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A082512
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a(n) = p is the smallest prime introducing a consecutive prime-difference pattern as follows: [2,2n,2], i.e., [p, p+2, p+2+2n, p+2+2n+2] are consecutive primes. Increasing middle prime gap in the immediate neighborhood of two small gaps(=2); a(n) = 0 if no such pattern exists.
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1
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0, 5, 0, 0, 137, 0, 0, 1931, 0, 0, 9437, 0, 0, 2969, 0, 0, 20441, 0, 0, 62987, 0, 0, 510401, 0, 0, 48677, 0, 0, 677471, 0, 0, 997811, 0, 0, 173357, 0, 0, 1134311, 0, 0, 3063287, 0, 0, 3591191, 0, 0, 4876511, 0, 0, 838247, 0, 0, 4297091, 0, 0, 15492437, 0, 0, 27458747
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OFFSET
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1,2
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COMMENTS
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It is conjectured that the twin primes in the neighborhood can be separated by an arbitrarily large gap.
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LINKS
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FORMULA
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EXAMPLE
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a(4) = 0 because no p can begin a [2,8,2] gap pattern since p mod 6 = 5 must hold and following 3 primes give modulo 6 residues 1, 3, and 5, so p + 2 + 8 is not prime; a(n)=0 if 2n congruent to 0 or 2 mod 6; a(n) has solution for n = 6k + 4;
For n=16, the 4 corresponding primes and 3 differences are {1931 [2] 1933 [16] 1949 [2] 1951}.
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MATHEMATICA
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d[x_] := Prime[x+1]-Prime[x]; h={k1=2, k2=82, k3=2}; de=Apply[Plus, h]; k=0; Do[If[Equal[d[n], k1]&&Equal[d[n+1], k2]&&Equal[d[n+2], k3], k=k+1; Print[k, n, h, {Prime[n], Prime[n+1], Prime[n+2], Prime[n+3]}]], {n, 1, 10000000}]
max = 20; v = Table[0, {max}]; p = Prime /@ Range[4]; count = 0; While[count < max, If[p[[2]] == p[[1]] + 2 && p[[4]] == p[[3]] + 2, d = ((p[[3]] - p[[2]])/2 - 2)/3 + 1; If[d <= max && v[[d]]==0, count++; v[[d]] = p[[1]]]]; p = Join[Rest[p], {NextPrime[p[[4]]]}]]; Riffle[Table[0, {2*max}], v, {2, -1, 3}] (* Amiram Eldar, Jan 21 2020 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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a(50) corrected and more terms added by Amiram Eldar, Jan 21 2020
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STATUS
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approved
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