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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.

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This question hasn't gotten much attention, but I will post a solution that I just found in case anyone is interested. It was easier than I had suspected.

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 $k=2m+1$.

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