## 1 Introduction and main results

In this paper, we study Turán inequalities $$p_n(x)^2 -p_{n-1}(x) \, p_{n+1}(x) \ge 0$$ for families of polynomials $$\{p_n(x)\}_n$$ attached to arithmetic functions.

Our work is motivated by a recent result by Ono et al. [8, 13]. Ono et al. proved the log-concavity conjecture ([9], Conjecture 1) for plane partitions $${{\textrm{pp}}}\left( n\right)$$ for $$n >11$$. Twenty-four years ago, Nicolas [12] had already proved the log-concavity property for the partition numbers p(n) for $$n >25$$. This result was reproved by DeSalvo and Pak [5]. For an introduction to partition numbers and plane partition numbers, we refer to Andrews’ book [1]. Further, to study the concept of log-concavity and related topics, Brenti [2] and Stanley [14, 15] are suitable references.

This paper is also a significant generalization of our previous result in [8]. Let $$a_d(n):= \frac{1}{n} \sum _{k=1}^n \left( \sum _{\ell \vert k} \ell ^d \right) \, a_d(n-k)$$, with $$a_d(0)=1$$. Let $$n \ge 6$$ be fixed. Then the sequence $$\{a_d(n)\}_n$$ is log-concave at n for almost all $$d\in {\mathbb {N}}$$ if and only if $$n \equiv 0 \pmod {3}$$. Note that $$p(n)= a_1(n)$$ and $${{\textrm{pp}}}(n)= a_2(n)$$. The quantities p(n) and $${{\textrm{pp}}}\left( n\right)$$ are induced by certain arithmetic functions. This leads to the following generalization.

### Definition

Let $${\mathbb {D}}$$ be the set of all double sequences $$\{g_d(n)\}_{d,n \ge 1}$$ with normalization $$g_d(1)=1$$, such that $$\sum _{n=1}^{\infty } g_d(n) \, q^{n-1}$$ is regular at $$q=0$$ with radius of convergence R, and

\begin{aligned} 0 \le g_{d}\left( n\right) -n^{d} \le g_{1} \left( n\right) \, \left( n-1\right) ^{d-1}, \end{aligned}

for all d and n.

We investigate sequences of polynomials $$\{P_n^{g_d}(x)\}_n$$, defined by the recurrence relation:

\begin{aligned} P_n^{g_d}(x) := \frac{x}{n} \, \sum _{k=1}^{n} g_d(k) \, P_{n-k}^{g_d}(x), \qquad \text {with } P_0^{g_d}(x):=1. \end{aligned}

We have the generating series

\begin{aligned} \sum _{n=0}^{\infty } P_n^{g_d}(x) \, q^n = {{\textrm{exp}}}\left( x \sum _{n=1}^{\infty } g_d(n) \frac{q^n}{n} \right) = \prod _{n=1}^{\infty } \left( 1 - q^n \right) ^{-x \, f_d(n)}, \end{aligned}

where $$n \, f_d(n) = \sum _{\ell \mid n } \mu (\ell ) \, g_d(n/ \ell )$$ with $$\mu$$ the Moebius function. Examples for $$g_d(n)$$ are $$\{\sigma _d(n)\}$$ and $$\{\psi _d(n)\}$$, where $$\sigma _d(n)= \sum _{\ell \mid n} \ell ^d$$ and $$\psi _d(n)= n^d$$. Turán’s inequality of $$\{P_n^{g_d}(x)\}$$ at n for a subset of $${\mathbb {R}}$$ is defined by

\begin{aligned} \Delta _{n}^{g_d}(x) := \left( P_n^{g_d}(x) \right) ^2 - P_{n-1}^{g_d}(x) \, P_{n+1}^{g_d}(x) \ge 0. \end{aligned}

Let $$x_0$$ be fixed, we call $$P_n^{g_d}(x_0)$$ log-concave at n if $$\Delta _{n}^{g_d}(x_0)\ge 0$$.

We note that the partition function and the plane partition function satisfy $$p(n)= P_{n}^{\sigma _1}(1)$$ and $${{\textrm{pp}}}\left( n\right) = P_n^{\sigma _2}(1)$$. Let $$E^{g_d}$$ be the set of all $$n \in {\mathbb {N}}$$ with $$\Delta _{n}^{g_d}(1)<0$$, denoted as strictly log-convex.

Nicolas [12] proved that the partition function p(n) is log-concave for almost all n. The set of exceptions is given by $$E^{\sigma _{1}}= \left\{ 2k +1 \, : \, 0 \le k \le 12\right\}$$. Ono et al. [13] proved that $$E^{\sigma _2}= \{1,3,5,7,9,11\}$$. Numerical investigations [8] for $$n \le 10^5$$ indicate that $$E^{\sigma _3}= \{1,3,5,7\}$$, $$E^{\sigma _4}= E^{\sigma _5}=\{1,5\}$$. Surprisingly, $$E^{\sigma _{20}}$$ has at least 10 elements. We believe that the general and clean patterns associated to double sequences in $${\mathbb {D}}$$ are displayed by $$g_d(n)= \psi _d(n)$$ (see Table 1).

In our main result, we capture the impact of the residue of n divided by 3 and the range of the argument of the $$\Delta _n^{g_d}(x)$$.

### Theorem 1.1

Let $$\{g_d(n)\}$$ be a double sequence in $${\mathbb {D}}$$. Let $$n \ge 6$$. Moreover let

\begin{aligned} \Delta _{n}^{g_d}(x)= \left( P_n^{g_d}(x) \right) ^2 - P_{n-1}^{g_d}(x) \, P_{n+1}^{g_d}(x) \ge 0, \end{aligned}
(1.1)

be the Turán inequality.

1. (a)

Let $$0 \le x < 2- \frac{12}{n+4}$$. Then (1.1) holds true for almost all d if and only if n is divisible by 3.

2. (b)

Let $$n \not \equiv 2 \pmod {3}$$ and $$x \ge 0$$. Then (1.1) holds true for almost all d if and only if n is divisible by 3.

The case $$g_{d}\left( n\right) = \sigma _{d} \left( n\right)$$ and $$x=1$$ leads to the results obtained in [8], Theorem 1.2 and Theorem 1.3. An explicit analysis of the bounds obtained in the proof of Theorem 1.1 leads to the following:

### Theorem 1.2

Let $$\left\{ g_{d} \left( n\right) \right\}$$ be a double sequence in $${\mathbb {D}}$$. Let $$n \ge 3$$ and $$n \ne 5$$. Let R be the radius of convergence of $$\sum _{n=1}^{\infty } g_1(n) \, \frac{q^n}{n}$$. For each x, let r(x) be chosen with $$0< r(x) < R$$ and $$P_n^{g_{1} }(x) \le r(x)^{-n}$$ for all n. Then we have the following properties.

1. (i)

Let $$n \equiv 0 \pmod {3}$$ and $$x >0$$. Then $$\Delta _n^{g_d}(x) \ge 0$$ for $$d \ge d_{0}\left( n,x\right)$$, where

\begin{aligned} d_{0}\left( n,x\right) =1+\frac{2n}{3\ln \left( 9/8\right) } \left( \ln \left( n/3\right) - \ln \left( x\right) -3\ln \left( r(x) \right) \right) . \end{aligned}
2. (ii)

Let $$n \equiv 1 \pmod {3}$$ and $$x >0$$. Then $$\Delta _n^{g_d}(x) < 0$$ for $$d \ge d_{0 }\left( n,x\right)$$, where

\begin{aligned} d_{0} \left( n,x\right) =1+\frac{2n}{3\ln \left( 9/8\right) } \left( \ln \left( \frac{n-1}{3}\right) -\frac{ 2n+1}{ 2n} \ln \left( x\right) -3\ln \left( r(x) \right) \right) . \end{aligned}
3. (iii)

Let $$n \equiv 2 \pmod {3}$$ and $$0<x<2-\frac{12}{n+4}$$. Then $$\Delta _n^{g_d}(x) < 0$$ for $$d \ge d_{0 }\left( n,x\right)$$, where

\begin{aligned} d_{0} \left( n,x\right){} & {} = 1+\frac{1}{ \ln \left( 9/8\right) }\left( -\min \left\{ 0,\ln \left( \frac{n-2}{3n+3} \left( \frac{1}{x}+\frac{1}{2}\right) -\frac{1}{3}\right) \right\} \right. \\{} & {} \left. \quad + \, \frac{n-2}{3}\ln \left( \frac{n-2}{3}\right) - \frac{n+1}{3}\ln \left( x\right) -n\ln \left( r(x) \right) \right) . \end{aligned}

### Remark

1. (a)

The positive real number r(x) always exists due to Cauchy–Hadamard’s theorem.

2. (b)

Let $$g_d(n)= n^d$$. Then $$\Delta _4^{g_5}(x)$$ has sign changes for positive real x, since there are two positive, real zeros $$\alpha _1 < \alpha _2$$.

## 2 Records

Let $$g_d(n)= \sigma _d(n)$$. Then for $$d=1$$ and $$d=2$$, complete results for the log-concavity $$\Delta _n^{g_d}(1) \ge 0$$, including the explicit $$E^{\sigma _d}$$, are provided by Nicolas [12] and Ono–Pujahari–Rolen [13]. For $$n\le 10^{5}$$ and $$d \le 8$$, further results have been obtained by Heim–Neuhauser [8].

Let $$g_d(n)= \psi _d(n)$$. In Table 2, we have displayed the results for $$1 \le d \le 9$$. In this paper, we prove the analogue to Nicolas’ result and give some numerical evidence for the case $$d=2$$, which is for $$\sigma _{2}\left( n \right)$$ the log-concavity for plane partitions.

More generally, let $$\{g_d(n)\}$$ be a double sequence in $${\mathbb {D}}$$. Then $$\Delta _1^{g_d}(1)$$ is always negative, since

\begin{aligned} \Delta _1^{g_d}(x) = \frac{x}{2} \, \left( x - g_d(2)\right) , \end{aligned}

and $$g_d(2) \ge 2$$. This explains the results for $$n=1$$ at $$x=1$$.

For $$x >0$$, we have $$\Delta _{1}^{\psi _{d}}\left( x\right) \ge 0$$ if and only if $$x \ge 2^d$$. Let $$n \ge 2$$. Table 3 records our results for Turán inequalities for small d.

In the case $$d=5$$, a new feature appears (see Fig. 2). Let $$3 \le n \le 100$$ then there are exactly two simple positive zeros. Their position implies $$\Delta _3^{\psi _5}(1)>0$$, $$\Delta _4^{\psi _5}(1)<0$$, and $$\Delta _n^{\psi _5}(1)>0$$ for $$5 \le n \le 100$$. We expect that this holds true for all $$n \ge 5$$.

## 3 Basic formulas

Let g be a normalized arithmetic function. Let

\begin{aligned} P_n^g(x):= \frac{x}{n} \sum _{k=1}^n g(k) \, P_{n-k}^g(x), \qquad \text {with } P_{0}^{g}\left( x\right) =1. \end{aligned}

Then $$P_n^g(x)$$ are polynomials of degree n. We refer to [10] for a detailed study of these polynomials. For example $$P_1^g(x)=x$$ and $$P_2^g(x) = x/2 \, (x +g(2))$$.

### 3.1 Coefficients of $$P_n^g(x)$$

Let

\begin{aligned} P_n^g(x) = \sum _{k=0}^n A_{n,k}^g \,\, x^k. \end{aligned}

Then $$A_{0,0}^g=1$$. Let $$n \ge 1$$ then $$A_{n,0}^g=0$$, $$A_{n,1}^g= g(n)\, / \, n$$ and $$A_{n,n}^g = 1 \, / \, n!$$. We also have [10] for $$1 \le m <n$$ and $$n-m=1,2,3$$:

\begin{aligned} A_{n,n-1}^{g}= & {} \frac{1}{n!} \,\, g\left( 2\right) \left( {\begin{array}{c}n\\ 2\end{array}}\right) ,\\ A_{n,n-2}^{g}= & {} \frac{1}{n!} \,\, \left[ 3\left( g\left( 2\right) \right) ^{2}\left( {\begin{array}{c}n\\ 4\end{array}}\right) +2g\left( 3\right) \left( {\begin{array}{c}n\\ 3\end{array}}\right) \right] ,\\ A_{n,n-3}^{g}= & {} \frac{1}{n!} \,\, \left[ 15\left( g\left( 2\right) \right) ^{3}\left( {\begin{array}{c}n\\ 6\end{array}}\right) +20g\left( 2\right) g\left( 3\right) \left( {\begin{array}{c}n\\ 5\end{array}}\right) +6g\left( 4\right) \left( {\begin{array}{c}n\\ 4\end{array}}\right) \right] . \end{aligned}

### Lemma 3.1

Let g be a normalized arithmetic function and

\begin{aligned} \Delta _{n}^g(x):= P_n^g(x)^2 - P_{n-1}^g(x) \, P_{n+1}^g(x). \end{aligned}

Then $$\Delta _1^g(x) = \frac{x}{2} \left( x-g(2) \right)$$ and $$\Delta _2^g(x) = \frac{x^2}{12} \left( x^2 + 3 \, g(2)^2 - 4 \, g(3)\right)$$. Further,

\begin{aligned} \Delta _3^g(x)= & {} \frac{ x^{2}}{144} \bigg (x^{ 4} + 3 g\left( 2\right) x^{ 3} + \left( 9 \left( g\left( 2\right) \right) ^{2} - 8 g\left( 3\right) \right) x^{ 2} + ( -9 \left( g\left( 2\right) \right) ^{3}\\{} & {} + 24 g\left( 3\right) g\left( 2\right) - 1 8 g\left( 4\right) ) x + \left( - 1 8 g\left( 4\right) g\left( 2\right) + 16 \left( g\left( 3\right) \right) ^{2}\right) \bigg ). \end{aligned}

This follows from the explicit form of the polynomials. We have

\begin{aligned} P_3^g(x)= & {} \frac{x}{6} \left( x^2 + 3 \, g(2)\, x + 2 \, g(3) \right) ,\\ P_4^g(x)= & {} \frac{x}{24} \left( x^3 + 6 \, g(2) \, x^2 + \left( 8 \, g(3) + 3 \, g(2)^2 \right) \, x + 6 \, g(4) \right) . \end{aligned}

### 3.2 Properties of $$\Delta _n^g(x)$$

Let us establish the following notation:

\begin{aligned} \Delta _n^g(x) = \sum _{k=0}^{2n } D_{n,k}^g \,\, x^k. \end{aligned}

In contrast to $$P_n^g(x)$$, the coefficients of $$\Delta _n^g(x)$$ are not always non-negative in general. Nevertheless, we have $$\Delta _n^g(0)=0$$ and the important asymptotic property

\begin{aligned} \lim _{x \rightarrow \infty } \Delta _n^g(x) = \infty . \end{aligned}

This follows from $$D_{n,2n}^g = \frac{1}{(n!)^2 \, (n+1)}$$. Let $$n \ge 2$$. We can always factor out $$x^2$$ and still have polynomials, since $$D_{n,0}^g = D_{n,1}^g =0$$. The new constant term is given by $$D_{n,2}^g$$, which does not need to be non-negative:

\begin{aligned} D_{n,2}^g= & {} \frac{1}{n^2} \, \left[ g(n)^2 - \frac{n^2}{n^2-1} \, g(n-1) \, g(n+1) \right] ,\\ D_{n,3}^g= & {} 2\frac{g\left( n\right) }{n^{2}}\sum _{k=1}^{n-1}\frac{g\left( k\right) g\left( n-k\right) }{k}-\frac{g\left( n-1\right) }{n -1}\sum _{k=1}^{n}\frac{g\left( n+1-k\right) g\left( k\right) }{2\left( n+1-k\right) k}\\{} & {} -\frac{g\left( n+1\right) }{n +1}\sum _{k=1}^{n-2}\frac{g\left( n-1-k\right) g\left( k\right) }{2\left( n-1-k\right) k}. \end{aligned}

### 3.3 Special cases

Let $$\{g_d(n)\}$$ be a double sequence in $${\mathbb {D}}$$. We have $$\Delta _1^{g_d}(x) = x \, (x-g_d(2))/2$$. Thus, $$\Delta _1^{g_d}(x) = 0$$ if $$x=0$$ or $$x= g_d(2)$$. Thus, $$\Delta _1^{g_d}(x)>0$$ if and only if $$x \not \in [0, g_d(2)]$$. This implies that $$\Delta _1^{g_d}(x)<0$$ for $$x\in \left( 0,g_{d}\left( 2\right) \right)$$ and all $$d \in {\mathbb {N}}$$. The case $$n=2$$ is still directly accessible. We have $$\Delta _2^{g_d}(x) = 0$$ if $$x=0$$ or $$x^2= 4\, g_d(3)-3 \,\left( g_{d}\left( 2\right) \right) ^2$$. We consider $$4\, g_d(3)-3 \,\left( g_{d}\left( 2\right) \right) ^2 \ge 0$$ and $$x \ne 0$$. Let $$g_{d} = \psi _{d}$$ or $$g_{d}=\sigma _{d}$$. Then $$\Delta _2^{\psi _d}(x) >0$$ for $$d \in {\mathbb {N}}$$, especially $$\Delta _2^{\psi _1}(x) =x^4/12$$.

## 4 Proof of Theorem 1.1

Our strategy is to utilize the well-known formula ([11, Sect. 4.7]):

\begin{aligned} P_{n}^{g_{d}}\left( x\right) = \sum _{k\le n}\sum _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+\ldots +m_{k}=n \end{array}} \frac{1}{k!} \, \frac{g _{d}\left( m_{1}\right) \cdots g _{d}\left( m_{k}\right) }{m_{1}\cdots m_{k}} \,\, x^k. \end{aligned}

### 4.1 Lower and upper bounds

In [8, Sect. 3], we have obtained lower and upper bounds for $$P_n^{\sigma _d}(1)$$. The invented proof method can be generalized in a straightforward manner to obtain the following result for all double sequences in $${\mathbb {D}}$$ and the associated polynomials for $$x >0$$.

### Proposition 4.1

Let the double sequence $$\{g_d(n)\}_{d,n \in {\mathbb {N}}}$$ be an element of $${\mathbb {D}}$$. Let $$n \ge 3$$ and $$x >0$$. Then we have for all $$d\ge 1$$ the following upper and lower bounds.

Let $$n \equiv 0 \pmod {3}$$ and $$n^{\prime }:= n/3$$. Then

\begin{aligned} \frac{3^{\left( d-1\right) n'}}{\left( n^{\prime } \right) !} \,\, x^{n^{\prime }} \, < \, P_n^{g_d}(x) \, \le \, 3^{\left( d-1\right) n^{\prime }} P_n^{g_1}(x). \end{aligned}

Further, let $$n \equiv 1 \pmod {3}$$ and $$n^{\prime }:= (n-4)/3$$. Then

\begin{aligned} \frac{\left( 4 \cdot 3^{n'}\right) ^{d-1}}{\left( n^{\prime } \right) !} \, \left( x^{n^{\prime }+1} + \frac{x^{n^{\prime }+2}}{2} \right) \, < \, P_n^{g_d}(x) \, \le \, \left( 4 \cdot 3^{n^{\prime }}\right) ^{d-1} \,\, P_n^{g_1}(x). \end{aligned}

Further, let $$n \equiv 2 \pmod {3}$$ and $$n' := (n-2)/3$$. Then

\begin{aligned} \frac{\left( 2\cdot 3^{n'}\right) ^{d-1}}{\left( n' \right) !} \,\, x^{n'+1} \, < \, P_n^{g_d}(x) \, \le \, \left( 2\cdot 3^{n'}\right) ^{d-1} P_n^{g_1}(x). \end{aligned}

Additionally, let $$n \equiv 2 \pmod {3}$$ and $$n \ge 8$$. Let $$n':= (n-2)/3$$. Then

\begin{aligned} P_n^{g_d}(x) \le \frac{\left( 2\cdot 3^{n'}\right) ^{d-1}}{\left( n'\right) !} \, x^{n'+1}+ \left( 16\cdot 3^{n'-2 }\right) ^{d-1} \, P_n^{g_1}(x). \end{aligned}

### 4.2 Proof of Theorem 1.1

For $$x=0$$, the inequality (1.1) holds certainly true. Therefore, let $$x>0$$. We apply Proposition 4.1.

#### 4.2.1 The case $$n\equiv 0 \pmod {3}$$

In the first step, we apply Proposition 4.1. This leads to

\begin{aligned} \frac{\left( P_{n}^{g_{d} }\left( x\right) \right) ^{2}}{P_{n-1}^{g_{d} }\left( x\right) P_{n+1}^{g_{d} }\left( x\right) } > \frac{x^{2n/3}}{\left( \left( n/3\right) !\right) ^{2}P_{n-1}^{g_{1} }\left( x\right) P_{n+1}^{g_{1} }\left( x\right) }\left( \frac{9}{8}\right) ^{d-1}. \end{aligned}

We choose $$r(x)>0$$, such that $$P_n^{g_1}(x) \le r(x)^{-n}$$ for all n. Let

\begin{aligned} d_{0}=d_{0}\left( n,x\right) =1+\frac{2n}{3\ln \left( 9/8\right) } \left( \ln \left( n/3\right) -\ln \left( x\right) -3\ln \left( r\left( {x}\right) \right) \right) . \end{aligned}

Then $$\left\{ P_{n}^{g_{d}}\left( x\right) \right\} _{n}$$ is strictly log-concave at n for $$d \ge d_0$$, since

\begin{aligned} \frac{\left( P_{n}^{g_{d}}\left( x\right) \right) ^{2}}{P_{n-1}^{g_{d}}\left( x\right) P_{n+1}^{g_{d}}\left( x\right) }> \frac{x^{2n/3}}{\left( n/3\right) ^{2n/3} \left( r\left( {x}\right) \right) ^{-2n}}\left( \frac{9}{8}\right) ^{d-1} \ge 1. \end{aligned}

#### 4.2.2 The case $$n\equiv 1 \pmod {3}$$

In the first step, we apply Proposition 4.1. This leads to

\begin{aligned} \frac{\left( P_{n}^{g_{d} }\left( x\right) \right) ^{2}}{P_{n-1}^{g_{d} }\left( x\right) P_{n+1}^{g_{d} }\left( x\right) } \, <\, \left( \left( \frac{n-1}{3} \right) ! \right) ^{2}\,\, \left( P_{n}^{g_{1}}\left( x\right) \right) ^{2} \,\, x^{- \left( 2n+ 1\right) /3} \,\, \left( \frac{8}{9}\right) ^{d-1}. \end{aligned}

We choose $$r(x)>0$$, such that $$P_n^{g_1}(x) \le r(x)^{-n}$$ for all n. Let

\begin{aligned} d_{0} =d_{0} \left( n,x\right) =1+\frac{ 2n}{3\ln \left( 9/8\right) }\left( \ln \left( \frac{n-1}{3}\right) -3 \ln \left( r\left( {x}\right) \right) -\frac{ 2n+1}{ 2n} \ln \left( x\right) \right) . \end{aligned}

Then the sequence $$\left\{ P_{n}^{g_{d}}\left( x\right) \right\} _{n}$$ is strictly log-convex at n for $$d \ge d_{0 }$$, since

\begin{aligned} \frac{\left( P_{n}^{g_{d}}\left( x\right) \right) ^{2}}{P_{n-1}^{g_{d}}\left( x\right) P_{n+1}^{g_{d}}\left( x\right) } < \left( \frac{n-1}{3}\right) ^{ 2n /3}\left( r\left( {x}\right) \right) ^{-2n} x ^{- \left( 2n+1\right) /3 }\left( \frac{8 }{9 }\right) ^{d-1} \le 1. \end{aligned}

#### 4.2.3 The case $$n\equiv 2 \pmod {3}$$

This final case involves some additional considerations. Again we first apply Proposition 4.1 and obtain

\begin{aligned}{} & {} \left( P_{n}^{g_{d} }\left( x\right) \right) ^{2} -P_{n-1}^{g_{d} }\left( x\right) P_{n+1}^{g_{d} }\left( x\right) \\{} & {} \quad < \left( \frac{\left( 2\cdot 3^{\left( n-2\right) /3}\right) ^{d-1} x^{\left( n+1\right) /3}}{\left( \left( n-2\right) /3\right) !}+ \left( 16\cdot 3^{\left( n-8\right) /3}\right) ^{d-1}P_{n}^{g_{1} }\left( x\right) \right) ^{2}\\{} & {} \qquad {}-\frac{ \left( 4\cdot 3^{\left( n-5\right) /3}\right) ^{d-1}\left( x^{\left( n-2\right) /3}+x^{\left( n+1\right) /3}/2\right) }{ \left( \left( n-5 \right) /3\right) !}\frac{3^{\left( d-1\right) \left( n+1\right) /3} x^{\left( n+1\right) /3}}{\left( \left( n+1\right) /3\right) !}\\{} & {} \quad \le \left( \frac{\left( 2\cdot 3^{\left( n-2\right) /3 }\right) ^{d-1}x^{\left( n+1\right) /3}}{\left( \left( n-2\right) /3\right) !}\right) ^{2} \left( 1-\frac{\left( n-2\right) /3}{\left( n+1\right) /3}\left( \frac{1}{x}+\frac{1}{2}\right) \right. \\{} & {} \qquad + 2\left( 8 \cdot 3^{ -2 }\right) ^{d-1}x^{-\left( n+1\right) /3}P_{n}^{g_{1}}\left( x\right) \left( \left( n-2\right) /3\right) !\\{} & {} \qquad \left. +\left( \left( \left( n-2\right) /3\right) !\left( 8 \cdot 3^{ -2 }\right) ^{d-1}x^{-\left( n+1\right) /3}P_{n}^{g_{1}}\left( x\right) \right) ^{2}\right) . \end{aligned}

The last inequality can only be not larger than zero if $$0<x<2-\frac{12}{n+4}$$. We choose $$r(x)>0$$ such that $$P_n^{g_1}(x) \le r(x)^{-n}$$ for all n. Then

\begin{aligned}{} & {} \left( \frac{n-2}{3}\right) !\left( \frac{8 }{9 }\right) ^{d-1}x^{-\left( n+1\right) /3}P_{n}^{g_{1}}\left( x\right) \\{} & {} < \left( \frac{n-2}{3}\right) ^{\left( n-2\right) /3}\left( \frac{8 }{9 }\right) ^{d-1}x^{-\left( n+1\right) /3} \left( r\left( {x }\right) \right) ^{-n} \le 1, \end{aligned}

for

\begin{aligned} d\ge 1+\frac{1}{\ln \left( 9/8\right) }\left( \frac{n-2}{3} \ln \left( \frac{n-2}{3}\right) -\frac{n+1}{3}\ln \left( x\right) -n\ln \left( r\left( {x}\right) \right) \right) . \end{aligned}

Let $$d_{0} =d_{0} \left( n,x\right)$$ be defined by

\begin{aligned}{} & {} 1+\frac{1}{\ln \left( 9/8\right) }\left( -\min \left\{ 0,\ln \left( \frac{n-2}{3n+3}\left( \frac{1}{x}+\frac{1}{2}\right) -\frac{1}{3}\right) \right\} \right. \\{} & {} \quad \left. {}+\frac{n-2}{3} \ln \left( \frac{n-2}{3}\right) -\frac{n+1}{3}\ln \left( x\right) -n\ln \left( r\left( {x}\right) \right) \right) . \end{aligned}

Then the sequence $$\{P_n^{g_d}(x)\}_d$$ is strictly log-convex for all $$d \ge d_{0}$$.

## 5 Turán inequalities

Let $$\{g_d(n)\}$$ be a double sequence in $${\mathbb {D}}$$. We are interested in finding the set of positive real numbers, such that $$\Delta _n^{g_d}(x) \ge 0$$, with special emphasis on the behavior at $$x=1$$.

In [7], a conjecture for $$\Delta _n^{\sigma _1}(x)$$ was stated, which generalized a conjecture of Chern–Fu–Tang [4] related to integers $$x \ge 2$$. The Chern–Fu–Tang conjecture was proven by Bringmann et al. [3]. Recently, a second conjecture [9] was proposed for $$\Delta _n^{\sigma _2}(x)$$. We have shown for $$x=1$$, the case of plane partitions, that $$\Delta _n^{\sigma _2}(1)>0$$ for almost all n. Finally, Ono et al. [13] have proven that $$\Delta _n^{\sigma _2}(1)>0$$ for all $$n \ge 12$$. We now show that $$\Delta _{n}^{\psi _{1}}\left( x\right) \ge 0$$ for all $$x \in {\mathbb {R}}$$. This is the first case where a full result on Turán inequalities is obtained for a double sequence in $${\mathbb {D}}$$ with d fixed.

### 5.1 $$\Delta _n^{\psi _1}(x) \ge 0$$

We have $$\psi _1(n)=1$$. The polynomials $$P_n^{\psi _1}(x)$$ had been studied in [6] and had been found intimately related with the $$\alpha$$-associated Laguerre polynomials. We have $$P_n^{\psi _1}(x)= \frac{x}{n} L_{n-1}^{(1)}(-x)$$, where

\begin{aligned} \sum _{n=0}^{\infty } L_n^{(\alpha )}(x) \, t^n = \frac{1}{(1-t)^{ \alpha +1 }}\,\, \textrm{e}^{-x \frac{t}{1-t}}, \qquad \alpha > -1. \end{aligned}

The Laguerre polynomials of degree n are given by $$L_n(x)= L_n^{(0)}(x)$$. It is known that $$\alpha$$-associated Laguerre polynomials for $$\alpha \ge 0$$ satisfy:

\begin{aligned} \left( L_n^{(\alpha )}(x)\right) ^2 - L_{n-1}^{(\alpha )}(x) \, L_{n+1}^{(\alpha )}(x) = \sum _{k=0}^{n-1} \frac{ \left( {\begin{array}{c}\alpha + n-1\\ n-k\end{array}}\right) }{ n \, \left( {\begin{array}{c}n\\ k\end{array}}\right) } \left( L_k^{(\alpha -1)}(x)\right) ^2 >0. \end{aligned}

These Turán inequalities are not sufficient to prove $$\Delta _n^{\psi _1}(x) \ge 0$$. We have to show for all $$x \in {\mathbb {R}}$$ that

\begin{aligned} \Delta _{n}^{\psi _1}\left( x\right) = \frac{1}{n^2} \left( L_{n-1}^{(1)}(-x) \right) ^2 - \frac{1}{n^2-1} L_{n-2}^{(1)}(-x) \,\, L_{n}^{(1)}(-x)\ge 0. \end{aligned}

Szegő [16] proved in 1948, that $$L_n^{(\alpha )}(x) / L_n^{(\alpha )}(0)$$ satisfies Turán inequalities, where $$L_n^{(1)}(0)= n+1$$. This proves our claim.

### 5.2 Challenges

We propose three open questions.

#### 5.2.1 Log-concavity of partition and plane partition numbers

Reprove the results of Nicolas [12] and Ono et al. [13] on the log-concavity of the partition numbers and the plane partition numbers utilizing the zero distribution of the polynomials $$\left\{ P_{n}^{\sigma _{d}}\left( x\right) \right\}$$ for $$d=1$$ and $$d=2$$.

#### 5.2.2 Turán inequalities $$\Delta _{n}^{\psi _d}(x) \ge 0$$

Based on our numerical investigations on the zeros of $$\Delta _n^{\psi _d}(x)$$, we believe that it is very likely for $$2 \le d \le 4$$ that $$\Delta _n^{\psi _d}(x) \ge 0$$ for $$n \ge 2$$ and $$x \in {\mathbb {R}}$$. Prove this observation.

#### 5.2.3 The case $$n \equiv 0 \pmod {3}$$

The following problem was presented at the Conference: 100 Years of Mock Theta Functions at Vanderbilt University in 2022 (organized by Rolen and Wagner). Prove that $$\Delta _n^{\psi _d}\left( x\right) \ge 0$$ for all $$n \equiv 0 \pmod {3}$$ and all $$d \in {\mathbb {N}}$$.