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# Do the Faber partition polynomials have integer coefficients?

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The Online Encyclopedia of Integer Sequences includes A263916: Coefficients of the Faber partition polynomials. Perhaps the clearest definition given is

-log(1 + b(1) x + b(2) x^2 + ...) = Sum_{n>=1} F(n,b(1),...,b(n)) * x^n/n

which in better notation is

$$\sum_{n \ge 1} F_n(b_1, \ldots, b_n) \frac{x^n}n = -\log(1 + b_1 x + b_2 x^2 + \cdots)$$

E.g. $$F_4(b_1,b_2,b_3,b_4) = -4b_4 + 4b_1 b_3 + 2b_2^2 - 4b_1^2 b_2 + b_1^4$$

The fact that it's listed in OEIS suggests that all of the coefficients should be integers, but none of the formulae or comments in the page obviously tells me that they are. What's the most straightforward way to see this?

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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.

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