Unusual class of polynomials related to partitions

In this paper, we study, for a given arithmetic function f, the sequence of polynomials (Pnf(t))n=0∞\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(P_{n}^{f}(t))_{n=0}^{\infty }$$\end{document}, defined by the recurrence P0f(x)=1,P1f(x)=x,Pnf(x)=xn∑k=1nf(k)Pn-kf(x),n≥2.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \left\{ \begin{array}{ll} P_{0}^{f}(x)=1, &{} \\ P_{1}^{f}(x)=x, &{} \\ P_{n}^{f}(x)=\frac{x}{n}\sum _{k=1}^{n}f(k)P_{n-k}^{f}(x), &{} n\ge 2. \end{array}\right. \end{aligned}$$\end{document}Using the ideas from the paper by Heim, Luca, and Neuhauser, we prove, under some assumptions on Pnf(t)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P_{n}^{f}(t)$$\end{document} for 1≤n≤10\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$1\le n\le 10$$\end{document}, that no root of unity can be a root of any polynomial Pnf(t)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$P_{n}^{f}(t)$$\end{document} for n∈N\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\in {\mathbb {N}}$$\end{document}. Then we specify the result to some functions f related to colored partitions.

Using the ideas from the paper by Heim, Luca, and Neuhauser, we prove, under some assumptions on P f n (t) for 1 ≤ n ≤ 10, that no root of unity can be a root of any polynomial P f n (t) for n ∈ N. Then we specify the result to some functions f related to colored partitions.

Introduction
A partition of a positive integer n is an expression of the form where n 1 , . . ., n s are positive integers.The history of partitions goes back to Euler, who proved (among other results) that for every positive integer n, the number of partitions of n into odd parts is equal to the number of partitions of n into distinct parts.For more information, see for example [1].
Instead of studying partitions themselves, it is sometimes useful to consider some classes of polynomials related to them.A usual way of defining polynomials related to partitions is the following one.They are defined as the coefficients in the product of the form a∈A 1 1 − t x a , for some non-empty set A ⊆ N, see for example [9].In this paper, we place the variable t in the exponent and define the sequence of polynomials P σ n (t) ∞ n=0 as coefficients of the power series expansion of the following product: Here σ (n) denotes the sum of all divisors of n.
In fact, we will consider a more general situation.Let f : N → Z be an arithmetic function satisfying f (1) = 1.Let us consider the following sequence of polynomials: ( In particular, if f (n) = σ (n), then we get the sequence of polynomials P σ n (t) ∞ n=0 .We will prove this fact later as a special case of Lemma 10.
Studying the products of the form (1), t ∈ N, has a long history.For example, for t = 1, we get the classical generating function for the sequence of ordinary partitions that were found by Euler.His pentagonal number theorem deals with the case of t = −1.The same product with t = −3 is considered in the well-known result by Jacobi, see, for example, [2].The general case was considered, for example, by Serre in [8].
Observe that if t ∈ N, then P σ n (t) is equal to the number of partitions of the number n such that every part can be colored with one among t colors.Interestingly, another type of interpretation of the numbers P σ n (t) can be also given for t ∈ Z, t < 0 through the Nekrasov-Okounkov hook-length formula, see [4].
An important unsolved problem in number theory is to describe the roots of polynomials P σ n (x).Probably, the first non-trivial result in this direction was obtained by Kostant in [7] by methods from the theory of Lie algebras.More precisely, he proved that if r ≥ max{4, n} is a natural number, then P σ n (1 − r 2 ) = 0. Han [4] used the Nekrasov-Okounkov formula in order to extend Kostant's result to all real numbers r ≥ max{4, n}.
A general result for complex roots of polynomials P σ n (t) was obtained by Heim and Neuhauser in [6].They showed that if a polynomial 1  z P σ n (z) is stable (that is, all its roots have negative real parts), then P σ n (z) = 0 for all z with real part less than − 3n(n−1) 2 .
Here we are interested in the problem of determining whether a root of unity can be a root of P f n (x) or not.This was previously studied by Heim et al. in [5] in the case of f (n) = σ (n).We will extend their result to the case of a general arithmetic function f and then we show the connection with partitions.
We can compute exact expressions for P f n (x) for small values of n.

Example 1
We have: We will also need the ordinary generating function of the sequence (P As long as f is fixed, we will write P n (x) instead of P f n (x), and F(x, q) instead of F f (x, q) for simplicity of notation.

Polynomials P f n (x) at roots of unity
Let ζ be a root of unity of order at least 3.The aim of this section is to prove, under some mild assumption, that for every n, we have P n (ζ ) = 0.More precisely, the main result of this section is the following.

Theorem 1 Let f be an arithmetic function such that f (1) = 1 and for every n, the polynomial P n (x) takes integer values at integer arguments. Assume moreover that if
A n (x) := n!P n (x), then 123 1. modulo 5: none of the polynomials A 3 (x) and A 4 (x) is divisible by a monic irreducible polynomial of degree 2 over F 5 .2. modulo 7: none of the polynomials A r (x) for 2 ≤ r ≤ 6 is divisible by a monic irreducible polynomial of degree 4 over F 7 .3. modulo 11: none of the polynomials A r (x) for 2 ≤ r ≤ 10 is divisible by a monic irreducible polynomial over F 11 that divides x 11 6 −1 − 1 and does not divide Then P n (ζ ) = 0 for all roots of unity ζ of order at least 3.
Before we move to the proof of Theorem 1 let us present some results that we will use in the sequel.We follow the idea from [5].
Proof The presented results were showed in [5, Lemma 8] for p ≥ 13.Remaining cases can be checked by hand.

Lemma 4
Let p be a prime and D = {d 1 , . . ., d s } ⊆ {1, . . ., p − 1} be a set of s numbers for some 1 ≤ s ≤ p − 1.For g ∈ N let W g ( p, D) denotes the set of the polynomials h(x) ∈ F p (x) such that: Then we get for p = 5: and Proof This is a long and tedious computation using Mathematica [10].
Lemma 5 Let f be an arithmetic function such that f (1) = 1 and for every n,the polynomial P n (x) takes integer values at integer arguments and denote A n (x) := n!P n (x).Then for every prime p, non-negative integer n, and r ∈ {1, . . ., p}, we have Proof From (2), we get In particular, A n (x) is a monic polynomial of degree n.
Observe that for every prime p, we have: Indeed, we have A p (x) = p!P p (x), and by the assumption, P p (x) is an integer-valued polynomial.Hence, every integer is a root modulo p of A p (x).Moreover, A p (x) is monic of degree p. Consequently, (5) holds.
In order to prove (3), we proceed by induction on m = np + r .Congruence (3) is obviously true for m ∈ {0, . . ., p − 1}.Now, if it is true for all m < np + r , then from (4) and ( 5) we have The result follows.
For primes p ≥ 5, we perform the following reasoning.Let us fix p and assume that we know that q | N for all primes q < p.We want to show that then p | N using Lemma 2 in the same way as in the previous Claims.Let us consider A np+r (x) for n ∈ N 0 and 0 ≤ r ≤ p − 1.By Lemma 5, we can see that it is enough to show that we have modulo p: where the numbers a, K 1 , . . ., K s satisfy the conditions from Lemma 2.
For every 0 ≤ r ≤ p − 1, let us write: where A r ,i (x) are distinct irreducible factors of A r (x) modulo p. Denote d r ,i := deg A r ,i (x).We want to apply Lemma 2 with The only condition that we need to check is that N p d i,r − 1 for all 1 ≤ i ≤ s r .We assumed that q< p q− prime q | N .Thus, if p ≥ 13, then Lemma 3 implies N p d i,r −1 for all 1 ≤ i ≤ s r .If p ∈ {5, 7, 11} and N | p d − 1, then Lemma 3 implies that ( p, d) ∈ {(5, 2), (5, 4), (7,4), (11, 6)}.However, by Lemma 4, we get that A r (x) has to be divisible modulo p by one of the following polynomials over F p : • a monic irreducible polynomial of degree 2 if p = 5, • a monic irreducible polynomial of degree 4 if p = 7, • a monic irreducible polynomial that divides x 11 6 −1 −1 and does not divide All these cases were excluded in the statement.Therefore, if A n (ζ ) = 0 then p | N .We have showed that if ζ is a root of unity of order N ≥ 3 such that A n (ζ ) = 0 for some n then N has to be divisible by all prime numbers.This is an obvious contradiction and thus the proof of Theorem 1 is finished.
Instead of the condition about divisibility of polynomials A r (x) modulo 5, 7, and 11, we could consider only roots of unity of degree N such that 385 = 5 • 7 • 11 | N .Using exactly the same method, we would get the following fact.
Corollary 8 Let f be an arithmetic function such that for every n, the polynomial P n (x) takes integer values at integer arguments.Let N be a number divisible by 385 and let ζ be a root of unity of order N .Then P n (ζ ) = 0 for all n ∈ N 0 .

Application to multicolored partitions
It is proved in [5] that the assumptions of Theorem 1 are satisfied in the case when for all n.Moreover, in this case,the generating function F σ (x, q) has strict connection with the generating function for the sequence of partitions as was explained in the introduction.
Here we prove that the conditions in Theorem 1 are satisfied in the following more general setting.Let C = (c n ) ∞ n=1 be a sequence of non-negative integers such that c 1 = 1.Let us consider If x ∈ N, then P C n (x) is equal to the number of partitions of n such that if k appears as a part, it can be colored with one among c k x colors.This generalizes the case (1) because now the number of colors can be different for different parts.Even more, choosing c k = 0, we restrict ourselves to count only those partitions, in which the number k does not appear as a part.
First of all, we need to find a function f such that F f (x, q) = F C (x, q), that is, In order to do so, we need to find a compact expression for the generating function F f (x, q).
Proof From (2), we get Hence, after dividing both sides by F(x, q) = ∞ n=0 P n (x)q n , we get This finished the proof of the Lemma.

Lemma 10
Let Proof From Lemma 9, we know the expression for the function F σ C (x, q).Hence, This finishes the proof.

123
Lemma 10 implies that the sequence of polynomials P C n (x) ∞ n=0 satisfies the following recurrence: P C 0 (x) = 1, P C 1 (x) = x, and for all n ≥ 2. Note that if x ∈ N, then P C n (x) ∈ N.This follows from the combinatorial interpretation of the numbers P C n (x) for arguments being positive integers.Therefore, we can apply Theorem 1 and Corollary 8 to get the following theorem.
Theorem 11 Let C be fixed and A n (x) := n!P C n (x).Assume that 1. modulo 5: none of the polynomials A 3 (x) and A 4 (x) is divisible by a monic irreducible polynomial of degree 2 over F 5 .2. modulo 7: none of the polynomials A r (x) for 2 ≤ r ≤ 6 is divisible by a monic irreducible polynomial of degree 4 over F 7 .3. modulo 11: none of the polynomials A r (x) for 2 ≤ r ≤ 10 is divisible by a monic irreducible polynomial over F 11 that divides x 11 6 −1 − 1 and does not divide x 11 d −1 − 1 for 1 ≤ d ≤ 10, d = 6.
Then P C n (ζ ) = 0 for all roots of unity ζ of order at least 3.Moreover, if ζ is a root of unity of order divisible by 385 then we can skip the above assumptions on A r (x) for 2 ≤ r ≤ 10.
Proof Follows immediately from Theorem 1 and Corollary 8. Now we take a look at the conditions on polynomials A r (x) for 2 ≤ r ≤ 10.They can be expressed (in general case) as conditions on values f (n) modulo 5, 7, and 11 for a finite number of n's.In the case of f (n) = σ C (n), we can change these conditions further into conditions on a finite number of terms of the sequence C. Unfortunately, it is difficult to list all of these conditions due to the fact that the number of them is very large.For example, there are 1 4 7 4 − 7 2 = 1176 monic irreducible polynomials of degree 4 over F 7 .This is a consequence of a general result that for every n ∈ N and a prime p the number of all monic irreducible polynomials of degree n is equal to This result can be found, for example, in [3, Theorem 4.2.2].
The proof is finished.The proof is finished.

Theorem 1
At first observe that it is enough to show for all roots of unity ζ of order N ≥ 3,the condition A n (ζ ) = 0 instead of P n (ζ ) = 0. Let us assume to the contrary that ζ is a root of unity of order N ≥ 3 and A n (ζ ) = 0 for some n.Claim 6 2 | N .
) p n/d , where μ(d) is the Möbius function defined as