TITLE: "inversion" of a convolution
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.