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 it seems (Warning: I didn't prove this yet!) thatwe have $$F_n(x)=L(n)+\sum_{1< d|n} L_{n/d}(x)\sum_{1=d_0|d_1|\dots|d_{k+1}=d}(-1)^{k+1}\prod_{i=0}^k F_n(x)=\sum_{1=d_0|d_1|\dots|d_k|n}L_{n/d_k}(x^{d_k}) (-1)^k\prod_{i=0}^{k-1} Z_{d_{i+1}/d_i}(x^{d_i}),$$
where in the second sum all divisors are proper, i.e., $d_i d_0 < d_{i+1}$d_1 < \dots < d_k \leq n$. In other words, we are summing over all chains from starting at$1$to 1$, below $d$.n$. The formula is easily shown by induction. 2 added 18 characters in body At least computing$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 it seems (Warning: I didn't prove this yet!) that $$F_n(x)=L(n)+\sum_{1< d|n} L_{n/d}(x)\sum_{1=d_0|d_1|\dots|d_{k+1}=d}(-1)^{k+1}\prod_{i=0}^k Z_{d_{i+1}/d_i}(x^{d_i}),$$ where in the second sum all divisors are proper, i.e.,$d_i < d_{i+1}$. In other words, we are summing over all chains from$1$to$d$. 1 At least computing$F_k(x)$turned out not to be that hard after all. Slightly more generally, consider $$\sum_{d|k} Z_d(x) F_{n/d}(x^d) = L_n(x).$$ Then it seems (Warning: I didn't prove this yet!) that $$F_n(x)=L(n)+\sum_{1< d|n} L_{n/d}(x)\sum_{1=d_0|d_1|\dots|d_{k+1}=d}(-1)^{k+1}\prod_{i=0}^k Z_{d_{i+1}/d_i}(x^{d_i}),$$ where in the second sum all divisors are proper, i.e.,$d_i < d_{i+1}$. In other words, we are summing over all chains from$1$to$d\$.