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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.
3
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
1,2
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.
LINKS
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Hugo Pfoertner, Nov 09 2019
STATUS
approved