JAMPJournal of Applied Mathematics and Physics2327-4352Scientific Research Publishing10.4236/jamp.2023.1111234JAMP-129474ArticlesPhysics&Mathematics On Prime Numbers between <i>kn</i> and (<i>k</i> + 1) <i>n</i> WingK. Yu1*Independent Researcher, Arlington, USA0311202311113712373413, October 202327, November 2023 30, November 2023© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

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.

Bertrand’s Postulate-Chebyshev’s Theorem The Prime Number Theorem Landau Problems Legendre’s Conjecture Prime Number Distribution1. Introduction

The Bertrand-Chebyshev’s theorem states that for any positive integer n, there always exists a prime number p such that n < p ≤ 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 2 n < p ≤ 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 ( λ n n ) where λ is a positive integer. In that

paper, the author proved that when n ≥ λ − 2 ≥ 25 , i.e., when n ≥ λ − 2 and λ ≥ 27 , there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n . In this article, we will use the same method to complete the entire work on this problem. We will prove that when n ≥ λ − 2 and λ ≥ 3 , there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n . Then, by converting λ to k, we prove that for two positive integers n and k, when n ≥ k − 1 , there is always at least a prime number p such that k n < p ≤ ( k + 1 ) 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 ≥ λ − 2 , there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n . In section 4, we will convert λ to k, and complete this article.

Definition: Γ a ≥ p > b { ( λ n n ) } denotes the prime number factorization operator of the integer expression ( λ n n ) . It is the product of the prime numbers in the decomposition of ( λ n n ) in the range of a ≥ p > b . In this operator, p is a prime number, a and b are real numbers, and λ n ≥ a ≥ p > b ≥ 1 .

It has some properties:

It is always true that Γ a ≥ p > b { ( λ n n ) } ≥ 1 (1.1)

If there is no prime number in ( λ n n ) within the range of a ≥ p > b , then Γ a ≥ p > b { ( λ n n ) } = 1 , or vice versa, if Γ a ≥ p > b { ( λ n n ) } = 1 , then there is no prime number in ( λ n n ) within the range of a ≥ p > b . (1.2)

For example, when λ = 5 and n = 4 , Γ 16 ≥ p > 10 { ( 20 4 ) } = 13 0 ⋅ 11 0 = 1 . No prime number 13 or 11 is in ( 20 4 ) within the range of 16 ≥ p > 10 .

If there is at least one prime number in ( λ n n ) in the range of a ≥ p > b , then Γ a ≥ p > b { ( λ n n ) } > 1 , or vice versa, if Γ a ≥ p > b { ( λ n n ) } > 1 , then there is at least one prime number in ( λ n n ) within the range of a ≥ p > b . (1.3)

For example, when λ = 5 and n = 4 , Γ 18 ≥ p > 16 { ( 20 4 ) } = 17 > 1 . A prime number 17 is in ( 20 4 ) within the range of 18 ≥ p > 16 .

Let v p ( n ) be the p-adic valuation of n, the exponent of the highest power of p that divides n.

We define R ( p ) by the inequalities p R ( p ) ≤ λ n < p R ( p ) + 1 , and determine the p-adic valuation of ( λ n n ) .

v p ( ( λ n n ) ) = v p ( ( λ n ) ! ) − v p ( ( ( λ − 1 ) n ) ! ) − v p ( n ! ) = ∑ i = 1 R ( p ) ( ⌊ λ n p i ⌋ − ⌊ ( λ − 1 ) n p i ⌋ − ⌊ n p i ⌋ ) ≤ R ( p )

because for any real numbers a and b , the expression of ⌊ a + b ⌋ − ⌊ a ⌋ − ⌊ b ⌋ is 0 or 1.

Thus, if p divides ( λ n n ) , then v p ( ( λ n n ) ) ≤ R ( p ) ≤ log p ( λ n ) , or p v p ( ( λ n n ) ) ≤ p R ( p ) ≤ λ n (1.4)

If n ≥ p > ⌊ λ n ⌋ , then 0 ≤ v p ( ( λ n n ) ) ≤ R ( p ) ≤ 1 . (1.5)

Let π ( n ) 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 ≡ 1 ( MOD   6 ) and p ≡ 5 ( MOD   6 ) . Thus, π ( n ) ≤ ⌊ n 3 ⌋ + 2 ≤ n 3 + 2 . (1.6)

From the prime number decomposition, when n > ⌊ λ n ⌋ ,

( λ n n ) = Γ λ n ≥ p > n { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ⋅ Γ n ≥ p > ⌊ λ n ⌋ { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! }       ⋅ Γ ⌊ λ n ⌋ ≥ p { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! }

When n ≤ ⌊ λ n ⌋ , ( λ n n ) ≤ Γ λ n ≥ p > n { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ⋅ Γ ⌊ λ n ⌋ ≥ p { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! }

Thus, ( λ n n ) ≤ Γ λ n ≥ p > n { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ⋅ Γ n ≥ p > ⌊ λ n ⌋ { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! }   ⋅ Γ ⌊ λ n ⌋ ≥ p { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! }

Γ λ n ≥ p > n { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } = Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } since all prime numbers in n ! do not appear in the range of λ n ≥ p > n .

If n ≥ λ − 2 , and there is a prime number p in Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } , then p ≥ n + 1 = ( n + 2 ) n + 1 > λ n . From (1.5), 0 ≤ v p ( Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ) ≤ R ( p ) ≤ 1 . Thus, if n ≥ λ − 2 , every prime number in Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } has a power of 0 or 1. (1.7)

Referring to (1.5), Γ n ≥ p > ⌊ λ n ⌋ { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ≤ ∏ n ≥ p p . It has been proven [6] that for n ≥ 3 , ∏ n ≥ p p < 2 2 n − 3 .

When n = 2 , ∏ n ≥ p p = 2 2 n − 3 , then for n ≥ 2 , Γ n ≥ p > ⌊ λ n ⌋ { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ≤ ∏ n ≥ p p ≤ 2 2 n − 3 .

Referring to (1.4) and (1.6), Γ ⌊ λ n ⌋ ≥ p { ( λ n ) ! n ! ⋅ ( ( λ − 1 ) n ) ! } ≤ ( λ n ) λ n 3 + 2 .

Thus, for λ ≥ 3 and n ≥ 2 , ( λ n n ) ≤ Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ⋅ 2 2 n − 3 ⋅ ( λ n ) λ n 3 + 2 (1.8)

2. Lemmas

Lemma 1: For λ ≥ 3 and n ≥ 2 , Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n − 1 ( λ n ) λ n 3 + 3 = f 8 ( n , λ ) (2.1)

Proof:

Let a real number x ≥ 3 , and f 1 ( x ) = 2 ( 2 x − 1 ) x − 1 ; then, f ′ 1 ( x ) = 2 ( x − 1 ) ( 2 x − 1 ) ′ − 2 ( 2 x − 1 ) ( x − 1 ) ′ ( x − 1 ) 2 = − 2 ( x − 1 ) 2 < 0 .

Thus, f 1 ( x ) is a strictly decreasing function for x ≥ 3 .

Since f 1 ( 3 ) = 5 , and lim x → ∞ f 1 ( x ) = 4 , for x ≥ 3 , we have 5 ≥ f 1 ( x ) = 2 ( 2 x − 1 ) x − 1 ≥ 4 .

Let f 2 ( x ) = ( x x − 1 ) x , then f ′ 2 ( x ) = ( ( x x − 1 ) x ) ′ = ( e x ⋅ ln x x − 1 ) ′ = e x ⋅ ln x x − 1 ⋅ ( x ⋅ ln x x − 1 ) ′

f ′ 2 ( x ) = ( x x − 1 ) x ⋅ ( ln x x − 1 + x ⋅ ( ln x x − 1 ) ′ ) = ( x x − 1 ) x ⋅ ( ln x x − 1 + x ⋅ x − 1 x ⋅ x − 1 − x ( x − 1 ) 2 )

f ′ 2 ( x ) = ( x x − 1 ) x ⋅ ( ln x x − 1 − 1 x − 1 ) (2.1.1)

In (2.1.1), for x ≥ 3 , 1 x − 1 = 1 x + 1 x 2 + 1 x 3 + 1 x 4 + 1 x 5 + 1 x 6 + ⋯

Using the formula: ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + x 5 5 − x 6 6 + ⋯ , we have

ln x x − 1 = ln 1 1 + − 1 x = − ln ( 1 + − 1 x ) = 1 x + 1 2 x 2 + 1 3 x 3 + 1 4 x 4 + 1 5 x 5 + 1 6 x 6 + ⋯

Thus, for x ≥ 3 , ln x x − 1 − 1 x − 1 < 0 .

Since ( x x − 1 ) x is a positive number for x ≥ 3 , f ′ 2 ( x ) = ( x x − 1 ) x ⋅ ( ln x x − 1 − 1 x − 1 ) < 0 .

Thus f 2 ( x ) is a strictly deceasing function for x ≥ 3 . Since f 2 ( 3 ) = 3.375 and lim x → ∞ f 2 ( x ) = e ≈ 2.718 , for x ≥ 3 , 3.375 ≥ f 2 ( x ) = ( x x − 1 ) x ≥ e (2.1.2)

Since for x ≥ 3 , f 1 ( x ) has a lower bound of 4 and f 2 ( x ) has an upper bound of 3.375,

f 1 ( x ) = 2 ( 2 x − 1 ) x − 1 > f 2 ( x ) = ( x x − 1 ) x

When x = λ ≥ 3 , we have 2 ( 2 λ − 1 ) λ − 1 > ( λ λ − 1 ) λ (2.1.3)

When λ ≥ 3 and n = 2 , ( λ n n ) = ( 2 λ 2 ) = 2 λ ( 2 λ − 1 ) ( 2 λ − 2 ) ! 2 ( 2 λ − 2 ) ! = λ ( 2 λ − 1 ) (2.1.4)

λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 = λ 2 λ − λ + 1 2 ( λ − 1 ) 2 ( λ − 1 ) − λ + 1 = λ ( λ − 1 ) 2 ⋅ ( λ λ − 1 ) λ (2.1.5)

Since λ ( λ − 1 ) 2 is a positive number for λ ≥ 3 , referring to (2.1.4) and (2.1.5), when λ ( λ − 1 ) 2 multiplies both sides of (2.1.3), we have

( λ ( λ − 1 ) 2 ) ( 2 ( 2 λ − 1 ) λ − 1 ) = λ ( 2 λ − 1 ) = ( λ n n ) > ( λ ( λ − 1 ) 2 ) ( λ λ − 1 ) λ = λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1

Thus, ( λ n n ) > λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 when λ ≥ 3 and n = 2 . (2.1.6)

By induction on n, when λ ≥ 3 , if ( λ n n ) > λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 is true for n, then for n + 1 ,

( λ ( n + 1 ) n + 1 ) = ( λ n + λ n + 1 ) = ( λ n + λ ) ( λ n + λ − 1 ) ⋯ ( λ n + 2 ) ( λ n + 1 ) ( λ n + λ − n − 1 ) ( λ n + λ − n − 2 ) ⋯ ( λ n − n + 1 ) ( n + 1 ) ⋅ ( λ n n )

( λ ( n + 1 ) n + 1 ) > ( λ n + λ ) ( λ n + λ − 1 ) ⋯ ( λ n + 2 ) ( λ n + 1 ) ( λ n + λ − n − 1 ) ( λ n + λ − n − 2 ) ⋯ ( λ n − n + 1 ) ( n + 1 ) ⋅ λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1

( λ ( n + 1 ) n + 1 ) > ( λ n + λ ) ( λ n + λ − 1 ) ⋯ ( λ n + 2 ) ( λ n + λ − n − 1 ) ( λ n + λ − n − 2 ) ⋯ ( λ n − n + 1 ) ⋅ λ n + 1 n ⋅ 1 n + 1 ⋅ λ λ n − λ + 1 ( λ − 1 ) ( λ − 1 ) n − λ + 1

Notice λ n + 1 n > λ , and ( λ n + λ ) ( λ n + λ − 1 ) ⋯ ( λ n + 2 ) ( λ n + λ − n − 1 ) ( λ n + λ − n − 2 ) ⋯ ( λ n − n + 1 ) > ( λ λ − 1 ) λ − 1 because λ n + λ λ n + λ − n − 1 = λ λ − 1 ; λ n + λ − 1 λ n + λ − n − 2 > λ λ − 1 ; ⋯ ; λ n + 2 λ n − n + 1 > λ λ − 1 . Thus,

( λ ( n + 1 ) n + 1 ) > λ λ − 1 ( λ − 1 ) λ − 1 ⋅ λ 1 ⋅ 1 n + 1 ⋅ λ λ n − λ + 1 ( λ − 1 ) ( λ − 1 ) n − λ + 1 = λ λ ( n + 1 ) − λ + 1 ( n + 1 ) ( λ − 1 ) ( λ − 1 ) ( n + 1 ) − λ + 1 (2.1.7)

From (2.1.6) and (2.1.7), we have for λ ≥ 3 and n ≥ 2 , ( λ n n ) > λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1

Referring to (1.8), for λ ≥ 3 and n ≥ 2 , ( λ n n ) ≤ Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ⋅ 2 2 n − 3 ⋅ ( λ n ) λ n 3 + 2 .

Then, Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } ⋅ 2 2 n − 3 ⋅ ( λ n ) λ n 3 + 2 > λ λ n − λ + 1 n ( λ − 1 ) ( λ − 1 ) n − λ + 1 .

Since when λ ≥ 3 and n ≥ 2 , then 2 2 n − 3 > 0 and ( λ n ) λ n 3 + 2 > 0 .

Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > λ λ n − λ + 1 ( λ n ) λ n 3 + 2 ⋅ 2 2 n − 3 ⋅ n ( λ − 1 ) ( λ − 1 ) n − λ + 1 = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n − 1 ( λ n ) λ n 3 + 3

Thus, Lemma 1 is proven.

Lemma 2: When n ≥ λ − 2 ≥ 9 , f 3 ( n , λ ) = 2 λ 2 ⋅ ( λ − 1 4 ⋅ e ) n − 1 ( λ n ) λ 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 λ ≥ 3 , ( λ λ − 1 ) λ ≥ e . From (2.1), when λ ≥ 3 and n ≥ 2 ,

Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > f 8 ( n , λ ) = 2 λ 2 ⋅ ( λ − 1 4 ⋅ ( λ λ − 1 ) λ ) n − 1 ( λ n ) λ n 3 + 3 , and f 8 ( n , λ ) ≥ 2 λ 2 ⋅ ( λ − 1 4 ⋅ e ) n − 1 ( λ n ) λ n 3 + 3 = f 3 ( n , λ ) > 0 .

Thus, Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > f 8 ( n , λ ) ≥ f 3 ( n , λ ) > 0 . (2.2.1)

Let x ≥ 2 and y ≥ 4 both be real numbers. When x = y − 2 ,

f 3 ( x , y ) = 2 ( x + 2 ) 2 ⋅ ( x + 1 4 ⋅ e ) x − 1 ( ( x + 2 ) ⋅ x ) x ⋅ ( x + 2 ) 3 + 3 > f 4 ( x ) = 2 ( x + 2 ) 2 ⋅ ( x + 1 4 ⋅ e ) x − 1 ( ( x + 2 ) ⋅ x ) x + 1 3 + 3 > 0 (2.2.2)

f ′ 4 ( x ) = f 4 ( x ) ⋅ ( 2 x + 2 + ln ( x + 1 4 ) + 4 3 − 2 x + 1 − 1 3 ln ( ( x + 2 ) ⋅ x ) − 10 3 x − 8 3 ( x + 2 ) ) = f 4 ( x ) ⋅ f 5 ( x )

where f 5 ( x ) = 2 x + 2 + ln ( x + 1 4 ) + 4 3 − 2 x + 1 − 1 3 ln ( ( x + 2 ) ⋅ x ) − 10 3 x − 8 3 ( x + 2 ) f ′ 5 ( x ) = 4 x + 6 ( x + 1 ) 2 ⋅ ( x + 2 ) 2 + x 2 + 2 x − 2 3 x ( x + 1 ) ( x + 2 ) + 10 3 x 2 + 8 3 ( x + 2 ) 2 > 0 when x ≥ 2 .

Thus, f 5 ( x ) is a strictly increasing function for x ≥ 2 .

When x = 9 , f 5 ( x ) = 2 9 + 2 + ln ( 9 + 1 4 ) + 4 3 − 2 9 + 1 − 1 3 ln ( 9 ) − 1 3 ln ( 9 + 2 ) − 10 27 − 8 33 > 0 . Thus, for x ≥ 9 , f 5 ( x ) > 0 . Then, f ′ 4 ( x ) = f 4 ( x ) ⋅ f 5 ( x ) > 0 .

Thus, f 4 ( x ) is a strictly increasing function for x ≥ 9 .

From (2.2.2), when x = y − 2 , f 3 ( x , y ) > f 4 ( x ) > 0 . Thus, when x = y − 2 ≥ 9 and x y ≥ 99 , then f 3 ( x , y ) is an increasing function respect to the product of xy. (2.2.3)

∂ f 3 ( x , y ) ∂ x = f 3 ( x , y ) ⋅ ( ln ( y − 1 4 ) + 1 − y 6 x ⋅ ln ( y x ) − y 3 x − 3 x ) = f 3 ( x , y ) ⋅ f 6 ( x , y ) (2.2.4)

where f 6 ( x , y ) = ln ( y − 1 4 ) + 1 − y 6 x ⋅ ln ( y x ) − y 3 x − 3 x

When x = y − 2 , then f 6 ( x , y ) = f 7 ( x ) = ln ( x + 1 4 ) + 1 − x + 2 6 x ⋅ ( ln ( x + 2 ) + ln ( x ) + 2 ) − 3 x

When x ≥ 2 , f ′ 7 ( x ) = 1 x + 1 − x + 2 6 x ⋅ ( 1 x + 2 + 1 x ) + ln ( x + 2 ) + ln ( x ) + 2 6 x x ( x + 2 ) + 3 x 2

f ′ 7 ( x ) = 1 x + 1 − x + 2 6 x ⋅ x + x + 2 x ( x + 2 ) + ln ( x + 2 ) + ln ( x ) + 2 6 x x ( x + 2 ) + 3 x 2 = 1 x + 1 − 1 3 x ⋅ x + 1 x x + 2 + 2 6 x x ( x + 2 ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 = 1 x + 1 − x 3 x x ( x + 2 ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 (2.2.5)

When x ≥ 2 , then 3 x ( x + 2 ) + ( x + 1 ) > 0

( 3 x ( x + 2 ) + ( x + 1 ) ) ⋅ ( 3 x ( x + 2 ) − ( x + 1 ) ) = ( 3 x ( x + 2 ) ) 2 − ( x + 1 ) 2 = 8 x 2 + 16 x − 1 > 0

Thus, ( 3 x ( x + 2 ) + ( x + 1 ) ) ⋅ ( 3 x ( x + 2 ) − ( x + 1 ) ) > 0

3 x ( x + 2 ) − ( x + 1 ) > 0

3 x ( x + 2 ) > x + 1 then 1 x + 1 > 1 3 x ( x + 2 )

When x ≥ 2 , 1 x + 1 − 1 3 x ( x + 2 ) > 0 , and from (2.2.5), ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 > 0 .

Then f ′ 7 ( x ) = ( 1 x + 1 − 1 3 x ( x + 2 ) ) + ln ( x + 2 ) + ln ( x ) 6 x x ( x + 2 ) + 3 x 2 > 0 .

Thus, when x ≥ 2 , f 7 ( x ) is a strictly increasing function.

When x = y − 2 ≥ 2 , since f 6 ( x , y ) = f 7 ( x ) , f 6 ( x , y ) is an increasing function respect to xy.

When x = y − 2 = 9 , f 6 ( x , y ) = ln ( 11 − 1 4 ) + 1 − 11 6 9 ⋅ ln ( 99 ) − 11 3 9 − 3 9 > 0 .

When x ≥ y − 2 ≥ 2 , ∂ f 6 ( x , y ) ∂ x = y 12 x x ⋅ ln ( y ) + y 12 x x ⋅ ln ( x ) + y 6 x x + y 6 x x + 3 x 2 > 0 .

Thus, when x ≥ y − 2 ≥ 9 , f 6 ( x , y ) > 0 , and it is an increasing function with respect to x and to the product of xy, then, from (2.2.4), ∂ f 3 ( x , y ) ∂ x = f 3 ( x , y ) ⋅ f 6 ( x , y ) > 0 .

Thus, when x ≥ y − 2 ≥ 9 , f 3 ( x , y ) is an increasing function with respect to x. (2.2.6)

Referring to (2.2.3) and (2.2.6), when x ≥ y − 2 ≥ 9 , then x y ≥ 99 , f 3 ( x , y ) is an increasing function with respect to the product of xy and with respect to x.

Let x = n and y = λ . Then when n ≥ λ − 2 ≥ 9 , f 3 ( n , λ ) 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 ≥ 24 and 26 ≥ λ ≥ 3 , f 8 ( n , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n − 1 ( λ n ) λ n 3 + 3 is a strictly increasing function respect to n. (2.3)

Proof:

Referring to (2.1), when λ ≥ 3 and n ≥ 2 , f 8 ( n , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n − 1 ( λ n ) λ n 3 + 3 > 0

Let a real number x ≥ 2 . When λ ≥ 3 , then f 8 ( x , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) x − 1 ( λ x ) λ x 3 + 3 > 0 (2.3.1)

When λ is an integer constant in the range of 10 ≥ λ ≥ 3 ,

∂ f 8 ( x , λ ) ∂ x = 2 λ 2 ⋅ ∂ ∂ x ( ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) x − 1 ) ⋅ ( ( λ x ) λ x 3 + 3 ) − ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) x − 1 ⋅ ∂ ∂ x ( ( λ x ) λ x 3 + 3 ) ( ( λ x ) λ x 3 + 3 ) 2

∂ f 8 ( x , λ ) ∂ x = f 8 ( x , λ ) ⋅ ( ln ( λ 4 ) + ln ( λ λ − 1 ) λ − 1 − λ ( ln ( x ) + ln ( λ ) + 2 ) 6 x − 3 x ) = f 8 ( x , λ ) ⋅ f 9 ( x , λ )

where

f 9 ( x , λ ) = ln ( λ 4 ) + ln ( λ λ − 1 ) λ − 1 − λ ( ln ( x ) + ln ( λ ) + 2 ) 6 x − 3 x

∂ f 9 ( x , λ ) ∂ x = λ ln ( λ ) + λ ln ( x ) 12 x x + 3 x 2 > 0 for x > 1 and λ > 2 ; then, f 9 ( x , λ ) is a strictly increasing function respect to x.

When x = 24 and 10 ≥ λ ≥ 3 , f 9 ( x , λ ) > 0 . The calculations are below.

When λ = 3 , f 9 ( x , λ ) = ln ( 3 4 ) + ln ( 3 3 − 1 ) 3 − 1 − 3 ( ln ( 24 ) + ln ( 3 ) + 2 ) 6 24 − 3 24 ≈ 0.0283 > 0 .

When λ = 4 , f 9 ( x , λ ) = ln ( 4 4 ) + ln ( 4 4 − 1 ) 4 − 1 − 4 ( ln ( 24 ) + ln ( 4 ) + 2 ) 6 24 − 3 24 ≈ 0.2814 > 0 .

When λ = 5 , f 9 ( x , λ ) = ln ( 5 4 ) + ln ( 5 5 − 1 ) 5 − 1 − 5 ( ln ( 24 ) + ln ( 5 ) + 2 ) 6 24 − 3 24 ≈ 0.4744 > 0 .

When λ = 6 , f 9 ( x , λ ) = ln ( 6 4 ) + ln ( 6 6 − 1 ) 6 − 1 − 6 ( ln ( 24 ) + ln ( 6 ) + 2 ) 6 24 − 3 24 ≈ 0.6113 > 0 .

When λ = 7 , f 9 ( x , λ ) = ln ( 7 4 ) + ln ( 7 7 − 1 ) 7 − 1 − 7 ( ln ( 24 ) + ln ( 7 ) + 2 ) 6 24 − 3 24 ≈ 0.7183 > 0 .

When λ = 8 , f 9 ( x , λ ) = ln ( 8 4 ) + ln ( 8 8 − 1 ) 8 − 1 − 8 ( ln ( 24 ) + ln ( 8 ) + 2 ) 6 24 − 3 24 ≈ 0.8044 > 0 .

When λ = 9 , f 9 ( x , λ ) = ln ( 9 4 ) + ln ( 9 9 − 1 ) 9 − 1 − 9 ( ln ( 24 ) + ln ( 9 ) + 2 ) 6 24 − 3 24 ≈ 0.8755 > 0 .

When λ = 10 , f 9 ( x , λ ) = ln ( 10 4 ) + ln ( 10 10 − 1 ) 10 − 1 − 10 ( ln ( 24 ) + ln ( 10 ) + 2 ) 6 24 − 3 24 ≈ 0.9347 > 0 .

From (2.3.1), when x ≥ 2 and λ ≥ 3 , f 8 ( x , λ ) > 0 .

Since ∂ f 8 ( x , λ ) ∂ x = f 8 ( x , λ ) ⋅ f 9 ( x , λ ) , when x = 24 and 10 ≥ λ ≥ 3 , ∂ f 8 ( x , λ ) ∂ x > 0 .

Thus, when x ≥ 24 and 10 ≥ λ ≥ 3 , f 8 ( x , λ ) is a strictly increasing function respect to x.

Let x = n ≥ 24 and 10 ≥ λ ≥ 3 , f 8 ( n , λ ) is a strictly increasing function respect to n. (2.3.2)

From (2.2.1), when λ ≥ 3 and n ≥ 2 , f 8 ( n , λ ) ≥ f 3 ( n , λ ) > 0 . Referring to (2.2), when λ ≥ 11 and n ≥ λ − 2 , f 3 ( n , λ ) is an increasing function with respect to the product of λn and with respect to n.

When n ≥ 24 and 26 ≥ λ ≥ 11 , since n ≥ λ − 2 and f 8 ( n , λ ) ≥ f 3 ( n , λ ) , f 8 ( n , λ ) 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 ≥ 24 and 26 ≥ λ ≥ 3 , f 8 ( n , λ ) is an increasing function with respect to n.

Thus, Lemma 3 is proven.

Lemma 4: When n ≥ n 0 ≥ 24 and 26 ≥ λ ≥ 3 , if f 8 ( n , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n − 1 ( λ n ) λ n 3 + 3 > 1 , then there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n . (2.4)

Proof:

Let integers m ≥ n ≥ n 0 ≥ 24 .

From (2.3), when n ≥ n 0 ≥ 24 and 26 ≥ λ ≥ 3 , f 8 ( n 0 , λ ) is an increasing function respect to n.

When 26 ≥ λ ≥ 3 , if f 8 ( n 0 , λ ) > 1 , then f 8 ( n , λ ) > 1 , and thus, f 8 ( m , λ ) > 1 ; then from (2.1), Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > f 8 ( n , λ ) > 1 , and Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } > f 8 ( m , λ ) > 1 . (2.4.1)

Note that m ≥ n ≥ n 0 ≥ 24 ≥ λ − 2 since 26 ≥ λ ≥ 3 .

From (1.7), when n ≥ λ − 2 , every prime number in Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } has a power of 0 or 1; when m ≥ λ − 2 , every prime number in Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } has a power of 0 or 1. (2.4.2)

Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = Γ λ m ≥ p > ( λ − 1 ) m { ( λ m ) ! ( ( λ − 1 ) m ) ! }   ⋅ ∏ i = 1 i = λ − 2 ( Γ ( λ − 1 ) m i ≥ p > λ m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ Γ λ m i + 1 ≥ p > ( λ − 1 ) m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } )

In ∏ i = 1 i = λ − 2 ( Γ ( λ − 1 ) m i ≥ p > λ m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) , for every distinct prime number p in these ranges, the numerator ( λ m ) ! has the product of p ⋅ 2 p ⋅ 3 p ⋯ i p = i ! ⋅ p i . The denominator ( ( λ − 1 ) m ) ! also has the same product of i ! ⋅ p i . Thus, they cancel each other in ( λ m ) ! ( ( λ − 1 ) m ) ! .

Referring to (1.2), ∏ i = 1 i = λ − 2 ( Γ ( λ − 1 ) m i ≥ p > λ m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) = 1 .

Thus,

Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = Γ λ m ≥ p > ( λ − 1 ) m { ( λ m ) ! ( ( λ − 1 ) m ) ! } ⋅ ∏ i = 1 i = λ − 2 ( Γ λ m i + 1 ≥ p > ( λ − 1 ) m i + 1 { ( λ m ) ! ( ( λ − 1 ) m ) ! } )

Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } = ∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) . (2.4.3)

∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) is the product of (λ − 1) sectors from i = 1 to i = λ − 1 .

Each of these sectors is the prime number factorization of the product of the consecutive integers between ( λ − 1 ) m i and λ m i .

Referring to (2.4.3), when Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } > 1 , then ∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) > 1 .

Referring to (1.1), Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ≥ 1 . Thus, when Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } > 1 , at least one of the sectors is greater than one in ∏ i = 1 i = λ − 1 ( Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } ) .

Let Γ λ m i ≥ p > ( λ − 1 ) m i { ( λ m ) ! ( ( λ − 1 ) m ) ! } > 1 be such a sector and let m = n i where λ − 1 ≥ i ≥ 1 from (2.4.3). Thus, when m = n i ≥ n ≥ λ – 2 ,

Γ λ n i i ≥ p > ( λ − 1 ) n i i { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } = Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } > 1 . (2.4.4)

( λ n i ) ! ( ( λ − 1 ) n i ) ! = ( λ n i ) ⋅ ( λ n i − 1 ) ⋯ ( λ n i − i ) ⋯ ( λ n i − 2 i ) ⋯ ( λ n i − ( n − 1 ) i ) ⋯ ( λ n i − n i + 1 ) ⋅ ( ( λ − 1 ) n i ) ! ( ( λ − 1 ) n i ) !

( λ n i ) ! ( ( λ − 1 ) n i ) ! = i ⋅ ( λ n ) ⋅ ( λ n i − 1 ) ⋯ i ⋅ ( λ n − 1 ) ⋯ i ⋅ ( λ n − 2 ) ⋯ i ⋅ ( λ n − n + 1 ) ⋯ ( λ n i − n i + 1 ) ⋅ ( ( λ − 1 ) n i ) ! ( ( λ − 1 ) n i ) !

Thus, ( λ n i ) ! ( ( λ − 1 ) n i ) ! contains all the factors of λ n , λ n − 1 , λ n − 2 , ⋯ , λ n − n + 1 in ( λ n ) ! ( ( λ − 1 ) n ) ! .

These factors make up all the consecutive integers in the range of λ n ≥ p > ( λ − 1 ) n in ( λ n ) ! ( ( λ − 1 ) n ) ! . Thus, ( λ n i ) ! ( ( λ − 1 ) n i ) ! contains ( λ n ) ! ( ( λ − 1 ) n ) ! .

Referring to the definition, all prime numbers in ( λ n i ) ! ( ( λ − 1 ) n i ) ! in the ranges of λ n i ≥ p > λ n and ( λ − 1 ) n > p do not contribute to Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } , nor does i for λ − 1 ≥ i ≥ 1 . Only the prime numbers in the prime factorization of ( λ n i ) ! ( ( λ − 1 ) n i ) ! in the range of λ n ≥ p > ( λ − 1 ) n present in Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } . Since ( λ n ) ! ( ( λ − 1 ) n ) ! is the product of all the consecutive integers in this range, Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } = Γ λ n ≥ p > ( λ − 1 ) n { ( λ n ) ! ( ( λ − 1 ) n ) ! } .

Referring to (2.4.4), Γ λ n ≥ p > ( λ − 1 ) n { ( λ n i ) ! ( ( λ − 1 ) n i ) ! } = Γ λ n ≥ p > ( λ − 1 ) n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 1 . (2.4.5)

Referring to (2.4.1), (2.4.3), (2.4.4), and (2.4.5), when 26 ≥ λ ≥ 3 and n ≥ n 0 ≥ 24 ,

if f 8 ( n , λ ) > 1 , then f 8 ( m , λ ) > 1 and Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } > 1 ;

when Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } > 1 , then Γ λ n ≥ p > ( λ − 1 ) n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 1 .

Thus, when 26 ≥ λ ≥ 3 and n ≥ n 0 ≥ 24 , if f 8 ( n , λ ) > 1 , then Γ λ n ≥ p > ( λ − 1 ) n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 1 , referring to (1.3), there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .

Thus, Lemma 4 is proven.

Lemma 5: When n ≥ λ − 2 ≥ 25 , there exists at least a prime number p such

that ( λ − 1 ) n < p ≤ λ n . (2.5)

Proof:

Referring to (2.2), when n ≥ λ − 2 ≥ 9 , f 3 ( n , λ ) = 2 λ 2 ⋅ ( λ − 1 4 ⋅ e ) n − 1 ( λ n ) λ 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 λ 1 = 27 . Then f 3 ( n 1 , λ 1 ) = 2 ⋅ 27 2 ⋅ ( 27 − 1 4 ⋅ e ) 25 − 1 ( 25 ⋅ 27 ) 675 3 + 3 ≈ 1.2495 E + 33 9.7843 E + 32 > 1 .

When m = n = n 1 = λ − 2 = λ 1 − 2 = 25 , f 3 ( m , λ ) = f 3 ( n , λ ) = f 3 ( n 1 , λ 1 ) > 1 .

From (2.2), when m = n = λ − 2 ≥ 25 , f 3 ( n , λ ) > 1 and f 3 ( m , λ ) > 1 since f 3 ( n , λ ) is an increasing function with respect to the product of λn. When m ≥ n ≥ λ − 2 ≥ 25 , then f 3 ( n , λ ) > 1 and f 3 ( m , λ ) > 1 since f 3 ( n , λ ) is also an increasing function with respect to n.

Referring to (2.2.1), when m ≥ n ≥ λ − 2 ≥ 25 , Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > f 3 ( n , λ ) ≥ f 3 ( n 1 , λ 1 ) > 1 ; and Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } > f 3 ( m , λ ) ≥ f 3 ( n , λ ) > 1 . (2.5.1)

Referring to (2.4.2), when m ≥ n ≥ λ − 2 ≥ 25 , every prime number in Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } and in Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } 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 ≥ n ≥ λ − 2 ≥ 25 , Γ λ n ≥ p > n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 1 , and Γ λ m ≥ p > m { ( λ m ) ! ( ( λ − 1 ) m ) ! } > 1 , then Γ λ n ≥ p > ( λ − 1 ) n { ( λ n ) ! ( ( λ − 1 ) n ) ! } > 1 ; referring to (1.3), there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ 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 λ = 3 , when n ≥ λ − 2 , there exists at least a prime number p such

that ( λ − 1 ) n < p ≤ λ n . (3.1)

Proof:

Referring to (2.3) for λ = 3 , when n ≥ n 0 ≥ 24 , f 8 ( n 0 , λ ) is a strictly increasing function on n 0 .

Let n 0 = 83 . When λ = 3 and n ≥ n 0 = 83 ,

f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 = 18 ⋅ 1.6875 83 − 1 249 249 3 + 3 ≈ 7.7493 E + 19 6.2000 E + 19 > 1 .

Thus, for λ = 3 , when n ≥ 83 , n > λ − 2 and f 8 ( n , λ ) ≥ f 8 ( n 0 , λ ) > 1 ; then, referring to (2.4), there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .

For λ = 3 , when 82 ≥ n ≥ 1 = λ − 2 , Table 1 shows that there exists a prime number p such that ( λ − 1 ) n < p ≤ λ n .

Thus, (3.1) is proven.

Proposition 2: For 5 ≥ λ ≥ 4 , when n ≥ λ − 2 , there exists at least a prime number p such that

( λ − 1 ) n < p ≤ λ n . (3.2)

For 82 ≥ n ≥ 1 and λ = 3 , there is a prime number p such that 2 n < p ≤ 3 n
n1234567891011121314
p3571113171919232329293131
n1516171819202122232425262728
p3737414143434747535359596161
n2930313233343536373839404142
p6767717173737979838389899797
n4344454647484950515253545556
p101101103103107107109109113113127127131131
n5758596061626364656667686970
p139139149149151151157157163163167167173173
n717273747576777879808182
p179179181181191191193193197197199199

Proof:

From (2.3), for 5 ≥ λ ≥ 4 , when n ≥ n 0 ≥ 24 , f 8 ( n 0 , λ ) is a strictly increasing function on n 0 .

Let n 0 = 40 . When n ≥ n 0 = 40 and 5 ≥ λ ≥ 4 , f 8 ( n , λ ) ≥ f 8 ( n 0 , λ ) > 1 . The calculations are below.

When λ = 4 , f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 32 ⋅ 2.3703 40 − 1 160 160 3 + 3 ≈ 1.3258 E + 16 8.0503 E + 15 > 1 .

When λ = 5 , f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 50 ⋅ 3.0518 40 − 1 200 200 3 + 3 ≈ 7.8968 E + 18 5.6253 E + 17 > 1 .

Thus, for 5 ≥ λ ≥ 4 , when n ≥ 40 , n > λ − 2 and f 8 ( n , λ ) > f 8 ( n 0 , λ ) > 1 ; then, referring to (2.4), there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .

For 5 ≥ λ ≥ 4 , when 39 ≥ n ≥ λ − 2 , Table 2 shows that there exists a prime number p such that ( λ − 1 ) n < p ≤ λ n .

Thus, (3.2) is proven.

Proposition 3: For 11 ≥ λ ≥ 6 , when n ≥ λ − 2 , there exists at least a prime number p such that

( λ − 1 ) n < p ≤ λ n . (3.3)

Proof:

Referring to (2.3), for 11 ≥ λ ≥ 6 , when n ≥ n 0 ≥ 24 , f 8 ( n 0 , λ ) is a strictly increasing function with respect to n 0 .

Let n 0 = 28 . When n ≥ n 0 = 28 and 11 ≥ λ ≥ 6 , f 8 ( n , λ ) ≥ f 8 ( n 0 , λ ) > 1 .

For 5 ≥ λ ≥ 4 and 39 ≥ n ≥ λ − 2 , there is a prime number p such that ( λ − 1 ) n < p ≤ λ
n234567891011121314
3n691215182124273033363942
p7111317192329313137374143
4n8121620242832364044485256
p131723293137414347535961
5n152025303540455055606570
n15161718192021222324252627
3n45485154576063666972757881
p47535959616167677173797983
4n60646872768084889296100104108
p6771737979838997101103107109113
5n7580859095100105110115120125130135
n282930313233343536373839
3n848790939699102105108111114117
p89899797101101103107109113127131
4n112116120124128132136140144148152156
p113127131137139149151157163173179191
5n140145150155160165170175180185190195

The calculations are below.

When λ = 6 , f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 72 ⋅ 3.7325 28 − 1 168 168 3 + 3 ≈ 2.0014 E + 17 1.9515 E + 16 > 1 .

When λ = 7 , f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 98 ⋅ 4.4128 28 − 1 196 196 3 + 3 ≈ 2.5033 E + 19 3.7501 E + 17 > 1 .

When λ = 8 , f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 128 ⋅ 5.0930 28 − 1 224 224 3 + 3 ≈ 1.5686 E + 21 5.9689 E + 18 > 1 .

When λ = 9 , f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 162 ⋅ 5.7730 28 − 1 252 252 3 + 3 ≈ 5.8527 E + 22 8.1510 E + 19 > 1 .

When λ = 10 , f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 200 ⋅ 6.4529 28 − 1 280 280 3 + 3 ≈ 1.4602 E + 24 9.7946 E + 20 > 1 .

When λ = 11 , f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 242 ⋅ 7.1328 28 − 1 308 308 3 + 3 ≈ 2.6414 E + 25 1.0560 E + 22 > 1 .

Thus, for 11 ≥ λ ≥ 6 , when n ≥ 28 , n > λ − 2 and f 8 ( n , λ ) ≥ f 8 ( n 0 , λ ) > 1 ; then, referring to (2.4), there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .

For 11 ≥ λ ≥ 6 , when 27 ≥ n ≥ λ − 2 , Table 3 shows that there exists a prime number p such that ( λ − 1 ) n < p ≤ λ n .

Thus, (3.3) is proven.

Proposition 4: For 26 ≥ λ ≥ 12 , when n ≥ λ − 2 , there exists at least a prime number p such that

( λ − 1 ) n < p ≤ λ n . (3.4)

Proof:

Referring to (2.3) for 26 ≥ λ ≥ 12 , when n ≥ n 0 ≥ 24 , f 8 ( n 0 , λ ) is a strictly increasing function with respect to n 0 . Let n 0 = 25 . When n ≥ n 0 = 25 and λ = 12 , then f 8 ( n , λ ) ≥ f 8 ( n 0 , λ ) and

f 8 ( n 0 , λ ) = 2 λ 2 ⋅ ( ( λ 4 ) ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 288 ⋅ 7.8126 25 − 1 300 300 3 + 3 ≈ 7.7000 E + 23 5.4078 E + 21 > 1 .

For n ≥ n 0 = 25 and 26 ≥ λ ≥ 13 , then f 8 ( n , λ ) ≥ f 8 ( n 0 , λ ) . It can be seen from Table 4 that the values of f 8 ( n 0 , λ ) as well as f 8 ( n , λ ) are all greater than 1.

(Detailed calculations are in Appendix.)

Thus, for 26 ≥ λ ≥ 12 , when n ≥ 25 > λ − 2 , f 8 ( n , λ ) ≥ f 8 ( n 0 , λ ) > 1 ; then, referring to (2.4), there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .

For 13 ≥ λ ≥ 12 and 24 ≥ n ≥ λ − 2 , Table 5 shows that there exists a prime number p such that ( λ − 1 ) n < p ≤ λ n .

For 26 ≥ λ ≥ 14 and 24 ≥ n ≥ λ − 2 , Table 6 shows that there exists a prime number p such that ( λ − 1 ) n < p ≤ λ n .

Thus, (3.4) is proven.

Combining (3.1), (3.2), (3.3), and (3.4), when 26 ≥ λ ≥ 3 and n ≥ λ − 2 , there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ 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 ≥ k − 1 , there exists at least a prime number p such that k n < p ≤ ( k + 1 ) n . (4.1)

For 11 ≥ λ ≥ 6 and 27 ≥ n ≥ λ − 2 , there is a prime number p such that ( λ − 1 ) n < p ≤ λ n
n456789101112131415
5n202530354045505560657075
λ = 6p232931374147535961677179
6n243036424854606672788490
λ = 7p3137475359616779838997
7n35424956637077849198105
λ = 8p4753596779838997101109
8n48566472808896104112120
λ = 9p596779838997107113127
9n6372819099108117126135
λ = 10p798397107113127137139
10n8090100110120130140150
λ = 11p97107113127137149157
11n99110121132143154165
n161718192021222324252627
5n80859095100105110115120125130135
λ = 6p838997101103107113127131137139149
6n96102108114120126132138144150156162
λ = 7p101107113127131137139149151157163167
7n112119126133140147154161168175182189
λ = 8p113127131137149149151163173179191199
8n128136144152160168176184192200208216
λ = 9p139137149163173179191199211223227229
9n144153162171180189198207216225234243
λ = 10p157163173179191199211223227229239251
10n160170180190200210220230240250260270
λ = 11p163179191199211223227239251257269281
11n176187198209220231242253264275286297
 When n ≥ n 0 = 25 and 26 ≥ λ ≥ 13 , then f 8 ( n , λ ) ≥ f 8 ( n 0 , λ ) > 1
λ1314151617181920212223242526
f 8 ( n 0 )15615714512510278.758.541.929.119.713.08.355.293.29
 For 13 ≥ λ ≥ 12 and 24 ≥ n ≥ λ − 2 , there is a prime number p such that ( λ − 1 ) n < p ≤ λ n
n101112131415161718192021222324
11n110121132143154165176187198209220231242253264
λ = 12p113127137149157167179191199211223241251269277
12n120132144156168180192204216228240252264276288
λ = 13p137149157179191199211223241251269277283293
13n143156169182195208221234247260273286299312
 For 26 ≥ λ ≥ 14 and 24 ≥ n ≥ λ − 2 , there is a prime number p such that ( λ − 1 ) n < p ≤ λ n
n12131415161718192021222324
13n156169182195208221234247260273286299312
λ = 14p157179191199211223241251269277293307317
14n168182196210224238252266280294308322336
λ = 15p191199211227241269277293307317331347
15n195210225240255270285300315330345360
λ = 16p211227241269277293307317331347373
16n224240256272288304320336352368384
λ = 17p241269277293307331347373389401
17n255272289306323340357374391408
λ = 18p277293307331347373389401419
18n288306324342360378396414432
λ = 19p307331347373389401419449
19n323342361380399418437456
λ = 20p347373389401419449461
20n360380400420440460480
λ = 21p389419433461467491
21n399420441462483504
λ = 22p433461467491521
22n440462484506528
λ = 23p467491521541
23n483506529552
λ = 24p521541563
24n528552576
λ = 25p563587
25n575600
λ = 26p613
26n624

Proof:

Referring to (2.5), when n ≥ λ − 2 ≥ 25 , there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n . This statement is the same as that when λ ≥ 27 and n ≥ λ − 2 , there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .

Referring to (3.5), when 26 ≥ λ ≥ 3 and n ≥ λ − 2 , there exists at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .

Thus, when λ ≥ 3 and n ≥ λ − 2 , there is at least a prime number p such that ( λ − 1 ) n < p ≤ λ n .

Let integer k = λ − 1 , then for n ≥ 1 and k ≥ 2 , when n ≥ k − 1 , there exists at least a prime number p such that k n < p ≤ ( k + 1 ) n . (4.2)

The Bertrand-Chebyshev’s theorem points out that for n ≥ 1 and k = 1 , there exists at least a prime number p such that k n < p ≤ ( k + 1 ) n . (4.3)

From (4.2) and (4.3), we can conclude that for n ≥ 1 and k ≥ 1 , when n ≥ k − 1 , there exists at least a prime number p such that k n < p ≤ ( k + 1 ) 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, π ( N ) ~ N ln ( N ) , where π ( N ) 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 ≥ 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] .

Conflicts of Interest

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

Cite this paper

Yu, W.K. (2023) On Prime Numbers between kn and (k + 1)n. Journal of Applied Mathematics and Physics, 11, 3712-3734. https://doi.org/10.4236/jamp.2023.1111234

Appendix

Calculation of f 8 ( n 0 , λ ) when n 0 = 25 and 26 ≥ λ ≥ 13 .

When λ = 13 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 338 ⋅ 8.4924 25 − 1 325 325 3 + 3 ≈ 6.6934 E + 24 4.2676 E + 22 ≈ 156 > 1 .

When λ = 14 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 392 ⋅ 9.1721 25 − 1 350 350 3 + 3 ≈ 4.9265 E + 25 3.1424 E + 23 ≈ 157 > 1 .

When λ = 15 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 450 ⋅ 9.8518 25 − 1 375 375 3 + 3 ≈ 3.1448 E + 26 2.1746 E + 24 ≈ 145 > 1 .

When λ = 16 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 512 ⋅ 10.5315 25 − 1 400 400 3 + 3 ≈ 1.7743 E + 27 1.4234 E + 25 ≈ 125 > 1 .

When λ = 17 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 578 ⋅ 11.2112 25 − 1 425 425 3 + 3 ≈ 8.9862 E + 27 8.8497 E + 25 ≈ 102 > 1 .

When λ = 18 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 648 ⋅ ( 11.8909 ) 25 − 1 450 450 3 + 3 ≈ 4.1374 E + 28 5.2574 E + 26 ≈ 78.7 > 1 .

When λ = 19 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 722 ⋅ 12.5705 25 − 1 475 475 3 + 3 ≈ 1.7498 E + 29 2.9904 E + 27 ≈ 58.5 > 1 .

When λ = 20 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 800 ⋅ 13.2502 25 − 1 500 500 3 + 3 ≈ 6.8616 E + 29 1.6366 E + 27 ≈ 41.9 > 1 .

When λ = 21 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 882 ⋅ 13.9298 25 − 1 525 525 3 + 3 ≈ 2.5127 E + 30 8.6291 E + 28 ≈ 29.1 > 1 .

When λ = 22 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 968 ⋅ 14.6094 25 − 1 550 550 3 + 3 ≈ 8.6507 E + 30 4.4015 E + 29 ≈ 19.7 > 1 .

When λ = 23 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 1058 ⋅ 15.2891 25 − 1 575 575 3 + 3 ≈ 2.8161 E + 31 2.1742 E + 30 ≈ 13.0 > 1 .

When λ = 24 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 1152 ⋅ 15.9687 25 − 1 600 600 3 + 3 ≈ 8.7081 E + 31 1.0425 E + 31 ≈ 8.35 > 1 .

When λ = 25 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 1250 ⋅ 16.6483 25 − 1 625 625 3 + 3 ≈ 2.5691 E + 32 4.8599 E + 31 ≈ 5.29 > 1 .

When λ = 26 , f 8 ( n 0 , λ ) = 2 λ ⋅ ( λ 4 ⋅ ( λ λ − 1 ) λ − 1 ) n 0 − 1 ( λ n 0 ) λ n 0 3 + 3 ≈ 1352 ⋅ 17.3279 25 − 1 650 650 3 + 3 ≈ 7.2594 E + 32 2.2080 E + 32 ≈ 3.29 > 1 .

ReferencesAigner, M. and Ziegler, G. (2014) Proofs from THE BOOK. 5th Edition, Springer, Berlin. https://doi.org/10.1007/978-3-662-44205-0El Bachraoui, M. (2006) Prime in the Interval [2n, 3n]. International Journal of Contemporary Mathematical Sciences, 1, 617-621. https://doi.org/10.12988/ijcms.2006.06065Loo, A. (2011) On the Prime in the Interval [3n, 4n]. https://arxiv.org/abs/1110.2377Shevelev, V., Greathouse IV, C. and Moses, P. (2013) On Intervals (kn, (kn + 1)n) Containing a Prime for All n > 1. Journal of Integer Sequences, 16, Article 13.7.3.Yu, W. (2023) The Proofs of Legendre’s Conjecture and Three Related Conjectures. Journal of Applied Mathematics and Physics, 11, 1319-1336. https://doi.org/10.4236/jamp.2023.115085Wikipedia. Proof of Bertrand’s Postulate. https://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulateWikipedia. Legendre’s Conjecture. https://en.wikipedia.org/wiki/Legendre%27s_conjectureWikipedia. Landau’s problems. https://en.wikipedia.org/wiki/Landau%27s_problems