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A329251
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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.
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3
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(PARI) 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]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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