%I #19 Nov 20 2017 05:31:42
%S 1,3,6,16,17,22,25,27,28,35,38,39,43,44,48,56,59,61,67,68,69,82,99,
%T 100,104,111,113,122,127,129,132,133,145,146,156,161,162,171,172,176,
%U 179,183,186,189,190,202,209,210,234,238,250,251,258,261,272,275,280,284
%N Values of n such that prime(2n+1) mod 12 = 5.
%H G. C. Greubel, <a href="/A116613/b116613.txt">Table of n, a(n) for n = 1..1000</a>
%F Equals the integer part of { odd terms in A160591 = A000720(A040117) } / 2. - _M. F. Hasler_, May 22 2009
%e 25 is in the sequence because the 51st prime is 233 and 233 mod 12 = 5.
%p a:=proc(n) if ithprime(2*n+1) mod 12 = 5 then n else fi end: seq(a(n),n=0..300);
%t Select[Range[0, 500], Mod[Prime[2*# + 1], 12] == 5 &] (* _G. C. Greubel_, Nov 19 2017 *)
%o (PARI) for(n=1,999, prime(2*n+1)%12==5 & print1(n",")) \\ _M. F. Hasler_, May 22 2009
%Y Cf. A116602, A116610, A116612-A116617, A160591-A160594. - _M. F. Hasler_, May 22 2009
%K nonn
%O 1,2
%A _Roger L. Bagula_, Mar 29 2006
%E Edited by _N. J. A. Sloane_, Apr 05 2006