# A Newton identity and the primes--the Faber partition polynomials and modular arithmetic

[Edit, July 6, 2022: Removed erroneous characterization of Faber polynomials as an Appell sequence.]

Dress and Siebeneicher in their tale of the Burnside family express an opinion (1.2) that, if I read it correctly, leads me to believe that the classic Faber partition polynomials which satisfy

$$\ln[A(x)] = \ln[1 + a_1 x + a_2 x^2 + \cdots ] = \sum_{\geq 1} -F_n(a_1,...,a_n)\; \frac{x^n}{n}$$ have the property

$$[F_n(a_1,a_2,...,a_n)- (-a_1)^n] \; mod(n) \; = 0$$

for $$n$$ prime and integral indeterminates $$a_n$$ .

I've been assured by a reputable authority that such is obvious. Can someone provide an 'obvious' proof or a least some other published hearsay on this?

Gessel and Ree in "Lattice paths and Faber polynomials" give (p. 4) a multinomial- coefficient type of expression for the coefficients of the Faber polynomials $$FP_n(u)$$ for which $$FP_n(0)=F_n[a_1,a_2,...,a_n]$$. (G & R use the notation $$F_n(u)$$ for what I denote as $$FP_n(u)$$.)

[Edit, July 6, 2022: Motivated by Peter's answer, I found two nice intros for the uninitiated to Kummer's theorem--"Legendre’s and Kummer’s Theorems Again" by Mihet and "Revisiting Kummer's and Legendre's Formulae" by Sury.]

On the 'ubiquity' of the Faber polynomials and Faber partition polynomials:

These Faber polynomials and their associated Faber partition polynomials crop up in multitude of discussions: in symmetric function theory in the Newton-Girard-Waring identities expressing the power symmetric polynomials in terms of the elementary symmetric polynomials; in operational calculus for a generic raising operator for Appell polynomials; in complex function theory in extending an analytic function defined on a closed curve to an analytic function within the curve (providing harmonic functions with prescribed boundary conditions); in an analog of Fourier series expansions of complex functions; in extracting the numerical values of the indeterminates of compositional partition polynomials given numerical evaluations of those polynomials; in relations between determinants and traces; in algebraic/geometric K-theory; in the combinatorial/analytic properties of random walks, lattice paths, noncrossing partitions, and associahedra; and in determining the compositional inverse of certain Laurent series--in fact, in an orgy with the family of reciprocal polynomials $$R_n$$ birthed by $$1/A(x) = \sum_{n \geq 0} R_n(a_1,...,a_n) x^n$$, the compositional inverse $$L^{(-1)}(z) = z + b_1 +b_2/z +b_3/z^2 + \cdots$$ of the formal Laurent series $$L(z) =z + a_1 +a_2/z +a_3/z^2 + \cdots$$ is given by the umbral recursion formula $$b_n(a_1,...,a_n) = \frac{1}{n}[R_n(b_1-F.(a_1,...),b_2,b_3,...,b_{n-1},0] - R_n(0,-b_2,-2b_3,...,-(n-2)b_{n-1},0)].$$

• If you consider $F_n(a_1,\ldots,a_n)$ as a polynomial in the $a_i$, is it obvious that the coefficients are integers? Jun 29, 2022 at 23:37
• @PeterTaylor, I'm being somewhat facetious in parodying a comment and the stated reason for the closing of my MO question mathoverflow.net/questions/425411/priming-for-the-primes by the thought police--think doublethink and disregard the opinionated editorial comments in the current Q. The math is not obvious to me nor most likely to anyone else. Jun 30, 2022 at 0:20
• I'd find it easier to disregard the opinionated comments in the question, and in your comment, if you would just delete them. (Delete the comments, not the question.) Jun 30, 2022 at 2:40
• Ok. If the coefficients are integers then the rest is easy, and since the coefficients are in OEIS it seems likely that someone has proven them to be integers; I was hoping that there's a combinatorial interpretation which makes that drop out. Jun 30, 2022 at 6:53

From $$\ln[A(x)] = \ln[1 + a_1 x + a_2 x^2 + \cdots ] = \sum_{n \geq 1} -F_n(a_1,...,a_n)\; \frac{x^n}{n}$$ and letting $$A'(x) = A(x) - 1$$ we have $$\begin{eqnarray*} F_n(a_1,...,a_n) &=& n [x^n] \sum_{i=1}^{\infty} \frac{(-A'(x))^i}{i} \\ &=& (-a_1)^n + n [x^n] \sum_{i=1}^{n-1} \frac{(-A'(x))^i}{i} \\ %&=& n \sum_{\lambda \, \vdash \, n} \frac{1}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} \prod_i (-a_i)^{f_i} \\ &=& (-a_1)^n + n \sum_{\lambda \, \vdash \, n,\; \lambda \neq 1^n} \frac{1}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} \prod_j (-a_j)^{f_j} \\ \end{eqnarray*}$$

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 $$\sum_i f_i$$, and the partition $$\lambda = 1^n$$ giving $$(-a_1)^n$$ is pulled out of the sum.

It remains to show that all of the multinomial coefficients in $$[x^n](-A'(x))^i$$ with $$i < n$$ must be divisible by $$i$$; they're certainly not divisible by any prime greater than $$i$$, which by hypothesis includes $$n$$.

Let $$p$$ be a prime factor of $$\operatorname{len}(\lambda) < n$$. There must be some $$c$$ for which $$f_c \neq 0 \pmod p$$, since otherwise $$p \mid n$$ contradicting the primality of $$n$$. But then $$\binom{\operatorname{len}(\lambda)}{f_1, \ldots, f_n} = \binom{\operatorname{len}(\lambda)}{f_c} \binom{\operatorname{len}(\lambda) - f_c}{\{f_j : j \neq c\}}$$ and by Kummer's theorem $$\nu_p \left( \binom{\operatorname{len}(\lambda)}{f_c} \right) \ge \nu_p \left(\operatorname{len}(\lambda)\right)$$. Therefore the multinomial coefficient is indeed divisible by $$\operatorname{len}(\lambda)$$, giving the desired result.

• Thanks, I guess that should be called Kummer's obvious theorem, which he deigned to prove in a paper in 1852. D & S extended these results. Jun 30, 2022 at 17:41
• @TomCopeland, as I observed in a comment on the question, the non-obvious thing is that the coefficients are integers. This argument just needs a tiny tweak to show that the coefficients are integers for non-prime $n$ too, although in that case not necessarily divisible by $n$; an argument could be made that it's obvious that a sequence in OEIS will only contain integers. It's also fair to say that things which are not obvious to one generation may be obvious to another which was taught a different syllabus. Jun 30, 2022 at 18:04
• KOT, something to sleep on. Jun 30, 2022 at 18:21
• Dold approaches similar questions in "Fixed point indices of iterated maps". Jul 1, 2022 at 0:24
• Another way to see that the coefficients are integers: setting $a_n$ equal to the $n$th complete homogeneous symmetric function, thereby identifying $-F_n$ with the $n$th power symmetric function, and then noting that the complete homogeneous symmetric functions are an integral basis for the ring of symmetric functions.
– AWO
Jul 8, 2022 at 14:46