

A219096


Indices of primes p such that the next two larger primes are p+6 and p+12.


2



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

1,1


COMMENTS

The primes themselves are given by A047948. Conjecture: if k == 0 mod 6 then there exists a prime p such that pk, p, p+k are consecutive primes. (This would follow from a proof of Dickson's conjecture; see the Comments and References at A186311.)


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000
Consecutive primes in arithmetic progression


FORMULA

a(n) = A000720(A047948(n)).  M. F. Hasler, Mar 11 2013


EXAMPLE

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.


MATHEMATICA

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*)
Prime[s] (*A047948*)
Flatten[Position[Partition[Prime[Range[1000]], 3, 1], _?(Differences[#] == {6, 6}&), {1}, Heads>False]] (* Harvey P. Dale, Jan 01 2015 *)


PROG

(PARI) n=0; p=2; q=3; forprime(r=5, 1e6, n++; if(rp==12&&qp==6, print1(n", ")); p=q; q=r) \\ Charles R Greathouse IV, Mar 05 2013


CROSSREFS

Cf. A047948, A128940, A000720.
Sequence in context: A280883 A241282 A249056 * A134509 A062712 A224719
Adjacent sequences: A219093 A219094 A219095 * A219097 A219098 A219099


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Mar 05 2013


STATUS

approved



