%I #26 Apr 03 2023 10:36:13
%S 15,36,39,54,55,73,102,107,110,118,129,160,164,184,187,194,199,218,
%T 271,272,291,339,358,387,419,426,464,465,508,520,553,605,621,629,667,
%U 682,683,702,709,710,733,761,791,813,821,822,829,830,882,896,952,962,988
%N Indices of primes p such that the next two larger primes are p+6 and p+12.
%C 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.)
%H Clark Kimberling, <a href="/A219096/b219096.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="https://t5k.org/top20/page.php?id=13">Consecutive primes in arithmetic progression</a>
%F a(n) = A000720(A047948(n)). - _M. F. Hasler_, Mar 11 2013
%e 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.
%t z = 1000; t = Differences[Prime[Range[z]]];
%t f[n_] := If[t[[n + 1]] - t[[n]] == 0, t[[n]], 0]
%t u = Table[f[n], {n, 1, z - 2}]; s = Flatten[Position[u, 6]] (*A219096*)
%t Prime[s] (*A047948*)
%t Flatten[Position[Partition[Prime[Range[1000]],3,1],_?(Differences[#] == {6,6}&),{1},Heads->False]] (* _Harvey P. Dale_, Jan 01 2015 *)
%o (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
%Y Cf. A047948, A128940, A000720.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Mar 05 2013