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%I #4 Dec 16 2013 09:54:06
%S 1,1,1,1,2,1,3,2,1,4,3,2,1,3,2,7,5,2,9,4,10,5,4,6,13,3,7,22,7,4,4,10,
%T 4,12,4,6,5,1,2,5,5,20,1,6,12,5,5,3,1,10,35,11,14,5,16,4,14,1,35,13,3,
%U 38
%N One sixth of gap g between 3 consecutive primes for the smallest k such that 6^n+k, 6^n+k+g, 6^n+k+2*g are consecutive primes in arithmetic progression
%C g is a multiple of 6 as otherwise 6^n+k, 6^n+k+g, or 6^n+k+2*g is divisible by 2 or 3.
%C Sequence starts at n=2 as no solution for n=1.
%C k=A233546(n).
%F a(n) = A233550(n)/6.
%Y Cf. A233546, A233550.
%K nonn
%O 2,5
%A _Jonathan Sondow_, Dec 16 2013
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