* Let $a(n)$ be [A204262](https://oeis.org/A204262) i.e. [permanent](https://en.wikipedia.org/wiki/Permanent_(mathematics)) of the matrix $n\times n$ with elements $\min(i,j)$.
* Let
$$
f\_{n,\ell}(x)=g\_{n,\ell}(x)+f\_{n,\ell-1}(\ell)-g\_{n,\ell}(\ell), \\
g\_{n,\ell}(x)=\int (n-\ell)^2 f\_{n-1,\ell}(x)\,dx, \\
f\_{n,0}(x)=n!x^n
$$
* Let
$$
R(n,q)=\sum\limits\_{j=0}^{q+1}\binom{q+1}{j}(j+1)R(n-1,j), \\
R(0,q)=1
$$


I conjecture that
$$R(n,0)=f\_{n+1,n+1}(0)=a(n+1).$$


Here is the PARI/GP prog to check it numerically:



```
a(n)=matpermanent(matrix(n, n, i, j, min(i, j)))
f_upto(n)=my(v1); v1=vector(n+1, i, i--; i!*x^i); for(i=1, n, for(j=i, n, my(A=intformal((j-i)^2*v1[j])); v1[j+1] = A + subst(v1[j+1] - A, x, i))); v1
R_upto(n)=my(v1, v2, v3, v4); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; v4=vector(n, i, vector(i+1, j, binomial(i, j-1)*j)); for(i=1, n, for(q=0, n-i, v2[q+1]=sum(j=0, q+1, v4[q+1][j+1]*v1[j+1])); v1=v2; v3[i+1]=v1[1];); v3
test1(n)=f_upto(n)==concat(1, R_upto(n-1))
test2(n)=f_upto(n)==vector(n+1, i, a(i-1))

```

Is there a way to prove it?



Remarkable recursions for the A204262
I can show the first identity $R(n,0) = f\_{n+1,n+1}(0)$, as a consequence of the more general identity


$$ R(n,q) = \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1)\tag{1}\label{1}$$
for $n,q \geq 0$. Indeed, note that $g\_{n+1,n+1} \equiv 0$ and thus $f\_{n+1,n+1}(0) = f\_{n+1,n}(n+1)$, so $R(n,0) = f\_{n+1,n+1}(0)$ is basically a special case of \eqref{1}.


One can prove \eqref{1} by induction on $n$. For $n=0$ we have
$$ R(0,q) = 1 = \frac{1}{(q+1)!} f\_{q+1,0}(1)$$
by hypothesis. For $n>0$, we see from Taylor expansion (and noting from induction that $f\_{n+q+1,n}$ is a polynomial of degree $q+1$) that
$$ \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1) = \sum\_{j=0}^{q+1} \frac{1}{(q+1)! j!} f\_{n+q+1,n}^{(j)}(n).$$
But from the recursive definition of $f\_{n,\ell}$ and the fundamental theorem of calculus one has
$$ f\_{n,\ell}'(x) = (n-\ell)^2 f\_{n-1,\ell}(x)$$
and hence on iterating
$$ f\_{n+q+1,n}^{(j)}(x) = \left(\frac{(q+1)!}{(q+1-j)!}\right)^2 f\_{n+q+1-j,n}(x)$$
so that
$$ \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1) = \sum\_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! (q+1-j)! j!} f\_{n+q+1-j,n}(n).$$
But from the recursive definition of $f\_{n,\ell}$ and the induction hypothesis we have
$$ f\_{n+q+1-j,n}(n) = f\_{n+q+1-j,n-1}(n) = (q+2-j)! R(n-1,q+1-j)$$
and hence
$$ \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1) = \sum\_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! j!} (q+2-j) R(n-1,q+1-j).$$
Replacing $j$ by $q+1-j$ we obtain
$$ \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1) = \sum\_{j=0}^{q+1} \binom{q+1}{j} (j+1) R(n-1,j)$$
recovering \eqref{1} by the recursive definition of $R$.


I discovered \eqref{1} by working backwards from the desired identity $R(n,0) = f\_{n+1,n+1}(0)$ to try to write $R(n,1) = \frac{1}{2} (R(n+1,0) - R(n,0))$ and then $R(n,2) = \frac{1}{3} (R(n+1,1) - R(n,0) - 4 R(n,1))$ in terms of the $f\_{n,\ell}$ in as simple a fashion as possible (using the ODE form $f\_{n,\ell}' = (n-\ell)^2 f\_{n-1,\ell}$, $f\_{n,\ell}(\ell) = f\_{n,\ell-1}(\ell)$ of the recursion for $f$ to eliminate the role of the indefinite integral $g$), at which point the general pattern became clear. Presumably a similar technique, using the multilinearity and symmetry properties of the permanent to express $R(n,1)$, then $R(n,2)$, etc., as a suitable permanent starting from the hypothesis $R(n,0) = a(n+1)$, will then give a similar identity relating $R(n,q)$ with the permanent of a suitable matrix which can then be established by induction to complete the proof, but I will leave this for someone else to try. EDIT: actually it may he easier to try to discover how $f\_{n,\ell}(x)$ can be expressed neatly as a permanent, as it has a simpler recursion.


SECOND EDIT: OK, I have found the permanental formula for $f\_{n,\ell}$:


$$ f\_{n,\ell}(x) = \mathrm{per}( A\_{n,\ell} + x B\_{n,\ell} ) \tag{2} \label{2}$$
where $A\_{n,\ell}$ is the $n \times n$ matrix with entries $\min(i,j) 1\_{\min(i,j) \leq \ell}$, and $B\_{n,\ell}$ is the $n \times n$ matrix with entries $1\_{\min(i,j) > \ell}$. (Guessing this formula for $\ell=n$ and $\ell = n-1$ was not too difficult, but it took a while to realize that $f\_{n,\ell}$ could be placed in this form for $\ell = n-2$. Again, at this point the pattern became clear.)


Indeed, one has
$$ f\_{n,0}(x) = x^n \mathrm{per}( B\_{n,0} ) = n! x^n $$
(since $B\_{n,0}$ is the all $1$'s matrix), and from the identity
$$ A\_{n,\ell} + \ell B\_{n,\ell} = A\_{n,\ell-1} + \ell B\_{n,\ell-1}$$
we have $f\_{n,\ell}(\ell) = f\_{n,\ell-1}(\ell)$. Finally, from multilinearity, cofactor expansion and the fact that the $n-1 \times n-1$ minor formed by deleting the $i^{th}$ row and $j^{th}$ column from $A\_{n,\ell} + x B\_{n,\ell}$ is equal to $A\_{n-1,\ell} + x B\_{n-1,\ell}$ when $\min(i,j) > \ell$, we have
$$ f'\_{n,\ell}(x) = (n-\ell)^2 f\_{n-1,\ell}(x)$$
(the factor $(n-\ell)^2$ coming from the number of 1's appearing in $B\_{n,\ell}$) and this is all we need to verify \eqref{2} recursively. Replacing $n,\ell$ by $n+1$ we obtain $f\_{n+1,n+1}(0) =a(n+1)$ as conjectured.


Combining \eqref{1}, \eqref{2} we also have the nice identity
$$ R(n,q) = \frac{1}{(q+1)!} \mathrm{per}( \min( i, j, n+1 ) )\_{1 \leq i,j \leq n+q+1}.$$


