# “inversion” of a convolution

I have the following relation:

$$\sum_{d|n} (1+1/x)^{d-1} F_{n/d}(x^d)=L_n(x)$$

where the right hand side is (for every $n$) a polynomial in $x$, which I have an expression for, but it's not extremely beautiful. The family of polynomials $F_k(x)$ is unknown, and is what I'm looking for.

Since this is close to Dirichlet convolution, I have not quite given up hope that there is something similar to Möbius inversion, that would give me $F_k(x)$ explicitely. Is this possible? Related instances of such a problem may also be interesting.

A possibly weaker, but still sufficient solution would be an expression in terms of $L_k$ and $R_k$ of the expression

$$\sum_{d|n} R_d(x) F_{n/d}(x^d)$$

where $R_k(x)$ is another family of polynomials, which is also unknown.

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A meta question: would a tag "inverse-relations" be appropriate? – Martin Rubey Aug 7 '10 at 10:16

At least computing $F_k(x)$ turned out not to be that hard after all. Slightly more generally, consider
$$\sum_{d|k} Z_d(x) F_{n/d}(x^d) = L_n(x),$$ with $Z_1(x)=1$.
Then we have $$F_n(x)=\sum_{1=d_0|d_1|\dots|d_k|n}L_{n/d_k}(x^{d_k}) (-1)^k\prod_{i=0}^{k-1} Z_{d_{i+1}/d_i}(x^{d_i}),$$
where in the sum $d_0 < d_1 < \dots < d_k \leq n$. In other words, we are summing over all chains starting at $1$, below $n$. The formula is easily shown by induction.