Consider the following recursive equation for $k\ge 1$: \begin{equation} a_k = \frac{1}{2} + \sum_{i=0}^{k-1} \frac{-1}{2k + 1 - 2i} \binom{2k}{2i} a_i \tag{1}. \end{equation} If $a_0 = 1$, then $a_k$ is the $2k$-th Bernoulli number $B_{2k}$. I want to prove that if $a_0 = 0$, then $a_k$ is $(2^{2k}-1)B_{2k}$. Does anybody have any idea? Thanks!
I checked the fact on the computer for big $k$. Every proof I attempt seems to lead to heavy computations.
Let's consider arbitrary $a_0$, rewrite the recurrence as $$\frac{1}{(2n)!}=\sum_{k=0}^n\frac{2}{(2n+1-2k)!}\frac{a_k}{(2k)!}\qquad(n>0)$$ and introduce the generating function $A(a_0,z)=\sum_{n=0}^\infty a_n z^{2n}/(2n)!$; then \begin{align*} \cosh z\underbrace{{}-1+2a_0}_{\text{to fix }n=0}&=\sum_{n=0}^\infty z^{2n}\sum_{k=0}^n\frac{2}{(2n+1-2k)!}\frac{a_k}{(2k)!}\\&=2\sum_{k=0}^\infty\frac{a_k}{(2k)!}\sum_{n=k}^\infty\frac{z^{2n}}{(2n+1-2k)!}\\\color{gray}{[n=m+k]}\quad&=2\sum_{k=0}^\infty\frac{a_k z^{2k}}{(2k)!}\sum_{m=0}^\infty\frac{z^{2m}}{(2m+1)!}=2A(a_0,z)\frac{\sinh z}{z}. \end{align*} Now, if we put $a_0=1$ and $a_0=0$, we obtain $$A(1,z)=\frac{z}{2}\frac{e^z+1}{e^z-1},\qquad A(0,z)=\frac{z}{2}\frac{e^z-1}{e^z+1},$$ and it remains to check $A(0,z)=A(1,2z)-A(1,z)$, which is done easily.