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Throughout this question $n$ is a positive integer greater than 1. Consider the following well-known identity by Euler, $$\sum_{k=1}^{n-1} \binom{2n}{2k}B_{2k}B_{2n-2k}=-(2n+1)B_{2n}.$$ Rather unconventionally, we can also write the above identity as $$\sum_{k=1}^{n} \binom{2n}{2k}B_{2k}B_{2n-2k}=-2n B_{2n}.$$

I was wondering if the following identity is also known. $$\sum_{k=1}^{n} \binom{2n}{2k}B_{2k}B_{2n-2k}=\sum_{k=1}^{n} 2^{2k} \binom{2n}{2k}B_{2k}B_{2n-2k}.$$

This would give the following nice identity $$\sum_{k=1}^{n}(2^{2k}-1) \binom{2n}{2k}B_{2k}B_{2n-2k}=0.$$

My proof involves Newton’s Identities and simple manipulations of the equation derived.

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This is well known. Equation (11) in is this Wolfram article implies that

$$\binom{n}{2}\frac{E_{n-2}(0)}{2}=\sum_{k=2}^n\binom{n}{k}(2^k-1)B_kB_{n-k}.$$

If $n>2$ is even, then $E_{n-2}(0)=0$, and in the sum all the $B_k$ are zero for odd $k$. This is your identity.

I think that the rule for Bernoulli numbers identities is: if it is a simple algebraic manipulation, it is surely known.

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Here's a simple proof of the identity. Let $$B(x) = \frac{x}{e^x-1} = \sum_{n} B_n \frac{x^n}{n!}.$$ Then $$\sum_n (2^n-1) B_n \frac{x^n}{n!} = B(2x) - B(x) = -\frac{x}{e^x+1}$$ and $$\bigl(B(2x) -B(x)\bigr) B(x)=-\frac{x^2}{e^{2x}-1}=-\frac{x}{2}B(2x).$$ So $$\sum_{k=0}^n \binom nk (2^k-1)B_k B_{n-k}=-2^{n-2}n B_{n-1}.$$ For $n$ even this reduces to the OP's identity since $B_n=0$ for $n$ odd and greater than 1.

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