Closed form for $\sum_{j=0}^{n}a^{-j}B_{j}B_{n-j}{n \choose n-j}$

Question

$n=2k+1$, $k\ge1$

Where $B_n$ ; Bernoulli number

$$\sum_{j=0}^{n}2^{-j}B_{j}B_{n-j}{n \choose n-j}=-\frac{2^{n-2}+1}{2^n}\cdot nB_{n-1}\tag1$$

We manage to figure the closed form for $(1)$

We are unable to work out $(2),$

$$\sum_{j=0}^{n}a^{-j}B_{j}B_{n-j}{n \choose n-j}=F(n,a)\tag2$$

Let $a\ge 2$

Does anyone know how to work out the general closed form for $(2)?$

Answer 1

Your notation should be something like $F(a,n)$.

Using exponential generating functions, I get for odd $n >2$ and integer $a$ $$ F(a,n) \equiv \sum_{j=0}^n a^{-j}B_jB_{n-j}\binom{n}{n-j} = -\left(\frac1{2a} + \frac12 a^{-(n-1)}\right) n B_{n-1} $$ Analytic continuation arguments say this should also be true for non-integer $a$, and indeed that is the case.

I did not try to also generalize to even $n$; the form has to be different since the closed form for odd $n$ comes out to zero when $n$ is even.

Answer 2

The Bernoulli numbers have a nice property that they are $0$ for all odd indices except for $B_1=-\frac12$. Therefore for odd $n $ all terms of the sum are $0$ except for $j=1$ and $j=n-1$: $$\begin{align} \sum_{j=0}^nB_jB_{n-j}\binom n {n-j}a^{-j}&=B_1B_{n-1}\binom n {n-1}a^{-1}+B_{n-1}B_1\binom n {1}a^{-(n-1)}\\ &=-\frac12 nB_{n-1}(a^{-1}+a^{1-n}). \end{align} $$