1 Introduction

The zero distribution of the D’Arcais polynomials [1, 3, 16]

$$\begin{aligned} P_{n}^{\sigma _1 }\left( x\right) =\frac{x}{n}\sum _{k=1}^{n} \sigma _1 \left( k\right) P_{n-k}^{\sigma _{1}}\left( x\right) ,\qquad n\ge 1, \end{aligned}$$

with initial value \(P_{0}^{\sigma _1 }\left( x\right) =1\), plays an important role in the study of the vanishing properties of the Fourier coefficients of complex powers of the Dedekind \(\eta \)-function [11, 17, 18]. Let \(\sigma _{d}\left( n \right) =\sum _{\ell \mid n}\ell ^{d}\) and \(q=\textrm{e}^{2\pi \textrm{i}z}\), where z is in the upper complex half plane. The identity

$$\begin{aligned} \sum _{n=0}^{\infty }P_{n}^{\sigma _{1}}\left( x\right) q^{n} =\prod _{n=1}^{\infty }\left( 1-q^{n}\right) ^{-x}=q^{x/24}\eta \left( z\right) ^{-x} \end{aligned}$$

encodes that for \(P_{n}^{\sigma _{1}}\left( x_{0}\right) =0\), the nth coefficient of \(q^{ x_{0}/24}\eta \left( z\right) ^{-x_{0}}\) vanishes. The famous conjecture of Lehmer [13] claims that \(x_{0}\) is never equal to \(-24\).

Let \(x_0 \in {\mathbb {Z}} {\setminus } \left\{ 0\right\} \) be even. Then, Serre [18] proved that the sequence \(\{P_n^{\sigma _{1}}(-x_0)\}_n\) is lacunary if and only if \(x_0 \in S_{\mathop {\textrm{even}}}:=\{2,4,6,8,10, 14,26\}\). Based on the records and results (connection to Maeda’s conjecture) provided in [11] we speculate that if \(n \in {\mathbb {N}}\) exists such that \(P_n^{\sigma _1}(-x_0) =0\) for \(x_0 \ne 0\) even, then \(x_0 \in S_{\mathop {\textrm{even}}}\).

Kostant [12], Theorem 4.28, by using techniques from simple complex Lie algebras, proves that \(P_{n}^{\sigma _{1}}\left( 1-m ^{2}\right) \ne 0\) for \(m\ge n\) and \(m >1\). Han [2] deduced from the Nekrasov–Okounkov hook length formula [15] that this already holds true for real \(\left| x\right| \ge n^{2}-1\).

Fig. 1
figure 1

Absolute values of the zeros of \(P_{n}^{\sigma _{1}}\left( x\right) \) (horizontal axis) with n on the vertical axis, real zeros in blue, non-real zeros in red

Full size image

Numerical computations (Fig. 1) indicate that a linear bound of the form \(\left| x\right| \ge \kappa \left( n-1\right) \) is actually in place. Suppose \(\kappa >0\) exists, such that for all complex x we have the following:

$$\begin{aligned} \left| x\right| \ge \kappa \left( n-1\right) \Rightarrow P_{n}^{\sigma _{1}}\left( x\right) \ne 0. \end{aligned}$$
(1)

Indeed, we have shown [5] that this holds true for \(\kappa \approx 10.82\). On the other hand, Fig. 1 and Table 1 in [8] give the obstruction that \(\kappa \) has to be larger than 9.72245.

In this paper, we prove a theorem on families of polynomials \(\{P_n^{g,h}(x)\}_n\) associated with two arithmetic functions g and h, controlling the growth (Theorem 2.1). As an application, we obtain upper bounds on the possible zeros. This includes Chebyshev polynomials of the second kind, 1-associated Laguerre polynomials, Hermite polynomials, and polynomials related to partitions’ and plane partitions’ numbers, and overpartitions’ numbers. As our most significant application and result we have that (1) holds true for \(\kappa \approx 9.7225\).

2 Statement of the main results

Let \(g:{\mathbb {N}}\rightarrow {\mathbb {C}}\) be normalized with \(g\left( 1\right) =1\) and \(\sum _{n=1}^{\infty } g(n) \, q^{n}\) regular at \( q=0\). Further, let \(h:{\mathbb {N}}\cup \left\{ 0\right\} \rightarrow {\mathbb {C}} \), \( h\left( 0 \right) =0 \) and \( h \left( n\right) \ne 0\) for all \(n\in {\mathbb {N}}\). Let

$$\begin{aligned} H\left( n\right) :=\max _{0\le j\le n}\left| h\left( j\right) \right| . \end{aligned}$$

We define

$$\begin{aligned} P_{n}^{g,h}\left( x\right) :=\frac{x}{h\left( n\right) }\sum _{k=1}^{n}g \left( k\right) P_{n-k}^{g,h}\left( x\right) \end{aligned}$$
(2)

for \(n\ge 1\) and initial value of \(P_0^{g,h}(x)=1\). Then, \(P_{n}^{\sigma _{1},\mathop {\textrm{id}}}\left( x\right) \) are the D’Arcais polynomials and \( P_{n}^{\sigma _{1},\mathop {\textrm{id}}} \left( x+1\right) \) the Nekrasov–Okounkov polynomials [15]. Further, note that \(\text {p}\left( n\right) =P_n^{\sigma _1, \mathop {\textrm{id}}}(1)\) provides the partition function and \(\mathop {\textrm{pp}}(n)=P_n^{\sigma _2, \mathop {\textrm{id}}}(1)\) the plane partition function (we refer to Sect. 5.2.2). In [10], it has been found that for \(g(n)=n \) and \(h(n) >0\) for \(n\ge 1\), we obtain orthogonal polynomials. The case \(h(n)=1\) leads to Chebyshev polynomials \(U_{n}\left( x\right) \) of the second kind and the case \(h(n)=n\) leads to 1-associated Laguerre polynomials \(L_{n}^{1}\left( x\right) \). We have \(P_{n}^{\mathop {\textrm{id}},\mathop {\textrm{id}}}\left( x\right) =\frac{x}{n}L_{n-1}^{1}\left( -x\right) \) and \(P_{n}^{\mathop {\textrm{id}},1}\left( x\right) =xU_{n-1}\left( \frac{x}{2}+1\right) \) (see also [4, 8]).

Theorem 2.1

Let \(c\in {\mathbb {C}}\) and let \(R>0\) be the radius of convergence of

$$\begin{aligned} G_{c }\left( q\right) :=\sum _{k=1 }^{\infty }\left| g\left( k+1\right) -c g\left( k\right) \right| q^{k}. \end{aligned}$$

Let \(q_{1}=\sup \left\{ 0\le q<R:G_{c}\left( q\right) <1\right\} \). For \(0<q<q_{1 }\le R\) let

$$\begin{aligned} \kappa =\kappa _{q}=\kappa _{c,q}^{g}:=\frac{1}{1-G_{c}\left( q\right) }\left( \frac{1}{q}+\left| c \right| \right) . \end{aligned}$$

Then for all \(n \in {\mathbb {N}}\) and \(x\in {\mathbb {C}}\) with \(\left| x \right| \ge {\kappa } \, H\left( n-1\right) \), we have:

$$\begin{aligned} \left| P_{n}^{g,h}\left( x\right) - \frac{x+c \, h\left( n-1\right) }{h\left( n\right) } P_{n-1}^{g,h}\left( x\right) \right| \le G_{c}\left( q\right) \frac{\left| x\right| }{H\left( n\right) }\left| P_{n-1}^{g,h}\left( x\right) \right| . \end{aligned}$$

An application of the triangle inequality shows the following:

Corollary 2.2

Assumptions as in Theorem 2.1. Then

$$\begin{aligned}{} & {} \left( \frac{\left| x+c \, h\left( n-1\right) \right| }{\left| h \left( n\right) \right| }-\frac{G_{c}\left( q\right) \left| x\right| }{H \left( n\right) }\right) \left| P_{n-1}^{g,h}\left( x\right) \right| \\{} & {} \quad \le \left| P_{n}^{g,h}\left( x\right) \right| \le \left( \frac{ \left| x+c h\left( n-1\right) \right| }{\left| h\left( n\right) \right| }+\frac{G_{c}\left( q\right) \left| x\right| }{ H\left( n\right) }\right) \left| P_{n-1}^{g,h}\left( x\right) \right| \end{aligned}$$

if \(\left| x \right| \ge {\kappa } H\left( n-1\right) \), \(n\ge 1\).

From Corollary 2.2 we obtain:

Corollary 2.3

Assumptions as in Theorem 2.1. We have \(P_{n}^{g,h}\left( x\right) \ne 0\) for \(\left| x\right| \ge \kappa \, H \left( n-1\right) \), \(n\ge 1\).

Proof

We have \(P_{1 }^{g,h}\left( x\right) = \frac{x}{h\left( 1\right) }\ne 0\) for \(x\ne 0\). By induction, and also as we have shown for \(\left| x\right| \ge \kappa H \left( n-1\right) \) in the preceding

$$\begin{aligned} \left| P_{n}^{g,h}\left( x\right) \right|\ge & {} \left( \frac{ \left| x +c \, h\left( n-1\right) \right| }{ H\left( n\right) }-G_{c}\left( q\right) \frac{\left| x\right| }{H\left( n\right) }\right) \left| P_{n-1}^{g,h}\left( x\right) \right| \\\ge & {} \left( \frac{ \left| x\right| -\left| c \, H\left( n-1\right) \right| }{ H\left( n\right) }-G_{c}\left( q\right) \frac{\left| x\right| }{H\left( n\right) }\right) \left| P_{n-1}^{g,h}\left( x\right) \right| \end{aligned}$$

and \(\left( 1-G_{c}\left( q\right) \right) \left| x\right| -\left| c \right| H \left( n-1\right) \ge \frac{H \left( n-1\right) }{q }>0\) for \(\left| x\right| \ge \kappa H \left( n-1\right) \), \(n\ge 2\). \(\square \)

Corollary 2.4

Let g, h be normalized non-negative arithmetic functions with h positive. Suppose that \(\sum _{n=1}^{\infty } g\left( n\right) \, q ^{n}\) is regular at \(q =0\). Suppose that \(\frac{1}{x} P_n^{g,h}(x)\) is a Hurwitz polynomial and let \(\kappa ^{g}\) be given (Corollary 2.3). Then, there exists a zero \(\alpha _n^{g,h}\) of \(P_n^{g,h}(x)\) such that

$$\begin{aligned} \frac{g\left( 2\right) }{n-1}\,\, \sum _{k=1}^{n-1} \,\,h(k) \, \le \, \left| \mathop {\textrm{Re}}\alpha _n^{g,h}\right| ,\qquad \left| \alpha _{n}^{g,h}\right| <\, \kappa ^{g}H\left( n-1\right) . \end{aligned}$$
(3)

A polynomial with real and non-negative coefficients is called Hurwitz polynomial if all zeros have negative real part. D’Arcais polynomials, up to the zero \(x=0\), possess this property for \(n \le 1500\) (we refer to [7], Section 5.3).

Proof

It follows from (2) that the sum of all zeros is equal to

$$\begin{aligned} -g(2) \, \, \sum _{k=1}^{n-1} \ h(k). \end{aligned}$$

Therefore, there exists a zero of \(P_n^{g,h}(x)\), due to the Hurwitz property and Corollary 2.3, satisfying (3). \(\square \)

If the D’Arcais polynomial \(P_n^{\sigma _1}(x)\) is Hurwitz this implies there exists a zero \(\alpha _n\) with

$$\begin{aligned} \frac{3 \, n}{2} \le \left| \mathop {\textrm{Re}}\alpha _n \right| ,\qquad \left| \alpha _{n}\right| <9.7225 \, (n-1). \end{aligned}$$

For the value of \(\kappa ^{\sigma _1}\) we refer to Corollary 4.1 and Corollary 4.6.

Remark 2.5

Note that for fixed c the function

$$\begin{aligned} q\mapsto \kappa _{ c,q}^{g}=\frac{1}{1-G_{c }\left( q\right) }\left( \frac{1}{q}+\left| c \right| \right) \end{aligned}$$

is convex on \(\left( 0,q_{1}\right) \), as

$$\begin{aligned} \kappa =\sum _{m=0}^{\infty }\left( G_{c}\left( q\right) \right) ^{m}\left( \frac{1}{q}+\left| c\right| \right) =\frac{1}{q}+F\left( q\right) \end{aligned}$$

where F is a power series in q whose coefficients are not negative. This implies that the second derivative \(\left( \frac{\partial }{\partial q}\right) ^{2}\kappa =\frac{2}{q^{3}}+ F^{\prime \prime }\left( q\right) >0\) on \(\left( 0,q_{1}\right) \).

Before providing the proof of Theorem 2.1, we illustrate how to utilize our main result towards the finding of upper bounds for the zeros with the largest absolute value.

Example 2.6

Let \(b>0\), \(g\left( n\right) =b^{n-1}\), and \(h\left( n\right) =1\) for all n (except 0). We choose \(c=b\) and for arbitrary \(q>0 \) we obtain

$$\begin{aligned} \left| P_{n}^{g,1}\left( x\right) -\left( x+b\right) P_{n-1}^{g,1}\left( x\right) \right| =0 \end{aligned}$$
(4)

for \(\left| x\right| \ge \kappa =\kappa _{q }\) where \(G_{b}\left( q\right) =0\) and \(\kappa =\frac{1}{1-G_{b}\left( q\right) }\left( \frac{1}{q }+c \right) =\frac{1}{1-0 }\left( \frac{1}{q }+b\right) \).

Therefore, q can be chosen arbitrarily large and \(\kappa \rightarrow b \). As \(P_{n}^{g,1}\left( x\right) =\left( x+b\right) ^{n-1}x\) for all \(n\ge 1\) equality (4) certainly holds always true.

From the theorem we then also find that \(P_{n}^{g,1}\left( x\right) \ne 0\) for all \(\left| x\right| \ge \kappa _{q}\) without knowing the explicit form of \(P_{n}^{g,1}\left( x\right) \). Therefore, in this case, the estimate is sharp as \(\kappa _{q} \rightarrow b\) for \(q \rightarrow \infty \).

Example 2.7

Let \(g\left( k\right) =3-2\delta _{1,k}\). We choose \(c=1\). Therefore, \(G_{1}\left( q\right) =2q\) and \(\kappa =\frac{1}{1-2q}\left( \frac{1}{q}+1\right) \) for any \(0<q<1/2=q_{1}\). Then

$$\begin{aligned} \frac{\partial }{\partial q}\kappa =\frac{2}{\left( 1-2q\right) ^{2}}\left( \frac{1}{q}+1\right) -\frac{1}{1-2q}\frac{1}{q^{2}}=0, \end{aligned}$$

if and only if \(0=2q^{2}+4q-1\). Thus, \(q=\frac{-4+\sqrt{16+8 }}{4 }=-1 +\sqrt{3/2 }\) and

$$\begin{aligned} \kappa =5 +2 \sqrt{6} \approx 9.89898. \end{aligned}$$

Note, that this value is actually the best possible for \(h\left( n\right) =1\), compare [8], Corollary 4.3 and Remark 4.4.

Example 2.8

Let \(g\left( k\right) =k\). For positive \(h\left( n\right) >0\), the sequence of polynomials \(P^{\mathop {\textrm{id}},h}\left( x\right) \) is related to orthogonal polynomials [10]. Then for \(c=\frac{3}{2}\) we obtain

$$\begin{aligned} G_{3/2}\left( q\right) = \frac{q-2q^{2}+2q^{3}}{2\left( 1-q\right) ^{2}}. \end{aligned}$$

Therefore, \(\kappa = \frac{\left( 3q+2 \right) \left( q-1\right) ^{3}}{-\left( 2q^{3} - 4q^{2} + 5q - 2\right) q}\) and

$$\begin{aligned} \frac{\partial }{\partial q}\kappa =\frac{\left( q-1\right) \left( 6q^{5} - 10q^{4} - 15q^{3} + 21q^{2} - 16q + 4\right) }{\left( 2q^{4} - 4q^{3} + 5q^{2} - 2q\right) ^{2}}. \end{aligned}$$

The relevant zero of the numerator is located at \(q\approx 0.37609\) and yields a minimal value of \(\kappa \approx 5.5928\).

Remark 2.9

To verify our speculation and to determine also all other integer zeros of \(P_n^{\sigma _{1}}(x)\) the results of Kostant [12] and Han [2] demand to check the values of \(P_n^{\sigma _{1}}(-m)\) for \(1 \le m \le n^2\). The results from [5] and this paper reduce the amount of values to \(1 \le m \le 9.7225 \, (n-1)\).

3 Proof of Theorem 2.1

We provide the proof by induction on n. The base case \(n=1\) holds true:

$$\begin{aligned} 0 = \left| P_{1}^{g,h}\left( x\right) - \frac{x+c h\left( 0\right) }{h\left( 1\right) }P_{0}^{g,h}\left( x\right) \right| \le G_{c }\left( q\right) \frac{\left| x\right| }{H\left( 1\right) }\left| P_{0}^{g,h}\left( x\right) \right| \end{aligned}$$

for \(\left| x\right| \ge \kappa H \left( 0\right) \). Let now \(n\ge 2\). Then

$$\begin{aligned} P_{n}^{g,h}\left( x\right) -c \frac{h\left( n-1\right) }{h\left( n\right) }P_{n-1}^{g,h}\left( x\right) \!=\!\frac{x}{h\left( n\right) }\left( P_{n-1}^{g,h}\left( x\right) \!+\!\sum _{k= 1}^{n-1}\left( g\left( k+1\right) \!-\!c g\left( k\right) \right) P_{n-1-k}^{g,h}\left( x\right) \right) . \end{aligned}$$

The basic idea for the induction step is to use the inequality

$$\begin{aligned}{} & {} \left| P_{n}^{g,h}\left( x\right) -\frac{x+c h\left( n-1\right) }{h\left( n\right) }P_{n-1}^{g,h}\left( x\right) \right| \le \left| \frac{ x }{h\left( n\right) }\right| \sum _{k= 1}^{n-1}\left| g \left( k+1\right) -c g\left( k\right) \right| \left| P_{n-1-k}^{g,h}\left( x\right) \right| . \end{aligned}$$

To estimate the sum, we apply the induction hypothesis for \(1\le j\le n-1\) to show that

$$\begin{aligned} \left| P_{j}^{g,h}\left( x\right) \right|{} & {} \ge \left| \frac{x+c h\left( j-1\right) }{h\left( j\right) }\right| \left| P_{j-1}^{g,h}\left( x\right) \right| -\left| P_{j}^{g,h}-\frac{x+c h\left( j-1\right) }{h\left( j\right) }P_{j-1}^{g,h}\left( x\right) \right| \\{} & {} \ge \left( \frac{\left| x\right| -\left| c h\left( j-1\right) \right| }{\left| h\left( j\right) \right| }-G_{c }\left( q\right) \frac{\left| x\right| }{H\left( j\right) } \right) \left| P_{j-1}^{g,h}\left( x\right) \right| \\{} & {} \ge \left( \frac{\left| x\right| -\left| c \right| H\left( j\right) }{\left| h\left( j\right) \right| }-G_{c }\left( q\right) \frac{\left| x\right| }{H\left( j\right) }\right) \left| P_{j-1}^{g,h}\left( x\right) \right| \\{} & {} \ge \frac{\left( 1-G_{c }\left( q\right) \right) \left| x\right| -\left| c \right| H \left( j\right) }{H \left( j\right) }\left| P_{j-1}^{g,h}\left( x\right) \right| \end{aligned}$$

for \(\left| x\right| \ge \kappa H\left( j-1\right) \). Note, that for \(\left| x\right| \ge \kappa H \left( n-1\right) \) we have

$$\begin{aligned} \left( 1-G_{c }\left( q\right) \right) \left| x \right| - \left| c \right| H \left( j\right) \ge \left( \frac{1}{q }+\left| c \right| \right) H \left( n-1\right) - \left| c \right| H \left( j\right) >0. \end{aligned}$$

Therefore,

$$\begin{aligned} \left| P_{j-1}^{g,h}\left( x\right) \right| \le \frac{H\left( j\right) }{\left( 1-G_{ c}\left( q\right) \right) \left| x\right| - \left| c \right| H \left( j\right) }\left| P_{j}^{g,h}\left( x\right) \right| . \end{aligned}$$

Iterating this inequality leads to

$$\begin{aligned} \left| P_{n-k}^{g,h}\left( x\right) \right|\le & {} \left| P_{n-k+1}^{g,h}\left( x\right) \right| \frac{H\left( n-k+1\right) }{\left( 1-G_{c}\left( q\right) \right) \left| x\right| -\left| c \right| H \left( n-k+1\right) }\le \ldots \\\le & {} \left| P_{n-1}^{g,h}\left( x\right) \right| \prod _{j=1}^{k-1}\frac{H\left( n-j\right) }{\left( 1-G_{c }\left( q\right) \right) \left| x\right| -\left| c \right| H \left( n-j\right) } \\\le & {} \left| P_{n-1}^{g,h}\left( x\right) \right| \left( \frac{ H\left( n-1\right) }{\left( 1-G_{ c}\left( q\right) \right) \left| x \right| -\left| c \right| H \left( n-1\right) }\right) ^{k-1} \end{aligned}$$

for \(\left| x\right| \ge \kappa H\left( n-1\right) \ge \kappa H\left( n-k\right) \) for all \(k=2,\ldots ,n\) by definition. We estimate the sum by

$$\begin{aligned}{} & {} \sum _{k=1 }^{n-1}\left| g \left( k+1\right) -c g\left( k\right) \right| \left| P_{n-1-k}^{g,h}\left( x\right) \right| \\{} & {} \quad \le \left| P_{n-1}^{g,h}\left( x\right) \right| \sum _{k= 1}^{n-1}\left| g \left( k+1\right) -c g\left( k\right) \right| \left( \frac{H\left( n-1\right) }{\left( 1-G_{c }\left( q\right) \right) \left| x\right| -\left| c \right| H \left( n-1\right) }\right) ^{k} \end{aligned}$$

and obtain

$$\begin{aligned}{} & {} \left| P_{n}^{g,h}\left( x\right) -\frac{x+c h\left( n-1\right) }{h\left( n\right) }P_{n-1}^{g,h}\left( x\right) \right| \\{} & {} \quad \le \left| \frac{ x }{h\left( n\right) }\right| \left| P_{n-1}^{g,h}\left( x\right) \right| \sum _{k= 1}^{n-1}\left| g\left( k+1\right) -c g \left( k\right) \right| \left( \frac{H\left( n-1\right) }{\left( 1-G_{c }\left( q\right) \right) \left| x\right| -\left| c \right| H \left( n-1\right) }\right) ^{k}. \end{aligned}$$

By estimating the sum using the assumption from the theorem, we obtain

$$\begin{aligned}{} & {} \sum _{k=1 }^{n-1}\left| g\left( k+1\right) -c g \left( k\right) \right| \left( \frac{ H\left( n-1\right) }{\left( 1-G_{c }\left( q\right) \right) \left| x\right| -\left| c \right| H \left( n-1\right) }\right) ^{k} \\{} & {} \quad \le G_{ c} \left( \frac{H \left( n-1\right) }{\left( 1-G_{c }\left( q\right) \right) \left| x\right| -\left| c \right| H \left( n-1\right) }\right) \le G_{ c} \left( q \right) \end{aligned}$$

since

$$\begin{aligned} \left| x\right| \ge {\kappa }H \left( n-1\right) =\frac{H \left( n-1\right) }{1-G_{ c}\left( q\right) }\left( \frac{1}{q }+\left| c \right| \right) \end{aligned}$$

which is equivalent to \(\frac{\left( 1-G_{ c}\left( q\right) \right) \left| x\right| }{H \left( n-1\right) }-\left| c \right| \ge \frac{1}{q }\) and \(G_{c}\) is increasing on \(\left[ 0,R\right) \) as \(\left| g\left( k+1\right) -c g\left( k\right) \right| \ge 0\) for all \(k\in {\mathbb {N}}\). Finally, this proves the theorem.

4 D’Arcais polynomials

One of the main applications of Theorem 2.1 provides the following comparison of the growth of \(P_n^{\sigma _1, h}\left( x\right) \) and \(P_{n-1}^{\sigma _1, h}\left( x\right) \). Here, the function h can be considered as a deformation.

Corollary 4.1

Let \(g=\sigma _1\). Suppose \(\left| x \right| \ge \frac{107}{11} H \left( n-1 \right) \) for all \(n\ge 1\). Then

$$\begin{aligned} \left| P_{n}^{\sigma _{1},h}\left( x\right) -\frac{x+4 h\left( n-1\right) /3}{h\left( n\right) }P_{n-1}^{\sigma _{1},h}\left( x\right) \right| \le \frac{2}{5}\, \frac{\left| x\right| }{ H \left( n\right) } \, \left| P_{n-1}^{\sigma _{1},h}\left( x\right) \right| . \end{aligned}$$

In particular, \(P_{n}^{\sigma _{1},h}\left( x\right) \ne 0\) for \(\left| x\right| \ge \frac{107 }{11}H \left( n-1\right) \), \(n\ge 2\).

For the proof, the following lemma will be useful. Note, that we want to apply our Theorem 2.1 for \(c=4/3\).

Lemma 4.2

\(\left| \frac{4}{3}\sigma _{1}\left( k\right) -\sigma _{1}\left( k+1\right) \right| \le \frac{1}{2}\left( {\begin{array}{c}k+2\\ 2\end{array}}\right) \) for all \(k\ge 2\).

Proof

It can be checked that \(\left| \frac{4}{3}\sigma _{1}\left( k\right) -\sigma _{1}\left( k+1\right) \right| \le \frac{1}{2}\left( {\begin{array}{c}k+2\\ 2\end{array}}\right) \) for all \(2\le k\le 42\) (see Table 1 for \(k\le 1 3\)). For \(k\ge 43\) holds \(1+\ln \left( k+1\right) \le \frac{k+1}{44 }+\ln \left( 44\right) \le \frac{3}{7}\frac{k+2}{4}\) as \(\exp \left( 3\cdot 45/28-1\right) >44\). Therefore,

$$\begin{aligned} \left| \frac{4}{3}\sigma _{1}\left( k\right) -\sigma _{1}\left( k+1\right) \right|\le & {} \frac{7}{3}\left( 1+\ln \left( k+1\right) \right) \left( k+1\right) \\\le & {} \frac{7}{3}\left( \frac{k+1}{44 }+\ln \left( 44\right) \right) \left( k+1\right) \le \frac{1}{2}\left( {\begin{array}{c}k+2\\ 2\end{array}}\right) . \end{aligned}$$

\(\square \)

Remark 4.3

We can now easily demonstrate that \(\kappa <10\). We have

$$\begin{aligned} G_{4/3}\left( q\right)\le & {} \frac{5}{3}q+\frac{5}{3}q^{3}+\frac{1}{2}\sum _{k=4}^{\infty }\left( {\begin{array}{c}k+2\\ 2\end{array}}\right) q^{k} \\= & {} \frac{1}{2\left( 1-q\right) ^{3}}-\frac{1}{2}+\frac{1}{6}q-3q^{2}-\frac{10}{3}q^{3} \end{aligned}$$

for \(0<q <1\). For \(q=\frac{4}{19}<q_{1}\), we obtain

$$\begin{aligned} \kappa =\frac{1}{1-G_{4/3}\left( q\right) }\left( \frac{1}{q}+\frac{4}{3}\right) =\frac{563295375}{56745988}<10. \end{aligned}$$

Remark 4.4

In the proof of Corollary 4.1, we will use the following notation. For \(K\in {\mathbb {N}}\), let \(\gamma \left( k\right) =\left| \sigma _{1}\left( k+1\right) -4\sigma _{1}\left( k\right) /3\right| \) for \(1\le k\le K-1\) and \(\gamma \left( k\right) =\frac{1}{2}\left( {\begin{array}{c}k+2\\ 2\end{array}}\right) \) for \(k\ge K\). (Note that \(\gamma \left( 2\right) =0\) see Table 1.)

Then, \(\left| \sigma _{1}\left( k+1\right) -\frac{4}{3}\sigma _{1}\left( k\right) \right| \le \gamma \left( k\right) \) for all \(k\ge 1\) by Lemma 4.2. Let \(M_{K} \left( q\right) =\sum _{k=1}^{\infty } \gamma \left( k\right) q^{k}\). The power series \(M_{K}\) is almost (except for the first K terms) a multiple of the second derivative of the geometric series in q.

Proof of Corollary 4.1

We have

$$\begin{aligned} G_{4/3}\left( q \right) =\sum _{k=1 }^{\infty }\left| \sigma _{1}\left( k+1\right) -\frac{4}{3}\sigma _{1}\left( k\right) \right| q ^{k}. \end{aligned}$$

We use the notation from Remark 4.4 with \(K=9\) and estimate

$$\begin{aligned} G_{4/3}\left( q\right) \le M_{9} \left( q\right) \end{aligned}$$

for \(0\le q<q_{0}=\sup \left\{ 0\le q<R:M_{9}\left( q\right) <1\right\} \le q_{1}\le R\). Therefore,

$$\begin{aligned} G_{4/3} \left( q\right) \le \frac{1}{2\left( 1-q\right) ^{3}}-\frac{1}{2} +\frac{1}{6}q- 3q^{2}- \frac{10}{3}q^{3}-\frac{25}{6}q^{4}-\frac{13}{2}q^{5}-6q^{6}-\frac{41}{3}q^{7}-\frac{31}{2}q^{8}. \end{aligned}$$

For \(q= \frac{11 }{50 }\) we obtain

$$\begin{aligned} G_{4/3}\left( \frac{11}{50}\right) = \frac{1832317044472581491}{4634296875000000000} \end{aligned}$$

and

$$\begin{aligned}{} & {} \frac{1}{1-G_{4/3} \left( q \right) }\left( \frac{1}{q}+\frac{4}{3}\right) \\{} & {} \quad \le \frac{\frac{1}{q}+\frac{4}{3}}{\frac{1}{2\left( 1-q \right) ^{3}}-\frac{1}{2}+\frac{1}{6} q -3 q^{2}-\frac{10}{3}q^{3}-\frac{25}{6} q^{4}-\frac{13}{2} q^{5}-6 q^{6}-\frac{41}{3} q^{7}-\frac{31}{2}q^8} \\{} & {} \quad =\frac{299684531250000000000}{30821778135801603599}<\frac{107}{11}=9.{\overline{72}}. \end{aligned}$$

The claim now follows from Theorem 2.1. \(\square \)

Remark 4.5

Let us choose \(c=4/3\) and \( K=10 \). Then, \(q=0.22\) leads to \(\kappa \approx 9.723\) by applying the same method.

Corollary 4.6

(of Theorem 2.1) Let \(c=4/3\) and \( K=13 \). Then, \(q=0.22\) leads to \(\kappa \approx 9.7225\).

We leave the related calculations to the interested reader. Note, that the values of \(\kappa \) apply to all admissible functions h, in particular to \(h\left( n\right) =1\), compare [8], Table 1. Therefore, we can conclude that \(\kappa \) is very close to the best possible value.

Table 1 Values of \(\left| \sigma _{1}\left( k+1\right) -\frac{4}{3}\sigma _{1}\left( k\right) \right| \) and related binomial coefficients
Full size table

Remark 4.7

Consider the original problem with the sum of divisors function \(\sigma _{1}=g\) and h the identity. It is worth noting that this allows to improve Corollary 3 of [5] from \(\kappa =\frac{119}{11} \) to \(\kappa =\frac{107 }{11}<10 \) by Corollary 2.3 and by Corollary 4.6 to \(\kappa \approx 9.7225\).

5 Final remarks

5.1 The geometric case \(h(n)=1\)

A special case of Theorem 2.1 leads to the following non-vanishing result:

Corollary 5.1

Let g be a normalized arithmetic function. Suppose that \(\textrm{G}\left( q \right) := \sum _{n=1}^{\infty } g\left( n\right) \, q^{n}\) is regular at \(q =0\) with radius of convergence \(\textrm{R} \). Then,

$$\begin{aligned} \sum _{n=0}^{\infty } P_{n}^{g,1}\left( x\right) \, q ^{n} = \frac{1}{1 - x \, \textrm{G}\left( q\right) }. \end{aligned}$$
(5)

Let \(c \in {\mathbb {C}}\) and \(0<q < R\) with \(G_{c}\left( q\right) <1\). Then, \(P_{n}^{g,1}\left( x\right) \ne 0\) for all \(n\ge 1\) and complex numbers x with \(\vert x \vert \ge \kappa _{q}\), where

$$\begin{aligned} \kappa _{q }=\frac{1}{1-G_{ c}\left( q\right) }\left( \frac{1}{q }+\left| c\right| \right) . \end{aligned}$$

The identity (5) already appeared in [6].

5.1.1 Sign changes and reciprocal of Eisenstein series

In [6] we studied properties of the q-expansion of reciprocals of Eisenstein series \(E_m\) of weight m. We have determined for \(g_m(n):= \sigma _{m-1}(n)\) the associated \(\kappa _{m}\), m even, (here we follow the notation given in [6]) to examine if

$$\begin{aligned} \left| \frac{2m}{B_m} \right| > \kappa _m, \end{aligned}$$
(6)

where \(B_m\) is the mth Bernoulli number. Let (6) be satisfied, then all the coefficients of \(1/E_m\) are non-vanishing. If \(B_m\) is negative, then the coefficients of \(1/E_m\) have alternating signs. For example, this is the case for \(1/E_4\).

With the method provided in this paper, we can improve the \(\kappa _m\). These are recorded in Table 2.

Table 2 Values \(\kappa _{m}^{\textrm{new}}\) for \(g\left( k\right) =\sigma _{m-1}\left( k\right) \) and \(\kappa _{m}\) from [6]
Full size table

5.1.2 Chebyshev polynomials

Let \(U_n(x)\) be the Chebyshev polynomials of the second kind. Let \(g(1)=g(2)=1\) and let \(g(n)=0\) for \(n \ge 3\). Then [9], Section 3.4 shows that

$$\begin{aligned} \sum _{n=0}^{\infty }P_{n}^{g,1}\left( -x^{2}\right) q^{n}= & {} \left( 1 -2\left( x/2\right) \left( -xq\right) +\left( -xq\right) ^{2} \right) ^{-1} =\sum _{n=0}^{\infty }U_{n}\left( x/2\right) \left( -x\right) ^{n}q^{n}. \end{aligned}$$

Then \(G_{c}\left( q\right) =\left| 1-c\right| q+\left| c\right| q^{2}\) and \(\kappa =\frac{1}{1-G_{c}\left( q\right) }\left( \frac{1}{q}+\left| c\right| \right) \). Assume \(0\le c\le 1\) then

$$\begin{aligned} \kappa =\frac{1}{1-\left( 1-c\right) q-cq^{2}}\left( \frac{1}{q}+c\right) = \frac{1}{q-q^{2}}, \end{aligned}$$

which attains its minimum at \(q=\frac{1}{2}\). For this q, we obtain \(\kappa =4\). This implies that \(U_n(x/2) \ne 0\) for \(\left| -x^2 \right| \ge 4\) which implies, that \(U_n(x) \ne 0\) if \(\vert x \vert \ge 1\). This shows that \(\kappa =4\) is the best possible.

5.2 The exponential case \(h(n)=n\)

5.2.1 Hermite polynomials

Let \(H_0(x)=1\),   \(H_1(x)=2x\),   \( H_2(x) = 4x^2 -2\),   \(H_3(x) = 8x^3-12x\), and more generally,

$$\begin{aligned} H_n(x):= (-1)^n \, \textrm{e}^{x^2} \, \frac{\textrm{d}^n}{\textrm{d}x^n} \textrm{e}^{-x^2} \end{aligned}$$

denote the Hermite polynomials. Let \(g(1)=g(2)=1\) and \(g(n)=0\) for \(n \ge 3\). Then, \(P_n^{g,\mathop {\textrm{id}}}(-2x^2) = (-x)^n \, H_n(x)/(n!).\) This follows from [9], Section 3.4. We have already obtained \(\kappa =4\), which is optimal. This implies that

$$\begin{aligned} H_n(x)\ne 0 \text { for }\left| x \right| \ge \sqrt{2} \, \sqrt{n-1}. \end{aligned}$$

5.2.2 Plane partitions and overpartitions

Let \(g(n)=\sigma _2(n)\). Then, the polynomials \(P_n^{\sigma _2,\mathop {\textrm{id}}}(x)\) interpolate the plane partition function \(\mathop {\textrm{pp}}(n)\). We have \(P_n^{\sigma _2,\mathop {\textrm{id}}}(1)=\mathop {\textrm{pp}}(n)\). With the method presented in this paper, we can demonstrate that \(P_n^{\sigma _2, \mathop {\textrm{id}}}(x) \ne 0\) for \(\left| x \right| \ge \kappa \, (n-1)\), where \(\kappa = 16.022\).

Let \(\bar{\textrm{p}}\left( n\right) \) be the number of overpartitions of n (we refer to [14] for recent results). Let \(g(n) = \sigma _1(n) - \sigma _1(n/2)\), where \(\sigma _1(y)=0\) for \(y \not \in {\mathbb {N}}\). Then, \(P_n^{g, \mathop {\textrm{id}}}(1) = \bar{\textrm{p}}\left( n\right) \). With the method of this paper, we obtain \(P_n^{g, \mathop {\textrm{id}}}(x) \ne 0\) for \(\vert x \vert \ge \kappa \, (n-1)\), where \(\kappa =6. 2133\).