwise distinct $\lambda_j \in \overline{\mathbb{Q}}^{\times}$ and nonzero
  $p_j (n) \in \overline{\mathbb{Q}} [n]$. As before, $A_0 (n)$ is a sequence with
  finite support, corresponding to the characteristic root $0$. It follows from \cref{prop:charroots:rational} that all characteristic roots
  $\lambda_j$ are rational, and this further implies that $p_j (n) \in
  \mathbb{Q} [n]$.

  Let $c_0 = \operatorname{ct} [P (\boldsymbol{x})] \in \mathbb{Q}$. From
  \cref{lem:ct:pr:large} (with $r = 1$ and $n = 1$) it follows that
  \begin{equation*}
    A (p + n) \equiv A (n) \cdot c_0 \pmod{p}
  \end{equation*}
  for all $n \geq 0$ and all large enough primes $p$ (namely, $p > \deg
  (P^n Q)$ and large enough so that $c_0 \in \mathbb{Z}_p$). Combining this
  congruence with \eqref{eq:C:lincomb:poly} and applying Fermat's little
  theorem to reduce $\lambda_j^{p + n}$ and $p_j (p + n)$ modulo $p$ to
  $\lambda_j^{n + 1}$ and $p_j (n)$ respectively, we find that
  \begin{equation}
    \sum_{j = 1}^d \lambda_j^{n + 1} p_j (n) \equiv c_0 \left[ A_0 (n) +
    \sum_{j = 1}^d \lambda_j^n p_j (n) \right] \pmod{p}
    \label{eq:Apn:modp}
  \end{equation}
  for all large enough $p$ (in particular, so that $p$ is larger than any
  denominator occuring in the $p_j (n)$ and so that $A_0 (p + n) = 0$). Note
  that both sides of~\eqref{eq:Apn:modp} are independent of $p$. Since they
  agree modulo any large enough $p$, it follows that they must be equal (for
  each fixed value of $n$). Accordingly, we have the identity
  \begin{equation}
    \sum_{j = 1}^d \lambda_j^{n + 1} p_j (n) = c_0 \left[ A_0 (n) + \sum_{j =
    1}^d \lambda_j^n p_j (n) \right] \label{eq:Apn} \quad \text{for all $n \geq 0$.}
  \end{equation}
  Note that both sides of \eqref{eq:Apn} are
  C-finite sequences so that, because the representation
  \eqref{eq:C:lincomb:poly} is unique, we must have, in particular, $c_0 A_0
  (n) = 0$. If $c_0 = 0$ then it follows by comparison with the left-hand side
  of \eqref{eq:Apn} that $d = 0$ so that $A (n) = A_0 (n)$ with the single
  characteristic root $\lambda = 0$. In the other case, that is if $c_0 \neq 0$, we have $A_0 (n) = 0$, so $0$ is not a characteristic root. Further comparing both sides of
  \eqref{eq:Apn}, we find that $\lambda_j = c_0$ for all $j$. Since the
  $\lambda_j$ are distinct, we conclude that $d = 1$ so that $A (n) =
  \lambda_1^n p_1 (n)$ with the single characteristic root $\lambda_1 \in
  \mathbb{Q}^{\times}$.
\end{proof}

We now extend \cref{thm:C:ct:1:pf} to the case of $r$-term
$\mathbb{Q}$-linear combinations of constant terms, thus proving our main result \cref{thm:C:ct}. We recall that its statement is that a C-finite $A (n)$ sequence is an $r$-term $\mathbb{Q}$-linear combination of constant terms if and only if it has at most $r$ characteristic roots, all of which are rational.

\begin{proof}[Proof of \cref{thm:C:ct}]
  The case $r = 1$ is proved by \cref{thm:C:ct:1:pf}. With the same argument as in \eqref{eq:ct:exp} it follows that any C-finite sequence with $r$ characteristic roots, all of which are rational, can be
  represented as a linear combination of $r$ constant terms.

  Therefore, suppose that $r > 1$ and that
  \begin{equation*}
    A (n) = \operatorname{ct} [P_1 (\boldsymbol{x})^n Q_1 (\boldsymbol{x})] + \cdots +
     \operatorname{ct} [P_r (\boldsymbol{x})^n Q_r (\boldsymbol{x})]
  \end{equation*}
  is an $r$-term $\mathbb{Q}$-linear combination of constant terms with $P_j,
  Q_j \in \mathbb{Q} [\boldsymbol{x}^{\pm 1}]$. We need to show that $A (n)$ has
  at most $r$ characteristic roots, all of which are rational. As in the proof
  of \cref{thm:C:ct:1:pf}, we find that all characteristic roots of $A
  (n)$ are rational and that $A (n)$ can be represented in the form
  \eqref{eq:C:lincomb:poly} with $p_j (n) \in \mathbb{Q} [n]$.

  Let $c_j = \operatorname{ct} [P_j (\boldsymbol{x})] \in \mathbb{Q}$. It follows from
  \cref{lem:ct:pr:large} that
  \begin{equation*}
    A (p + n) \equiv c_1 \operatorname{ct} [P_1 (\boldsymbol{x})^n Q_1 (\boldsymbol{x})]
     + \cdots + c_r \operatorname{ct} [P_r (\boldsymbol{x})^n Q_r (\boldsymbol{x})] \pmod{p}
  \end{equation*}
  for all $n \geq 0$ and all large enough primes $p$. On the other hand, for large $p$,
  by Fermat's little theorem,
  \begin{equation*}
    A (p + n) \equiv \sum_{j = 1}^d \lambda_j^{n + 1} p_j (n) \pmod{p} .
  \end{equation*}
  Note that the right-hand sides of the last two congruences are independent
  of~$p$. Since the congruences hold modulo all large enough primes, we
  conclude that
  \begin{equation*}
    \sum_{j = 1}^d \lambda_j^{n + 1} p_j (n) = c_1 \operatorname{ct} [P_1
     (\boldsymbol{x})^n Q_1 (\boldsymbol{x})] + \cdots + c_r \operatorname{ct} [P_r
     (\boldsymbol{x})^n Q_r (\boldsymbol{x})] .
  \end{equation*}
  Note that the sequence
  \begin{equation*}
    B (n) \coloneqq \sum_{j = 1}^d \lambda_j^{n + 1} p_j (n) - c_1 A (n) =
     \sum_{j = 1}^d (\lambda_j - c_1) \lambda_j^n p_j (n) - c_1 A_0 (n)
  \end{equation*}
  is C-finite and is an $(r - 1)$-term $\mathbb{Q}$-linear combination of
  constant terms. By induction, we may conclude that $B (n)$ has at most $r -
  1$ characteristic roots, all of which are rational. By comparison with
  \eqref{eq:C:lincomb:poly}, we see that $A (n)$ has at most one more
  characteristic root than $B (n)$. Thus $A (n)$ has at most $r$
  characteristic roots, which is what we had to show.
\end{proof}

\cref{thm:C:ct} classifies those rational recursive sequences with constant coefficients which can be represented as a linear combination of $r$ constant terms. In particular, a rational C-finite sequence $A (n)$ is a linear combination of constant terms if and only if all of its characteristic roots are rational. It is natural to wonder whether we can restrict to integer sequences and conclude that all characteristic roots must be integral. This leads to the following general conjecture on C-finite sequences.

\begin{conjecture}\label{conj:new}
  Let $A (n)$ be a C-finite sequence with characteristic roots $\lambda_1,
  \ldots, \lambda_d \in \mathbb{Q}$. If $A (n)$ is an integer sequence, then
  $\lambda_1, \ldots, \lambda_d \in \mathbb{Z}$.
\end{conjecture}

Even though the statement of \cref{conj:new} seems to be quite plausible, we could find neither its statement nor its proof (even for particular cases) in the vast literature on C-finite sequences.

Note that even in the particular case $d=2$ and  $A(n) = \lambda_1^n - \lambda_2^n$, \cref{conj:new} is a non-trivial statement; in this case, it follows from the classical ``lifting-the-exponent lemma'' due to Carmichael~\cite[Thm.~V]{Carmichael1909}.
In many other cases with $d=2$,
\cref{conj:new} follows similarly from~\cite[Thm.~1.5]{Sanna16}. For arbitrary $d$, we expect that \cref{conj:new} can be tackled by using similar (analytic) $p$-adic arguments.

\section{An analog of Minton's theorem}\label{sec:minton:ct}

In this section, we record the following result which, though having a much simpler proof, is pleasingly similar to \cref{thm:minton} due to Minton \cite{minton-cong}. Moreover, this result gives a classification of constant
term sequences of the form $A (n) = \operatorname{ct} [P (\boldsymbol{x})^n]$ among all
constant term sequences $\operatorname{ct} [P (\boldsymbol{x})^n Q (\boldsymbol{x})]$.

\begin{proposition}
  \label{lem:minton:ct}Suppose $A (n) = \operatorname{ct} [P (\boldsymbol{x})^n Q
  (\boldsymbol{x})]$ with $P, Q \in \mathbb{Q} [\boldsymbol{x}^{\pm 1}]$. Then the
  following are equivalent:
  \begin{enumerate}
    \item For all large enough primes $p$ and for all $r\geq 1$, $A (n)$ satisfies the Gauss
    congruences~\eqref{eq:gauss}.

    \item For all large enough primes $p$, $A (n)$ satisfies the
    congruences~\eqref{eq:gauss:r1}.

    \item $A (n) = A (0) \operatorname{ct} [P (\boldsymbol{x})^n]$.
  \end{enumerate}
\end{proposition}

\begin{proof}
  We conclude from \cref{lem:ct:pr:large} with $r = 1$ and $k = 0$ that
  \begin{equation*}
    A (p n) \equiv A (0) \operatorname{ct} [P (\boldsymbol{x})^n] \pmod{p}
  \end{equation*}
  for large enough $p$ (namely, if $p > \deg (Q)$). If $A (n)$ satisfies the
  congruences~\eqref{eq:gauss:r1}, we find that, for large enough $p$,
  \begin{equation*}
    A (n) \equiv A (0) \operatorname{ct} [P (\boldsymbol{x})^n] \pmod{p}.
  \end{equation*}
  In that case, since this congruence holds modulo
  infinitely many $p$, we conclude the equality $A (n) = A (0) \operatorname{ct} [P
  (\boldsymbol{x})^n]$. Thus the third condition follows from the second.

  To complete the proof, we need to show that the third condition implies the
  first. This follows from \cref{lem:ct:pr:large} with $k = 0$ and $Q =
  1$.
\end{proof}

\begin{remark}
  Note that \cref{lem:minton:ct} does not imply that if $A (n) = \operatorname{ct}
  [P (\boldsymbol{x})^n Q (\boldsymbol{x})]$ satisfies the Gauss
  congruences~\eqref{eq:gauss} for large enough primes, then $Q$ must be constant. For instance, for any $P (x) \in \mathbb{Z} [x^{\pm 1}]$, the constant terms $\operatorname{ct} [P (x^2)^n (1 + x)] = \operatorname{ct} [P (x^2)^n] = \operatorname{ct} [P (x)^n]$ satisfy the Gauss congruences for all primes $p$, even though the
  first constant term has a non-constant $Q$. \cref{lem:minton:ct} rather shows that if (a) or (b) are fulfilled, then $Q$ can be replaced by $\operatorname{ct} [Q]$.
\end{remark}

\section{Hypergeometric constant terms}\label{sec:hyp}

Exiting the class of C-finite sequences, we find it natural to ask (\cref{Q5} in the introduction): Which hypergeometric sequences\footnote{Recall that a sequence $A (n)$ is hypergeometric if it satisfies a first-order recurrence $\alpha (n) A (n + 1) = \beta (n) A (n)$ for some
polynomials $\alpha (n), \beta (n) \in \mathbb{Q} [n]$. For our purposes we will assume $\alpha (n) \neq 0$ for all $n \geq 0$.} $A (n)$ are constant term sequences?

The reason for the specialization to hypergeometric sequences is threefold. First, it can be argued that it is the easiest P-recursive case. Second, similar to constant terms, hypergeometric sequences are not stable under addition. Finally, as we will see below in \cref{lem:ct:p:large:0:c}, the congruences proven in \cref{sec:cong} behave nicely with the hypergeometric assumption.\\

It follows from \cref{lem:ct:pr:large} (specialized to $n = 1$ and $r =
1$) that if $A (n) = \operatorname{ct} [P (\boldsymbol{x})^n Q (\boldsymbol{x})]$ with $P,
Q \in \mathbb{Q} [\boldsymbol{x}^{\pm 1}]$, then
\begin{equation*}
  A (p + k) \equiv A (k) \operatorname{ct} [P (\boldsymbol{x})] \pmod{p}
\end{equation*}
for all integers $k \geq 0$, provided that $p > \deg (P^k Q)$ and $P, Q
\in \mathbb{Z}_p [\boldsymbol{x}^{\pm 1}]$. In other words, if $A (n)$ is a
constant term, then there exists a constant $c \in \mathbb{Q}$ such that, for
each $k \in \mathbb{Z}_{\geq 0}$, the congruences
\begin{equation}
  A (p + k) \equiv A (k) \cdot c \pmod{p} \label{eq:ct:p:large:k:c}
\end{equation}
hold for all large enough $p$. We shall now show that, for hypergeometric
sequences, the congruences \eqref{eq:ct:p:large:k:c} follow from the base case
$k = 0$.

\begin{lemma}\label{lem:ct:p:large:0:c}
Let $A (n)$ be a hypergeometric sequence.
  Suppose that there exists a constant $c \in \mathbb{Q}$ such that
  \begin{equation}
    A (p) \equiv c \pmod{p} \label{eq:ct:p:large:0:c}
  \end{equation}
  for all large enough $p$. Then, for each $k \in \mathbb{Z}_{\geq 0}$,
  the congruence \eqref{eq:ct:p:large:k:c} holds for sufficiently large $p$.
\end{lemma}

\begin{proof}
  Since $A (n)$ is hypergeometric, we have $A (n + 1) = \rho (n) A (n)$ for a
  rational function $\rho (n) = \beta (n) / \alpha (n)$ with $\alpha (n), \beta
  (n) \in \mathbb{Z} [n]$. Fix $k \in \mathbb{Z}_{\geq 0}$ and suppose
  that the congruence \eqref{eq:ct:p:large:k:c} holds for all large enough
  $p$. By applying the hypergeometric recurrence twice, we obtain
  \begin{equation*}
    A (p + k + 1) = \rho (p + k) A (p + k) \equiv \rho (k) c A (k) = c A (k + 1)
     \pmod{p},
  \end{equation*}
  which is \eqref{eq:ct:p:large:k:c} with $k+1$ in place of $k$.
  Here we used that $\rho (p + k) \equiv \rho (k) \pmod{p}$,
  which holds true provided that $\alpha (k) \mathrel{\not\equiv} 0 \pmod{p}$. The latter is true for all sufficiently large $p$ since, by
  assumption, $\alpha (k) \neq 0$. The claim therefore follows by induction on $k$.
\end{proof}

\begin{remark}
  Note that \cref{lem:ct:p:large:0:c} does not hold for
  non-hypergeometric sequences in general. For instance, it does not hold for
  the Lucas numbers $L (n)$ as defined in \eqref{eq:lucas:L}. These form a
  trace sequence so that, by Minton's \cref{thm:minton}, the Gauss
  congruences \eqref{eq:gauss} are satisfied. In the case $n = 1$, these imply
  the congruences \eqref{eq:ct:p:large:0:c}. However, the Lucas numbers do not
  satisfy the congruences \eqref{eq:ct:p:large:k:c} for $k > 0$.
\end{remark}

\cref{lem:ct:pr:large} gives a necessary condition for $A (n)$ to be a constant term sequence. It is natural to wonder whether, or to
which degree, this condition is sufficient: Is an integer hypergeometric sequence $A(n)$ that satisfies the congruences \eqref{eq:ct:p:large:0:c} a constant term? Natural sources of potential counterexamples to this question are families of integer sequences that are quotients of binomial coefficients but cannot be written as products of those, for example $A(n) = \binom{8n}{4n} \binom{4n}{n}\binom{2n}{n}^{-1}$
(see~\cite[Thm. 1.2]{Bober09}).

We recall that the corresponding question for diagonals \eqref{eq:diag} (namely, to classify
which hypergeometric sequences $A (n)$ are coefficients of diagonals) also remains open. The following conjecture due to Christol~\cite[Conjecture 4, p.~55]{christol-glob} attempts such a classification.
In its statement, we call a sequence $(A(n))_{n\geq0}$ \emph{almost integral} if there exists a positive integer $K$ such that $K^{n+1} A(n) \in \mathbb{Z}$ for all integers $n\geq0$.  An almost integral sequence with (at most) geometric growth is called \emph{globally bounded}.

\begin{conjecture}[\cite{christol-glob}]
  \label{conj:christol} Let $(A(n))_{n\geq0}$ be sequence of rational numbers. The generating function $
  \sum_{n\geq0} A(n)t^n$ is the diagonal of a rational function if and only if $(A(n))_{n\geq0}$ is P-recursive and globally bounded.
\end{conjecture}

Any hypergeometric sequence is P-recursive since it satisfies, by definition, a recurrence with polynomial coefficients of order one. Moreover, thanks to a result of Christol \cite{christol-hyp-glob}, \cite{christol-glob} it is easy to check when a hypergeometric sequence is integral (in the case when $\alpha(n)$ and $\beta(n)$ in the definition split in $\mathbb{Q}[n]$). This makes hypergeometric sequences a natural source of potential counterexamples to \cref{conj:christol}. We refer to \cite{bbchm-diag}, \cite{akm-christol} and \cite{by-hyp-diag} for recent progress in this area. Here, we only mention that even for
\begin{equation} \label{eq:christolsexample}
A(n) = \frac{\left(\tfrac{1}{9} \right)_n \left(\tfrac{4}{9} \right)_n \left(\tfrac{5}{9} \right)_n}{n!^2 \left(\tfrac{1}{3} \right)_n}
\end{equation}
the conjecture is open. In other words, it is an open question whether the sequence \eqref{eq:christolsexample} is the diagonal of a rational function. On the other hand, we will show in this section that \eqref{eq:christolsexample} is not a constant term. Before doing so, we first prove the following result answering \cref{Q5} for a special family of hypergeometric sequences.
\begin{lemma}
  \label{lem:hyp:2f1:ct}Let $m \geq 2$ be an integer and consider the
  sequence
  \begin{equation}
    A_m (n) = \frac{\left(\tfrac{1}{m} \right)_n \left(1 - \tfrac{1}{m}
    \right)_n}{n!^2} . \label{eq:hyp:2f1:ct}
  \end{equation}
  \begin{enumerate}
    \item $A_m (n)$ is a diagonal for all $m \geq 2$.

    \item $A_m (n)$ is a constant term if and only if $m \in \{ 2, 3, 4, 6
    \}$.
  \end{enumerate}
\end{lemma}

Note that the classification in \cref{lem:hyp:2f1:ct} suggests that constant term
sequences are special among diagonals and often have significant additional
arithmetic properties. Indeed, the cases $m \in \{ 2, 3, 4, 6 \}$ (see \href{http://oeis.org/A002894}{A002894}, \href{http://oeis.org/A006480}{A006480}, \href{http://oeis.org/A000897}{A000897} and \href{http://oeis.org/A113424}{A113424} in the on-line encyclopedia of integer sequences~\cite{OEIS}) correspond
precisely to those special hypergeometric functions underlying Ramanujan's
theories of elliptic functions ($m = 2$ being the classical case and $m = 3,
4, 6$ corresponding the alternative bases). We refer to \cite{bbg-rama-ell}
for more information.

\begin{example}
  The hypergeometric sequence
  \begin{equation}
    B (n) = 5^{3 n} \frac{\left(\tfrac{1}{5} \right)_n \left(\tfrac{4}{5}
    \right)_n}{n!^2} = 1, 20, 1350, 115500, 10972500, \ldots
    \label{eq:hyp:2f1:5}
  \end{equation}
  is an integer sequence and grows at most exponentially. As suggested by
  Christol's \cref{conj:christol} and stated in \cref{lem:hyp:2f1:ct}, the sequence $B (n)$ is a diagonal. However, $B (n)$ is not a constant term. The
  proof of \cref{lem:hyp:2f1:ct} in this case proceeds by showing that we have the
  congruences
  \begin{equation*}
    B (p) \equiv \begin{cases}
       20, & \text{if $p \equiv \pm 1 \bmod{5}$},\\
       30, & \text{otherwise},
     \end{cases} \pmod{p} ,
  \end{equation*}
  which contradict \cref{lem:ct:p:large:0:c}.
\end{example}

\begin{proof}[Proof of \cref{lem:hyp:2f1:ct}]
  Part (a) follows from the fact
  that the generating function of $A_m (n)$ is the Hadamard (term-wise) product of $(1 -
  x)^{- 1 / m}$ and $(1 - x)^{1 / m - 1}$. The latter are algebraic functions
  and hence diagonals by a result of Furstenberg \cite{furstenberg-diag}.
  Since diagonals are closed under Hadamard products \cite{christol-diag88},
  it follows that $A_m (n)$ is a diagonal.

  That $A_m (n)$ is a constant term if $m \in \{ 2, 3, 4, 6 \}$ follows from
  the following alternative representations as products of binomial
  coefficients:
  \begin{eqnarray*}
    2^{4 n} A_2 (n) & = & \frac{(2 n) !^2}{n!^4} = \binom{2 n}{n}^2,\\
    3^{3 n} A_3 (n) & = & \frac{(3 n) !}{n!^3} = \binom{3 n}{2 n} \binom{2
    n}{n},\\
    4^{3 n} A_4 (n) & = & \frac{(4 n) !}{(2 n) !n!^2} = \binom{4 n}{2 n}
    \binom{2 n}{n},\\
    2^{4 n} 3^{3 n} A_6 (n) & = & \frac{(6 n) !}{(3 n) ! (2 n) !n!} = \binom{6
    n}{3 n} \binom{3 n}{n}.
  \end{eqnarray*}
  In the remainder, we will show that $A_m (n)$ is not a constant term if $m
  \not\in \{ 2, 3, 4, 6 \}$. If $m$ is coprime to $p$ (as it is for large enough $p$) then the right-hand side of
  \begin{equation*}
    m^p \left(\tfrac{1}{m} \right)_p = 1 \cdot (m + 1) (2 m + 1) \cdots ((p
     - 1) m + 1)
  \end{equation*}
  is a product of all the residues modulo $p$. In particular, exactly one
  factor is of the form $a p$ where $a \in \{ 1, 2, \ldots, m - 1 \}$ is
  characterized by $a p \equiv 1 \pmod{m}$. By Wilson's
  theorem, we therefore have
  \begin{equation*}
    m^p \left(\tfrac{1}{m} \right)_p \equiv - a p \pmod{p^2}
  \end{equation*}
  or, equivalently,
  \begin{equation*}
    \frac{m^p \left(\tfrac{1}{m} \right)_p}{p!} \equiv a \pmod{p} .
  \end{equation*}
  Similarly,
  \begin{equation*}
    m^p \left(1 - \tfrac{1}{m} \right)_p = (m - 1) (2 m - 1) \cdots (p m -
     1)
  \end{equation*}
  and, again, the right-hand side features a product of all residues modulo
  $p$. Exactly one factor is of the form $b p$ where $b \in \{ 1, 2, \ldots, m
  - 1 \}$ is characterized by $b p \equiv - 1 \pmod{m}$. It
  follows that $b = m - a$. Combined, we conclude that
  \begin{equation}
    m^{2 p} A_m (p) = \frac{m^{2 p} \left(\tfrac{1}{m} \right)_p \left(1 -
    \tfrac{1}{m} \right)_p}{p!^2} \equiv a (m - a) \pmod{p} .
    \label{eq:hyp:2f1:modp}
  \end{equation}
  Since $a \in \{ 1, 2, \ldots, m - 1 \}$ is characterized by $a \equiv
  1 / p \pmod{m}$ it, in particular, depends only on the
  residue class of $p$ modulo $m$. As $p$ ranges through all primes, it
  follows from Dirichlet's theorem on primes in
  arithmetic progressions, that each value
  $a \in \{ 1, 2, \ldots, m - 1 \}$ with $a$ coprime to $m$ appears infinitely
  many times. There are $\phi (m)$ many such values of $a$, where $\phi$ is
  Euler's totient function. Consequently, the quantity $a (m - a)$ on the
  right-hand side of \eqref{eq:hyp:2f1:modp} takes $\phi (m) / 2$ many
  different values as $p$ ranges through all primes $p > m$.

  On the other hand, if $A_m (n)$ is a constant term sequence, then by
  \eqref{eq:ct:p:large:0:c} there exists a constant $c \in \mathbb{Q}$ such
  that $m^{2 p} A_m (p) \equiv c \pmod{p}$ for all large
  enough primes. If \eqref{eq:hyp:2f1:modp} holds for infinitely many $p$, we
  necessarily have $c = a (m - a)$, which is only possible if $\phi (m) / 2 =
  1$.

  Thus, if $\phi (m) > 2$ then $m^{2 p} A_m (p)$ cannot satisfy the
  congruences \eqref{eq:ct:p:large:0:c} for all large enough primes and,
  hence, the sequences $m^{2 n} A_m (n)$ and $A_m (n)$ cannot be constant
  terms. Since $\phi (m) > 2$ for all integers $m \geq 2$ except for $m
  \in \{ 2, 3, 4, 6 \}$, the claim follows.
\end{proof}

For hypergeometric sequences, we therefore have the following inclusions
\begin{equation*}
  \left\{ \text{constant terms} \right\} \subsetneq \left\{ \text{diagonals}
   \right\} \subseteq \left\{ \text{P-recursive \& globally bounded seq's} \right\}.
\end{equation*}
We note that these inclusions are also true for C-finite as well as for P-recursive sequences. An example for the strictness of the first inclusion in the realm of hypergeometric sequences is given by the sequence~\eqref{eq:hyp:2f1:5} and in the class of C-finite sequences by the Fibonacci numbers. The second inclusion is a consequence of a result due to Lipshitz~\cite{Lipshitz88} and it is strict if and only if Christol's \cref{conj:christol}
(restricted to hypergeometric sequences) is false.
A potential candidate of a globally bounded hypergeometric sequence that is not a diagonal is sequence \eqref{eq:christolsexample}.
We now show that this sequence is not a constant term.

\begin{lemma} \label{lem:christol_hypergeom}
  The hypergeometric sequence $A (n)$ defined in \eqref{eq:christolsexample}
  is not a constant term sequence.
\end{lemma}

\begin{proof}
  Proceeding as in the proof of \cref{lem:hyp:2f1:ct}, we find
  \begin{equation*}
    m^p \left(\tfrac{r}{m} \right)_p = r (m + r) \cdots ((p - 1) m + r),
  \end{equation*}
  where the right-hand side is a product over all residues modulo $p$. Exactly
  one factor is of the form $a p$ where $a \in \{ 1, 2, \ldots, m - 1 \}$ is
  characterized by $a p \equiv r \pmod{m}$. In that case,
  \begin{equation*}
    \frac{m^p \left(\tfrac{r}{m} \right)_p}{p!} \equiv a \pmod{p} .
  \end{equation*}
  If $p \equiv 1 \pmod{9}$, we therefore find
  \begin{equation*}
    \frac{9^p \left(\tfrac{1}{9} \right)_p}{p!} \equiv \frac{3^p \left(\tfrac{1}{3} \right)_p}{p!} \equiv 1, \quad \frac{9^p \left(\tfrac{4}{9}
     \right)_p}{p!} \equiv 4, \quad \frac{9^p \left(\tfrac{5}{9}
     \right)_p}{p!} \equiv 5 \pmod{p},
  \end{equation*}
  which combine to
  \begin{equation*}
    3^{5 p} A (p) = 3^{5 p} \frac{\left(\tfrac{1}{9} \right)_p \left(\tfrac{4}{9} \right)_p \left(\tfrac{5}{9} \right)_p}{p!^2 \left(\tfrac{1}{3} \right)_p} \equiv \frac{1 \cdot 4 \cdot 5}{1} = 20 \pmod{p} .
  \end{equation*}
  On the other hand, if $p \equiv - 1 \pmod{9}$, then
  \begin{equation*}
    \frac{9^p \left(\tfrac{1}{9} \right)_p}{p!} \equiv 8, \quad \frac{3^p
     \left(\tfrac{1}{3} \right)_p}{p!} \equiv 2, \quad \frac{9^p \left(\tfrac{4}{9} \right)_p}{p!} \equiv 5, \quad \frac{9^p \left(\tfrac{5}{9}
     \right)_p}{p!} \equiv 4 \pmod{p},
  \end{equation*}
  which combine to
  \begin{equation*}
    3^{5 p} A (p) = 3^{5 p} \frac{\left(\tfrac{1}{9} \right)_p \left(\tfrac{4}{9} \right)_p \left(\tfrac{5}{9} \right)_p}{p!^2 \left(\tfrac{1}{3} \right)_p} \equiv \frac{8 \cdot 4 \cdot 5}{2} = 80 \pmod{p} .
  \end{equation*}
  As in the proof of \cref{lem:hyp:2f1:ct} we conclude that $3^{5 n} A
  (n)$ and, hence, $A (n)$ cannot be a constant term.
\end{proof}

\paragraph{Acknowledgments.} Warm thanks go to the \href{https://www-polsys.lip6.fr/}{PolSys team} of the Sorbonne University for providing a very nice atmosphere for the three authors to collaborate. The first and third authors were supported by ANR-19-CE40-0018 \href{https://specfun.inria.fr/chyzak/DeRerumNatura/}{De Rerum Natura} and the \href{https://oead.at/en/}{WTZ collaboration}/\href{https://www.campusfrance.org/}{Amadeus project} FR-09/2021 (46411YJ).
The second author gratefully acknowledges support through a Collaboration Grant (\#514645) awarded by the Simons Foundation.
The third author is funded by DOC scholarship P-26101 of the \href{https://www.oeaw.ac.at/en/1/austrian-academy-of-sciences}{ÖAW}.

\bibliographystyle{alphaabbr}

\newcommand{\etalchar}[1]{$^{#1}$}


\section{Introduction}

Investigations concerning extra dimensions have been had considerable activity recently,
and have been motivated by possible extra dimensions arising within a TeV scale for quantum gravity.  The idea that the universe
is trapped on a membrane in some higher-dimensional spacetime may explain why gravity is so weak, and could be
 tested at high-energy particle accelerators. In this context, the observable universe can be described by
a $1+3$  manifold, the so-called \emph{brane}, embedded in a spacetime with $1+3+\delta$ dimensions, denominated \emph{bulk}, where $\delta$ is the number of extra
dimensions.  The idea of extra dimensions dates back to the 1890s, when Heinrich R. Hertz \cite{hertz} and Luther P. Eisenhart \cite{eise} dextrously proposed
the trajectories of a general conservative system in classical dynamics can be put into a one-to-one correspondence with the geodesics of a suitable Riemannian manifold in
one higher dimension \cite{barut1}. Thereafter in the 1920s,
 the second attempt to include extra dimensions  in a physical theory was made by Theodor Kaluza and Oskar Klein \cite{kaluza, Kaluza}.
 They suggested a unification of electromagnetism and gravity into a geometrical formulation,
involving one extra dimension that, when compactified, makes a unique formalism that unravels
the Lorentz and the U(1) gauge symmetry of electromagnetism simultaneously.

In spite of the possible existence of extra dimensions, they still remain up to now unaccessible to experiments.
Kaluza-Klein (KK) and string theories
indicate the introduction of compact extra dimensions \cite{5} of small radii of size of the order of the
Planck scale, $l_P \sim 10^{-33}$ cm. Those unaccessed dimensions can also be as large as millimeters, implying
deviations of the Newton's law of gravity at these scales, and the highest energy particle accelerators extend our range of sight to
include the weak and strong forces down to small scales, around 10-15 mm. Below about 1 mm, objects may be gravitating in more
dimensions. The electromagnetic, weak and strong forces, as well as all the matter
in the universe, would be trapped on a brane with three spatial dimensions. Only gravitons would be able to leave the surface
 and move throughout the full volume \cite{gr, zwi, zwi1, zwi2}.

At low energies, gravity is localized on the brane and general relativity is recovered, but at high energies,
 significant changes are introduced in gravitational dynamics, forcing general relativity to break down
to be overcome by a quantum gravity theory \cite{rov}.
A plausible reason for the gravitational force appear to be so weak can be its dilution in possibly existing extra dimensions
related to a bulk, where $p$-branes \cite{gr, zwi, zwi1, zwi2, bs, Townsend} are embedded. $p$-branes \cite{8, 17}
are good candidates for brane-worlds because they possess gauge symmetries \cite{zwi, zwi1, zwi2} and automatically incorporate a quantum theory of gravity. The
gauge symmetry arises from open strings, which can collide to form a closed string that can leak into the higher-dimensional
bulk. The simplest excitation modes of these closed strings
correspond precisely to gravitons \cite{qfts}.

It has been suggested the existence of infinite extra dimensions \cite{ken, Rubakov}, where particles in the bulk can be detectable in low
energy as 4-dimensional fields \cite{Dvali}. Also, an alternative scenario can be achieved using 5-dimensional theories over manifolds endowed with
the Randall-Sundrum  metric (RS), which induces a volcano barrier-shaped effective potential for gravitons around the brane \cite{Likken}.
The corresponding spectrum of gravitational perturbations has a massless bound state on the brane,
and a continuum of bulk modes with suppressed couplings to brane fields. These bulk modes introduce small
corrections at short distances, and the introduction of more compact dimensions does not affect the localization of matter fields.
However, true localization takes place only for massless fields \cite{Gregory}, and in the massive case the bound state becomes metastable,
being able to leak into the extra space. This is shown to be exactly the case for astrophysical massive objects,
where highly energetic stars and the process of gravitational collapse, which can originate black holes, leads to deviations from the $4D$ general relativity problem.

We can explore the idea that, if black holes and specially the supermassive black holes (SMBHs) present in the nucleus of galaxies and quasars,
 do cause deviations from the $4D$ general relativity,  these corrections should cause a small deviation in all SMBH properties. Consequently,
corrections in the accretion theory that supports quasars radiative processes. The radiative flux observed from quasars can then indicate
 these deviations and the AdS$_5$ bulk curvature radius.

In order to the article to be as self-contained as possible, in the first three Sections we make a brief review about
field equations on the brane. Although the description of black hole theory on the brane is relatively well known,
there are still some inconsistencies. One of them is the apparent discrepancy between the perturbation of the Schwarzschild form,
coming from the correction in the gravitational potential that arises from KK modes in RS formalism,
and the large bulk curvature radius, which is computed via quasar observations.

It is also shown an observational constraint between the given
perturbation in the Schwarzschild metric and the Schwarzschild
radius on the brane, showing the magnitude of the AdS$_5$ bulk
curvature radius, and subsequently the extra dimension
compactification radius. By a \emph{gedanken} experiment concerning
quasar luminosity, we obtain as a result a large bulk curvature
radius, and consequently a compactification radius $k \approx
10^{-14}$cm. This can indicate remarkable guidelines in the future
compactification scale investigations.

This article  is organized as follows: in Section 2,  Einstein field equations, written in terms of the Weyl tensor,
the tensor splittings on the brane and Codazzi equations, are used to derive a field equation on the brane. In Section 3 we derive a suitable
expression for this field equations, supposing the conservation of energy in the brane and the Israel-Darmois junction conditions.
In Section 4,  a Schwarzschild black hole on the brane is investigated in this context, and in Section 5
 the range of the AdS$_5$ curvature radii due to theory of accretion in quasars is computed, together with the extra dimension compactification radius.
Finally we present in Appendix some geometrical features
concerning the Weyl tensor, after proposing in the Concluding Remarks, two alternative  perspectives, respectively cohomological, concerning Penrose's
twistor theory \cite{pe1, pe2, ro1}, and geometrical, using brane-world
scenarios over non-orientable manifolds in order to circumvent all the
formal prerequisites necessary to approach the arising questions of the present formalism.

\section{Field equations on the brane}

In what follows, we adopt natural units, where $c=1$ and $G=1$, where $c$ denotes the light speed and $G$ the gravitational constant.
In this section we present and discuss the mathematical preliminaries necessary to explore the cosmological theory of black holes and
quasars in brane-world scenarios constraining the AdS$_5$ universe radius.
For a complete exposition about arbitrary manifolds and fiber bundles, see, e.g, \cite{frankel, naka, koni, moro, rowa}.
Hereon $\{e_\mu\}, \mu = 0,1,2,3$ [$\{e_A\}, A=0,1,2,3,4$] denotes a basis for the tangent space $T_xM$ at a point $x$ in $M$, where $M$ denotes the manifold
modelling a brane embedded in a bulk.
Naturally the cotangent space at $x$ has an orthonormal basis $\{\theta^\mu\}$ [$\{\theta^A\}$] such that $\theta^\mu(e_\nu) = \delta^\mu_\nu$.
If we choose a local coordinate chart, it is possible to represent $e_A = \partial/\partial x^A \equiv \partial_A$ and $\theta^A = dx^A$.
Given the extrinsic curvature $K = K_{AB}\theta^A \wedge\theta^B$, it is possible to project the bulk curvature on the brane.

Take $n = n^Ae_A$ a  vector orthogonal to $T_xM$ and let $y$ be the Gaussian coordinate orthogonal to the $T_xM$, on the brane at $x$, indicating
how much an observer upheave out the brane into the bulk.
In particular we have $n_AdX^A = dy$.
A vector $v = v^Ae_A$ in the  bulk  is split in components in the brane and orthogonal to the brane, respectively as
 $v = v^\mu e_\mu + ye_4 = (v^{\mu},y)$. Since the bulk is endowed with a
non-degenerate bilinear symmetric form $g$ that can be written in a coordinate basis, where $e^A:= dx^A,\; e_A = \partial/\partial x^A$, as
 $g = g_{AB}dx^A\otimes dx^B$,  the components of the metric on the brane and on the bulk are hereon denoted respectively by  $g_{AB}$ and
 $^{(5)}g_{AB}$, and related by \cite{mis}
\begin{equation}\label{neo}
{}^{(5)}g_{AB} = g_{AB} + n_An_B.
\end{equation}

The Gauss equation, relating curvatures in a 5-dimensional spacetime projected in a 4-dimensional one, can be written as \cite{frankel, mis, man}:
\begin{eqnarray}\label{duvida}
\hspace{-0.3cm}R^A_{\;\; BCD} &=& {}^{(5)}R^E _ {\;\; FGH}g_E^{\;\;A}g_{\;\;B}^Fg_{\;\;C}^Gg_{\;\;D}^H + K^A _ {\;\;C}K_{BD} - K^A _{\;\;D}K_{BC}\nonumber\\
&=&{}^{(5)}R^E _ {\;\; FGH}({}^{(5)}g_E^{\;\;A} - n_En^A)({}^{(5)}g_{\;\;B}^F - n^Fn_B)({}^{(5)}g_{\;\;C}^G-n^Gn_C)({}^{(5)}g_{\;\;D}^H -n_Dn^H) + K^A _ {\;\;C}K_{BD} - K^A _{\;\;D}K_{BC}
\end{eqnarray} Now it arises, from the antisymmetry of the first two and the last two indices of the Riemann tensor, that all but the terms
\begin{eqnarray}\label{uvida}
 {}^{(5)}R^E _ {\;\; FGH}{}^{(5)}g_E^{\;\;A}{}^{(5)}g_{\;\;B}^F{}^{(5)}g_{\;\;C}^G{}^{(5)}g_{\;\;D}^H + K^A _ {\;\;C}K_{BD} - K^A _{\;\;D}K_{BC}
\end{eqnarray}\noindent survive in Eq.(\ref{duvida}) when we contract the Riemann tensor indices in order to obtain the Ricci scalar. We then consider, without loss of generality in what follows that
\begin{eqnarray}\
R^A_{\;\; BCD}&=& {}^{(5)}R^A _ {\;\; BCD} + K^A _ {\;\; C}K_{BD} - K^A _ {\;\; D}K_{BC}.\nonumber
\end{eqnarray}
\noindent
Contracting the indices and left multiplying by $g^C _ {\;\; A}$, the components of the Ricci form in the bulk are derived
\begin{equation}\label{ricci}
R_{BD} = (^{(5)}g^C _ {\;\; A} - n^Cn_A)^{(5)}R^A _ {\;\;BCD} + ({\rm Tr}\; K)K_{BD} - K^C _ {\;\; D}K_{BC} + g^C _ {\;\; A}R^A _ {\;\; BCD}n^An^B,
\end{equation}
\noindent and, in the brane, we have:
\begin{equation}\label{ricci1}
R_{\mu\nu} = {}^{(5)}R_{BD}g_{\;\;\mu} ^{ B}g_{\;\;\nu} ^{ D} - {}^{(5)}R^A _{\;\; BCD}n^Cn_Ag_{\;\;\mu} ^{ B}g_{\;\;\nu} ^{ D} + ({\rm Tr}\; K) K_{\mu\nu} - K^C _{ \nu}K_{\mu C}.
\end{equation}
Using  eq.(\ref{ricci}), the Ricci scalar is calculated from the following equation:
\begin{eqnarray}
R  = g^{BD}R_{BD} &=& {}^{(5)}R - {}^{(5)}R_{BD}n^Bn^D - {}^{(5)}R^A_{\;\;BCD}n_An^C{}^{(5)}g^{BD} + ({\rm Tr}\; K)^2 - K^{BC}K_{BC}
+ R^A_{\;\;BCD}n_An^Bn^Cn^D
\end{eqnarray}
\noindent
Expressing  $g^{BD} =  g_{\;\;\mu}^Bg_{\;\;\nu}^Dg^{\mu\nu}$ yields
\begin{equation}\label{escalar}
Rg_{\mu\nu} = {}^{(5)}Rg_{BD}g_{\;\;\mu}^Bg_{\;\;\nu}^D - {}^{(5)}R_{BD}n^Bn^Dg_{\mu\nu} - {}^{(5)}R^A_{\;\;BCD}n_An^Cg_{\;\;\mu}^Bg_{\;\;\nu}^D +
 [({\rm Tr}\; K)^2 - ({\rm Tr}\; K^2)]g_{\mu\nu}.
\end{equation}
\noindent
From eqs.(\ref{ricci1}) and (\ref{escalar}), the tensor $G_{\mu\nu}:= R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}$ is given by:
\begin{eqnarray}
G_{\mu\nu} &=& \left(^{(5)}R_{BD} -
\frac{1}{2}{}^{(5)}Rg_{BD}\right)g_{\;\;\mu}^Bg_{\;\;\nu}^D +
\frac{1}{2}{}^{(5)}R_{BD}n^Bn^Dg_{\mu\nu} +
\frac{1}{2}{}^{(5)}R^{\;\;A}_Cn_An^C g_{\mu\nu}
-{}^{(5)}R^A_{\;\;BCD}n_An^Cg_{\;\;\mu}^Bg_{\;\;\nu}^D \nonumber\\
&& - \frac{1}{2}{}^{(5)}R^A_{\;\;BCD}n^Bn^Cn^Dn_A g_{\mu\nu} + ({\rm Tr}\; K) K_{\mu\nu} - K^C _{ \nu}K_{\mu C}
 - \frac{1}{2}g_{\mu\nu}\left(({\rm Tr}\; K)^2 - ({\rm Tr}\; K^2)\right).
\end{eqnarray}
\noindent
Since $g_{BD} = {}^{(5)}g_{BD} - n_Bn_D$, then
\begin{eqnarray}
G_{\mu\nu} &=& \left(^{(5)}R_{BD} -
\frac{1}{2}{}^{(5)}R{}^{(5)}g_{BD}\right)g_{\;\;\mu}^Bg_{\;\;\nu}^D +
{}^{(5)}R_{BD}n^Bn^Dg_{\mu\nu} +
\frac{1}{2}{}^{(5)}R^{\;\;A}_Cn_An^C g_{\mu\nu}
-{}^{(5)}R^A_{\;\;BCD}n_An^Cg_{\mu}^Bg_{\nu}^D\nonumber\\
&& - \frac{1}{2}{}^{(5)}R^A_{\;\;BCD}n^Bn^Cn^Dn_A g_{\mu\nu} + ({\rm Tr}\; K) K_{\mu\nu} - K^C _{ \nu}K_{\mu C}
 - \frac{1}{2}g_{\mu\nu}(({\rm Tr}\; K)^2 - ({\rm Tr}\; K^2)).
\end{eqnarray}
Using Einstein equations \cite{mis}, it is possible to write
\begin{equation}
{}^{(5)}G_{AB} = - \frac{1}{2}{\Lambda}_5{}^{\; (5)}g_{AB} + \kappa_5^2{}^{(5)}T_{AB},
\end{equation}
\noindent
where $\Lambda_5$ denotes  the cosmological constant in the bulk, which
prevents gravity from leaking into the extra dimension at low energies. The field equations on the brane can be written as
\begin{eqnarray}\label{field}
G_{\mu\nu}  &=& -\frac{1}{2}{\Lambda}_5g_{\mu\nu} + \kappa_5^2{}^{(5)}T_{\mu\nu} +
{}^{(5)}R_{BD}n^Bn^Dg_{\mu\nu} +
\frac{1}{2}{}^{(5)}R^{\;\;A}_Cn_An^C g_{\mu\nu}\nonumber\\
&&-{}^{(5)}R^A_{\;\;BCD}n_An^Cg_{\;\;\mu}^Bg_{\;\;\nu}^D
 - \frac{1}{2}{}^{(5)}R^A_{\;\;BCD}n^Bn^Cn^Dn_A g_{\mu\nu} + ({\rm Tr}\; K) K_{\mu\nu} - K^C _{ \nu}K_{\mu C}\nonumber\\
&& - \frac{1}{2}g_{\mu\nu}(({\rm Tr}\; K)^2 - ({\rm Tr}\; K^2)).
\end{eqnarray}
\noindent where $\kappa_5 = 8\pi G_5$. The Riemann tensor can be expressed as
a combination of the Weyl tensor, the Ricci tensor and the Ricci scalar, as
$^{(5)}R^A_{\;\;BCD} = {}^{(5)}C^A_{\;\;BCD} + {}^{(5)}D^A_{\;\;BCD}$, where
$^{(5)}C^A_{\;\;BCD}$ denotes the Weyl tensor and $^{(5)}D^A_{\;\;BCD}$ can be written as \cite{Shiromizu}
\begin{eqnarray} \label{weyl}
{}^{(5)}D^A_{\;\;BCD} &=& \frac{2}{3}\left({}^{(5)}g^A_{\;\;C}{}^{(5)}R_{DB} - {}^{(5)}g^A_{\;\;D}
{}^{(5)}R_{CB} - {}^{(5)}g_{BC}{}^{(5)}R^A_{\;\;D} + {}^{(5)}g_{BD}{}^{(5)}R^A_{\;\;C}\right)
 \nonumber\\
&&- \frac{1}{6}\left({}^{(5)}g^A_{\;\;C}{}^{(5)}
g_{BD} - {}^{(5)}g^A_{\;\;D}{}^{(5)}g_{BC}\right){}^{(5)}R.
\end{eqnarray}

\section{Energy conservation and junction conditions}

From eqs.(\ref{field}) and (\ref{weyl}) it follows that
\begin{eqnarray}\label{einstein}
G_{\mu\nu}  &=& -\frac{1}{2}{\Lambda}_5g_{\mu\nu} +  \frac{2}{3}\kappa_5^2\left[^{(5)}T_{AB}g_{\;\;\mu}^{A}g_{\;\;\nu}^{B} + \left(^{(5)}T_{AB}n^An^B - \frac{1}{4}({\rm Tr}\; ^{\; (5)}T)\right)g_{\mu\nu}\right]\nonumber\\
&&+ ({\rm Tr}\; K) K_{\mu\nu} - K^C _{ \nu}K_{\mu C} + \frac{1}{2}g_{\mu\nu}[({\rm Tr}\; K)^2 - ({\rm Tr}\; K^2)] - E_{\mu\nu}.
\end{eqnarray}
\noindent
The term $E_{\mu\nu}$ is the projection of the bulk Weyl tensor on the brane:
\begin{equation}
E_{\mu\nu} = {}^{(5)}C_{ACBD}n^Cn^Dg_{\;\;\mu}^{A}g_{\;\;\nu}^{B}.
\end{equation}

Now, the curvature factors are to be expressed in terms of
momentum-energy tensors. This can be done using two conditions: the Israel-Darmois junction conditions and the $\mathbb{Z}_2$-symmetry.
This last one physically asserts that, when one approaches the brane from one side and go through it, she emerges into a bulk that looks the same,
but with the normal vector direction reversed,
$n \mapsto -  n$, meaning a mirror condition. In more formal terms, taking another basis $\{v_A\}$ of the bulk,  it is well known
that the exterior product among elements of each basis $\{e_A\}$ and $\{v_A\}$ are related by
$v_0\wedge v_1\wedge\cdots\wedge v_4 = {\rm det}(v_A^{\;\;B})\;e_0\wedge e_1\wedge \cdots\wedge e_4$, where $v_A = v_A^{\;\;B}e_B$.  The number
${\rm det}(v_A^{\;\;B})$ is positive [negative]
if the bases $\{e_A\}$ and $\{v_A\}$ have the same [opposite] orientation, where the orientation in  the bulk
is defined as a choice in $\mathbb{Z}_2$ of equivalence classes of bases  in the bulk
 (or also in the brane).
Any tangent vector space  has only two orientation possibilities, depending of the signal of  ${\rm det}(v_A^{\;\;B})$, and
any homomorphism given by $e_A\mapsto F\,e_A$, satisfying ${\rm det}(F_A^{\;\;B}) > 0$,
does not change the tangent space orientation. Then the mirror condition means the orientation over $T_xM$ is changed. It is worth to emphasize
that such mirror transformations are trivially achieved in bulks described by non-orientable manifolds, like the M\"obius strip, for instance.
In this case it is possible to achieve mirror transformations without puncturing the surface, by means of the parallel transport of a normal
vector field, over a closed path. In this sense, after parallel transporting the vector, normal to the surface, over a closed path,
it points in the opposite direction and, despite
reversing the normal vector direction, the related observer is on the another side of the brane.
The formalism of brane world scenarios on non-orientable manifolds are beyond the scope of the present paper.

The Israel-Darmois conditions \cite{is, dar, lanc, sen, ko} are obtained assuming that in the bulk field equations,
\begin{equation}
^{(5)}G_{AB} = - \frac{1}{2}{\Lambda}_5{}^{\; (5)}g_{AB} + \kappa_5^2{}^{(5)}T_{AB},
\end{equation}\noindent  the momentum-energy tensor $^{(5)}T_{AB}$ is a sum of a bulk intrinsic momentum-energy tensor
 and a brane momentum-energy tensor integrated in a brane neighbour region
(from $y = - \epsilon$ to $y = + \epsilon$, where $y$ is the coordinate representing the bulk from the brane). The Israel-Darmois conditions are given by
\begin{equation}
g^+_{\mu\nu} - g^-_{\mu\nu} = 0,\qquad\qquad K^+_{\mu\nu} - K^-_{\mu\nu} = -\kappa^2_5\left(T_{\mu\nu}^{\rm brane} - \frac{1}{3}T^{\rm brane}g_{\mu\nu}\right),
\end{equation}
\noindent
where $T^{\rm brane} = ({\rm Tr}\; T^{\rm brane}_{\mu\nu})$. The term $T_{\mu\nu}^{\rm brane}$
is the description of the momentum-energy tensor of the brane by an \emph{inhabitant} of the bulk. In order to write $T_{\mu\nu}^{\rm brane}$ using entities
realized by an inhabitant in the brane, we use
\begin{equation}
T_{\mu\nu}^{\rm brane} = T_{\mu\nu} - {\lambda}g_{\mu\nu},
\end{equation}
\noindent
where $\lambda$ is defined as the tension on the brane.
The second assumption is the $\mathbb{Z}_2$-symmetry, resulting in the conditions
 $K^+_{\mu\nu} = - K^-_{\mu\nu} = K_{\mu\nu}$. Substituting these expressions  in the Israel-Darmois conditions it follows that
\begin{equation}
K_{\mu\nu} = -\frac{1}{2}\kappa^2_5\left(T_{\mu\nu} + \frac{1}{3}(\lambda - T)g_{\mu\nu}\right).
\end{equation}
Another step is to use the results above in order to develop the extrinsic curvatures terms in eq.(\ref{einstein}):
\begin{equation}\label{adriana}
({\rm Tr}\; K) K_{\mu\nu} - K^C _{ \nu}K_{\mu C} = \frac{1}{4}\kappa^4_5\left(TT_{\mu\nu} - T^C _{ \nu}T_{\mu C}\right),
\end{equation}\noindent obviously implying that
\begin{equation}\label{tamara}
({\rm Tr}\; K)^2 - ({\rm Tr}\; K^2) = \frac{1}{4}\kappa^4_5\left(({\rm Tr}\; T)^2 - ({\rm Tr}\; T^2)\right).
\end{equation} From eqs.(\ref{adriana}) and (\ref{tamara}) it follows that
\begin{eqnarray}\label{curvaturas}
({\rm Tr}\; K) K_{\mu\nu} - K^C _{ \nu}K_{\mu C} + \frac{1}{2}g_{\mu\nu}(({\rm Tr}\; K)^2 - ({\rm Tr}\; K^2))= \frac{1}{4}\kappa_5^4\left[TT_{\mu\nu} - T^C _{ \nu}T_{\mu C} + \frac{1}{2}g_{\mu\nu}(({\rm Tr}\; T)^2 - ({\rm Tr}\; T^2))\right].&&
\end{eqnarray}
\noindent
Substituting eq.(\ref{curvaturas}) in eq.(\ref{einstein}), we obtain the following equation for the fields in the brane:
\begin{eqnarray}
G_{\mu\nu}  &=& -\frac{1}{2}{\Lambda}_5g_{\mu\nu} +  \frac{2}{3}\kappa_5^2\left[^{\; (5)}T_{AB}g_{\;\mu}^{A}g_{\;\nu}^{B} + \left(^{\; (5)}T_{AB}n^An^B - \frac{1}{4}({\rm Tr}\; ^{\; (5)}T)\right)g_{\mu\nu}\right]\nonumber\\
&& + \frac{1}{4}\kappa_5^4\left[TT_{\mu\nu} - T^C _{ \nu}T_{\mu C} + \frac{1}{2}g_{\mu\nu}(({\rm Tr}\; T)^2 - ({\rm Tr}\; T^2))\right] - E_{\mu\nu}.
\end{eqnarray}
This equation can be further simplified if the momentum-energy conservation law is valid on the brane, i.e.,
\begin{equation}
T_{\mu\nu}^{\; \; \; ;\nu} = 0.
\end{equation}
\noindent
An expression for $T_{\mu\nu}^{\; \; \;  ;\nu}$ in terms of $^{(5)}T_{AB}$ can be found by combining  the following equations, respectively
describing the field equations in the bulk, the Israel-Darmois junction conditions and the Gauss-Codazzi equations:
\begin{equation}\label{fields}
^{(5)}G_{AB} = - \Lambda_5 ^{\; (5)}g_{AB} + \kappa^2_5 {}^{\; (5)}T_{AB},
\end{equation}
\begin{equation}\label{israel}
K_{\mu\nu} = -\frac{1}{2}\kappa^2_5\left[T_{\mu\nu} + \frac{1}{3}(\lambda - T)g_{\mu\nu}\right],
\end{equation}
\noindent
\begin{equation}\label{codazzi}
K^{\;\; B}_{A\; \; ;B} - K_{;A} = {}^{(5)}R_{BC}g_{\;\;A}^{B}n^C.
\end{equation}
\noindent
Performing the covariant derivative of the curvatures and using eq.(\ref{israel}) it follows that
\begin{equation}\label{facial}
K_{\mu\nu}^{ \; \; ; \nu} - K^{;\nu} = - \frac{1}{2}\kappa_5^2T_{\mu\nu}^{\; \;  ; \nu},
\end{equation}
\noindent
and using eq.(\ref{codazzi}) it follows that
\begin{equation}
^{\; (5)}R_{AB}g^A_{\;\; \nu}n^B = K^{\; A}_{\nu\; \; ; A} - K_{;\nu}.
\end{equation}
\noindent
But from eq.(\ref{fields}), the expression
\begin{equation}
^{(5)}R_{AB} = - \Lambda_5 ^{(5)}g_{AB} + \kappa^2_5 {}^{(5)}T_{AB} + \frac{1}{2} {}^{(5)}g_{AB}{}^{(5)}R
\end{equation}
\noindent
holds, which implies that
\begin{equation}\label{desenvolvimento}
K^{\; A}_{\nu\; \;  ; A} - K_{;\nu} = \left(- \Lambda_5 ^{\; (5)}g_{AB} + \kappa_5 ^{2\; (5)}T_{AB} + \frac{1}{2}{}^{(5)}g_{AB}{}^{(5)}R\right)g_{A\nu}n_B.
\end{equation}
\noindent
Now, introducing eq.(\ref{neo}) in eq.(\ref{desenvolvimento}) and attempting for the fact that the projections of the Ricci tensors in the $n^A$ direction are zero,
we get from eq.(\ref{facial})
\begin{equation}
T^{\; A}_{\mu \; \; ;A} = - 2^{\; (5)}T_{AB}g_{\mu}^{\; A}n^B.
\end{equation}
\noindent
Assuming that $T^{\; A}_{\mu \; \; ;A} = 0$ in the brane, it is immediate that $^{(5)}T_{AB} = 0$.
This means the bulk is in complete vacuum and the particles are in fact on the brane. Now, the field equations reduce to
\begin{eqnarray}\label{123}
G_{\mu\nu}  &=& -\frac{1}{2}{\Lambda}_5g_{\mu\nu}
+ \frac{1}{4}\kappa_5^4\left[TT_{\mu\nu} - T^C _{ \nu}T_{\mu C} + \frac{1}{2}g_{\mu\nu}(({\rm Tr}\; T)^2 - ({\rm Tr}\; T^2))\right] - E_{\mu\nu},
\end{eqnarray}
\noindent
showing the contribution of the bulk on the brane is only due to the Weyl tensor.

\section{Black holes on the brane}
The Bianchi identities, applied to the Einstein field equations found in the last section, are expressed as
$G_{\mu\nu}^{\; \; \; ;\nu} = 0$, which implies that $E_{\mu\nu}^{\; \; \; ;\nu} - S_{\mu\nu}^{\; \; \; ;\nu}= 0,$
where
\begin{equation}
S_{\mu\nu} := \frac{1}{4}\kappa_5^4\left[TT_{\mu\nu} - T^C _{ \nu}T_{\mu C} + \frac{1}{2}g_{\mu\nu}(({\rm Tr}\; T)^2 - ({\rm Tr}\; T^2))\right].
\end{equation}
A vacuum on the brane, where $T_{\mu\nu} = 0$ outside a black hole, implies that
\begin{equation}\label{21}
E_{\mu\nu}^{\; \; \; ;\nu} = 0.
\end{equation}
\noindent
Eqs.(\ref{21}) are referred to the non-local conservation equations. Other useful equations for the black hole case are
\begin{equation}\label{ricci2}
G_{\mu\nu} = - \frac{1}{2}\Lambda_ 5g_{\mu\nu} - E_{\mu\nu}, \qquad\qquad R = R^{\mu}_{\;\; \mu} = 0 = E^{\mu}_{\;\; \mu}.
\end{equation}
\noindent
The Weyl term $E_{\mu\nu}$  carries an imprint of high-energy effects sourcing KK modes. It means that highly energetic
stars and the process of gravitational collapse, and naturally  black holes, lead to deviations from the $4D$ general
relativity problem. This occurs basically because the gravitational collapse unavoidably produces energies high enough to make
these corrections significant. From the brane-observer viewpoint, the KK corrections in $E_{\mu\nu}$ are nonlocal, since they
incorporate $5D$ gravity wave modes. These nonlocal corrections cannot be determined purely from data on the brane \cite{Maartens}.
In the perturbative analysis of Randall-Sundrum (RS) brane, KK modes generate a correction in the gravitational potential $V(r) = GM/r$, which is
given by \cite{Randall}
\begin{equation}\label{potential}
V(r) \approx \frac{GM}{r}\left(1 + \frac{2l^2}{3r^2}\right).
\end{equation}
\noindent
The KK modes that generate this correction are responsible for a nonzero $E_{\mu\nu}$. This term carries the modification to the weak-field field equations, as we have
already seen.
The RS metric is in general expressed as
\begin{equation}
^{(5)}ds^2 = e^{-2k|y|}g_{\mu\nu}dx^{\mu}dx^{\nu} + dy^2,\qquad 0\leq y\leq \pi r_c
\end{equation}
\noindent
where $k^2 = 3/(2l^2)$, $l$ denotes the bulk curvature radius, $k$ is a scale of order the Planck scale,
 $r_c$ is the compactification radius and the term $e^{-2|y|/l}$ is called the \emph{the warp factor} \cite{Randall}, which
reflects the confinement role of the bulk cosmological constant $\Lambda_5$, preventing gravity from leaking
into the extra dimension at low energies.

Concerning the anti-de Sitter (AdS$_5$) bulk, the cosmological constant can be written as $\Lambda_5 = -6/l^2$,
where $l$ is the curvature radius of AdS$_5$.
The brane is localized
at $y = 0$, where the metric recovers the usual aspect. Therefore, a particular manner to express the vacuum field equations in the brane given by eq.(\ref{ricci2}) is
\begin{equation}
E_{\mu\nu} = - R_{\mu\nu},
\end{equation}
\noindent
where the bulk cosmological constant is incorporated to the warp factor in the metric.
One can use a Taylor expansion in order to probe properties of a static black hole on the brane \cite{Dadhich}, and for a vacuum brane metric,
\begin{eqnarray}\label{metrica}
g_{\mu\nu}(x,y) &=& g_{\mu\nu}(x,0) - E_{\mu\nu}(x,0)y^2 - \frac{2}{l}E_{\mu\nu}(x,0)y^3
+\frac{1}{12}\left({\Box}E_{\mu\nu} - \frac{32}{l^2}E_{\mu\nu} + 2R_{\mu\alpha\nu\beta}E^{\alpha\beta} + 6E_{\mu}^{\; \alpha}E_{\alpha\nu}\right)_{y=0}{\; \; }y^4 + \cdots
\nonumber\end{eqnarray}
\noindent where $\Box$ denotes the usual d'Alembertian.
It shows in particular that the propagating effect of $5D$ gravity arises only at the fourth order of the expansion. For a static spherical metric on the brane
given by \begin{equation}\label{124}
g_{\mu\nu}dx^{\mu}dx^{\nu} = - F(r)dt^2 + \frac{dr^2}{H(r)} + r^2d\Omega^2,
\end{equation}
\noindent
 the projected Weyl term on the brane is given by the expressions
\begin{equation}
E_{00} = \frac{F}{r}\left(H' - \frac{1 - H}{r}\right),\qquad
E_{rr} = -\frac{1}{rH}\left(\frac{F'}{F} - \frac{1 - H}{r}\right),
\qquad E_{\theta\theta} = -1 + H +\frac{r}{2}H\left(\frac{F'}{F} + \frac{H'}{H}\right).
\end{equation}
\noindent Note that in eq.(\ref{124}) the metric is led to the Schwarzschild one, if $F(r)$ equals $H(r)$.
The exact determination of these radial functions remains an open problem in black hole theory on the brane \cite{Maartens}.

These components allow one to evaluate the metric coefficients in eq.(\ref{metrica}). The area of the
$5D$ horizon is determined by $g_{\theta\theta}$. Defining $\psi(r)$ as the deviation from a Schwarzschild form $H$, i.e.,
\begin{equation}\label{h}
H(r) = 1 - \frac{2M}{r} + \psi(r),
\end{equation}
\noindent
where $M$ is constant, yields
\begin{equation}\label{gtheta}
g_{\theta\theta}(r,y) = r^2  + \psi'\left(1 + \frac{2}{l}y\right)y^2 + \frac{1}{6r^2}\left[\psi' + \frac{1}{2}(1 + \psi')(r\psi' - \psi)'\right]y^4 + \cdots
\end{equation}
\noindent
It shows how $\psi$ and its derivatives determine the change in the area of the horizon along the extra dimension.
 For a large black hole, with horizon scale $r \gg l$, we have from eq.(\ref{potential}) that
\begin{equation}\label{psi}
\psi \approx -\frac{4Ml^2}{3r^3},
\end{equation}
\noindent
that shall be used to estimate the AdS$_5$ bulk curvature radius due to
 observational luminosities of quasars, that is the same as to the supermassive black holes accreting gas.
\section{Luminosity in quasars and AdS curvature radius}
The observation of quasars (QSOs) in $X$-ray band can constrain the measure of the AdS$_5$ bulk curvature radius $l$, and indicate
 how the bulk is curled, from its geometrical and topological features.
QSOs are astrophysical objects that can be found at large astronomical distances (redshifts $z > 1$).
For \emph{gedanken} experiment involving a static black hole being accreted, in a simple model, the accretion eficiency $\eta$ is given by
\begin{equation}\label{eta}
\eta = \frac{M}{6R_{{\rm Sbrane}}},
\end{equation}
\noindent
where $R_{{\rm Sbrane}}$ is the Schwarzschild radius corrected for the case of brane-world. The luminosity $L$ due to accretion in a black hole, that generates a quasar,
 is given by
\begin{equation}
L = \eta \dot{M},
\end{equation}
\noindent
where $\dot{M}$ denotes the accretion rate and depends on some specific model of accretion.
In order to estimate $R_{{\rm Sbrane}}$, fix $H(r) = 0$ in  eq.(\ref{h}), resulting in
\begin{equation}
1 - \frac{2M}{R_{{\rm Sbrane}}} - \frac{4Ml^2}{3R_{{\rm Sbrane}}^3} = 0.
\end{equation}
\noindent
The above solution for $R_{{\rm Sbrane}}$ is relatively straightforward to be found in terms of the curvature radius $l$.
It is then possible to find an expression for the luminosity $L$  in terms of the radius of curvature
\begin{equation}
L(l) = \eta(l) \dot{M}.
\end{equation}
\noindent
Having observational values for the luminosity $L$, it is  possible to estimate a value for $l$, given a black hole accretion model.
As usual,  denoting $M_{\odot} = 10^{33}$ g  the mass of the sun, for a tipical supermassive black hole of $10^9 M_{\odot}$ in a massive quasar,
a spherical accretion model as the Bondi-Hoyle model \cite{Bondi} gives $\dot{M} \sim 800 M_{\odot}\;{\rm yr}^{-1}.$
Supposing the quasar radiates in Eddington limit, given by (see, e.g., \cite{shapiro})
\begin{equation}
L(l) = L_{Edd} = 1.26 \times 10^{45}\left(\frac{M}{10^7 M_{\odot}}\right)\; {\rm erg\,  s^{-1}}
\end{equation}
\noindent
for a quasar with a supermassive black hole of $10^9 M_{\odot}$, the luminosity is given by $L \sim 10^{47}\, {\rm erg\, s^{-1}}$.
The accretion efficiency, in this case, is given by
$\eta(l) \sim 10^{-2}$ or, from eq.(\ref{eta}),
$R_{{\rm Sbrane}}(l) \sim 10M.
$ Estimating $R_{{\rm Sbrane}}(l)$ in terms of $l$ is equivalent to estimate the curvature radius of the AdS$_5$ bulk.
Substituting the result above in eq.(\ref{h}) gives $\psi \approx -0.8$, with respect to a SMBH with mass of magnitude $10^9 M_\odot$.
In the first approximation given by eq.(\ref{psi}) the AdS$_5$ curvature radius is $l \sim 25M$.
It accords to the WMAP observational results \cite{wmap}, since it agrees with the possibility of an approximately flat
universe at large scales.

Now we are able to determine the order of the extra dimension compactification scale $k$, since it is closely related to
the AdS$_5$ bulk curvature radius \cite{Randall} by the expression $k^2 = 3/(2l^2)$, leading to  $k \approx 10^{-2}/M$, where
it must be emphasized $M$ denotes a SMBH mass.
This can be useful for the future accelerators experiments in order to impose limits on the investigation range of the extra dimension
compactification radius, among other branches of applications.

\section{Concluding Remarks}

The outcomes presented in the last Sections points basically for two possible conclusions. On the one hand, the AdS$_5$
bulk curvature radius, and consequently the extra dimension compactification radius, is determinated from quasar luminosity observations.
On the other hand, the approximation given by eq.(\ref{psi}), proposed by a lot of authors, is not a so good solution in order to describe a Schwarzschild black hole on the brane.
In fact, for the approximation adopted, $g_{\theta\theta}$ in eq.(\ref{gtheta})  decreases as one moves off the brane. This is consistent with a
pancake-like shape of the horizon. However, the correct horizon shape is the tubular one, in Gaussian normal coordinates (see, e.g.,  \cite{Gian}).  Thus
a simple brane-based approach, while giving useful insights, does not lead to a very realistic black hole solution. Indeed, up to this moment, there is no known
 solution representing a realistic black hole localized on the brane, which is stable and does not present any naked singularity \cite{Maartens}.
This remains a key open question on nonlinear brane-world gravity. Some applications in astrophysics \cite{darocha3} and particle physics \cite{coimbra}, involving extensions of the present paper, are presented \cite{darocha3,coimbra}.

However, the points discussed in the present paper sheds new light on the theory of supermassive black holes in brane-world scenarios,
since it exhibits a constraint involving quasars luminosity and the AdS$_5$ curvature radius, in order to find a
suitable black hole solution. If the brane-world models are
to be correct,  the quasars luminosity does contain a smoke gun signature to estimate the right correction of the term $\psi$ in eq.(\ref{h}),
 and the correct fixed value to the AdS$_5$ curvature radius $l$.

One interesting remark is that the AdS$_5$ influence on the brane is totally due to the Weyl tensor, related to the bulk geometry. From the
 geometric viewpoint, it carries intrinsic modifications in the black hole accretion mechanism.
 Some alternative directions to approach the present formalism can be pointed out, trying to encompass the questions related to this article.
The first one arises from the fact that, since after Witten proved
general relativity is a renormalizable quantum system in (1+2) dimensions \cite{Witten},
a lot of interesting motivations to investigate the AdS$_5$ spacetime have been arising, in the light of the generalization
of the gravity gauge theory in (1+2) dimensions to higher ones \cite{darocha}. The first attempt was to enlarge the Poincar\'e group of symmetries,
 supposing an AdS group symmetry \cite{tron}, which contains the Poincar\'e group. Also, the AdS/CFT correspondence
 asserts that a  maximal supersymmetric Yang-Mills theory in 4-dimensional Minkowski-spacetime is equivalent to a type IIB closed
superstring theory \cite{gr, hoker, mald1, mald2, mald3, mald4, mald5, mald15}.
The 10-dimensional arena for the type IIB superstring theory is then described by the product manifold
$S^5\times$ AdS$_5$, and so our future investigations are directed towards brane-world scenarios and AdS$_5$ spacetime and their
possible relationships with supersymmetric theories.

Other two possible directions to find black hole solutions in a brane-world scenario involve respectively the cohomological
and geometrical aspects of the brane and the bulk.
On the one hand, nowadays the search for a brane-world scenario is based on an orientable manifold describing the bulk. The $\mathbb{Z}_2$-symmetry in the brane,
known as the mirror symmetry, describes a change in the orientation when one passes through the brane, from one side to the other side
of the brane. It can be interpreted by a non-homomorphic `puncture' map on the brane, which can be led to a homomorphic map, if one supposes that the brane
is \emph{non-orientable} \cite{darocha}. For instance, considering the non-orientable M\"obius strip,
the parallel transport of a normal vector on a closed path makes the vector reverse the sign after
returning to the initial point over the strip. Up to our knowledge, there is no investigation about branes in non-orientable manifolds, which certainly can
bring some novelty in this open research field.
On the another hand, the Weyl tensor
 has been more formally investigated by its increasing importance concerning symplectic
geometry and twistor fibrations, which are closely related to open problems in quantum gravity \cite{giu, gu1}.
It is known that  the set of all compatible complex structures on the
tangent spaces of the bulk $M$ has twistor structure, by considering a canonical almost complex structure induced by a symplectic connection on $M$. The
integrability equation of the almost complex structure appears in terms of the curvature and it agrees with the vanishing
 of the Weyl curvature tensor, related to conformal flatness. It proposes a close relation between the investigations on symplectic connections and that of twistor spaces \cite{beng},
and the recent developments in string theory \cite{b,b1}.
More comments concerning such structures are beyond the scope of the present paper.

\section{Acknowledgements}
The authors are grateful to Prof. Paul K. Townsend for his comments and suggestions, to Prof.
Roy Maartens for his patience and clearing up expositions concerning branes, to Julio Hoff M. da Silva for pointing out some missing points,
and specially to Dr. Ricardo A. Mosna for his careful reading of the manuscript
and enlightening  suggestions.

\section*{Appendix: some useful properties of the Weyl tensor}

 The curvature tensor can be decomposed into the part which depends on the Ricci curvature, and the Weyl tensor.
Considering a manifold of dimension $n$, when $n > 3$ the second part can be non-zero.
In a conformal change of the spacetime metric $g$, the Weyl tensor is invariant, and it is also called conformal tensor.
A necessary condition for a Riemannian manifold to be conformally flat is that the Weyl tensor vanish.
 Moreover, the Weyl tensor is identically null if and only if
the metric is locally conformal to the standard Euclidean metric.

The Weyl curvature tensor is defined as being the traceless component of the Riemann curvature tensor.
 In other words, it is a tensor that has the same symmetries as the Riemann curvature tensor with the extra condition that its Ricci curvature must vanish \cite{mis}.

In dimensions two and three the Weyl curvature tensor vanishes identically.
The Weyl tensor can be obtained from the full curvature tensor by subtracting out various traces.
This is most easily done by writing the Riemann tensor as a (0,4) valent tensor, by contracting with the metric. The (0,4) valent Weyl tensor is then written as
$$
W = R - \frac{1}{n-2}(Ric - \frac{s}{n}g)\circ g - \frac{s}{2n(n-1)}g\circ g
$$\noindent
where $Ric$ denotes the Ricci tensor, $R$ denotes the scalar curvature, and $g\circ h$  denotes the
Kulkarni-Nomizu product of two arbitrary metrics $g$ and $h$:
$$(g\circ h) (u,v,w,z):= g(u,w)h(v,z) + h(u,w)g(v,z) - g(u,z)h(v,w) - g(v,w)h(u,z).$$

The ordinary (1,3) valent Weyl tensor is then given by contracting the above with the inverse of the metric.
The Ricci tensor describes the local gravitational field of the nearby matter. The long-range gravitational field,
namely gravitational waves and tidal forces, propagates through the Weyl conformal curvature tensor. It is well known that spacetime, at least locally, can be split
in spacelike surfaces plus time, if a timelike vector field $n$,  tangent to the worldline of a given observer, is chosen.
 The splitting of the gravitational field into its local and nonlocal components is shown in the following decomposition of the Riemann tensor
\begin{equation}
 R_{abcd} = C_{abcd} + \frac{1}{2} (g_{ac} R_{bd} + g_{bd} R_{ac} -  g_{bc} R_{ad} -   g_{ad} R_{bc}) + \frac{1}{6} R ( g_{ac} g_{bd} -   g_{ad} g_{bc}),
\end{equation}\noindent where $C_{abcd}$ denotes the components of the  Weyl tensor, which presents all the symmetries of the Riemann tensor and also $C^c_{\;\;acb} = 0$.
Relative to the fundamental observers, the Weyl tensor decomposes further into its irreducible parts according to
\begin{equation}
C_{abcd} = (g_{abqp} g_{cdsr} - \eta_{abqp}\eta_{cdsr}) n_q n_s E^{pr} +  (\eta_{abqp} g_{cdsr} + g_{abqp}\eta_{cdsr}) u_q u_s H^{pr}
\end{equation}\noindent where $g_{abcd} := g_{ac}g_{bd} - g_{ad} g_{bc}$ \cite{scheilamelo, tsagas}. The symmetric and trace-free components $E_{ab}$  and $H_{ab}$  are known as
 the electric and magnetic Weyl components and they are given by $E_{ab} = C_{acbd} u^c u^d$ and $H_{ab} = \frac{1}{2}\epsilon_a^{cd} C_{cdbe} u^e$,
with $E_{ab} u^b = 0 = H_{ab} u^b$. The electric part of the Weyl tensor is associated with the tidal field.
The magnetic component is associated with gravitational waves \cite{anegrini}. Of course, both tensors are required if gravitational waves are to exist.
 The Weyl tensor represents the part of the curvature that is not determined locally by matter.
The dynamics of the Weyl field is not entirely arbitrary because the Riemann tensor satisfies the Bianchi identities, that
when contracted take the form \cite{scheilamelo}
\begin{equation}
\nabla\label{1} C_{abcd} =  \nabla_{[b} R_{a]c} + \frac{1}{6} g_{c[b}\nabla_{a]} R.
\end{equation}\noindent Then, field equations for the Weyl tensor are simulated by Bianchi identitites,
determining the part of the spacetime curvature that depends on the matter distribution at other points \cite{scheilamelo}.
 Eq.(\ref{1}) splits into a set of two propagation and two constraint equations, which gives the evolution of the electric and magnetic Weyl components
\cite{1}.


\section*{Acknowledgments}
The author would like to thank Yukawa Institute for Theoretical
Physics (YITP), Kyoto University.
This work was motivated by discussions during the workshop on
{\it New Frontier of QCD 2010 (NFQCD2010)}, Jan. 18 -- Mar. 19, 2010,
YITP, Kyoto University, Kyoto, Japan. He also would like to
appreciate Professor S.~Takeda for informing a recent analysis in
nucleon $\sigma$ term from lattice QCD.