Crossposted from


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 $9!! =$

  • $B_i$ 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 $1,3,5,.., M-2$.


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



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

$$ \binom{5}{1}[(k-7)+1]B_4[(k)(k-2)(k-4)] + \binom{5}{2}[(k-7)(k-5)-1.3]B_3[k(k-2)] + \binom{5}{3}[(k-7)(k-5)(k-3)+1.3.5]B_2[k] + \binom{5}{4}[(k-7)(k-5)(k-3)(k-1)-]B_1 + \binom{5}{5}[(k-7)(k-5)(k-3)(k-1)(k+1)+]\frac{B_0}{k+2}$$

Putting $k=7$, using that $B_3=0$ and factoring out, 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.

  • 3
    $\begingroup$ You should definitely put the whole question into the text here, and not make people visit the other site to see it. $\endgroup$ May 4, 2021 at 22:11
  • 1
    $\begingroup$ @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. $\endgroup$ May 4, 2021 at 22:54
  • 1
    $\begingroup$ @David, Sam: thanks! $\endgroup$
    – Fox Mulder
    May 5, 2021 at 0:41
  • 1
    $\begingroup$ 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). $\endgroup$ May 7, 2021 at 16:52

1 Answer 1


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

  • $\begingroup$ 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! $\endgroup$
    – Fox Mulder
    May 7, 2021 at 13:58
  • $\begingroup$ @FoxMulder: You seem to be correct, and there was a typo in my code causing miscalculation. I'm taking my answer down for now. $\endgroup$ May 7, 2021 at 16:12
  • $\begingroup$ @FoxMulder: I've added a proof of your conjecture. $\endgroup$ May 8, 2021 at 16:05
  • $\begingroup$ 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! $\endgroup$
    – Fox Mulder
    May 9, 2021 at 1:52
  • $\begingroup$ Sure, you are welcome to email me to maxal@gwu.edu $\endgroup$ May 9, 2021 at 2:24

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