A268867 Number of integers of the form m^2+1 between two consecutive pairs of primes of the same form.

0, 7, 7, 27, 67, 77, 177, 77, 167, 7, 67, 377, 47, 27, 67, 27, 37, 57, 187, 47, 57, 7, 277, 87, 27, 7, 307, 47, 77, 127, 167, 87, 207, 167, 227, 217, 17, 247, 127, 17, 187, 237, 7, 117, 47, 7, 157, 57, 37, 197, 217, 87, 17, 137, 147, 287, 67, 547, 37, 187, 787

Offset: 1

Author

Michel Lagneau, Feb 15 2016

Keywords

• nonn

Comments

Or number of integers of the form m^2+1 between two consecutive twin k^2+1 primes.

a(n)==7 mod 10 for n>1.

The primes of the sequence are 7, 17, 37, … (A030432), subsequence of A017353.

Conjecture 1: the sequence is infinite.

Conjecture 2: for n>1, the sequence sorted in ascending order with distinct values generates the set B = {b(k)} = {7, 17, 27, 37, …} = {7 + 10k}, k = 0,1,2,… and {a(n)}/qZ = B/qZ with q = 2^m, m = 1, 2,… is a multiplicative group.

This has been verified for n up to 10^8.

A remarkable simple way to perceive properties of invariability in this sequence (or other particular sequences) is the use of finite groups of integers modulo q. This study can provide interesting interpretations for some regularities which describe properties in other mathematical spaces.

However, this concept is based on a conjecture if it is impossible to prove the infinity character of a sequence. This requires, for the calculations, a large number of elements in the sequence.

The groups of integers modulo 2^m are:

q = 2 => B/2Z = {1}, the trivial group.

q = 4 => B/4Z = {1,3}, the cyclic group of order 2 with two elements.

q = 8 => B/8Z = {1,3,5,7}, the group of order 4 with generating set {3,7} => the Klein four-group of order 2. The square of each element of B/8Z is 1. The group is not cyclic.

q = 16 => B/16Z = {1,3,5,7,9,11,13,15}, the group of order 8 with generating set {3,15}. The powers of 3 {1,3,9,11} are a subgroup of order 4, as are the powers of 5, {1,5,9,13}. The group B/16Z is not cyclic.

For higher powers q = 2^k, k>2, B/(2^k)Z = {1,3,5,…,2^k-1}, with generating set {3, 2^k-1}. The group B/(2^k)Z is not cyclic.

The order of the group is given by Euler’s totient function (A000010): this is the product of the orders of the cyclic groups in the direct product (see the links).

Examples

			a(1)=0 because there exists 0 number of the form m^2+1 between the two consecutive pairs of primes(2^2+1, 4^2+1) and (4^2+1, 6^2+1);
a(2) = 7 because there exists 7 numbers of the form m^2+1 between the two consecutive pairs of primes(4^2+1, 6^2+1) and (14^2+1, 16^2+1): 50, 65, 82, 101, 122, 145 and 170.
		

Links

• Michel Lagneau, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Finite Group
• Wikipedia, Finite group
• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A005574, A030432, A017353, A096012, A206328.

Programs

Maple

nn:=10000:T:=array(1..200):kk:=0:
for n from 4 by 2 to nn do:
   p1:=n^2+1:p2:=(n+2)^2+1:
    if isprime(p1) and isprime(p2)
     then
     kk:=kk+1:T[kk]:=n:
     else
    fi:
  od:
    for m from 1 to kk-1 do:
      q:=T[m+1]-T[m]-3:printf(`%d, `,q):
    od:

Mathematica

lst={};Do[If[PrimeQ[n^2+1],AppendTo[lst,n]],{n,1,10000}];Module[{tr=Transpose[Select[Partition[lst,2,1],#[[2]]-#[[1]]==2&]],fir,las},fir=Rest[tr[[1]]];las=Most[tr[[2]]];Flatten[Abs[Differences/@Thread[{fir,las}]]]-1/.{-1->0}]