A329251 -id:A329251 - cp's OEIS frontend

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

Hugo Pfoertner, Nov 09 2019

nonn

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.

			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.

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

Cf. A329164, A329165, A329251, A329252.

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

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.

Original entry on oeis.org

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

Hugo Pfoertner, Nov 10 2019

nonn

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.

			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.

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

Cf. A329160, A329161, A329250, A329251.

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]

cp's OEIS Frontend

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.

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