Ask for a proof of an identity involving the product of two Bernoulli numbers It is well known that the Bernoulli numbers $B_{k}$ for $k\in\{0,1,2,\dotsc\}$ can be generated by
\begin{equation*}
\frac{z}{\textrm{e}^z-1}=\sum_{k=0}^\infty B_k\frac{z^k}{k!}=1-\frac{z}2+\sum_{k=1}^\infty B_{2k}\frac{z^{2k}}{(2k)!}, \quad \vert z\vert<2\pi.
\end{equation*}
Could you please find or give a proof for the identity
\begin{equation}\label{Qi-EFinat-S-ID}\tag{BQQID}
\sum_{j=1}^{k-1}\binom{2k}{2j}\bigl(1-2^{2j-1}-2^{2k-2j-1}\bigr)B_{2j}B_{2k-2j}
=\bigl(2^{2k}-1\bigr)B_{2k}
\end{equation}
for $k\ge2$?
 A: More generally, let $B_k(x)$, $k\ge0$, be the Bernoulli polynomials defined by the exponential generating function
\begin{equation*}
\frac{ze^{xz}}{e^z-1}=\sum_{k=0}^\infty B_k(x)\frac{z^k}{k!},
\end{equation*}
thus $B_k=B_k(0)$.
These satisfy the general identity (which can be proven using the definition above, simple manipulations, and comparing coefficients)
\begin{equation*}
\sum_{i=0}^n \binom{n}{i} B_i(x)B_{n-i}(y)=(1-n)B_n(x+y)+n(x+y-1)B_{n-1}(x+y),
\end{equation*}
that for $x=y=0$ becomes
\begin{equation*}
\sum_{i=0}^n \binom{n}{i} B_iB_{n-i}=(1-n)B_n-nB_{n-1},
\end{equation*}
and  for $x=y=1/2$ becomes
\begin{equation*}
\sum_{i=0}^n \binom{n}{i} B_i(1/2)B_{n-i}(1/2)=(1-n)B_n(1).
\end{equation*}
Since $B_i(1/2)=(2^{1-i}-1)B_i$, the LHS of the last equation is
\begin{align*}
\sum_{i=0}^n \binom{n}{i} B_i(1/2)B_{n-i}(1/2)&=
\sum_{i=0}^n \binom{n}{i} (2^{1-i}-1)B_i\times(2^{1-n+i}-1)B_{n-i}\\
&=\sum_{i=0}^n \binom{n}{i} (1-2^{1-i}-2^{1-n+i}+2^{2-n})B_iB_{n-i}\\
&=\sum_{i=0}^n \binom{n}{i}B_iB_{n-i}
+2^{2-n}\sum_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B_iB_{n-i},
\end{align*}
hence
\begin{equation*} 
(1-n)B_n(1)=(1-n)B_n-nB_{n-1}+2^{2-n}\sum_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B_iB_{n-i}.
\end{equation*}
If $n\ge4$ is even, then $B_{n}(1)=B_n$ and $B_{n-1}=0$, and it follows that
\begin{align*}
0&=\sum_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B_iB_{n-i}\\
&=(1-2^{n})B_n+\sum_{i=2}^{n-2} \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B_iB_{n-i},
\end{align*}
which proves your identity.
