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

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

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 $$(2)?$$

## closed as off-topic by user21820, RRL, Lee David Chung Lin, YuiTo Cheng, Xander HendersonMay 15 at 13:37

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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.
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}