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A219096
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Indices of primes p such that the next two larger primes are p+6 and p+12.
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2
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15, 36, 39, 54, 55, 73, 102, 107, 110, 118, 129, 160, 164, 184, 187, 194, 199, 218, 271, 272, 291, 339, 358, 387, 419, 426, 464, 465, 508, 520, 553, 605, 621, 629, 667, 682, 683, 702, 709, 710, 733, 761, 791, 813, 821, 822, 829, 830, 882, 896, 952, 962, 988
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OFFSET
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1,1
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COMMENTS
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The primes themselves are given by A047948. Conjecture: if k == 0 mod 6 then there exists a prime p such that p-k, p, p+k are consecutive primes. (This would follow from a proof of Dickson's conjecture; see the Comments and References at A186311.)
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LINKS
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FORMULA
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EXAMPLE
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a(1) = 15 since p(15), p(16), p(17) are consecutive primes (47, 53, 59) with common difference 6: 53 - 47 = 6, and 59 - 53 = 6.
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MATHEMATICA
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z = 1000; t = Differences[Prime[Range[z]]];
f[n_] := If[t[[n + 1]] - t[[n]] == 0, t[[n]], 0]
u = Table[f[n], {n, 1, z - 2}]; s = Flatten[Position[u, 6]] (*A219096*)
Flatten[Position[Partition[Prime[Range[1000]], 3, 1], _?(Differences[#] == {6, 6}&), {1}, Heads->False]] (* Harvey P. Dale, Jan 01 2015 *)
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PROG
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(PARI) n=0; p=2; q=3; forprime(r=5, 1e6, n++; if(r-p==12&&q-p==6, print1(n", ")); p=q; q=r) \\ Charles R Greathouse IV, Mar 05 2013
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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