Crossposted from


<https://math.stackexchange.com/questions/4116414/conjecture-on-bernoulli-numbers-and-binomial-coefficients>


In playing around with some formulas, I have come up with the following conjecture. I have checked it for a lot of cases, and have good reason to believe it to be true. If anyone could help, I would sincerely appreciate it, and I would be happy to include as co-author.


Some notation:


* The exponent $R(i)$ means the formal raising by $2$s for $i$ terms. For example, $(k-7)^{R(3)} = (k-7)(k-5)(k-3)$.
* The exponent $D(i)$ means the formal descending by $2$s for $i$ terms. For example, $k^{D(4)} = (k)(k-2)(k-4)(k-6)$.
* $k^{D(1)}:=k$, $k^{D(0)}:=1$ and $k^{D(-1)}:=\frac{1}{k+2}$.
* !! denotes the double factorial, for example $9!! = 1.3.5.7.9$
* $B\_i$ denotes the ith Bernoulli number, with the convention that $B\_1 = 1/2$.


In what follows, $M$ is a positive odd integer bigger than or equal to $3$, $k$ is a formal variable. For a given odd $M\geq 3$, the following is a polynomial in $k$. My conjecture is that it has roots $1,3,5,.., M-2$.


$$\sum\limits\_{i=1}^{\frac{M+1}{2}}\binom{\frac{M+1}{2}}{i}\left[(k-(M-2))^{R(i)}+(-1)^{i+1}(2i-1)!!\right]B\_{\frac{M+1}{2}-i}\left[k^{D(\frac{M+1}{2}-i-1)}\right]$$


For example, when $M=3$, we get


$$\binom{2}{1}[(k-1)^{R(1)}+1!!]B\_{1}k^{D(0)}+\binom{2}{2}[(k-1)^{R(2)}-3!!]B\_{0}[k^{D(-1)}]
$$


$$=2[k]B\_1+[(k-1)(k+1)-3]\frac{B\_0}{k+2},
$$ which has a root of $1$.


When $M=5$ we get


$$\binom{3}{1}[(k-3)^{R(1)}+1!!]B\_2[k^{D(1)}]+\binom{3}{2}[(k-3)^{R(2)}-3!!]B\_1[k^{D(0)}]+\binom{3}{3}[(k-3)^{R(3)}+5!!]B\_0[k^{D(-1)}]$$


$$=3[k-3+1]B\_2[k]+3[(k-3)(k-1)-1.3]B\_1+[(k-3)(k-1)(k+1)+1.3.5]\frac{B\_0}{k+2}.$$


Putting $k=1$ gives


$$-3B\_2-9B\_1+5B\_0 = 0,$$


putting $k=3$ gives


$$9B\_2-9B\_1+3B\_0 = 0.$$


As another examples, when $M=9$ we get


$$ \binom{5}{1}[(k-7)+1]B\_4[(k)(k-2)(k-4)]
+ \binom{5}{2}[(k-7)(k-5)-1.3]B\_3[k(k-2)]
+ \binom{5}{3}[(k-7)(k-5)(k-3)+1.3.5]B\_2[k]
+ \binom{5}{4}[(k-7)(k-5)(k-3)(k-1)-1.3.5.7]B\_1
+ \binom{5}{5}[(k-7)(k-5)(k-3)(k-1)(k+1)+1.3.5.7.9]\frac{B\_0}{k+2}$$


Putting $k=7$, using that $B\_3=0$ and factoring out 1.3.5.7, we get


$$\binom{5}{1}B\_4+\binom{5}{3}B\_2-\binom{5}{4}B\_1+\binom{5}{5}B\_0 = -\frac{5}{30}+\frac{10}{6}-\frac{5}{2}+1 = 0$$


In special cases the conjecture boils down to well known identities involving Bernoulli numbers and binomial coefficients. I would appreciate an elementary proof, but really anything will do. Perhaps someone into combinatorics or generating functions can help? Thanks again.



Conjecture on bernoulli numbers and binomial coefficients
**UPDATE.** My earlier answer was incorrect due to miscalculation. Below I give a proof of the conjecture.


First, we notice that $k^{R(i)} = (k/2+i-1)\_i\cdot 2^i = \binom{k/2+i-1}i\cdot i!\cdot 2^i$. Similarly, $k^{D(i)} = (k/2)\_i\cdot 2^i = \binom{k/2}i\cdot i!\cdot 2^i$ and $(2i-1)!!=\frac{(2i)!}{i!2^i}=\binom{2i}{i}\cdot \frac{i!}{2^i}$. Then the proposed conjecture follows from the following polynomial (in $x$) identity:
$$\sum\_{i=0}^m \big(\binom{x+\frac12-m+i}i + (-1)^{i+1}\binom{2i}{i} \frac{1}{4^i}\big) B^+\_{m-i} \binom{x+1}{m-i} = (x+1) \binom{x-\frac12}{m-1}$$
by taking $m = \frac{M+1}2$ and $x=\frac{k}2$. Noticing that $\binom{x+\frac12-m+i}i = (-1)^i \binom{m-x-\frac32}i$ and $B^+\_{m-i} = (-1)^{m-i} B^-\_{m-i}$, we rewrite the above identity as
$$(\star)\qquad \sum\_{i=0}^m \big(\binom{m-x-\frac32}i - \binom{2i}{i} \frac{1}{4^i}\big) B^-\_{m-i} \binom{x+1}{m-i} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$


**Proof.** Since for a fixed $m$ the l.h.s. and r.h.s. of $(\star)$ are polynomials in $x$, it is enough to prove $(\star)$ for $x$ being a nonnegative integer. We notice that
$$\binom{m-x-\frac32}i - \binom{2i}{i}\frac{1}{4^i} = [z^i]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big)$$
while
$$B^-\_{m-i} \binom{x+1}{m-i} = [z^{m-i}]\ z^{x+1}{\cal B}\_{x+1}(\frac1z),$$
where ${\cal B}\_{x+1}(t)$ is the $(x+1)$-st [Bernoulli polynomial](https://en.wikipedia.org/wiki/Bernoulli_polynomials) and $[z^n]$ is the operator extracting the coefficient of $z^n$.


It follows that the l.h.s. of $(\star)$ equals
$$[z^m]\ \big( (1+z)^{m-x-\frac32} - (1-z)^{-\frac12}\big) z^{x+1}{\cal B}\_{x+1}(\frac1z).$$


Using [Lagrange–Bürmann formula](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem#Lagrange%E2%80%93B%C3%BCrmann_formula), we conclude that
$$[z^m]\ (1+z)^{m-x-\frac32} z^{x+1}{\cal B}\_{x+1}(\frac1z) = [z^m]\ (1-z)^{-\frac12} z^{x+1} {\cal B}\_{x+1}(\frac1z-1).$$
Then, by the properties of Bernoulli polynomials,
$${\cal B}\_{x+1}(\frac1z-1) = {\cal B}\_{x+1}(\frac1z) - (x+1) (\frac1z-1)^x.$$


So, the Bernoulli polynomials in the l.h.s. of $(\star)$ cancel out, and it reduces to
$$-(x+1)\cdot [z^m]\ z (1-z)^{x-\frac12} = (-1)^m (x+1) \binom{x-\frac12}{m-1}.$$
**QED**


