On the Consecutive Integers n+<em>i</em>-1=(<em>i</em>+1) <em>P<sub>i</sub></em> | OMICS International
ISSN: 1736-4337
Journal of Generalized Lie Theory and Applications
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# On the Consecutive Integers n+i-1=(i+1) Pi

Jiang CX*

Institute for Basic Research, Palm Harbor, P.O. Box 3924, Beijing 100854, P.R. China

*Corresponding Author:
Jiang CX
Institute for Basic Research
Palm Harbor, P.O. Box 3924
Beijing 100854, P.R. China
Tel: +1-727-688 3992
E-mail: [email protected]

Received Date: February 23, 2017; Accepted Date: May 16, 2017; Published Date: May 22, 2017

Citation: Jiang CX (2017) On the Consecutive Integers n+i-1=(i+1) Pi. J Generalized Lie Theory Appl 11: 267. doi: 10.4172/1736-4337.1000267

Copyright: © 2017 Jiang CX. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

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#### Abstract

By using the Jiang’s function J2 (ω) we prove that there exist infinitely many integers n such that n=2P1, n+1=3P2, n+k−1=(k+1) Pk are all composites for arbitrarily long k, where P1, P2,…, Pk are all primes. This result has no prior occurrence in the history of number theory.

#### Keywords

Consecutive integers; Jiang’s function

#### Introduction

Theorem 1

There exist infinitely many integers n such that the consecutive integers n=2P1, n+1=3P2,…, n+k−1=(k+1) Pk are all composites for arbitrarily long k, where P1, P2,…, Pk are all primes.

Proof: Suppose that . We define the prime equations:

, (1)

Where i=1, 2,…, k

The Jiang’s function [1] is:

(2)

Where (P)=−k if P2 m; χ(P)=−k+1 if Pm; χ(P)=0 otherwise, .

Since J2 (ω)→∞ as ω→, there exist infinitely many integers x such that P1, P2,…, Pk are all primes.

We have the asymptotic formula of the number of integers xN [1]

(3)

Where, .

From (1) we have,

Example 1: Let k=5, we have n=2 × 53281, n+1=3 × 35521, n+2=4 × 26641, n+3=5 × 21313, n+4=6 × 17761.

Theorem 2

There exist infinitely many integers n such that the consecutive integers n=(1+2b) P1, n+1=(2+2b) P2,…, n+k−1=(k+2b) Pk are all composites for arbitrarily long k, where P1, P2,, Pk are all primes [2].

Proof: Suppose that . We define the prime equations:

(4)

Where i=1, 2,…, k.

The Jiang’s function [1] is:

(5)

Where χ(P)=−k if P2m; χ(P)=−k+1 if Pm; χ(P)=0 otherwise.

Since, J2 (ω)→∞ asω→∞, there exist infinitely many integers x such that P1, P2,…, Pk are all primes.

We have the asymptotic formula of the number of integers xN [1]

(6)

From (4) we have:

Example 2: Let b=1 and k=4, we have n=3 × 27361, n+1=4 × 20521, n+2=5 × 16417, n+3=6 × 13681.

Theorem 3

There exist infinitely many integers n such that the consecutive integers n=3P1, n+2=5P2,…, n+2=5P2,…,n+2 (k−1)=(2k+1) Pk are all composites for arbitrarily long k, where P1, P2,, Pk are all primes [3].

Proof: Suppose that . We define the prime equations:

(7)

Where i=1, 2,…, k.

The Jiang’s function [1] is:

(8)

Where χ(P)=−k if P2 m; χ(P)=−k+1 if P m; χ(P)=0 otherwise.

Since, J2 (ω)→∞ asω→∞, there exist infinitely many integers x such that P1, P2,…, Pk are all primes.

We have the asymptotic formula of the number of integers xN [1]

(9)

From (7) we have:

Example 3: Let k=4, we have n=3 × 631, n+2=5 × 379, n+4=7 × 271, n+6=9 × 211.

Theorem 4

There exist infinitely many integers n such that the consecutive integers n+2=3P2,…,n+2=3P2,…,n+2 (k−1)=(2k+1) Pk are all composites for arbitrarily long k, where P1, P2,, Pk are all primes [4].

Proof: Suppose that . We define the prime equations:

(10)

Where i=1, 2,…, k.

The Jiang’s function [1] is:

(11)

Where χ(P)=−k if P2 m; χ(P)=−k+1 if P m; χ(P)=0 otherwise.

Since, J2 (ω)→∞ asω→∞, there exist infinitely many integers x such that P1, P2,…, Pk are all primes.

We have the asymptotic formula of the number of integers xN [1]

(12)

From (10) we have:

Example 4: Let k=4, we have n=9661, n+2=3 × 3221, n+4=5 × 1933, n+6=7 × 1381.

Theorem 5

There exist infinitely many integers n such that the consecutive integers n=3P1,…,n+4=7P2,…, n+4 (k−1)=(4k+1) Pk are all composites for arbitrarily long k, where P1, P2,…, Pk are all primes [5].

Example 5: Let k=4, we have n=3 × 2311, n+4=7 × 991, n+8=11 × 631, n+12=15 × 463.

Theorem 6

There exist infinitely many integers n such that the consecutive integers n=5P1,…, n+4=9P2,…,n+4 (k−1)=(4k+1) Pk are all composites for arbitrarily long k, where P1, P2,…, Pk are all primes [6].

#### Conclusion

Jiang’s function J2 (ω) prove that there exist infinitely many integers n such that n=2P1, n+1=3P2,…, n+k−1=(k+1) Pk are all composites for arbitrarily long k, where P1, P2,, Pk are all primes which result has no prior occurrence in the history of number theory.

#### References

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