# An Identity Involving Bernoulli Numbers and Stirling Numbers

I am trying to prove the following identity involving the Bernoulli numbers $B_n$: $$\sum_{i=0}^m\sum_{t=0}^{m-i}B_{2t}2^{2t}{4m+4\choose 2t,2i+1,4m-2t-2i+3}=(2m+2)\left(2^{4m+2}-{4m+2\choose 2m+1}\right).$$ One strategy I have tried is to find a combinatrial interpretation for the LHS since the RHS is so combinatorially simple. I am not sure how to do this, mainly because I can't give a direct combinatorial interpretation to the Bernoulli numbers. Using the known identity $$B_{2t}=\sum_{r=0}^{2t}\frac{(-1)^r}{r+1}r!S(2t,r),$$ where $S(2t,r)$ denotes a Stirling number of the second kind, I have rewritten the LHS of my identity as $$\sum_{r\geq 0}\frac{(-1)^r}{r+1}U(r),$$ where $$U(r)=\sum_{i=0}^m\sum_{t=0}^{m-i}r!S(2t,r)2^{2t}{4m+4\choose 2i+1,2t,4m-2i-2t+3}.$$ I don't know if transforming the LHS in this way is at all helpful, but it does lend itself to some sort of a combinatorial interpretation since we can think of $U(r)$ as the number of ways to do the following:

First, choose disjoint subsets $S_1,S_2,S_3$ of $\{1,2,\ldots,4m+4\}$ with $\vert S_1\vert$ odd, $\vert S_2\vert$ even, $\vert S_1\cup S_2\vert<2m+2$, and $S_1\cup S_2\cup S_3=\{1,2,\ldots,4m+4\}$ (think that $\vert S_1\vert=2i+1$ and $\vert S_2\vert=2t$ in the sum). Then, choose a subset $T_2$ of $S_2$, and choose an ordered partition of $S_2$ into $r$ blocks.

I don't yet know how to push this any further. Perhaps I'm on the wrong track. Maybe there is an easy analytic proof that I'm missing. Any help would be greatly appreciated.

We use the fact that if $m\geq 3$ is odd, then $B_m=0$. The equation $$\sum_{m=0}^{n-1}B_m{n\choose m}=0 \tag{1}$$ holds for any positive integer $n$. If $n$ is odd, then $$\sum_{m=0}^{n}B_m2^m{n\choose m}=0. \tag{2}$$
Because $\sum_{i=0}^{k-t}{2(k-t)+2\choose 2i+1}=2^{2k-2t+1}$, we have $$\sum_{i+t\leq k}B_{2t}2^{2t}{\textstyle{2k+2\choose 2t,2i+1,2k-2t-2i+1}}=\sum_{t=0}^k B_{2t}2^{2t}{\textstyle{2k+2\choose 2t}}\sum_{i=0}^{k-t}{\textstyle{2(k-t)+2\choose 2i+1}}=2^{2k+1}\sum_{t=0}^{k}B_{2t}{\textstyle{2k+2\choose 2t}}.$$ Since $B_m=0$ for all odd $m\geq 3$,
$$\sum_{i+t\leq k}B_{2t}2^{2t}{\textstyle{2k+2\choose 2t,2i+1,2k-2t-2i+1}}=2^{2k+1}\left(\sum_{\ell=0}^{2k+1}B_\ell{\textstyle{2k+2\choose \ell}}-B_1{\textstyle{2k+2\choose 1}}\right)=2^{2k+1}(k+1). \tag{3}$$ Note that we used $(1)$ along with the fact that $B_1=-\frac 12$ to deduce the last equality above.
Next, $$\sum_{\left\lfloor k/2\right\rfloor<i+t\leq k}B_{2t}2^{2t}{\textstyle{2k+2\choose 2t,2i+1,2k-2t-2i+1}}=\sum_{m=\left\lfloor k/2\right\rfloor+1}^k{\textstyle{2k+2\choose 2m+1}}\sum_{t=0}^m B_{2t}2^{2t}{\textstyle{2m+1\choose 2t}}$$ $$=\sum_{m=\left\lfloor k/2\right\rfloor+1}^k{\textstyle{2k+2\choose 2m+1}}\left(\sum_{\ell=0}^{2m+1} B_\ell2^\ell{\textstyle{2m+1\choose \ell}}-2B_1{\textstyle{2m+1\choose 1}}\right)=\sum_{m=\left\lfloor k/2\right\rfloor+1}^k{\textstyle{2k+2\choose 2m+1}}(2m+1),$$ where we have used $(2)$ to see that $\sum_{\ell=0}^{2m+1} B_\ell2^\ell{\textstyle{2m+1\choose \ell}}=0$. Therefore, $$\sum_{\left\lfloor k/2\right\rfloor<i+t\leq k}B_{2t}2^{2t}{\textstyle{2k+2\choose 2t,2i+1,2k-2t-2i+1}}=\sum_{m=\left\lfloor k/2\right\rfloor+1}^k(2k+2)\left[{\textstyle{2k\choose 2m}}+{\textstyle{2k\choose 2m-1}}\right]$$ $$=(k+1)\left(\sum_{m=\left\lfloor k/2\right\rfloor+1}^k\left[{\textstyle{2k\choose 2m}}+{\textstyle{2k\choose 2m-1}}\right]+\sum_{j=0}^{k-\left\lfloor k/2\right\rfloor-1}\left[{\textstyle{2k\choose 2j}}+{\textstyle{2k\choose 2j+1}}\right]\right)=(k+1)(2^{2k}-(-1)^k{\textstyle{2k\choose k}}). \tag{4}$$
If we subtract $(4)$ from $(3)$, we find that $$\sum_{i+t\leq\left\lfloor k/2\right\rfloor}B_{2t}2^{2t}\textstyle{2k+2\choose 2t,2i+1,2k-2t-2i+1}=(k+1)\left(2^{2k}+(-1)^k{2k\choose k}\right).$$ The desired identity now follows by setting $k=2m+1$.