Log-concavity of infinite product generating functions

Abstract

In the 1970s Nicolas proved that the coefficients $p_d(n)$ defined by the generating function

$$\begin{aligned} \sum _{n=0}^{\infty } p_d(n) \, q^n = \prod _{n=1}^{\infty } \left( 1- q^n\right) ^{-n^{d-1}} \end{aligned}$$

are log-concave for $d=1$. Recently, Ono, Pujahari, and Rolen have extended the result to $d=2$. Note that $p_1(n)=p(n)$ is the partition function and $p_2(n)=\mathrm{pp}\left( n\right)$ is the number of plane partitions. In this paper, we invest in properties for $p_d(n)$ for general d. Let $n \ge 6$. Then $p_d(n)$ is almost log-concave for n divisible by 3 and almost strictly log-convex otherwise.

1 Introduction and main results

Many partition functions have infinite products as generating function. Numerous researchers have investigated their asymptotic behaviors and combinatorial properties. Let $\alpha := (\alpha _n)_n$ be a sequence of positive integers. In this paper, we investigate the log-concavity of the coefficients $p_{\alpha }(n)$ defined by [1], chapter 6:

$$\begin{aligned} 1+ \sum _{n=1}^{\infty } p_{\alpha }(n) \, q^n := \prod _{n \in S} \left( 1- q^n \right) ^{-\alpha _n}, \qquad S\subset \mathbb {N}, \,\, \vert q \vert < 1. \end{aligned}$$

We focus on $S=\mathbb {N}$ and the sequences $\alpha _n(d):= n^{d-1}$, $d \in \mathbb {N}$, and denote the coefficients by $p_d(n)$. A sequence of non-negative integers $\left( a_{n}\right) _{n}$ is considered to be log-concave at n ([2, 14] and [12] introduction) if

$$\begin{aligned} \Delta (n):= a_n^2-a_{n-1} \, a_{n+1} \ge 0. \end{aligned}$$

Most prominent is the partition function p(n), counting all non-ordered decompositions of a natural number n as a sum of positive integers. Euler proved the identity $p(n)=p_1(n)$, the famous Eulerian infinite product identity

$$\begin{aligned} \sum _{n=0}^{\infty } p(n) \, q^n = \prod _{n=1}^{\infty } \frac{1}{1 - q^n}. \end{aligned}$$

Building on the celebrated work of Hardy and Ramanujan [5] on the asymptotic formula

$$\begin{aligned} p(n) \sim \frac{1}{4 n \sqrt{3}}\,\, e^{\pi \sqrt{2n/3}} \end{aligned}$$

and Lehmer [8] and Rademacher [13], Nicolas [11], in 1978, proved that p(n) is log-concave if and only if n is an even integer or n is odd and $n >25$. We refer to recent work by DeSalvo and Pak [3]. Another prominent partition function is provided by the plane partitions ${\mathrm{pp}}\left( n\right)$. For further information we refer to Andrews, Krattenthaler, and Stanley [1, 7, 14] and Sect. 2. A century ago, MacMahon [9, 10] proved the fundamental identity ${\mathrm{pp}}\left( n\right) = p_2(n)$. Improving on the works of E. M. Wright [16] on asymptotic formulas of ${\mathrm{pp}}\left( n\right)$, Ono, Pujahari, and Rolen [12] finally proved that ${\mathrm{pp}}\left( n\right)$ is log-concave if and only if n is an even integer or n is odd and $n>11$. Previously it was known [4], that ${\mathrm{pp}}\left( n\right)$ is log-concave for almost all n and tested up to $10^5$.

Supplied with these results we analyze the log-concavity patterns for all $d\ge 1$. Table 1 displays the results for $1 \le d \le 8$.

Table 1 Log-concavity patterns for $1 \le d \le 8$

Let d be fixed. We call a natural number n an exception if $\Delta _d(n)=\left( p_{d}\left( n\right) \right) ^{2}-p_{d}\left( n-1\right) p_{d}\left( n+1\right) <0$. This indicates that $p_d(n)$ is strictly log-convex at n. It is obvious that $n=1$ is an exception for all d. Moreover, Table 1 indicates that $n=4$ and $n=5$ are also candidates for $\Delta _d(n) <0$ for $d>4$. It turns out that this is true for $n \ge 6$ coprime to 3 and almost all d. The bounds depend on n. Thus, let us first study $\Delta _d(n)$ for small n.

Theorem 1.1

Let $1 \le n \le 9$. Then for all $d \in \mathbb {N}$ we have:

$$\begin{aligned} \begin{array}{|c|c|c|} \hline n &{} \text { strictly log-convex if and only if } \\ \hline 1 &{} d \ge 1 \\ 2 &{} \emptyset \\ 3 &{} 1 \le d \le 3 \\ 4 &{} d \ge 6 \\ 5 &{} 1 \le d \le 9\\ 6 &{} \emptyset \\ 7 &{} 1 \le d \le 3, d \ge 11 \\ 8 &{} d \ge 9 \\ 9 &{} 1 \le d \le 2\\ \hline \end{array} \end{aligned}$$

Obviously, Theorem 1.1 illustrates some irregular behavior of $\Delta _d(n)$. Nevertheless, there is a dominant pattern for each $n \ge 6$, coprime to 3, forcing $\Delta _d(n)$ to be negative for almost all d. We also have an interesting result for n divisible by 3.

Theorem 1.2

Let $n \ge 6$ be divisible by 3. Then $\Delta _d(n)>0$ for almost all d.

Remarks

Let $n \ge 6$ be divisible by 3. Nicolas states that partitions p(n) are not log-concave if and only if $n=9, 15$, and $n=21$ (the case $d=1$). Ono, Pujahari, and Rolen’s result states, that plane partitions ${\mathrm{pp}}\left( n\right)$ are not log-concave if and only if $n=9$ (the case $d=2$). Further, for $n \le 6$ we have only $\Delta _3(3)< 0$. Moreover, it is much more likely that $\Delta _d(n)>0$ for $n \ge 6$, divisible by 3 and $d \ge 3$. For each $n\ge 6$ not divisible by 3, we have an explicit bound C(n), such that $d \ge C(n)$ implies $\Delta _d(n)<0$.

Theorem 1.3

Let $n \ge 6$. Let $n \equiv 1$ or $n \equiv 2 \mod 3$. Then $\Delta _d(n) < 0$ for almost all d. In particular let

$$\begin{aligned} {\tilde{C}}_1(n):= & {} 1 + 6\, \big ( 1 + \ln (n-1) \big ) \, n, \\ {\tilde{C}}_2(n):= & {} 1 + 3\, \big ( 3 + \ln (n+1) \big ) \, n. \end{aligned}$$

Then for $r=1$ or $r=2$, $\Delta _d(n)<0$ for $n \equiv r \mod 3$ and $d \ge {\tilde{C}}_r(n)$.

In this paper we study the n and d aspects of $p_d(n)$. For the rest of this section we consider $p_d(n)$ as a double indexed sequence. Table 4 gives a decent description of the underlying landscape.

Note that the results of column $d=1$ are due to Nicolas [11]. He proved that for $n \ge 26$, there are no further exceptions. Column $d=2$ was conjectured by Heim, Neuhauser, and Tröger [4] and proved by Ono, Pujahari, and Rolen [12]. They proved that for $n \ge 12$, there are no further exceptions. We also increased the values of the parameters d and n.

Observation 1

Let $n\ge 6$ and $n \equiv 0 \mod 3$. Then we expect

$$\begin{aligned} \Delta _d(n)>0 \Longrightarrow \Delta _{d+1}(n)>0. \end{aligned}$$

Observation 2

Moreover, let ${D}_{n}:= \min \left\{ d > 3\, : \, \Delta _d(n) <0 \right\}$. Let $n \ge 6$ and $n \equiv 1 \mod 3$. Then Table 4 indicates ${D}_{n} > {D}_{n+1}$.

2 Ordinary and plane partition functions

According to Stanley, plane partitions are fascinating generalizations of partitions of integers ([15], Section 7.20). Andrews [1] gave an excellent introduction of plane partitions in the context of higher-dimensional partitions. The most recent survey is presented by Krattenthaler on plane partitions in the work by Stanley and his school [7].

A plane partition $\pi$ of n is an array $\pi = \left( \pi _{ij} \right) _{i,j \ge 1}$ of non-negative integers $\pi _{ij}$ with finite sum $\vert \pi \vert := \sum _{i,j\ge 1} \pi _{ij}=n$, which is weakly decreasing in rows and columns.

Plane partitions are also displayed by a filling of a Ferrers diagram with weakly decreasing rows and columns, where the sum of all these numbers is equal to n. Let the numbers in the filling represent the heights for stacks of blocks placed on each cell of the diagram (Fig. 1).

Fig. 1

Representation of plane partitions

Let p(n) denote the number of partitions of n and ${\mathrm{pp}}(n)$ denote the number of plane partitions of n. As usual, ${p}\left( 0\right) :=1$ and ${\mathrm{pp}} \left( 0 \right) : =1$. In Table 2 we have listed the first values of the partition and plane partition function, which are equal to $p_1(n)$ and $p_2(n)$ and related to $d=1$ and $d=2$. There had been speculations about the sequence $p_3(n)$. To indicate that this is still an open topic we labeled $p_3(n)$ with $X_n$.

Table 2 Values for $0 \le n \le 10$

In the context of higher partitions [1] one has the so-called solid partitions as certain 3-dimensional arrays. These are different from $p_3(n)$. MacMahon proposed possible generating series, which did not work out finally.

3 Proofs of Theorems 1.2 and 1.3

We divide the proof into several parts. Let $\sigma _{d}(n):= \sum _{ t \mid n} t^d$. Our strategy is to utilize the following formula ([6], Sect. 4.7):

$$\begin{aligned} p_{d}(n) =\sum _{k\le n}\sum _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+\ldots +m_{k}=n \end{array}}\frac{1}{k!}\frac{\sigma _{d}\left( m_{1}\right) \cdots \sigma _{d}\left( m_{k}\right) }{m_{1}\cdots m_{k}}. \end{aligned}$$

3.1 Partitions $n=\sum _{j=1}^k m_{ j}$ and the maximum of $\prod _{ j=1}^k m_{ j}$

We start with the following well known property. For completeness, we include a proof.

Lemma 3.1

Let $m_{1},\ldots ,m_{k}\ge 1$ and $m_{1}+\ldots +m_{k}=n\ge 2$. Then the largest values for $m_{1}\cdots m_{k}$ are

$$\begin{aligned} \max _{k\le n}\max _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+ \ldots +m_{k}=n \end{array}}m_{1}\cdots m_{k}= & {} 3^{n/3} \text { for } n \equiv 0\mod 3, \\ \max _{k\le n}\max _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+ \ldots +m_{k}=n \end{array}}m_{1}\cdots m_{k}= & {} 4\cdot 3^{\left( n-4\right) /3} \text { for } n \equiv 1\mod 3,\\ \max _{k\le n}\max _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+\ldots +m_{k}=n \end{array}}m_{1}\cdots m_{k}= & {} 2\cdot 3^{\left( n-2\right) /3} \text { for } n\equiv 2\mod 3. \end{aligned}$$

Proof

Let $k\le n$ and $m_{1},\ldots ,m_{k}\ge 1$ so that $m_{1}+\ldots +m_{k}=n$. We can assume that $m_{j}\ge 2$ for all $j\le k$. Otherwise, if $m_{j}=1$, we can take the partition with $m_{j-1}$ replaced by $m_{j-1}+1$, or $m_{j+1}$ replaced by $m_{j+1}+1$ and drop $m_{j}$ and obtain a larger product. We can also assume, that $m_{j}\le 4$ for all $j\le k$. Otherwise, if $m_{j} >4$, we split it as $m_{j}=2+\left( m_{j}-2\right)$ and $2\left( m_{j}-2\right) >m_{j}$, resulting in a larger product.

Therefore, if $m_{1}\cdots m_{k}$ is maximal, then it follows that $2\le m_{j}\le 4$ for all $j\le k$. Assume there are at least three $m_{j}=2$. But $2+2+2=6=3+3$ and $2^{3}<3^{2}$. So, if we replace the three 2s by two 3s we obtain a larger product. Therefore, in the maximal product there can be at most two 2s. Similarly, if there were to be one $m_{j}=4$ and another $=2$, we obtain $4+2=6=3+3$ and again $4\cdot 2<3^{2}$. Again, the product becomes larger if 4 and 2 are replaced by two 3s. Therefore, in the maximal product there can only be either one 4 and no 2, or no 4 if there is a 2.

In total, we obtain that in the maximal product we have only 3s except for either precisely one 4 or at most two 2s. Note, that if there is either one 4 or two 2s, it results in the same product as $4=2^{2}$. If $n\equiv 0\mod 3$, this yields only one possibility, when $n=\frac{n}{ 3} \cdot 3$ resulting in $3^{n/3}$. If $n\equiv 1\mod 3$, we have two possible partitions: $n=4+\frac{n-4}{3}\cdot 3=2+2+\frac{n-4}{3}\cdot 3$ where both result in the product $4\cdot 3^{\left( n-4\right) /3}$. Note that we assumed $n\ge 2$. If $n\equiv 2\mod 3$, we again have only one possible partition $n=2+\frac{n-2}{3}\cdot 3$ given the restrictions, resulting in the product $2\cdot 3^{\left( n-2\right) /3 }$. $\square$

Lemma 3.2

For $n\ge 8$, $n\equiv 2\mod 3$ the second largest product is $16\cdot 3^{\left( n-8\right) /3}$.

Proof

In this case in the corresponding partition $n=m_{1}+\ldots +m_{k}$ is an $m_{j}=1$, an $m_{j}\ge 5$, more than one $m_{j}=4$, at least three $m_{j}=2$, or a 4 and a 2 together.

First suppose there is an $m_{j}=1$. Our argument is similar to that for the proof of Lemma 3.1 but in addition, we have to ensure that the result is not maximal. We replace either $m_{j-1}$ by $m_{j-1}+1$ or $m_{j+1}$ by $m_{j+1}+1$ and drop the $m_{j}=1$. This can only result in $n=2+\frac{n-2}{3}\cdot 3$ if either there were two 2s in the partition of n, or another 1. If it was $n=1+2+2+\frac{n-5}{3}\cdot 3$, then since $n\ge 8$ we can change it to $n=2+2+4+\frac{n-8}{3}\cdot 3$ and obtain a larger product. In the second case, $n=1+1+\frac{n-2}{3}\cdot 3$. Since $n\ge 8$ we can replace it by $n=1+4+\frac{n-5}{3}\cdot 3$ resulting in a larger product.

In the second case we assume there is an $m_{j}\ge 5$. If we again replace it by $2+\left( m_{j}-2\right)$ this will yield the partition $n=2+\frac{n-2}{3}\cdot 3$, if and only if it was $n=5+\frac{n-5}{3}\cdot 3$ before. As $n\ge 8$ we can increase this to $n=4+4+\frac{n-8}{3}\cdot 3$, since $4^{2}>5\cdot 3$.

In the third case we assume there are more than one 4. We replace $4+4$ by $2+3+3$ and obtain a larger product $2\cdot 3^{2}>4^{2}$. The product of the parts of such a partition would only be maximal if the partition was $n=4+4+\frac{n-8}{3}\cdot 3$.

The fourth case is that there are at least three 2s. The product can be increased by replacing $2+2+2=3+3$. This would only be maximal if the partition was $n=4\cdot 2+\frac{n-8}{3}\cdot 3$.

In the fifth case we assume that there is at least one 2 and at least one 4 together in the partition. This could be increased by replacing $2+4=3+3$ as $2\cdot 4<3^{2}$. This will yield the maximal product only if the partition was $n=2+2+4+\frac{n-8}{3}$.

In total, the second largest product $m_{1}\cdots m_{k}$ among all partitions $\left( m_{1},\ldots ,m_{k}\right)$ of n since $n\ge 8$ is

$$\begin{aligned} 16\cdot 3^{\left( n-8\right) /3} \end{aligned}$$

(3.1)

from the partitions

$$\begin{aligned} n=4+4+\frac{n-8}{3}\cdot 3=2+2+4+\frac{n-8}{3}\cdot 3=4\cdot 2+\frac{n-8}{3}\cdot 3. \end{aligned}$$

$\square$

3.2 Upper and lower bounds for $p_d(n)$

Let $n\ge 2$ and $d \ge 1$. Then we have the inequalities

$$\begin{aligned} \frac{3^{\left( d-1\right) n/3}}{\left( n/3\right) !}< & {} p_{d}(n)\le 3^{\left( d-1\right) n/3}p_{1}(n)\text { for }n\equiv 0\mod 3,\\ \frac{3 \left( 4\cdot 3^{\left( n-4\right) /3}\right) ^{d-1} }{2\left( \left( n-4\right) /3\right) !}< & {} p_{d}(n) \le \left( 4\cdot 3^{\left( n-4\right) /3}\right) ^{d-1} p_{1}(n)\text { for }n\equiv 1\mod 3,\\ \frac{\left( 2\cdot 3^{\left( n-2\right) /3}\right) ^{d-1} }{\left( \left( n-2\right) /3\right) !}< & {} p_{d}(n) \le \left( 2\cdot 3^{\left( n-2\right) /3}\right) ^{d-1}p_{1}(n)\text { for }n\equiv 2\mod 3. \end{aligned}$$

Moreover, let $n \equiv 2 \mod 3$ and $n\ge 8$. Then

$$\begin{aligned} p_{d}(n)\le \frac{\left( 2\cdot 3^{\left( n-2\right) /3}\right) ^{d-1}}{\left( \left( n-2\right) /3\right) !}+ \left( 16\cdot 3^{\left( n-8\right) /3 }\right) ^{d-1}p_{1}(n). \end{aligned}$$

(3.2)

To show these inequalities we consider

$$\begin{aligned} p_{d}\left( n\right)= & {} \sum _{k\le n}\frac{1}{k!}\sum _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+ \ldots +m_{k}=n \end{array}} \frac{\sigma _{d}\left( m_{1}\right) \cdots \sigma _{d}\left( m_{k}\right) }{m_{1}\cdots m_{k}}\nonumber \\= & {} \sum _{k\le n}\frac{1}{k!}\sum _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+ \ldots +m_{k}=n \end{array}} \frac{1}{m_{1}\cdots m_{k}}\left( \sum _{t_{1}\mid m_{1}}t_{1}^{d}\right) \cdots \left( \sum _{t_{k}\mid m_{k}}t_{k}^{d}\right) . \end{aligned}$$

(3.3)

For the upper bounds we use the following estimate

$$\begin{aligned} p_{d}\left( n\right)\le & {} \sum _{k\le n}\frac{1}{k!}\sum _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+ \ldots +m_{k}=n \end{array}} \frac{1}{m_{1}\cdots m_{k}}\left( \sum _{t_{1}\mid m_{1}}t_{1}\right) \cdots \left( \sum _{t_{k}\mid m_{k}}t_{k}\right) \left( m_{1}\cdots m_{k}\right) ^{d-1}\\\le & {} p_{1}\left( n\right) \max _{k\le n}\max _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+\ldots +m_{k}=n \end{array}}\left( m_{1}\cdots m_{k}\right) ^{d-1}. \end{aligned}$$

Using Lemma 3.1 we obtain the upper bounds. Similarly, we obtain the lower bounds by taking only those growth terms from the previous sum (3.3) where the product is maximal. This we can obtain in the following way. Let $M=\max _{k\le n}\max _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+\ldots +m_{k}=n \end{array}}m_{1}\cdots m_{k}$. Continuing (3.3) we then obtain

$$\begin{aligned} p_{d}\left( n\right) \ge \sum _{k\le n}\frac{1}{k!}\sum _{\begin{array}{c} m_{1},\ldots ,m_{k}\ge 1 \\ m_{1}+\ldots +m_{k}=n \\ m_{1}\cdots m_{k}=M \end{array}}M^{d-1}. \end{aligned}$$

Note that for $n\equiv 1\mod 3$, we obtain the maximal value for $m_{1}=4$, $m_{2}=\ldots =m_{\left( n-1\right) /3}=3$ and $m_{1}=m_{2}=2$, $m_{3}=\ldots =m_{\left( n+2\right) /3}=3$, which contribute

$$\begin{aligned} \frac{3}{2\left( \left( n-4\right) /3\right) !}\left( 4\cdot 3^{\left( n-4\right) /3}\right) ^{d-1} \end{aligned}$$

to the lower bound.

Now assume $n\equiv 2\mod 3$. Then we obtain the improved upper bound (3.2). For this recall that in the proof of Lemma 3.1, we have shown that $m_{1}\cdots m_{k}$ is maximal only if in the partition of $n=m_{1}+\ldots +m_{k}$ there are only 3s except for at most either precisely one 4 or at most two 2s. Since presently $n\equiv 2\mod 3$, the only possibility is $n=2+\frac{n-2}{3}\cdot 3$.

The rest follows using Lemma 3.2.

3.3 Final step in the proofs of Theorems 1.2 and 1.3

In this section we prove Theorem 1.2 (first inequality in Theorem 3.3) and Theorem 1.3 (second and third inequality in Theorem 3.3).

Theorem 1.2

Let $n \ge 6$ be divisible by 3. Then $\Delta _d(n)>0$ for almost all d.

Theorem 1.3

Let $n \ge 6$. Let $n \equiv 1$ or $n \equiv 2 \mod 3$. Then $\Delta _d(n) < 0$ for almost all d. In particular let

$$\begin{aligned} {\tilde{C}}_1(n):= & {} 1 + 6\, \big ( 1 + \ln (n-1) \big ) \, n , \\ {\tilde{C}}_2(n):= & {} 1 + 3\, \big ( 3 + \ln (n+1) \big ) \, n. \end{aligned}$$

Then for $r=1$ or $r=2$, $\Delta _d(n)<0$ for $n \equiv r \mod 3$ and $d \ge {\tilde{C}}_r(n)$.

Note that in Theorem 1.3, we can take the least upper integer bound $\left\lceil C\right\rceil$ of the constants since d is integer. We show the following improvement of Theorems 1.2 and 1.3.

Theorem 3.3

Let $n\ge 6$. Then:

$$\begin{aligned} \left( p_{d }\left( n\right) \right) ^{2}> & {} p_{d }\left( n-1\right) p_{d }\left( n+1\right) \text { for }n\equiv 0\mod 3,\,d\ge C_{0}\left( n\right) ,\\ \left( p_{d }\left( n\right) \right) ^{2}< & {} p_{d }\left( n-1\right) p_{d }\left( n+1\right) \text { for }n\equiv 1\mod 3,\, d \ge C_{1}\left( n\right) ,\\ \left( p_{d }\left( n\right) \right) ^{2}< & {} p_{d }\left( n-1\right) p_{d }\left( n+1\right) \text { for }n\equiv 2\mod 3, \, d\ge C_{2}\left( n\right) \end{aligned}$$

with

$$\begin{aligned} C_{0}\left( n\right):= & {} 1+2\frac{ \ln \left( 2 \right) +\ln \left( n /3\right) /3 }{\ln \left( 9/8\right) }n,\\ C_{1}\left( n\right):= & {} 1+2\frac{ \ln \left( 2 \right) +\ln \left( \left( n-1\right) /3\right) /3 }{\ln \left( 9/8\right) }n, \\ C_{2}\left( n\right):= & {} 1+\frac{\ln \left( 3 \right) + \ln \left( \left( n+1\right) /3\right) /3 }{\ln \left( 9/8\right) } n. \end{aligned}$$

Proof

We apply the upper and lower bounds from Sect. 3.2. Let $n\ge 3$.

Case 1: $n\equiv 0\mod 3$. Then

$$\begin{aligned}&\frac{\left( p_{d }\left( n\right) \right) ^{2}}{p_{d }\left( n-1\right) p_{d }\left( n+1\right) }\\&\quad \ge \frac{3^{\left( d-1\right) 2n/3}}{\left( \left( n/3\right) !\right) ^{2}p_{1 }\left( n-1\right) \left( 2\cdot 3^{\left( n-3\right) /3}\right) ^{d-1}p_{1 }\left( n+1\right) \left( 4\cdot 3^{ \left( n-3\right) /3}\right) ^{d-1} }\\&\quad =\frac{1}{\left( \left( n/3\right) !\right) ^{2}p_{1 }\left( n-1\right) p_{1 }\left( n+1\right) }\left( \frac{9}{8}\right) ^{d-1}\\&\quad \ge \left( n/3\right) ^{-2n/3}2^{-2n}\left( \frac{9}{8}\right) ^{d-1} >1 \end{aligned}$$

for $d\ge 1+2\frac{ \ln \left( 2 \right) +\ln \left( n /3\right) /3 }{\ln \left( 9/8\right) }n$.

Case 2: $n\equiv 1\mod 3$. Then

$$\begin{aligned} \frac{\left( p_{d }\left( n\right) \right) ^{2}}{p_{d }\left( n-1\right) p_{d }\left( n+1\right) }\le & {} \frac{\left( \left( n-1\right) /3\right) !\left( \left( n-1\right) /3\right) !\left( \left( 4\cdot 3^{\left( n-4\right) /3}\right) ^{d-1}p_{1 }\left( n\right) \right) ^{2} }{\left( 2\cdot 3^{\left( n-1\right) /3}\right) ^{d-1}3^{\left( n-1\right) \left( d-1\right) /3}}\\= & {} \left( \left( \left( n-1\right) /3\right) !p_{1 }\left( n\right) \right) ^{2}\left( \frac{8}{9}\right) ^{d-1}\\\le & {} \left( \left( n-1\right) /3\right) ^{2n/3}2^{2n}\left( \frac{8}{9}\right) ^{d-1} <1 \end{aligned}$$

for $d\ge 1+2\frac{ \ln \left( 2\right) +\ln \left( \left( n-1\right) /3\right) /3 }{\ln \left( 9/8\right) }n$.

Case 3: $n\equiv 2\mod 3$ and $n\ge 11$. Then

$$\begin{aligned}&\frac{\left( p_{d }\left( n\right) \right) ^{2}}{p_{d }\left( n-1\right) p_{d }\left( n+1\right) }\\&\quad \le \frac{\left( \frac{\left( 2\cdot 3^{\left( n-2\right) /3}\right) ^{d-1}}{\left( \left( n-2\right) /3\right) !}+ \left( 16\cdot 3^{\left( n-8\right) /3}\right) ^{d-1}p_{1 }\left( n\right) \right) ^{2}}{\frac{3\left( 4\cdot 3^{\left( n-5\right) /3}\right) ^{d-1}}{2\left( \left( n-5 \right) /3\right) !}\frac{3^{\left( d-1\right) \left( n+1\right) /3}}{\left( \left( n+1\right) /3\right) !}}\\&\quad =\frac{2\left( n+1\right) /3}{3\left( n-2\right) / 3}+\frac{ \frac{2}{\left( \left( n-2\right) /3\right) !}+ \left( \frac{8}{9}\right) ^{d-1} p_{1 }\left( n\right) }{\frac{3}{2\left( \left( n-5 \right) /3\right) !\left( \left( n+1\right) /3\right) !}}\left( \frac{8 }{ 9 }\right) ^{d-1}p_{1 }\left( n\right) \\&\quad <\frac{8}{9}+ \frac{2}{3}\left( 2+\left( \left( n+1\right) /3\right) ^{n/3}\left( \frac{8}{9}\right) ^{d-1}2^{n}\right) \left( \left( n+1\right) /3\right) ^{n/3} \left( \frac{8}{9}\right) ^{d-1}2^{n}. \end{aligned}$$

If we suppose $d\ge C_{2}\left( n\right)$ we obtain $\left( \left( n+1\right) /3\right) ^{n/3}\left( {8}/{9}\right) ^{d-1}2^{n}\le \left( 2/3\right) ^{ n}$ and therefore,

$$\begin{aligned} \frac{\left( p_{d}\left( n\right) \right) ^{2}}{p_{d}\left( n-1\right) p_{d}\left( n+1\right) }< \frac{8}{9}+\frac{2}{3}\left( 2+ \left( \frac{2}{3}\right) ^{ n}\right) \cdot \left( \frac{2}{3}\right) ^{ n}<1. \end{aligned}$$

Case 4: $n=8$. Here the previous argument does not apply, since $\frac{2\left( n+1\right) }{3\left( n-2\right) }=1$. Therefore, this case has to be treated separately. To estimate $p_{d}\left( 8\right)$, we have to determine the third largest products $m_{1}\cdots m_{k}$ when $m_{1}+\ldots +m_{k}=8$. The largest product is $18=2\cdot 3\cdot 3$. From (3.1) we obtain 16 from $4\cdot 4=4\cdot 2\cdot 2= 2^{4}$ as the second largest value. So, the third largest value could be 15, which is indeed obtained for $5\cdot 3$. Therefore, we obtain

$$\begin{aligned} p_{d}\left( 8\right) <\frac{1}{2}18^{d-1}+\frac{25}{24}16^{d-1}+21 \cdot 15^{d-1} . \end{aligned}$$

In Lemma 4.1 we show that

$$\begin{aligned} p_{d}\left( 7\right) >\frac{3}{2}12^{d-1} . \end{aligned}$$

In a similar way we can obtain that

$$\begin{aligned} p_{d}\left( 9\right) >\frac{1}{6}27^{d-1}+\frac{7}{6}24^{d-1} \end{aligned}$$

from $3\cdot 3\cdot 3$, the largest product, and $2\cdot 2\cdot 2\cdot 3=2\cdot 4\cdot 3$ the second largest products. Therefore,

$$\begin{aligned}&p_{d}\left( 7\right) p_{d}\left( 9\right) -\left( p_{d}\left( 8\right) \right) ^{2}\\&\quad>\frac{17}{24}288^{d-1} -21 \cdot 270^{d-1}-\frac{625}{576}256^{d-1}-\frac{ 175}{4 }240^{d-1} -441 \cdot 225^{d-1}\\&\quad>\frac{17}{24}288^{d-1} -508 \cdot 270^{d-1}>0 \end{aligned}$$

for $d\ge 103$. For $9\le d\le 102$ it can be checked that also $\frac{\left( p_{d}\left( 8\right) \right) ^{2}}{p_{d}\left( 7\right) p_{d}\left( 9\right) }<1$.

Remarks

With minor simplifications (see below) and using numerical values, we obtain also the following lower bounds for d:

$$\begin{aligned} \left( p_{d }\left( n\right) \right) ^{2}> & {} p_{d }\left( n-1\right) p_{d }\left( n+1\right) \text { for }n\equiv 0\mod 3,\, d\ge C_{0}^{*}\left( n\right) ,\\ \left( p_{d }\left( n\right) \right) ^{2}< & {} p_{d }\left( n-1\right) p_{d }\left( n+1\right) \text { for }n\equiv 1\mod 3,\, d\ge C_{1}^{*}\left( n\right) ,\\ \left( p_{d }\left( n\right) \right) ^{2}< & {} p_{d }\left( n-1\right) p_{d }\left( n+1\right) \text { for }n\equiv 2\mod 3, \, d\ge C_{2}^{*}\left( n\right) \end{aligned}$$

where we have

$$\begin{aligned} C_{0}^{*}\left( n\right):= & {} 1+ 5.67\left( 1+\ln \left( n\right) \right) n,\\ C_{1}^{*}\left( n\right):= & {} 1+ 5.67\left( 1+\ln \left( n-1\right) \right) n, \\ C_{2}^{*}\left( n\right):= & {} 1+2.84\left( 2.2 +\ln \left( n+1\right) \right) n. \end{aligned}$$

Proof

(Simplifications) Note that $\ln \left( n/3\right) /3=-\ln \left( 3\right) /3+\ln \left( n\right) /3$ and

$$\begin{aligned} 2\frac{\ln \left( 2\right) -\ln \left( 3\right) /3}{\ln \left( 9/8\right) }\approx 5.55<5.67 \end{aligned}$$

and $2\frac{1/3}{\ln \left( 9/8\right) }\approx 5.66<5.67$. Therefore, $5.67\left( 1+\ln \left( n\right) \right) >2\frac{\ln \left( 2\right) +\ln \left( n/3\right) /3}{\ln \left( 9/8\right) }$ which implies $C_{r }^{*}\left( n\right) >C_{r }\left( n\right)$ for $r =0,1$. On the other hand $\frac{\ln \left( 3 \right) -\ln \left( 3\right) /3}{\ln \left( 9/8\right) }\approx 6.22< 6.248$ and $\frac{1/3}{\ln \left( 9/8\right) }\approx 2.83<2.84$. Therefore, $2.84\left( 2.2 +\ln \left( n+1\right) \right) >\frac{\ln \left( 3 \right) +\ln \left( \left( n+1\right) /3\right) /3}{\ln \left( 9/8\right) }$ which implies $C_{2 }^{*}\left( n\right) >C_{ 2}\left( n\right)$.

4 Proof of Theorem 1.1

We first express $p_d(n)$ by $p_d(n-1)$. This may be interesting in its own way.

Lemma 4.1

Let $d \ge 1$. Then $p_{d}\left( 0 \right) = p_{d}\left( 1 \right) =1$ and $p_{d} (2) = 2^{d-1}+ p_{d}(1)$. Moreover, $p_{d}(3) = 3^{d-1}+p_{d}(2)$, $p_{d}(4) = \frac{3}{2}\cdot 4^{d-1}+\frac{1}{2}2^{d-1}+p_{d}(3)$, and

$$\begin{aligned} p_{d}(5)= & {} 5^{d-1}+ 6^{d-1}+p_{d}(4),\\ p_{d}(6)= & {} \frac{1}{2}9^{d-1}+\frac{7}{6}8^{d-1}+6^{d-1}+\frac{1}{2}4 ^{ d-1 }+\frac{1}{2}3^{d-1} +\frac{1 }{3}2^{d-1}+p_{d}(5),\\ p_{d}(7)= & {} \frac{3 }{2 }12^{d-1}+10^{d-1}+7^{d-1}+\frac{1}{2}6^{d-1}+p_{d}(6),\\ p_{d}\left( 8\right)= & {} \frac{1}{2 }18^{d-1}+\frac{25}{ 24}16^{d-1}+15^{d-1}+12^{d-1}+\frac{7 }{4 }8^{d-1}+\frac{1}{2}6^{d-1} \\&{}+\frac{23 }{24}4^{d-1}+\frac{1}{4}2^{d-1}+p_{d}\left( 7\right) . \end{aligned}$$

Corollary 4.2

For $1\le n \le 7$ we have the following relations between $\left( p_{d }\left( n\right) \right) ^{2}$ and $p_{d }\left( n-1\right) p_{d }\left( n+1\right)$:

$$\begin{aligned} \left( p_{d}\left( 1\right) \right) ^{2}< & {} p_{d}\left( 0\right) p_{d}\left( 2\right) \text { for all }d\ge 1, \\ \left( p_{d}\left( 2\right) \right) ^{2}> & {} p_{d}\left( 1\right) p_{d}\left( 3\right) \text { for all }d\ge 1, \\ \left( p_{d}\left( 3\right) \right) ^{2}> & {} p_{d}\left( 2\right) p_{d}\left( 4\right) \text { if and only if }d\ge 4, \\ \left( p_{d}\left( 4\right) \right) ^{2}< & {} p_{d}\left( 3\right) p_{d}\left( 5\right) \text { if and only if }d\ge 6, \\ \left( p_{d}\left( 5\right) \right) ^{2}> & {} p_{d}\left( 4\right) p_{d}\left( 6\right) \text { if and only if }d\ge 10, \\ \left( p_{d}\left( 6\right) \right) ^{2}> & {} p_{d}\left( 5\right) p_{d}\left( 7\right) \text { for all }d\ge 1, \\ \left( p_{d}\left( 7\right) \right) ^{2}< & {} p_{d}\left( 6\right) p_{d}\left( 8\right) \text { if and only if }d\le 3\text { or }d\ge 11. \end{aligned}$$

To complete the proof of Theorem 1.1, Lemma 4.1 could be extended to $n=9$ and $n=10$. Since the bounds $C_{2}\left( 8\right)$ and $1+18 \frac{\ln \left( 2\right) +\ln \left( 3\right) /3}{\ln \left( 9/8\right) }$ are not too large, we can use the results of Theorem 3.3.

Assume $n=8$. Then $\left\lceil C_{2}\left( 8\right) \right\rceil = 101$. It can be checked that $\frac{\left( p_{d}\left( 8\right) \right) ^{2}}{p_{d}\left( 7\right) p_{d}\left( 9\right) }< 1$ for $9 \le d \le 100$ and that $\frac{\left( p_{d}\left( 8\right) \right) ^{2}}{p_{d}\left( 7\right) p_{d}\left( 9\right) }>1$ for $d\le 8$. For $n=9$ we obtain $\left\lceil 1+18 \frac{\ln \left( 2\right) +\ln \left( 9/3\right) /3}{\ln \left( 9/8\right) }\right\rceil =163$. Again, it can be checked that $\frac{\left( p_{d}\left( 9\right) \right) ^{2}}{p_{d}\left( 8\right) p_{d}\left( 10\right) }>1$ for all $3\le d \le 162$ and that $\frac{\left( p_{d}\left( 9\right) \right) ^{2}}{p_{d}\left( 8\right) p_{d}\left( 10\right) }<1$ for $d\le 2$.

Appendices

Appendix A: bounds on d

Fig. 2

Graphical representation of the bounds on d from Table 3: left $C_{1}\left( n\right) ,C_{2}\left( n\right)$, right ${\tilde{C}}_{1}\left( n\right) ,{\tilde{C}}_{2}\left( n\right)$

Table 3 Comparison of the bounds on d

Appendix B: proofs of Lemma 4.1 and Corollary 4.2

Proof of Lemma 4.1

We have $p_d(0)= p_d(1)=1$. Moreover $p_d(2) = \frac{\sigma _{d}\left( 2\right) +1}{2}=2^{d-1}+1$. Further, $p_{d}(3)=\frac{\sigma _{d}\left( 3\right) }{3}+\frac{ \sigma _{d}\left( 2\right) }{2}+\frac{1}{6}=3^{d-1}+2^{d-1}+1=3^{d-1}+p_{d}(2)$. Moreover,

$$\begin{aligned} p_{d}(4)= & {} \frac{\sigma _{d}\left( 4\right) }{4}+\frac{\sigma _{d}\left( 3\right) }{3}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{2}}{8}+\frac{\sigma _{d}\left( 2\right) }{4}+\frac{1}{24} =\frac{3}{2}\cdot 4^{d-1}+\frac{1}{2}2^{d-1}+p_{d}(3),\\ p_{d}(5)= & {} \frac{\sigma _{d}\left( 5\right) }{5}+\frac{\sigma _{d}\left( 4\right) }{4}+\frac{\sigma _{d}\left( 3\right) \sigma _{d}\left( 2\right) }{6}+\frac{\sigma _{d}\left( 3\right) }{6}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{2}}{8}+\frac{\sigma _{d}\left( 2\right) }{12}+\frac{1}{120}\\= & {} 6^{d-1}+5^{d-1}+\frac{3}{2}\cdot 4 ^{ d-1 }+ 3^{d-1}+\frac{3}{2}\cdot 2^{d-1 } +1=6^{d-1}+5^{d-1}+ p_{d}(4),\\ p_{d}(6)= & {} \frac{\sigma _{d}\left( 6\right) }{6}+\frac{\sigma _{d}\left( 5\right) }{5}+\frac{\sigma _{d}\left( 4\right) \sigma _{d}\left( 2\right) }{8}+\frac{\sigma _{d}\left( 4\right) }{8}+\frac{\left( \sigma _{d}\left( 3\right) \right) ^{2}}{18}+\frac{\sigma _{d}\left( 3\right) \sigma _{d}\left( 2\right) }{6}\\&{}+\frac{\sigma _{d}\left( 3\right) }{18}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{3}}{48}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{2}}{16}+\frac{\sigma _{d}\left( 2\right) }{48}+\frac{1}{720}\\= & {} \frac{1}{2 }9^{d-1}+\frac{7}{6 }8^{d-1}+2\cdot 6^{d-1}+5^{d-1}+2\cdot 4^{d-1}+\frac{3}{2}3^{d-1}+\frac{11}{6}2^{d-1}+1\\= & {} \frac{1}{2 }9^{d-1}+\frac{7}{6 }8^{d-1}+6^{d-1}+\frac{1}{2}4 ^{ d-1 }+\frac{1}{2}3^{d-1} +\frac{1 }{3 }2^{d-1}+p_{d}(5),\\ p_{d}(7)= & {} \frac{\sigma _{d}\left( 7\right) }{7}+\frac{\sigma _{d}\left( 6\right) }{6}+\frac{\sigma _{d}\left( 5\right) \sigma _{d}\left( 2\right) }{10}+\frac{\sigma _{d}\left( 5\right) }{10}+\frac{\sigma _{d}\left( 4\right) \sigma _{d}\left( 3\right) }{12}\\&{}+\frac{\sigma _{d}\left( 4\right) \sigma _{d}\left( 2\right) }{8}+\frac{\sigma _{d}\left( 4\right) }{24}+\frac{\left( \sigma _{d}\left( 3\right) \right) ^{2}}{18}+\frac{\sigma _{d}\left( 3\right) \left( \sigma _{d}\left( 2\right) \right) ^{2}}{24}\\&{}+\frac{\sigma _{d}\left( 3\right) \sigma _{d}\left( 2\right) }{12}+\frac{\sigma _{d}\left( 3\right) }{72}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{3}}{48}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{2}}{48}+\frac{\sigma _{d}\left( 2\right) }{240}+\frac{1}{5040}\\= & {} \frac{3 }{2 }12^{d-1}+10^{d-1}+\frac{1}{2}9^{d-1}+\frac{7}{6}8^{d-1}+7^{d-1}+\frac{5}{2}6^{d-1}+5^{d-1}+2\cdot 4^{d-1}\\&{}+\frac{3 }{2}3^{d-1}+\frac{11}{6 }2^{d-1}+1 = \frac{3}{2}12^{d-1}+10^{d-1}+7^{d-1}+\frac{1}{2}6^{d-1}+p_{d}(6), \end{aligned}$$

and finally

$$\begin{aligned} p_{d}\left( 8\right)= & {} \frac{\sigma _{d}\left( 8\right) }{8}+\frac{\sigma _{d}\left( 7\right) }{7}+\frac{\sigma _{d}\left( 6\right) \sigma _{d}\left( 2\right) }{12}+\frac{\sigma _{d}\left( 6\right) }{12}+\frac{\sigma _{d}\left( 5\right) \sigma _{d}\left( 3\right) }{15}\\&{}+\frac{\sigma _{d}\left( 5\right) \sigma _{d}\left( 2\right) }{10}+\frac{\sigma _{d}\left( 5\right) }{30}+\frac{\left( \sigma _{d}\left( 4\right) \right) ^{2}}{32}+\frac{\sigma _{d}\left( 4\right) \sigma _{d}\left( 3\right) }{12}\\&{}+\frac{\sigma _{d}\left( 4\right) \left( \sigma _{d}\left( 2\right) \right) ^{2}}{32}+\frac{\sigma _{d}\left( 4\right) \sigma _{d}\left( 2\right) }{16}+\frac{\sigma _{d}\left( 4\right) }{96}+\frac{\left( \sigma _{d}\left( 3\right) \right) ^{2}\sigma _{d}\left( 2\right) }{36}\\&{}+\frac{\left( \sigma _{d}\left( 3\right) \right) ^{2}}{36}+\frac{\sigma _{d}\left( 3\right) \left( \sigma _{d}\left( 2\right) \right) ^{2}}{24}+\frac{\sigma _{d}\left( 3\right) \sigma _{d}\left( 2\right) }{36}+\frac{\sigma _{d}\left( 3\right) }{360}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{4}}{384}\\&{}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{3}}{96}+\frac{\left( \sigma _{d}\left( 2\right) \right) ^{2}}{192}+\frac{\sigma _{d}\left( 2\right) }{1440}+\frac{1}{40320}\\= & {} \frac{1}{2}18^{d-1}+\frac{25}{ 24}16^{d-1}+15^{d-1}+\frac{5}{2}12^{d-1}+10^{d-1} +\frac{1}{2}9^{d-1}+\frac{35}{12 }8^{d-1}\\&{} +7^{d-1} +3\cdot 6^{d-1}+5^{d-1}+\frac{71}{24}4^{d-1}+\frac{3}{2}3^{d-1}+\frac{25}{12 }2^{d-1}+1\\= & {} \frac{1}{2 }18^{d-1}+\frac{25}{ 24}16^{d-1} +15^{d-1}+12^{d-1}+\frac{7}{4 }8^{d-1}+\frac{1}{2}6^{d-1}+\frac{23 }{24 }4^{d-1}\\&+\frac{1}{4 }2^{d-1} +p_{d}\left( 7\right) .\\ \end{aligned}$$

$\square$

Proof of Corollary 4.2

The property $\Delta _d(1)< 0$ follows from $p_d(0)=p_d(1)=1$ and $p_d(2)= 2^{d-1} +1 >1$ for $d \ge 1$. We use the results of Lemma 4.1 to estimate the $p_{d}\left( n\right)$. Therefore, $p_{d}(2)\ge 2^{d-1}$ and $p_{d}(3) \le 3^{d}$. Thus,

$$\begin{aligned} \frac{p_{d}(2)^{2}}{p_{d}(1) \, p_{d}(3) }\ge \frac{4^{d-1}}{3^{d}}>1 \end{aligned}$$

for $d\ge 5$. For the remaining cases $1\le d\le 4$ we checked directly that $\left( p_{d}\left( 2 \right) \right) ^{2} >p_{d}\left( 1 \right) p_{d}\left( 3 \right)$.

We have $p_{d}\left( 3\right) \ge 3^{d-1}$, $p_{d}\left( 2\right) \le 2^{d}$, and $p_{d}\left( 4 \right) \le 5\cdot 4^{d-1}$. Therefore,

$$\begin{aligned} \frac{\left( p_{d }\left( 3\right) \right) ^{2}}{p_{d }\left( 2\right) p_{d }\left( 4\right) }\ge \frac{9^{d-1}}{10\cdot 8^{d-1}}>1 \end{aligned}$$

for $d\ge 21$. For $4 \le d\le 20$ it can be checked that $\left( p_{d}\left( 3 \right) \right) ^{2}>p_{d}\left( 2 \right) p_{d}\left( 4 \right)$ and for $d\le 3$ that $\left( p_{d}\left( 3\right) \right) ^{2}<p_{d}\left( 2\right) p_{d}\left( 4\right)$.

From Lemma 4.1, we obtain $p_{d}\left( 3 \right) p_{d}\left( 5 \right) \ge 18^{d-1}$ and $p_{d}\left( 4\right) ^{2}\le \left( 5 \cdot 4^{d-1}\right) ^{2}=25\cdot 16^{d-1}$. Then

$$\begin{aligned} \frac{p_{d}\left( 3\right) p_{d}\left( 5\right) }{ \left( p_{d}\left( 4\right) \right) ^2}\ge \frac{18^{d-1}}{ 25\cdot 16^{d-1}}>1 \end{aligned}$$

for $d\ge 29$. For $6\le d\le 28$ we can check that $\frac{p_{d}\left( 3 \right) p_{d}\left( 5\right) }{\left( p_{d}\left( 4\right) \right) ^{2}}>1$ and for $d\le 5$ that $\frac{p_{d}\left( 3 \right) p_{d}\left( 5 \right) }{\left( p_{d}\left( 4\right) \right) ^{2}}<1$.

We have $p_{d}\left( 5 \right) \ge 6^{d-1}$, $p_{d}\left( 4\right) \le \frac{3}{2}4^{d-1}+\frac{7}{2}3^{d-1}$, and $p_{d}\left( 6\right) \le \frac{1}{2}9^{d-1}+\frac{21}{2}8^{d-1}$. Therefore,

$$\begin{aligned} \frac{\left( p_{d}\left( 5\right) \right) ^{2}}{p_{d}\left( 4\right) p_{d}\left( 6\right) }\ge & {} \frac{36^{d-1}}{\frac{3}{4}36^{d-1}+\frac{63 }{4}32^{d-1}+\frac{7}{4}27^{d-1}+\frac{147 }{4}24^{d-1}}\\> & {} \frac{36^{d-1}}{\frac{3}{4}36^{d-1}+55 \cdot 32^{d-1}}\ge 1 \end{aligned}$$

for $d\ge 47$. For $10 \le d\le 46$ it can be checked that $\left( p_{d}\left( 5 \right) \right) ^{2} >p_{d}\left( 4\right) p_{d}\left( 6\right)$ and for $d\le 9$ that $\left( p_{d}\left( 5\right) \right) ^{2}< p_{d}\left( 4\right) p_{d}\left( 6\right)$.

We have $p_{d}\left( 6\right) \ge \frac{1}{2}9^{d-1}$, $p_{d}\left( 5\right) \le 7\cdot 6^{d-1}$, and $p_{d}\left( 7\right) \le 15\cdot 12^{d-1}$. Therefore,

$$\begin{aligned} \frac{\left( p_{d}\left( 6\right) \right) ^{2}}{p_{d}\left( 5\right) p_{d}\left( 7\right) }\ge \frac{81^{d-1}}{420\cdot 72^{d-1}}>1 \end{aligned}$$

for $d\ge 53$. For $d\le 52$ it can be checked that $\left( p_{d}\left( 6\right) \right) ^{2}>p_{d}\left( 5\right) p_{d}\left( 7\right)$.

We have $p_{d}\left( 7\right) \le 15\cdot 12^{d-1}$, $p_{d}\left( 6\right) >\frac{1}{2}9^{d-1}$, and $p_{d}\left( 8\right) >\frac{1}{2}18^{d-1}$. Therefore,

$$\begin{aligned} \frac{\left( p_{d}\left( 7\right) \right) ^{2}}{p_{d}\left( 6\right) p_{d}\left( 8\right) }<\frac{225\cdot 144^{d-1}}{\frac{1}{4}162^{d-1}}<1 \end{aligned}$$

for $d\ge 59$. For $d\le 3$ and $11\le d \le 58$ it can also be checked that $\frac{\left( p_{d}\left( 7\right) \right) ^{2}}{p_{d}\left( 6\right) p_{d}\left( 8\right) }<1$ and for $4\le d\le 10$ that $\frac{\left( p_{d}\left( 7\right) \right) ^{2}}{p_{d}\left( 6\right) p_{d}\left( 8\right) }>1$. $\square$

Appendix C: patterns in the n and d directions

Table 4 Exceptions for $1\le d\le 20$ and $1\le n\le 26$