A329252 Let P1 >= 5, P2, P3 be consecutive primes, with P2 - P1 = 2. a(n) = (P1 + P2)/12 for the first occurrence of (P3 - P2)/2 = n.

1, 5, 0, 23, 33, 0, 322, 87, 0, 325, 278, 0, 495, 1293, 0, 2027, 4725, 0, 3468, 2690, 0, 27177, 14438, 0, 4245, 6773, 0, 13283, 24938, 0, 104283, 92067, 0, 28893, 60015, 0, 119362, 46905, 0, 44270, 106323, 0, 90713, 67475, 0, 266618, 207107, 0

Offset: 2

Author

Hugo Pfoertner, Nov 10 2019

Keywords

• nonn

Comments

Position of first occurrence of a gap of length P3 - P2 = 2*n containing no primes, immediately following the twin primes (P1,P2). To indicate impossible gaps of lengths 8, 14, 20, ..., a(3k+1) is set to 0 for all k >= 1.

Examples

			a(5) = 23 because the prime gap following P1 = 6*23 - 1 = 137, P2 = 6*23 + 1 = 139 is the first such gap with length 2*n = 10. P3 - P2 = 149 - 139 = 10.
		

Links

• Hugo Pfoertner, Table of n, a(n) for n = 2..209

Crossrefs

Cf. A329160, A329161, A329250, A329251.

Programs

PARI

my(v=vector(60), p1=5, p2=7, d); forprime(p3=11, 5e6, if(p2-p1==2, d=(p3-p2)/2; if(v[d]==0, v[d]=(p1+p2)/12)); p1=p2; p2=p3); v[2..49]