Prime gap subsequence repeats twice then infinitely times

twice then often

Cody Luo (cody@ustc.edu)
Enter Additional Author Here

Any prime gap subsequence s, Length(s)=len,  if s occurs twice in the prime gap sequence, then s will continue to appear,  until infinitely times. if s recently twice occurs in prime gap sequence at index (Subscript[i, k])and (Subscript[i, k+1]) , the next time:

(Subscript[i, k+2]) ⩽((((Subscript[i, k])+ (Subscript[i, k+1])))^len)

How to predict .

Prime Gap Subsequence Theorem

Denote p[i] as the i-th prime. Let g[i]=p[i+1]-p[i] is the i-th prime gap, for i=1,2,3,...
{g[i]}={1,2,2,4,2,4,...} is the prime gap sequence.

p=Prime
g[i_Integer]:=p[i+1]-p[i];
M=1000
Table[p[i],{i,1,M+1}]
Table[g[i],{i,1,M}]

Theorem 1(Prime Gap Subsequence Theorem)

Any prime gap subsequence s, Length(s)=len,  if s occurs twice in the prime gap sequence, then s will continue to appear,  until infinitely times.To be a little accurate,if s recently twice occurs in prime gap sequence at index (Subscript[i, k])and (Subscript[i, k+1])
s={g[(Subscript[i, k])], g[1+(Subscript[i, k])], g[2+(Subscript[i, k])], ..., g[len-1+(Subscript[i, k])]}={g[(Subscript[i, k+1])], g[1+(Subscript[i, k+1])], g[2+(Subscript[i, k+1])], ..., g[len-1+(Subscript[i, k+1])]}
then, the next appearance of s will be at the before of index ((((Subscript[i, k])+ (Subscript[i, k+1])))^len).

({{(Subscript[i, k+2]) ⩽((((Subscript[i, k])+ (Subscript[i, k+1])))^len)}})

Proof: (to be complete) Firstly consider the case of s={2*m}, Length(s)=1. Observing the process of prime Sieve of Eratosthenes, ..., it will have

({{({{(Subscript[i, k+2]) ⩽((Subscript[i, k])+ (Subscript[i, k+1]))}})}})

...

{2,2} only occurs once at the prime gap sequence. While {2*m}, {2,4,2}, {6,2,6}... repeat infinitely times. However, such {2,6,2} never appears in the prime gap sequence.

Twin primes conjecture and Polignac’s conjecture

When take the prime gap subsequence s={2}, Length(s)=1, this is the content of the twin prime conjecture, which states that there are infinitely many primes p such that p + 2 is also prime. 
Below verify (Subscript[i, k+2]) <((Subscript[i, k])+ (Subscript[i, k+1])), for s={2} and i >3

M=10^6
seq=Table[g[i],{i,1,M}];
pos=Flatten@Position[seq,2]
i=4;length=Length[pos]
While[ pos[[i]]<pos[[i-2]]+pos[[i-1]],
i++;If[i>=length,Break[]]]
{i,length}

In above code, one line is pos=Flatten@Position[seq,2], replace “2” as 4 or 6,8,10,... 
Polignac’s conjecture from 1849 states that for every positive even natural number 2m, there are infinitely many consecutive prime pairs p and p′ such that p′ − p = 2m (i.e. there are infinitely many prime gaps of size 2m). 
Below test (Subscript[i, k+2]) <((Subscript[i, k])+ (Subscript[i, k+1]))until to the M-th prime, for prime gap subsequence s={12}

M=10^5
seq=Table[g[i],{i,1,M}];
pos=Flatten@Position[seq,12]
i=3;length=Length[pos]
While[ pos[[i]]<pos[[i-2]]+pos[[i-1]],
i++;If[i>=length,Break[]]]
{i,length}

Prime quadruplet , the prime gap subsequence s={2,4,2}

When take the prime gap subsequence s={2,4,2}, Length(s)=3, this is the case of prime quadruplet. A prime quadruplet (sometimes called prime quadruple) is a set of four primes of the form {p, p+2, p+6, p+8}.
Below verify (Subscript[i, k+2]) <((Subscript[i, k])+ (Subscript[i, k+1]))^3, for s={2,4,2} and i =3,4,5,...

M=10^5
seq=Table[g[i],{i,1,M}];
sub={2,4,2}
len=Length[sub]
pos=SequencePosition[seq,sub]
i=3;posLength=Length[pos]
While[ pos[[i,1]]<(pos[[i-2,1]]+pos[[i-1,1]])^len,
i++;If[i>=posLength,Break[]]]
{i,posLength}

Summary

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Conclusion

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Appendix

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∫x dx+(Sqrt[z])

Acknowledgments

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References

https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

https://en.wikipedia.org/wiki/Prime_quadruplet#Prime_quintuplets

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