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.



Bilinear recurrence relation between even Bernoulli numbers
This is well known. Equation (11) in is this [*Wolfram article*](http://mathworld.wolfram.com/EulerPolynomial.html) 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. 


