A329252 -id:A329252 - 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

A329251 Let P1 >= 3, P2, P3 be consecutive primes, with P3 - P2 = 2. a(n) = (P2 + P3)/12 for the first occurrence of (P2 - P1)/2 = n.

Original entry on oeis.org

1, 2, 5, 0, 25, 87, 0, 325, 213, 0, 192, 758, 0, 500, 1158, 0, 1668, 5383, 0, 4217, 13130, 0, 15180, 4713, 0, 5955, 19583, 0, 66642, 17127, 0, 48108, 49485, 0, 28905, 171005, 0, 175530, 61838, 0, 314192, 76967, 0, 192637, 96147, 0, 812768, 708780, 0

Offset: 1

Hugo Pfoertner, Nov 10 2019

nonn

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

			a(5) = 25 because the prime gap immediately before P2 = 25*6 - 1 = 149, P3 = 25*6 + 1 = 151 is the first such gap with length 2*n = 2*5 = 10. P2 - P1 = 149 - 139 =10.

Hugo Pfoertner, Table of n, a(n) for n = 1..224

Cf. A329158, A329159, A329250, A329252.

Module[{nn=500000,lst},lst={(#[[2]]-#[[1]])/2,(#[[2]]+#[[3]])/12}&/@ Select[ Partition[Prime[Range[2,nn]],3,1],#[[3]]-#[[2]]==2&];Table[ SelectFirst[ lst,#[[1]]==n&],{n,50}]/.Missing["NotFound"]->{0,0}] [[All,2]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Nov 04 2020 *)

my(v=vector(70), p1=3, p2=5, d); forprime(p3=7, 5e6, if(p3-p2==2, d=(p2-p1)/2; if(v[d]==0, v[d]=(p2+p3)/12)); p1=p2; p2=p3); v[1..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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