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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

%I #11 Nov 11 2019 13:53:04

%S 1,23,322,1573,495,3407,10498,85067,8113,112912,166302,28893,189052,

%T 510548,598532,812752,139708,716182,2582073,4576458,2497092,5130198,

%U 5761777,25381573,7315173,20200532,40629683,33185292,69948743,38771927,13194622

%N 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.

%C 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.

%H Hugo Pfoertner, <a href="/A329250/b329250.txt">Table of n, a(n) for n = 1..63</a>

%e 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.

%o (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

%Y Cf. A329164, A329165, A329251, A329252.

%K nonn

%O 1,2

%A _Hugo Pfoertner_, Nov 09 2019

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Last modified August 13 01:27 EDT 2024. Contains 375113 sequences. (Running on oeis4.)