Let $a_k =0 \ \forall k \le 0$ be an integer sequence, and define
$$f(n,s):=\sum_{k=0}^n k^s a_k a_{n-k}$$
We can clearly express $f(n,1)$ in terms of $f(n,0)$ as: $$f(n,1)=\frac{n}{2} f(n,0)$$
My question is whether there is a way to express $f(n,2)$ in terms of $f(n,1)$ and, if there is not any, how to disprove it.
More generally, when I was working with this type of discrete convolutions, I noticed that it is possible to express $f(n,2l+1)$ in terms of combinations of $f(n,2l), \dots ,f(n,0)$ just by making equal:
$$\sum_{k=0}^n k^s a_k a_{n-k} = \sum_{k=0}^n (n-k)^s a_k a_{n-k}$$
And developing Newton's Binomial. An alternative proof can be found by taking its Fourier transform and working with its properties.
However, this method does not work for even values of $s$, hence my question of wether there exists an alternative method for finding such an expression (while my question if focused on the case $f(n,2)$, there may exist a generalization for $n=2l$). If it were not possible, it would be great to have a proof of it.
Thank you in advance.
If I understand correctly, you want a formula of the form $f(n,2) = g\big( n, f(n,1) \big)$ for some function $g$ (which, in particular, does not take the invididual $a_j$ into account).
Such a function cannot exist, as we can see just by taking $n=2$: \begin{align*} f(2,2) &= a_1^2 + 4a_0a_2 \\ f(2,1) &= a_1^2 + 2a_0a_2. \end{align*} There are certainly triples $(a_0,a_1,a_2)$ for which $a_1^2 + 2a_0a_2$ gives the same value but for which $a_1^2 + 4a_0a_2$ does not; for example, $(a_0,a_1,a_2)=(1,1,5)$ and $(a_0,a_1,a_2)=(3,1,1)$ have this property.