A329250 Let P1, P2, P3, P4 be consecutive primes, with P2 - P1 = P4 - P3 = 2. a(n) = (P1 + P2)/12 for the first occurrence of (P3 - P1)/6 = n.

1, 23, 322, 1573, 495, 3407, 10498, 85067, 8113, 112912, 166302, 28893, 189052, 510548, 598532, 812752, 139708, 716182, 2582073, 4576458, 2497092, 5130198, 5761777, 25381573, 7315173, 20200532, 40629683, 33185292, 69948743, 38771927, 13194622

Offset: 1

Author

Hugo Pfoertner, Nov 09 2019

Keywords

• nonn

Comments

Position of first occurrence of a gap of length P3 - P2 = 6*n - 2 containing no primes, bounded by twin primes (P1,P2) below and (P3,P4) above.

Examples

			a(4) = 1573, because the 4 primes P1 = 6*1573 - 1 = 9437, P2 = 6*1573 + 1 = 9439, P3 = P1 + 6*4 = 9461, P4 = 9463 produce the first occurrence of the gap P3 - P2 = 9461 - 9439 = 6*4 - 2 = 22. See also example in A329164.
		

Links

• Hugo Pfoertner, Table of n, a(n) for n = 1..63

Crossrefs

Cf. A329164, A329165, A329251, A329252.

Programs

PARI

my(v=vector(31),p1=3,p2=5,p3=7,r=0,d);forprime(p4=11,5e8,if(p2-p1==2&&p4-p3==2,d=(p3-p1)/6;if(v[d]==0,v[d]=(p1+p2)/12));p1=p2;p2=p3;p3=p4);v