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 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. 3

%I #12 Nov 11 2019 11:51:29

%S 1,5,0,23,33,0,322,87,0,325,278,0,495,1293,0,2027,4725,0,3468,2690,0,

%T 27177,14438,0,4245,6773,0,13283,24938,0,104283,92067,0,28893,60015,0,

%U 119362,46905,0,44270,106323,0,90713,67475,0,266618,207107,0

%N 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.

%C 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.

%H Hugo Pfoertner, <a href="/A329252/b329252.txt">Table of n, a(n) for n = 2..209</a>

%e 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.

%o (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]

%Y Cf. A329160, A329161, A329250, A329251.

%K nonn

%O 2,2

%A _Hugo Pfoertner_, Nov 10 2019

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Last modified September 12 14:45 EDT 2024. Contains 375851 sequences. (Running on oeis4.)