I have an arithmetic function $f$ defined by the following recurrence relation:
$$f(n+1)=f(n)+\sum _{a=1}^n \sum _{b=1}^n e^{n-a b}$$
Is it possible to create a closed-form expression for $f$? I've been noodling around with generating functions, but I'm not sure if that's the right way to go about this. There are so many possibilities, and I don't know how to proceed.
I am aware that there are lots of questions on this forum about this general subject (recurrence $\rightarrow$ closed form), but I haven't found one that meets my specific needs. The issue, I suspect, is the double summation.
*BTW, I'm teaching myself all this stuff from scratch. Hints are appreciated, of course - but please be aware that I may not even understand the hint. My A-Level was in 1982, and since then, I've done no maths at all...
Hint.
$$ \frac{f_{n+1}-f_n}{e^n}-\frac{f_{n}-f_{n-1}}{e^{n-1}}=2\frac{e^{-n(n+1)}-1}{e^{-n}-1}-e^{-n^2}-2 $$
giving
$$ f_n = c_1+c_2 e^n+e^n \sum _{k=0}^{n-1} \frac{e^{-(k+1)^2} \left(2 e^{(k+1)^2}-e^{k+1}-1\right)}{(e-1) \left(e^{k+1}-1\right)}-\sum _{j=0}^{n-1}\frac{e^{-(j+1)^2-j (j+1)+j+1} \left(2 e^{(j+1)^2+j (j+1)}-e^{j (j+1)}-e^{(j+1)^2}\right)}{(e-1) \left(e^{j+1}-1\right)} $$
NOTE
$$ \sum_j^n\sum_k^n e^{-j k}-\sum_j^{n-1}\sum_k^{n-1} e^{-j k} = \sum_{j=1}^n e^{-nj}+\sum_{k=1}^n e^{-nk}-e^{-n^2} $$
where
$$ \sum_{j=1}^n e^{-nj} = \frac{e^{-n(n+1)}-1}{e^{-n}-1}-1 $$
hence
$$ \sum_j^n\sum_k^n e^{-j k}-\sum_j^{n-1}\sum_k^{n-1} e^{-j k} = 2\frac{e^{-n(n+1)}-1}{e^{-n}-1}-e^{-n^2}-2 $$
after that, consider $g_n = \frac{f_{n+1}-f_n}{e^n}$