Sum of function over even divisors and Mobius inversion

Question

I have a function of the form $$f(n) = \sum_{i,j}(-1)^i a_{i,j}^n$$ (where the $a_{i,j}$ are complex numbers and the sum over $i$ and $j$ is a finite sum for all $n$ whose bounds depend on $n$). Moreover, I know $f$ satisfies the relation $$ f(n) = \sum_{d\mid n}d g(d), $$ but I know very little about $g$ besides this. I need to work with $$ \sum_{\substack{d\mid n \\ d\text{ even}}}g(d). $$ Via Mobius inversion, I know that $$ g(d) = \frac{1}{d}\sum_{k\mid d} \mu(k)f(d/k), $$ so $$ \sum_{\substack{d\mid n \\ d\text{ even}}}g(d) = \sum_{\substack{d\mid n \\ d\text{ even}}} \frac{1}{d}\sum_{k\mid d} \mu(k)f(d/k) = \sum_{\substack{d\mid n \\ d\text{ even}}} \frac{1}{d} \sum_{k\mid d} \mu(k) \sum_{i,j}(-1)^i a_{i,j}^{d/k}. $$ But this is rather unpleasant to work with. Can I hope to get a nicer or more approachable expression for this? Or is this the best I could reasonably expect? (To be a little more precise: I would especially like to somehow eliminate the nested sums over divisors, perhaps by using the fact that the Mobius function is $0$ on numbers with repeated prime factors… but I couldn’t quite see how to go about doing this.)