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Let $B(x) = b_1 x + b_2 x^2 + \cdots$. Then
$$\begin{eqnarray*}
F_n(b_1, \ldots, b_n) &=& - [x^n] n \log(1 + B(x)) \\\\
&=& [x^n] n \sum_{i \ge 1} \frac{(-B(x))^i}{i} \\\\
&=& \sum_{\lambda \\, \vdash \\, n} \frac{n}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} \prod_j (-b_j)^{f_j} \\
\end{eqnarray*}$$
where the sum in the last line is over partitions $\lambda = 1^{f_1}2^{f_2} \cdots n^{f_n}$ with $\sum_i if_i = n$ and the length $\operatorname{len}(\lambda)$ defined as $\sum_i f_i$.

So the question is whether $n \binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n}$ is divisible by $\operatorname{len}(\lambda)$. Consider a prime $p$ which divides $\operatorname{len}(\lambda)$ and look at $p$-adic valuations. Specifically, choose $c$ such that $\nu_p(f_c) = \min_i(\nu_p(f_i))$. We can split the multinomial as $$\binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} = \binom{\operatorname{len}(\lambda)}{f_c} \binom{\operatorname{len}(\lambda) - f_c}{ \\{ f_i : i \neq c \\}}$$ By Kummer's theorem, $$\nu_p\left(\binom{\operatorname{len}(\lambda)}{f_c}\right) \ge \nu_p(\operatorname{len}(\lambda)) - \nu_p(f_c) \tag{1}$$

Since $\sum_i if_i = n$ we have $\nu_p(n) \ge \nu_p(f_c)$ and we can combine that with $(1)$ to get $$\nu_p\left(\binom{\operatorname{len}(\lambda)}{f_c}\right) + \nu_p(n) \ge \nu_p(\operatorname{len}(\lambda))$$
Therefore $$\nu_p\left( \frac{n}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} \right) \ge 0$$ and since this holds for every prime divisor of the denominator, we have an integer.