Manipulations with Bernoulli numbers

Question

Is there a simple way to rewrite the expression $$ c(g):= \sum_{k=0}^{2g} \frac{B_k B_{2g-k}}{k! (2g-k)!} (-1)^k $$ involving a sum of Bernoulli numbers, as a product of just one or two Bernoulli numbers, times possibly a numerical coefficient depending on $g$?

Answer 1

Actually, I think I found the answer by myself: it is $c_0=1$ and $$ c_{g \geq 1} = - \frac {B_{2g}}{2g (2g-2)!}$$ and it is obtained by differentiating the generating function, $$ \frac {d}{dt} \frac{1}{e^t-1}$$