# Conjecture on bernoulli numbers and binomial coefficients

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.

• You should definitely put the whole question into the text here, and not make people visit the other site to see it. – Sam Hopkins May 4 at 22:11
• @Fox you can hit 'edit' on the other question and copy and paste the source (then cancel the edit), rather than re-type the thing. – David Roberts May 4 at 22:54
• @David, Sam: thanks! – Fox Mulder May 5 at 0:41
• Strengthening a conjecture usually makes it easier to prove, so it is worth noting that empirically the given polynomial for $M = 2m+1$ appears to be $(m+1) (k - 1)^{D(m)}$ (and, FWIW, this holds for $M=1$ too). – Peter Taylor May 7 at 16:52

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 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, 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

• Hi Max, thanks for your answer, maybe I'm doing something incorrect, but it seems like when M=9, k=7 is indeed an integer root, I put the calculation in the body of the question at the end, it was too long for a comment. Thanks again for looking into this, greatly appreciated! – Fox Mulder May 7 at 13:58
• @FoxMulder: You seem to be correct, and there was a typo in my code causing miscalculation. I'm taking my answer down for now. – Max Alekseyev May 7 at 16:12
• @FoxMulder: I've added a proof of your conjecture. – Max Alekseyev May 8 at 16:05
• Thanks Max, greatly appreciated! If it's ok with you, I'll email you to chat a bit about this and the wider problem, thanks again! – Fox Mulder May 9 at 1:52
• Sure, you are welcome to email me to maxal@gwu.edu – Max Alekseyev May 9 at 2:24