show/hide this revision's text 3 correct typo, now proved

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.
show/hide this revision's text 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 $d$.
show/hide this revision's text 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 $d$.