# Sum of function over even divisors and Mobius inversion

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.)