On Prime Numbers between kn and (k + 1) n

Wing K. Yu

Independent Researcher, Arlington, USA.

DOI: 10.4236/jamp.2023.1111234                

Abstract

In this paper along with the previous studies on analyzing the binomial coefficients, we will complete the proof of a theorem. The theorem states that for two positive integers n and k, when n ≥ k - 1, there always exists at least a prime number p such that kn < p ≤ (k +1)n. The Bertrand-Chebyshev’s theorem is a special case of this theorem when k = 1. In the field of prime number distribution, just as the prime number theorem provides the approximate number of prime numbers relative to natural numbers, while the new theory indicates that prime numbers exist in the specific intervals between natural numbers, that is, the new theorem provides the approximate positions of prime numbers among natural numbers.

Keywords

Bertrand’s Postulate-Chebyshev’s Theorem, The Prime Number Theorem, Landau Problems, Legendre’s Conjecture, Prime Number Distribution

1. Introduction

The Bertrand-Chebyshev’s theorem states that for any positive integer n, there always exists a prime number p such that $n<p\le \text{2}n$ . Pafnuty Chebyshev proved this in 1850 [1] . In 2006, M. El Bachraoui [2] extended the theorem and proved that for any positive integer n, there exists a prime number p such that $\text{2}n<p\le \text{3}n$ . In 2011, Andy Loo [3] proved that when n ≥ 2, there are prime numbers in the interval (3n, 4n). In 2013, Vladimir Shevelev et al. [4] proved that when the integer k ≤ 100,000,000, only k = 1, 2, 3, 5, 9, 14, for all n ≥ 1, the interval (kn, (k + 1)n) contains prime numbers. This raises the question: for all k ≥ 1, under what conditions does the interval (kn, (k + 1)n) contain prime numbers? Previously, the author partially answered this question in the paper [5] by

analyzing the binomial coefficients $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ where λ is a positive integer. In that

paper, the author proved that when $n\ge \lambda -\text{2}\ge \text{25}$ , i.e., when $n\ge \lambda -\text{2}$ and $\lambda \ge \text{27}$ , there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ . In this article, we will use the same method to complete the entire work on this problem. We will prove that when $n\ge \lambda -\text{2}$ and $\lambda \ge 3$ , there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ . Then, by converting λ to k, we prove that for two positive integers n and k, when $n\ge k-\text{1}$ , there is always at least a prime number p such that $kn<p\le \left(k+1\right)n$ . The Bertrand-Chebyshev’s theorem is a special case of this theorem when k = 1.

We will use the same definition and concepts from [5] in this section and in section 2. In section 3, we will prove that for λ from 3 to 26, when $n\ge \lambda -\text{2}$ , there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ . In section 4, we will convert λ to k, and complete this article.

Definition: ${\Gamma }_{a\ge p>b}\left\{\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right\}$ denotes the prime number factorization operator of the integer expression $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ . It is the product of the prime numbers in the decomposition of $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ in the range of $a\ge p>b$ . In this operator, p is a prime number, $a$ and $b$ are real numbers, and $\lambda n\ge a\ge p>b\ge 1$ .

It has some properties:

It is always true that ${\Gamma }_{a\ge p>b}\left\{\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right\}\ge 1$ (1.1)

If there is no prime number in $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ within the range of $a\ge p>b$ , then ${\Gamma }_{a\ge p>b}\left\{\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right\}=1$ , or vice versa, if ${\Gamma }_{a\ge p>b}\left\{\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right\}=1$ , then there is no prime number in $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ within the range of $a\ge p>b$ . (1.2)

For example, when $\lambda =5$ and $n=4$ , ${\Gamma }_{16\ge p>10}\left\{\left(\begin{array}{c}20\\ 4\end{array}\right)\right\}={13}^0\cdot {11}^0=1$ . No prime number 13 or 11 is in $\left(\begin{array}{c}20\\ 4\end{array}\right)$ within the range of $16\ge p>10$ .

If there is at least one prime number in $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ in the range of $a\ge p>b$ , then ${\Gamma }_{a\ge p>b}\left\{\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right\}>1$ , or vice versa, if ${\Gamma }_{a\ge p>b}\left\{\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right\}>1$ , then there is at least one prime number in $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ within the range of $a\ge p>b$ . (1.3)

For example, when $\lambda =5$ and $n=4$ , ${\Gamma }_{18\ge p>16}\left\{\left(\begin{array}{c}20\\ 4\end{array}\right)\right\}=17>1$ . A prime number 17 is in $\left(\begin{array}{c}20\\ 4\end{array}\right)$ within the range of $18\ge p>16$ .

Let $v_p\left(n\right)$ be the p-adic valuation of n, the exponent of the highest power of p that divides n.

We define $R\left(p\right)$ by the inequalities $p^{R\left(p\right)}\le \lambda n<p^{R\left(p\right)+1}$ , and determine the p-adic valuation of $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ .

$\begin{array}{c}{v}_p\left(\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right)=v_p\left(\left(\lambda n\right)!\right)-v_p\left(\left(\left(\lambda -1\right)n\right)!\right)-v_p\left(n!\right)\\ ={\displaystyle {\sum }_{i=1}^{R\left(p\right)}\left(\lfloor \frac{\lambda n}{p^i}\rfloor -\lfloor \frac{\left(\lambda -1\right)n}{p^i}\rfloor -\lfloor \frac{n}{p^i}\rfloor \right)}\le R(\; p\; )\end{array}$

because for any real numbers $a$ and $b$ , the expression of $\lfloor a+b\rfloor -\lfloor a\rfloor -\lfloor b\rfloor$ is 0 or 1.

Thus, if p divides $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)$ , then $v_p\left(\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right)\le R\left(p\right)\le {\log }_p\left(\lambda n\right)$ , or $p^{v_p\left(\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right)}\le p^{R\left(p\right)}\le \lambda n$ (1.4)

If $n\ge p>\lfloor \sqrt{\lambda n}\rfloor$ , then $0\le v_p\left(\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\right)\le R\left(p\right)\le 1$ . (1.5)

Let $\pi \left(n\right)$ be the number of distinct prime numbers less than or equal to 𝑛. Among the first six consecutive natural numbers are three prime numbers 2, 3, and 5. Then, for each additional six consecutive natural numbers, at most one

can add two prime numbers, $p\equiv 1\left(\text{MOD}6\right)$ and $p\equiv 5\left(\text{MOD}6\right)$ . Thus, $\pi \left(n\right)\le \lfloor \frac{n}{3}\rfloor +2\le \frac{n}{3}+2$ . (1.6)

From the prime number decomposition, when $n>\lfloor \sqrt{\lambda n}\rfloor$ ,

$\begin{array}{c}\left(\begin{array}{c}\lambda n\\ n\end{array}\right)={\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\cdot {\Gamma }_{n\ge p>\lfloor \sqrt{\lambda n}\rfloor }\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\\ \cdot {\Gamma }_{\lfloor \sqrt{\lambda n}\rfloor \ge p}\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\end{array}$

When $n\le \lfloor \sqrt{\lambda n}\rfloor$ , $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\le {\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\cdot {\Gamma }_{\lfloor \sqrt{\lambda n}\rfloor \ge p}\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}$

Thus, $\begin{array}{c}\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\le {\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\cdot {\Gamma }_{n\ge p>\lfloor \sqrt{\lambda n}\rfloor }\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\\ \cdot {\Gamma }_{\lfloor \sqrt{\lambda n}\rfloor \ge p}\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\end{array}$

${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}={\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}$ since all prime numbers in $n!$ do not appear in the range of $\lambda n\ge p>n$ .

If $n\ge \lambda -2$ , and there is a prime number p in ${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}$ , then $p\ge n+1=\sqrt{\left(n+2\right)n+1}>\sqrt{\lambda n}$ . From (1.5), $0\le v_p\left({\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}\right)\le R\left(p\right)\le 1$ . Thus, if $n\ge \lambda -2$ , every prime number in ${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}$ has a power of 0 or 1. (1.7)

Referring to (1.5), ${\Gamma }_{n\ge p>\lfloor \sqrt{\lambda n}\rfloor }\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\le {\displaystyle {\prod }_{n\ge p}p}$ . It has been proven [6] that for $n\ge 3$ , ${\displaystyle {\prod }_{n\ge p}p}<2^{2n-3}$ .

When $n=2$ , ${\displaystyle {\prod }_{n\ge p}p}=2^{2n-3}$ , then for $n\ge 2$ , ${\Gamma }_{n\ge p>\lfloor \sqrt{\lambda n}\rfloor }\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\le {\displaystyle {\prod }_{n\ge p}p}\le 2^{2n-3}$ .

Referring to (1.4) and (1.6), ${\text{Γ}}_{\lfloor \sqrt{\lambda n}\rfloor \ge p}\left\{\frac{\left(\lambda n\right)!}{n\text{!}\cdot \left(\left(\lambda -1\right)n\right)!}\right\}\le {\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+2}$ .

Thus, for $\lambda \ge 3$ and $n\ge 2$ , $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\le {\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}\cdot 2^{2n-3}\cdot {\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+2}$ (1.8)

2. Lemmas

Lemma 1: For $\lambda \ge 3$ and $n\ge 2$ , ${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}=f_8\left(n,\lambda \right)$ (2.1)

Proof:

Let a real number $x\ge 3$ , and $f_1\left(x\right)=\frac{2\left(2x-1\right)}{x-1}$ ; then, ${f^{\prime }}_1\left(x\right)=\frac{2\left(x-1\right){\left(2x-1\right)}^{\prime }-2\left(2x-1\right){\left(x-1\right)}^{\prime }}{{\left(x-1\right)}^2}=\frac{-2}{{\left(x-1\right)}^2}<0$ .

Thus, $f_1\left(x\right)$ is a strictly decreasing function for $x\ge 3$ .

Since $f_1\left(3\right)=5$ , and ${\lim }_{x\to \infty }{f}_1\left(x\right)=4$ , for $x\ge 3$ , we have $5\ge f_1\left(x\right)=\frac{2\left(2x-1\right)}{x-1}\ge 4$ .

Let $f_2\left(x\right)={\left(\frac{x}{x-1}\right)}^x$ , then ${f^{\prime }}_2\left(x\right)={\left({\left(\frac{x}{x-1}\right)}^x\right)}^{\prime }={\left({\text{e}}^{x\cdot \ln \frac{x}{x-1}}\right)}^{\prime }={\text{e}}^{x\cdot \ln \frac{x}{x-1}}\cdot {\left(x\cdot \ln \frac{x}{x-1}\right)}^{\prime }$

$\begin{array}{c}{f^{\prime }}_2\left(x\right)={\left(\frac{x}{x-1}\right)}^x\cdot \left(\ln \frac{x}{x-1}+x\cdot {\left(\ln \frac{x}{x-1}\right)}^{\prime }\right)\\ ={\left(\frac{x}{x-1}\right)}^x\cdot \left(\ln \frac{x}{x-1}+x\cdot \frac{x-1}{x}\cdot \frac{x-1-x}{{\left(x-1\right)}^2}\right)\end{array}$

${f^{\prime }}_2\left(x\right)={\left(\frac{x}{x-1}\right)}^x\cdot \left(\ln \frac{x}{x-1}-\frac{1}{x-1}\right)$ (2.1.1)

In (2.1.1), for $x\ge 3$ , $\frac{1}{x-1}=\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+\frac{1}{x^6}+\cdots$

Using the formula: $\ln \left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\cdots$ , we have

$\ln \frac{x}{x-1}=\ln \frac{1}{1+\frac{-1}{x}}=-\ln \left(1+\frac{-1}{x}\right)=\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{3x^3}+\frac{1}{4x^4}+\frac{1}{5x^5}+\frac{1}{6x^6}+\cdots$

Thus, for $x\ge 3$ , $\ln \frac{x}{x-1}-\frac{1}{x-1}<0$ .

Since ${\left(\frac{x}{x-1}\right)}^x$ is a positive number for $x\ge 3$ , ${f^{\prime }}_2\left(x\right)={\left(\frac{x}{x-1}\right)}^x\cdot \left(\ln \frac{x}{x-1}-\frac{1}{x-1}\right)<0$ .

Thus $f_2\left(x\right)$ is a strictly deceasing function for $x\ge 3$ . Since $f_2\left(3\right)=3.375$ and ${\lim }_{x\to \infty }{f}_2\left(x\right)=e\approx 2.718$ , for $x\ge 3$ , $3.375\ge f_2\left(x\right)={\left(\frac{x}{x-1}\right)}^x\ge e$ (2.1.2)

Since for $x\ge 3$ , $f_1\left(x\right)$ has a lower bound of 4 and $f_2\left(x\right)$ has an upper bound of 3.375,

$f_1\left(x\right)=\frac{2\left(2x-1\right)}{x-1}>f_2\left(x\right)={\left(\frac{x}{x-1}\right)}^x$

When $x=\lambda \ge 3$ , we have $\frac{2\left(2\lambda -1\right)}{\lambda -1}>{\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda }$ (2.1.3)

When $\lambda \ge 3$ and $n=2$ , $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)=\left(\begin{array}{c}2\lambda \\ 2\end{array}\right)=\frac{2\lambda \left(2\lambda -1\right)\left(2\lambda -2\right)!}{2\left(2\lambda -2\right)!}=\lambda \left(2\lambda -1\right)$ (2.1.4)

$\frac{{\lambda }^{\lambda n-\lambda +1}}{n{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}=\frac{{\lambda }^{2\lambda -\lambda +1}}{2{\left(\lambda -1\right)}^{2\left(\lambda -1\right)-\lambda +1}}=\frac{\lambda \left(\lambda -1\right)}{2}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda }$ (2.1.5)

Since $\frac{\lambda \left(\lambda -1\right)}{2}$ is a positive number for $\lambda \ge 3$ , referring to (2.1.4) and (2.1.5), when $\frac{\lambda \left(\lambda -1\right)}{2}$ multiplies both sides of (2.1.3), we have

$\left(\frac{\lambda \left(\lambda -1\right)}{2}\right)\left(\frac{2\left(2\lambda -1\right)}{\lambda -1}\right)=\lambda \left(2\lambda -1\right)=\left(\begin{array}{c}\lambda n\\ n\end{array}\right)>\left(\frac{\lambda \left(\lambda -1\right)}{2}\right){\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda }=\frac{{\lambda }^{\lambda n-\lambda +1}}{n{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}$

Thus, $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)>\frac{{\lambda }^{\lambda n-\lambda +1}}{n{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}$ when $\lambda \ge 3$ and $n=2$ . (2.1.6)

By induction on n, when $\lambda \ge 3$ , if $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)>\frac{{\lambda }^{\lambda n-\lambda +1}}{n{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}$ is true for n, then for $n+1$ ,

$\begin{array}{l}\left(\begin{array}{c}\lambda \left(n+1\right)\\ n+1\end{array}\right)=\left(\begin{array}{c}\lambda n+\lambda \\ n+1\end{array}\right)\\ =\frac{\left(\lambda n+\lambda \right)\left(\lambda n+\lambda -1\right)\cdots \left(\lambda n+2\right)\left(\lambda n+1\right)}{\left(\lambda n+\lambda -n-1\right)\left(\lambda n+\lambda -n-2\right)\cdots \left(\lambda n-n+1\right)\left(n+1\right)}\cdot \left(\begin{array}{c}\lambda n\\ n\end{array}\right)\end{array}$

$\left(\begin{array}{c}\lambda \left(n+1\right)\\ n+1\end{array}\right)>\frac{\left(\lambda n+\lambda \right)\left(\lambda n+\lambda -1\right)\cdots \left(\lambda n+2\right)\left(\lambda n+1\right)}{\left(\lambda n+\lambda -n-1\right)\left(\lambda n+\lambda -n-2\right)\cdots \left(\lambda n-n+1\right)\left(n+1\right)}\cdot \frac{{\lambda }^{\lambda n-\lambda +1}}{n{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}$

$\left(\begin{array}{c}\lambda \left(n+1\right)\\ n+1\end{array}\right)>\frac{\left(\lambda n+\lambda \right)\left(\lambda n+\lambda -1\right)\cdots \left(\lambda n+2\right)}{\left(\lambda n+\lambda -n-1\right)\left(\lambda n+\lambda -n-2\right)\cdots \left(\lambda n-n+1\right)}\cdot \frac{\lambda n+1}{n}\cdot \frac{1}{n+1}\cdot \frac{{\lambda }^{\lambda n-\lambda +1}}{{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}$

Notice $\frac{\lambda n+1}{n}>\lambda$ , and $\frac{\left(\lambda n+\lambda \right)\left(\lambda n+\lambda -1\right)\cdots \left(\lambda n+2\right)}{\left(\lambda n+\lambda -n-1\right)\left(\lambda n+\lambda -n-2\right)\cdots \left(\lambda n-n+1\right)}>{(\frac{\lambda }{\lambda -1})}^{\lambda -1}$ because $\frac{\lambda n+\lambda }{\lambda n+\lambda -n-1}=\frac{\lambda }{\lambda -1};\frac{\lambda n+\lambda -1}{\lambda n+\lambda -n-2}>\frac{\lambda }{\lambda -1};\cdots ;\frac{\lambda n+2}{\lambda n-n+1}>\frac{\lambda }{\lambda -1}$ . Thus,

$\left(\begin{array}{c}\lambda \left(n+1\right)\\ n+1\end{array}\right)>\frac{{\lambda }^{\lambda -1}}{{\left(\lambda -1\right)}^{\lambda -1}}\cdot \frac{\lambda }{1}\cdot \frac{1}{n+1}\cdot \frac{{\lambda }^{\lambda n-\lambda +1}}{{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}=\frac{{\lambda }^{\lambda \left(n+1\right)-\lambda +1}}{\left(n+1\right){\left(\lambda -1\right)}^{\left(\lambda -1\right)\left(n+1\right)-\lambda +1}}$ (2.1.7)

From (2.1.6) and (2.1.7), we have for $\lambda \ge 3$ and $n\ge 2$ , $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)>\frac{{\lambda }^{\lambda n-\lambda +1}}{n{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}$

Referring to (1.8), for $\lambda \ge 3$ and $n\ge 2$ , $\left(\begin{array}{c}\lambda n\\ n\end{array}\right)\le {\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}\cdot 2^{2n-3}\cdot {\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+2}$ .

Then, ${\text{Γ}}_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}\cdot 2^{2n-3}\cdot {\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+2}>\frac{{\lambda }^{\lambda n-\lambda +1}}{n{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}$ .

Since when $\lambda \ge 3$ and $n\ge 2$ , then $2^{2n-3}>0$ and ${\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+2}>0$ .

${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>\frac{{\lambda }^{\lambda n-\lambda +1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+2}\cdot 2^{2n-3}\cdot n{\left(\lambda -1\right)}^{\left(\lambda -1\right)n-\lambda +1}}\text{=}\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}$

Thus, Lemma 1 is proven.

Lemma 2: When $n\ge \lambda -2\ge 9$ , $f_3\left(n,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda -1}{4}\cdot e\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}$ is an increasing function with respect to the product of λn and with respect to n. (2.2)

Proof:

Referring to (2.1.2), when $\lambda \ge 3$ , ${\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda }\ge e$ . From (2.1), when $\lambda \ge 3$ and $n\ge 2$ ,

${\text{Γ}}_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>f_8\left(n,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda -1}{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda }\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}$ , and $f_8\left(n,\lambda \right)\ge \frac{2{\lambda }^2\cdot {\left(\frac{\lambda -1}{4}\cdot e\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}=f_3\left(n,\lambda \right)>0$ .

Thus, ${\text{Γ}}_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>f_8\left(n,\lambda \right)\ge f_3\left(n,\lambda \right)>0$ . (2.2.1)

Let $x\ge 2$ and $y\ge 4$ both be real numbers. When $x=y-2$ ,

$f_3\left(x,y\right)=\frac{2{\left(x+2\right)}^2\cdot {\left(\frac{x+1}{4}\cdot e\right)}^{x-1}}{{\left(\left(x+2\right)\cdot x\right)}^{\frac{\sqrt{x\cdot \left(x+2\right)}}{3}+3}}>f_4\left(x\right)=\frac{2{\left(x+2\right)}^2\cdot {\left(\frac{x+1}{4}\cdot e\right)}^{x-1}}{{\left(\left(x+2\right)\cdot x\right)}^{\frac{x+1}{3}+3}}>0$ (2.2.2)

$\begin{array}{c}{f^{\prime }}_4\left(x\right)=f_4\left(x\right)\cdot \left(\frac{2}{x+2}+\ln \left(\frac{x+1}{4}\right)+\frac{4}{3}-\frac{2}{x+1}-\frac{1}{3}\ln \left(\left(x+2\right)\cdot x\right)-\frac{10}{3x}-\frac{8}{3\left(x+2\right)}\right)\\ =f_4\left(x\right)\cdot f_5(\; x\; )\end{array}$

where $f_5\left(x\right)=\frac{2}{x+2}+\ln \left(\frac{x+1}{4}\right)+\frac{4}{3}-\frac{2}{x+1}-\frac{1}{3}\ln \left(\left(x+2\right)\cdot x\right)-\frac{10}{3x}-\frac{8}{3\left(x+2\right)}$ ${f^{\prime }}_5\left(x\right)=\frac{4x+6}{{\left(x+1\right)}^2\cdot {\left(x+2\right)}^2}+\frac{x^2+2x-2}{3x\left(x+1\right)\left(x+2\right)}+\frac{10}{3x^2}+\frac{8}{3{\left(x+2\right)}^2}>0$ when $x\ge 2$ .

Thus, $f_5\left(x\right)$ is a strictly increasing function for $x\ge 2$ .

When $x=9$ , $f_5\left(x\right)=\frac{2}{9+2}+\ln \left(\frac{9+1}{4}\right)+\frac{4}{3}-\frac{2}{9+1}-\frac{1}{3}\ln \left(9\right)-\frac{1}{3}\ln \left(9+2\right)-\frac{10}{27}-\frac{8}{33}>0$ . Thus, for $x\ge 9$ , $f_5\left(x\right)>0$ . Then, ${f^{\prime }}_4\left(x\right)=f_4\left(x\right)\cdot f_5\left(x\right)>0$ .

Thus, $f_4\left(x\right)$ is a strictly increasing function for $x\ge 9$ .

From (2.2.2), when $x=y-2$ , $f_3\left(x,y\right)>f_4\left(x\right)>0$ . Thus, when $x=y-2\ge 9$ and $xy\ge 99$ , then $f_3\left(x,y\right)$ is an increasing function respect to the product of xy. (2.2.3)

$\frac{\partial f_3\left(x,y\right)}{\partial x}=f_3\left(x,y\right)\cdot \left(\ln \left(\frac{y-1}{4}\right)+1-\frac{\sqrt{y}}{6\sqrt{x}}\cdot \ln \left(yx\right)-\frac{\sqrt{y}}{3\sqrt{x}}-\frac{3}{x}\right)=f_3\left(x,y\right)\cdot f_6\left(x,y\right)$ (2.2.4)

where $f_6\left(x,y\right)=\ln \left(\frac{y-1}{4}\right)+1-\frac{\sqrt{y}}{6\sqrt{x}}\cdot \ln \left(yx\right)-\frac{\sqrt{y}}{3\sqrt{x}}-\frac{3}{x}$

When $x=y-2$ , then $f_6\left(x,y\right)=f_7\left(x\right)=\ln \left(\frac{x+1}{4}\right)+1-\frac{\sqrt{x+2}}{6\sqrt{x}}\cdot \left(\ln \left(x+2\right)+\ln \left(x\right)+2\right)-\frac{3}{x}$

When $x\ge 2$ , ${f^{\prime }}_7\left(x\right)=\frac{1}{x+1}-\frac{\sqrt{x+2}}{6\sqrt{x}}\cdot \left(\frac{1}{x+2}+\frac{1}{x}\right)+\frac{\ln \left(x+2\right)+\ln \left(x\right)+2}{6x\sqrt{x\left(x+2\right)}}+\frac{3}{x^2}$

$\begin{array}{c}{f^{\prime }}_7\left(x\right)=\frac{1}{x+1}-\frac{\sqrt{x+2}}{6\sqrt{x}}\cdot \frac{x+x+2}{x\left(x+2\right)}+\frac{\ln \left(x+2\right)+\ln \left(x\right)+2}{6x\sqrt{x\left(x+2\right)}}+\frac{3}{x^2}\\ =\frac{1}{x+1}-\frac{1}{3\sqrt{x}}\cdot \frac{x+1}{x\sqrt{x+2}}+\frac{2}{6x\sqrt{x\left(x+2\right)}}+\frac{\ln \left(x+2\right)+\ln \left(x\right)}{6x\sqrt{x\left(x+2\right)}}+\frac{3}{x^2}\\ =\frac{1}{x+1}-\frac{x}{3x\sqrt{x\left(x+2\right)}}+\frac{\ln \left(x+2\right)+\ln \left(x\right)}{6x\sqrt{x\left(x+2\right)}}+\frac{3}{x^2}\end{array}$ (2.2.5)

When $x\ge 2$ , then $3\sqrt{x\left(x+2\right)}+\left(x+1\right)>0$

$\begin{array}{l}\left(3\sqrt{x\left(x+2\right)}+\left(x+1\right)\right)\cdot \left(3\sqrt{x\left(x+2\right)}-\left(x+1\right)\right)\\ ={\left(3\sqrt{x\left(x+2\right)}\right)}^2-{\left(x+1\right)}^2=8x^2+16x-1>0\end{array}$

Thus, $\left(3\sqrt{x\left(x+2\right)}+\left(x+1\right)\right)\cdot \left(3\sqrt{x\left(x+2\right)}-\left(x+1\right)\right)>0$

$3\sqrt{x\left(x+2\right)}-\left(x+1\right)>0$

$3\sqrt{x\left(x+2\right)}>x+1$ then $\frac{1}{x+1}>\frac{1}{3\sqrt{x\left(x+2\right)}}$

When $x\ge 2$ , $\frac{1}{x+1}-\frac{1}{3\sqrt{x\left(x+2\right)}}>0$ , and from (2.2.5), $\frac{\ln \left(x+2\right)+\ln \left(x\right)}{6x\sqrt{x\left(x+2\right)}}+\frac{3}{x^2}>0$ .

Then ${f^{\prime }}_7\left(x\right)=\left(\frac{1}{x+1}-\frac{1}{3\sqrt{x\left(x+2\right)}}\right)+\frac{\ln \left(x+2\right)+\ln \left(x\right)}{6x\sqrt{x\left(x+2\right)}}+\frac{3}{x^2}>0$ .

Thus, when $x\ge 2$ , $f_7\left(x\right)$ is a strictly increasing function.

When $x=y-2\ge 2$ , since $f_6\left(x,y\right)=f_7\left(x\right)$ , $f_6\left(x,y\right)$ is an increasing function respect to xy.

When $x=y-2=9$ , $f_6\left(x,y\right)=\ln \left(\frac{11-1}{4}\right)+1-\frac{\sqrt{11}}{6\sqrt{9}}\cdot \ln \left(99\right)-\frac{\sqrt{11}}{3\sqrt{9}}-\frac{3}{9}>0$ .

When $x\ge y-2\ge 2$ , $\frac{\partial f_6\left(x,y\right)}{\partial x}=\frac{\sqrt{y}}{12x\sqrt{x}}\cdot \ln \left(y\right)+\frac{\sqrt{y}}{12x\sqrt{x}}\cdot \ln \left(x\right)+\frac{\sqrt{y}}{6x\sqrt{x}}+\frac{\sqrt{y}}{6x\sqrt{x}}+\frac{3}{x^2}>0$ .

Thus, when $x\ge y-2\ge 9$ , $f_6\left(x,y\right)>0$ , and it is an increasing function with respect to x and to the product of xy, then, from (2.2.4), $\frac{\partial f_3\left(x,y\right)}{\partial x}=f_3\left(x,y\right)\cdot f_6\left(x,y\right)>0$ .

Thus, when $x\ge y-2\ge 9$ , $f_3\left(x,y\right)$ is an increasing function with respect to x. (2.2.6)

Referring to (2.2.3) and (2.2.6), when $x\ge y-2\ge 9$ , then $xy\ge 99$ , $f_3\left(x,y\right)$ is an increasing function with respect to the product of xy and with respect to x.

Let $x=n$ and $y=\lambda$ . Then when $n\ge \lambda -2\ge 9$ , $f_3\left(n,\lambda \right)$ is an increasing function with respect to the product of λn and with respect to n.

Thus, Lemma 2 is proven.

Lemma 3: When $n\ge 24$ and $26\ge \lambda \ge 3$ , $f_8\left(n,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}$ is a strictly increasing function respect to n. (2.3)

Proof:

Referring to (2.1), when $\lambda \ge 3$ and $n\ge 2$ , $f_8\left(n,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}>0$

Let a real number $x\ge 2$ . When $\lambda \ge 3$ , then $f_8\left(x,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{x-1}}{{\left(\lambda x\right)}^{\frac{\sqrt{\lambda x}}{3}+3}}>0$ (2.3.1)

When λ is an integer constant in the range of $10\ge \lambda \ge 3$ ,

$\frac{\partial f_8\left(x,\lambda \right)}{\partial x}=2{\lambda }^2\cdot \frac{\frac{\partial }{\partial x}\left({\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{x-1}\right)\cdot \left({\left(\lambda x\right)}^{\frac{\sqrt{\lambda x}}{3}+3}\right)-{\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{x-1}\cdot \frac{\partial }{\partial x}\left({\left(\lambda x\right)}^{\frac{\sqrt{\lambda x}}{3}+3}\right)}{{\left({\left(\lambda x\right)}^{\frac{\sqrt{\lambda x}}{3}+3}\right)}^2}$

$\begin{array}{c}\frac{\partial f_8\left(x,\lambda \right)}{\partial x}=f_8\left(x,\lambda \right)\cdot \left(\ln \left(\frac{\lambda }{4}\right)+\ln {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}-\frac{\sqrt{\lambda }\left(\ln \left(x\right)+\ln \left(\lambda \right)+2\right)}{6\sqrt{x}}-\frac{3}{x}\right)\\ =f_8\left(x,\lambda \right)\cdot f_9\left(x,\lambda \right)\end{array}$

where

$f_9\left(x,\lambda \right)=\ln \left(\frac{\lambda }{4}\right)+\ln {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}-\frac{\sqrt{\lambda }\left(\ln \left(x\right)+\ln \left(\lambda \right)+2\right)}{6\sqrt{x}}-\frac{3}{x}$

$\frac{\partial f_9\left(x,\lambda \right)}{\partial x}=\frac{\sqrt{\lambda }\ln \left(\lambda \right)+\sqrt{\lambda }\ln \left(x\right)}{12x\sqrt{x}}+\frac{3}{x^2}>0$ for $x>1$ and $\lambda >2$ ; then, $f_9\left(x,\lambda \right)$ is a strictly increasing function respect to x.

When $x=24$ and $10\ge \lambda \ge 3$ , $f_9\left(x,\lambda \right)>0$ . The calculations are below.

When $\lambda =3$ , $f_9\left(x,\lambda \right)=\ln \left(\frac{3}{4}\right)+\ln {\left(\frac{3}{3-1}\right)}^{3-1}-\frac{\sqrt{3}\left(\ln \left(24\right)+\ln \left(3\right)+2\right)}{6\sqrt{24}}-\frac{3}{24}\approx 0.0283>0$ .

When $\lambda =4$ , $f_9\left(x,\lambda \right)=\ln \left(\frac{4}{4}\right)+\ln {\left(\frac{4}{4-1}\right)}^{4-1}-\frac{\sqrt{4}\left(\ln \left(24\right)+\ln \left(4\right)+2\right)}{6\sqrt{24}}-\frac{3}{24}\approx 0.2814>0$ .

When $\lambda =5$ , $f_9\left(x,\lambda \right)=\ln \left(\frac{5}{4}\right)+\ln {\left(\frac{5}{5-1}\right)}^{5-1}-\frac{\sqrt{5}\left(\ln \left(24\right)+\ln \left(5\right)+2\right)}{6\sqrt{24}}-\frac{3}{24}\approx 0.4744>0$ .

When $\lambda =6$ , $f_9\left(x,\lambda \right)=\ln \left(\frac{6}{4}\right)+\ln {\left(\frac{6}{6-1}\right)}^{6-1}-\frac{\sqrt{6}\left(\ln \left(24\right)+\ln \left(6\right)+2\right)}{6\sqrt{24}}-\frac{3}{24}\approx 0.6113>0$ .

When $\lambda =7$ , $f_9\left(x,\lambda \right)=\ln \left(\frac{7}{4}\right)+\ln {\left(\frac{7}{7-1}\right)}^{7-1}-\frac{\sqrt{7}\left(\ln \left(24\right)+\ln \left(7\right)+2\right)}{6\sqrt{24}}-\frac{3}{24}\approx 0.7183>0$ .

When $\lambda =8$ , $f_9\left(x,\lambda \right)=\ln \left(\frac{8}{4}\right)+\ln {\left(\frac{8}{8-1}\right)}^{8-1}-\frac{\sqrt{8}\left(\ln \left(24\right)+\ln \left(8\right)+2\right)}{6\sqrt{24}}-\frac{3}{24}\approx 0.8044>0$ .

When $\lambda =9$ , $f_9\left(x,\lambda \right)=\ln \left(\frac{9}{4}\right)+\ln {\left(\frac{9}{9-1}\right)}^{9-1}-\frac{\sqrt{9}\left(\ln \left(24\right)+\ln \left(9\right)+2\right)}{6\sqrt{24}}-\frac{3}{24}\approx 0.8755>0$ .

When $\lambda =10$ , $f_9\left(x,\lambda \right)=\ln \left(\frac{10}{4}\right)+\ln {\left(\frac{10}{10-1}\right)}^{10-1}-\frac{\sqrt{10}\left(\ln \left(24\right)+\ln \left(10\right)+2\right)}{6\sqrt{24}}-\frac{3}{24}\approx 0.9347>0$ .

From (2.3.1), when $x\ge 2$ and $\lambda \ge 3$ , $f_8\left(x,\lambda \right)>0$ .

Since $\frac{\partial f_8\left(x,\lambda \right)}{\partial x}=f_8\left(x,\lambda \right)\cdot f_9\left(x,\lambda \right)$ , when $x=24$ and $10\ge \lambda \ge 3$ , $\frac{\partial f_8\left(x,\lambda \right)}{\partial x}>0$ .

Thus, when $x\ge 24$ and $10\ge \lambda \ge 3$ , $f_8\left(x,\lambda \right)$ is a strictly increasing function respect to x.

Let $x=n\ge 24$ and $10\ge \lambda \ge 3$ , $f_8\left(n,\lambda \right)$ is a strictly increasing function respect to n. (2.3.2)

From (2.2.1), when $\lambda \ge 3$ and $n\ge 2$ , $f_8\left(n,\lambda \right)\ge f_3\left(n,\lambda \right)>0$ . Referring to (2.2), when $\lambda \ge 11$ and $n\ge \lambda -2$ , $f_3\left(n,\lambda \right)$ is an increasing function with respect to the product of λn and with respect to n.

When $n\ge 24$ and $26\ge \lambda \ge 11$ , since $n\ge \lambda -2$ and $f_8\left(n,\lambda \right)\ge f_3\left(n,\lambda \right)$ , $f_8\left(n,\lambda \right)$ is an increasing function with respect to the product of λn and with respect to n. (2.3.3)

From (2.3.2) and (2.3.3), when $n\ge 24$ and $26\ge \lambda \ge 3$ , $f_8\left(n,\lambda \right)$ is an increasing function with respect to n.

Thus, Lemma 3 is proven.

Lemma 4: When $n\ge n_0\ge 24$ and $26\ge \lambda \ge 3$ , if $f_8\left(n,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}>1$ , then there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ . (2.4)

Proof:

Let integers $m\ge n\ge n_0\ge 24$ .

From (2.3), when $n\ge n_0\ge 24$ and $26\ge \lambda \ge 3$ , $f_8\left(n_0,\lambda \right)$ is an increasing function respect to n.

When $26\ge \lambda \ge 3$ , if $f_8\left(n_0,\lambda \right)>1$ , then $f_8\left(n,\lambda \right)>1$ , and thus, $f_8\left(m,\lambda \right)>1$ ; then from (2.1), ${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>f_8\left(n,\lambda \right)>1$ , and ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}>f_8\left(m,\lambda \right)>1$ . (2.4.1)

Note that $m\ge n\ge n_0\ge 24\ge \lambda -2$ since $26\ge \lambda \ge 3$ .

From (1.7), when $n\ge \lambda -2$ , every prime number in ${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}$ has a power of 0 or 1; when $m\ge \lambda -2$ , every prime number in ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}$ has a power of 0 or 1. (2.4.2)

$\begin{array}{l}{\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}={\Gamma }_{\lambda m\ge p>\left(\lambda -1\right)m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\\ \cdot {\displaystyle {\prod }_{i=1}^{i=\lambda -2}\left({\Gamma }_{\frac{\left(\lambda -1\right)m}{i}\ge p>\frac{\lambda m}{i+1}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\cdot {\Gamma }_{\frac{\lambda m}{i+1}\ge p>\frac{\left(\lambda -1\right)m}{i+1}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\right)}\end{array}$

In ${\displaystyle {\prod }_{i=1}^{i=\lambda -2}\left({\Gamma }_{\frac{\left(\lambda -1\right)m}{i}\ge p>\frac{\lambda m}{i+1}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\right)}$ , for every distinct prime number p in these ranges, the numerator $\left(\lambda m\right)!$ has the product of $p\cdot 2p\cdot 3p\cdots ip=i\text{!}\cdot p^i$ . The denominator $\left(\left(\lambda -1\right)m\right)!$ also has the same product of $i\text{!}\cdot p^i$ . Thus, they cancel each other in $\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}$ .

Referring to (1.2), ${\displaystyle {\prod }_{i=1}^{i=\lambda -2}\left({\Gamma }_{\frac{\left(\lambda -1\right)m}{i}\ge p>\frac{\lambda m}{i+1}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\right)}=1$ .

Thus,

$\begin{array}{l}{\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\\ ={\Gamma }_{\lambda m\ge p>\left(\lambda -1\right)m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\cdot {\displaystyle {\prod }_{i=1}^{i=\lambda -2}\left({\Gamma }_{\frac{\lambda m}{i+1}\ge p>\frac{\left(\lambda -1\right)m}{i+1}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\right)}\end{array}$

${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}={\displaystyle {\prod }_{i=1}^{i=\lambda -1}\left({\Gamma }_{\frac{\lambda m}{i}\ge p>\frac{\left(\lambda -1\right)m}{i}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\right)}$ . (2.4.3)

${\displaystyle {\prod }_{i=1}^{i=\lambda -1}\left({\Gamma }_{\frac{\lambda m}{i}\ge p>\frac{\left(\lambda -1\right)m}{i}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\right)}$ is the product of (λ − 1) sectors from $i=1$ to $i=\lambda -1$ .

Each of these sectors is the prime number factorization of the product of the consecutive integers between $\frac{\left(\lambda -1\right)m}{i}$ and $\frac{\lambda m}{i}$ .

Referring to (2.4.3), when ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}>1$ , then ${\displaystyle {\prod }_{i=1}^{i=\lambda -1}\left({\Gamma }_{\frac{\lambda m}{i}\ge p>\frac{\left(\lambda -1\right)m}{i}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\right)}>1$ .

Referring to (1.1), ${\Gamma }_{\frac{\lambda m}{i}\ge p>\frac{\left(\lambda -1\right)m}{i}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\ge 1$ . Thus, when ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}>1$ , at least one of the sectors is greater than one in ${\displaystyle {\prod }_{i=1}^{i=\lambda -1}\left({\Gamma }_{\frac{\lambda m}{i}\ge p>\frac{\left(\lambda -1\right)m}{i}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}\right)}$ .

Let ${\Gamma }_{\frac{\lambda m}{i}\ge p>\frac{\left(\lambda -1\right)m}{i}}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}>1$ be such a sector and let $m=ni$ where $\lambda -1\ge i\ge 1$ from (2.4.3). Thus, when $m=ni\ge n\ge \lambda –2$ ,

${\Gamma }_{\frac{\lambda ni}{i}\ge p>\frac{\left(\lambda -1\right)ni}{i}}\left\{\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\right\}={\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\right\}>1$ . (2.4.4)

$\begin{array}{l}\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\\ =\frac{\left(\lambda ni\right)\cdot \left(\lambda ni-1\right)\cdots \left(\lambda ni-i\right)\cdots \left(\lambda ni-2i\right)\cdots \left(\lambda ni-\left(n-1\right)i\right)\cdots \left(\lambda ni-ni+1\right)\cdot \left(\left(\lambda -1\right)ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\end{array}$

$\begin{array}{l}\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\\ =\frac{i\cdot \left(\lambda n\right)\cdot \left(\lambda ni-1\right)\cdots i\cdot \left(\lambda n-1\right)\cdots i\cdot \left(\lambda n-2\right)\cdots i\cdot \left(\lambda n-n+1\right)\cdots \left(\lambda ni-ni+1\right)\cdot \left(\left(\lambda -1\right)ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\end{array}$

Thus, $\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}$ contains all the factors of $\lambda n,\lambda n-1,\lambda n-2,\cdots ,\lambda n-n+1$ in $\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}$ .

These factors make up all the consecutive integers in the range of $\lambda n\ge p>\left(\lambda -1\right)n$ in $\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}$ . Thus, $\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}$ contains $\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}$ .

Referring to the definition, all prime numbers in $\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}$ in the ranges of $\lambda ni\ge p>\lambda n$ and $\left(\lambda -1\right)n>p$ do not contribute to ${\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\right\}$ , nor does i for $\lambda -1\ge i\ge 1$ . Only the prime numbers in the prime factorization of $\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}$ in the range of $\lambda n\ge p>\left(\lambda -1\right)n$ present in ${\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\right\}$ . Since $\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}$ is the product of all the consecutive integers in this range, ${\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\right\}={\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}$ .

Referring to (2.4.4), ${\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda ni\right)!}{\left(\left(\lambda -1\right)ni\right)!}\right\}={\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>1$ . (2.4.5)

Referring to (2.4.1), (2.4.3), (2.4.4), and (2.4.5), when $26\ge \lambda \ge 3$ and $n\ge n_0\ge 24$ ,

if $f_8\left(n,\lambda \right)>1$ , then $f_8\left(m,\lambda \right)>1$ and ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}>1$ ;

when ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}>1$ , then ${\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>1$ .

Thus, when $26\ge \lambda \ge 3$ and $n\ge n_0\ge 24$ , if $f_8\left(n,\lambda \right)>1$ , then ${\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>1$ , referring to (1.3), there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Thus, Lemma 4 is proven.

Lemma 5: When $n\ge \lambda -2\ge 25$ , there exists at least a prime number p such

that $\left(\lambda -1\right)n<p\le \lambda n$ . (2.5)

Proof:

Referring to (2.2), when $n\ge \lambda -2\ge 9$ , $f_3\left(n,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda -1}{4}\cdot e\right)}^{n-1}}{{\left(\lambda n\right)}^{\frac{\sqrt{\lambda n}}{3}+3}}$ is an increasing function with respect to the product of λn and with respect to n.

Let integers $n_1=25$ , and ${\lambda }_1=27$ . Then $f_3\left(n_1,{\lambda }_1\right)=\frac{2\cdot {27}^2\cdot {\left(\frac{27-1}{4}\cdot e\right)}^{25-1}}{{\left(25\cdot 27\right)}^{\frac{\sqrt{675}}{3}+3}}\approx \frac{1.2495E+33}{9.7843E+32}>1$ .

When $m=n=n_1=\lambda -2={\lambda }_1-2=25$ , $f_3\left(m,\lambda \right)=f_3\left(n,\lambda \right)=f_3\left(n_1,{\lambda }_1\right)>1$ .

From (2.2), when $m=n=\lambda -2\ge 25$ , $f_3\left(n,\lambda \right)>1$ and $f_3\left(m,\lambda \right)>1$ since $f_3\left(n,\lambda \right)$ is an increasing function with respect to the product of λn. When $m\ge n\ge \lambda -2\ge 25$ , then $f_3\left(n,\lambda \right)>1$ and $f_3\left(m,\lambda \right)>1$ since $f_3\left(n,\lambda \right)$ is also an increasing function with respect to n.

Referring to (2.2.1), when $m\ge n\ge \lambda -2\ge 25$ , ${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>f_3\left(n,\lambda \right)\ge f_3\left(n_1,{\lambda }_1\right)>1$ ; and ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}>f_3\left(m,\lambda \right)\ge f_3\left(n,\lambda \right)>1$ . (2.5.1)

Referring to (2.4.2), when $m\ge n\ge \lambda -2\ge 25$ , every prime number in ${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}$ and in ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}$ has a power of 0 or 1. (2.5.2)

Referring to (2.5.1), (2.4.3), (2.4.4), and (2.4.5), when $m\ge n\ge \lambda -2\ge 25$ , ${\Gamma }_{\lambda n\ge p>n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>1$ , and ${\Gamma }_{\lambda m\ge p>m}\left\{\frac{\left(\lambda m\right)!}{\left(\left(\lambda -1\right)m\right)!}\right\}>1$ , then ${\Gamma }_{\lambda n\ge p>\left(\lambda -1\right)n}\left\{\frac{\left(\lambda n\right)!}{\left(\left(\lambda -1\right)n\right)!}\right\}>1$ ; referring to (1.3), there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Thus, Lemma 5 is proven. It was proven in [ [5] , pp 1324-1329] with more details.

3. A Prime Number between (λ − 1)n and λn When 26 ≥ λ ≥ 3 and n ≥ λ − 2

Proposition 1: For $\lambda =3$ , when $n\ge \lambda -2$ , there exists at least a prime number p such

that $\left(\lambda -1\right)n<p\le \lambda n$ . (3.1)

Proof:

Referring to (2.3) for $\lambda =3$ , when $n\ge n_0\ge 24$ , $f_8\left(n_0,\lambda \right)$ is a strictly increasing function on $n_0$ .

Let $n_0=83$ . When $\lambda =3$ and $n\ge n_0=83$ ,

$f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}=\frac{18\cdot {1.6875}^{83-1}}{{249}^{\frac{\sqrt{249}}{3}+3}}\approx \frac{7.7493E+19}{6.2000E+19}>1$ .

Thus, for $\lambda =3$ , when $n\ge 83$ , $n>\lambda -2$ and $f_8\left(n,\lambda \right)\ge f_8\left(n_0,\lambda \right)>1$ ; then, referring to (2.4), there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

For $\lambda =3$ , when $82\ge n\ge 1=\lambda -2$ , Table 1 shows that there exists a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Thus, (3.1) is proven.

Proposition 2: For $\text{5}\ge \lambda \ge \text{4}$ , when $n\ge \lambda -2$ , there exists at least a prime number p such that

$\left(\lambda -1\right)n<p\le \lambda n$ . (3.2)

Table 1. For $\text{82}\ge n\ge \text{1}$ and $\lambda =\text{3}$ , there is a prime number p such that $\text{2}n<p\le \text{3}n$ .

Proof:

From (2.3), for $\text{5}\ge \lambda \ge \text{4}$ , when $n\ge n_0\ge 24$ , $f_8\left(n_0,\lambda \right)$ is a strictly increasing function on $n_0$ .

Let $n_0=40$ . When $n\ge n_0=40$ and $\text{5}\ge \lambda \ge \text{4}$ , $f_8\left(n,\lambda \right)\ge f_8\left(n_0,\lambda \right)>1$ . The calculations are below.

When $\lambda =4$ , $f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{32\cdot {2.3703}^{40-1}}{{160}^{\frac{\sqrt{160}}{3}+3}}\approx \frac{1.3258E+16}{8.0503E+15}>1$ .

When $\lambda =5$ , $f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{50\cdot {3.0518}^{40-1}}{{200}^{\frac{\sqrt{200}}{3}+3}}\approx \frac{7.8968E+18}{5.6253E+17}>1$ .

Thus, for $\text{5}\ge \lambda \ge \text{4}$ , when $n\ge 40$ , $n>\lambda -2$ and $f_8\left(n,\lambda \right)>f_8\left(n_0,\lambda \right)>1$ ; then, referring to (2.4), there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

For $\text{5}\ge \lambda \ge \text{4}$ , when $39\ge n\ge \lambda -2$ , Table 2 shows that there exists a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Thus, (3.2) is proven.

Proposition 3: For $11\ge \lambda \ge 6$ , when $n\ge \lambda -2$ , there exists at least a prime number p such that

$\left(\lambda -1\right)n<p\le \lambda n$ . (3.3)

Proof:

Referring to (2.3), for $11\ge \lambda \ge 6$ , when $n\ge n_0\ge 24$ , $f_8\left(n_0,\lambda \right)$ is a strictly increasing function with respect to $n_0$ .

Let $n_0=28$ . When $n\ge n_0=28$ and $11\ge \lambda \ge 6$ , $f_8\left(n,\lambda \right)\ge f_8\left(n_0,\lambda \right)>1$ .

Table 2. For $\text{5}\ge \lambda \ge \text{4}$ and $39\ge n\ge \lambda -2$ , there is a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$

The calculations are below.

When $\lambda =6$ , $f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{72\cdot {3.7325}^{28-1}}{{168}^{\frac{\sqrt{168}}{3}+3}}\approx \frac{2.0014E+17}{1.9515E+16}>1$ .

When $\lambda =7$ , $f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{98\cdot {4.4128}^{28-1}}{{196}^{\frac{\sqrt{196}}{3}+3}}\approx \frac{2.5033E+19}{3.7501E+17}>1$ .

When $\lambda =8$ , $f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{128\cdot {5.0930}^{28-1}}{{224}^{\frac{\sqrt{224}}{3}+3}}\approx \frac{1.5686E+21}{5.9689E+18}>1$ .

When $\lambda =9$ , $f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{162\cdot {5.7730}^{28-1}}{{252}^{\frac{\sqrt{252}}{3}+3}}\approx \frac{5.8527E+22}{8.1510E+19}>1$ .

When $\lambda =10$ , $f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{200\cdot {6.4529}^{28-1}}{{280}^{\frac{\sqrt{280}}{3}+3}}\approx \frac{1.4602E+24}{9.7946E+20}>1$ .

When $\lambda =11$ , $f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{242\cdot {7.1328}^{28-1}}{{308}^{\frac{\sqrt{308}}{3}+3}}\approx \frac{2.6414E+25}{1.0560E+22}>1$ .

Thus, for $11\ge \lambda \ge 6$ , when $n\ge \text{28}$ , $n>\lambda -2$ and $f_8\left(n,\lambda \right)\ge f_8\left(n_0,\lambda \right)>1$ ; then, referring to (2.4), there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

For $11\ge \lambda \ge 6$ , when $27\ge n\ge \lambda -2$ , Table 3 shows that there exists a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Thus, (3.3) is proven.

Proposition 4: For $26\ge \lambda \ge 12$ , when $n\ge \lambda -2$ , there exists at least a prime number p such that

$\left(\lambda -1\right)n<p\le \lambda n$ . (3.4)

Proof:

Referring to (2.3) for $26\ge \lambda \ge 12$ , when $n\ge n_0\ge 24$ , $f_8\left(n_0,\lambda \right)$ is a strictly increasing function with respect to $n_0$ . Let $n_0=25$ . When $n\ge n_0=25$ and $\lambda =12$ , then $f_8\left(n,\lambda \right)\ge f_8\left(n_0,\lambda \right)$ and

$f_8\left(n_0,\lambda \right)=\frac{2{\lambda }^2\cdot {\left(\left(\frac{\lambda }{4}\right)\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{288\cdot {7.8126}^{25-1}}{{300}^{\frac{\sqrt{300}}{3}+3}}\approx \frac{7.7000E+23}{5.4078E+21}>1$ .

For $n\ge n_0=25$ and $26\ge \lambda \ge 13$ , then $f_8\left(n,\lambda \right)\ge f_8\left(n_0,\lambda \right)$ . It can be seen from Table 4 that the values of $f_8\left(n_0,\lambda \right)$ as well as $f_8\left(n,\lambda \right)$ are all greater than 1.

(Detailed calculations are in Appendix.)

Thus, for $26\ge \lambda \ge 12$ , when $n\ge 25>\lambda -2$ , $f_8\left(n,\lambda \right)\ge f_8\left(n_0,\lambda \right)>1$ ; then, referring to (2.4), there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

For $13\ge \lambda \ge 12$ and $24\ge n\ge \lambda -2$ , Table 5 shows that there exists a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

For $26\ge \lambda \ge 14$ and $24\ge n\ge \lambda -2$ , Table 6 shows that there exists a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Thus, (3.4) is proven.

Combining (3.1), (3.2), (3.3), and (3.4), when $26\ge \lambda \ge 3$ and $n\ge \lambda -2$ , there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ . (3.5)

4. A Prime Number between kn and (k + 1)n When n ≥ k − 1

Proposition 5: For two positive integers n and k, when $n\ge k-1$ , there exists at least a prime number p such that $kn<p\le \left(k+1\right)n$ . (4.1)

Table 3. For $11\ge \lambda \ge 6$ and $27\ge n\ge \lambda -2$ , there is a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Table 4. When $n\ge n_0=25$ and $26\ge \lambda \ge 13$ , then $f_8\left(n,\lambda \right)\ge f_8\left(n_0,\lambda \right)>1$ .

Table 5. For $13\ge \lambda \ge 12$ and $24\ge n\ge \lambda -2$ , there is a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Table 6. For $26\ge \lambda \ge 14$ and $24\ge n\ge \lambda -2$ , there is a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Proof:

Referring to (2.5), when $n\ge \lambda -2\ge 25$ , there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ . This statement is the same as that when $\lambda \ge \text{27}$ and $n\ge \lambda -2$ , there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Referring to (3.5), when $26\ge \lambda \ge 3$ and $n\ge \lambda -2$ , there exists at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Thus, when $\lambda \ge 3$ and $n\ge \lambda -2$ , there is at least a prime number p such that $\left(\lambda -1\right)n<p\le \lambda n$ .

Let integer $k=\lambda -1$ , then for $n\ge 1$ and $k\ge \text{2}$ , when $n\ge k-1$ , there exists at least a prime number p such that $kn<p\le \left(k+1\right)n$ . (4.2)

The Bertrand-Chebyshev’s theorem points out that for $n\ge 1$ and $k=\text{1}$ , there exists at least a prime number p such that $kn<p\le \left(k+1\right)n$ . (4.3)

From (4.2) and (4.3), we can conclude that for $n\ge 1$ and $k\ge \text{1}$ , when $n\ge k-1$ , there exists at least a prime number p such that $kn<p\le \left(k+1\right)n$ . Thus, Proposition 5 becomes a theorem, Theorem 4.1, and Bertrand-Chebyshev’s theorem is a special case of this theorem.

In the field of prime number distribution, an important theorem is the prime number theorem, $\pi \left(N\right)~\frac{N}{\ln \left(N\right)}$ , where $\pi \left(N\right)$ is the number of distinct

prime numbers less than or equal to a natural number N. The prime number theorem provides the approximate number of prime numbers relative to the natural numbers, while Theorem 4.1 shows that when $n\ge k-1$ , the prime number exists in the interval between kn and (k + 1)n, that is, Theorem 4.1 provides the approximate locations of prime numbers among natural numbers. Using this theorem, Legendre’s conjecture [7] and several other conjectures can be easily proven. The method of proving Theorem 4.1 can also help to study and solve some difficult problems in number theory such as the other three Landau problems [8] .

Appendix

Calculation of $f_8\left(n_0,\lambda \right)$ when $n_0=25$ and $26\ge \lambda \ge 13$ .

When $\lambda =13$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{338\cdot {8.4924}^{25-1}}{{325}^{\frac{\sqrt{325}}{3}+3}}\approx \frac{6.6934E+24}{4.2676E+22}\approx 156>1$ .

When $\lambda =14$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{392\cdot {9.1721}^{25-1}}{{350}^{\frac{\sqrt{350}}{3}+3}}\approx \frac{4.9265E+25}{3.1424E+23}\approx 157>1$ .

When $\lambda =15$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{450\cdot {9.8518}^{25-1}}{{375}^{\frac{\sqrt{375}}{3}+3}}\approx \frac{3.1448E+26}{2.1746E+24}\approx 145>1$ .

When $\lambda =16$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{512\cdot {10.5315}^{25-1}}{{400}^{\frac{\sqrt{400}}{3}+3}}\approx \frac{1.7743E+27}{1.4234E+25}\approx 125>1$ .

When $\lambda =17$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{578\cdot {11.2112}^{25-1}}{{425}^{\frac{\sqrt{425}}{3}+3}}\approx \frac{8.9862E+27}{8.8497E+25}\approx 102>1$ .

When $\lambda =18$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{648\cdot {\left(11.8909\right)}^{25-1}}{{450}^{\frac{\sqrt{450}}{3}+3}}\approx \frac{4.1374E+28}{5.2574E+26}\approx 78.7>1$ .

When $\lambda =19$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{722\cdot {12.5705}^{25-1}}{{475}^{\frac{\sqrt{475}}{3}+3}}\approx \frac{1.7498E+29}{2.9904E+27}\approx 58.5>1$ .

When $\lambda =20$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{800\cdot {13.2502}^{25-1}}{{500}^{\frac{\sqrt{500}}{3}+3}}\approx \frac{6.8616E+29}{1.6366E+27}\approx 41.9>1$ .

When $\lambda =21$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{882\cdot {13.9298}^{25-1}}{{525}^{\frac{\sqrt{525}}{3}+3}}\approx \frac{2.5127E+30}{8.6291E+28}\approx 29.1>1$ .

When $\lambda =22$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{968\cdot {14.6094}^{25-1}}{{550}^{\frac{\sqrt{550}}{3}+3}}\approx \frac{8.6507E+30}{4.4015E+29}\approx 19.7>1$ .

When $\lambda =23$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{1058\cdot {15.2891}^{25-1}}{{575}^{\frac{\sqrt{575}}{3}+3}}\approx \frac{2.8161E+31}{2.1742E+30}\approx 13.0>1$ .

When $\lambda =24$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{1152\cdot {15.9687}^{25-1}}{{600}^{\frac{\sqrt{600}}{3}+3}}\approx \frac{8.7081E+31}{1.0425E+31}\approx 8.35>1$ .

When $\lambda =25$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{1250\cdot {16.6483}^{25-1}}{{625}^{\frac{\sqrt{625}}{3}+3}}\approx \frac{2.5691E+32}{4.8599E+31}\approx 5.29>1$ .

When $\lambda =26$ , $f_8\left(n_0,\lambda \right)=\frac{2\lambda \cdot {\left(\frac{\lambda }{4}\cdot {\left(\frac{\lambda }{\lambda -1}\right)}^{\lambda -1}\right)}^{n_0-1}}{{\left(\lambda n_0\right)}^{\frac{\sqrt{\lambda n_0}}{3}+3}}\approx \frac{1352\cdot {17.3279}^{25-1}}{{650}^{\frac{\sqrt{650}}{3}+3}}\approx \frac{7.2594E+32}{2.2080E+32}\approx 3.29>1$ .

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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