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%I #15 Aug 01 2019 04:08:58
%S 1,1,1,1,1,1,1,1,1,1,2,1,1,1,2,2,1,1,1,1,4,1,2,1,1,2,1,1,1,3,1,1,1,1,
%T 1,1,2,1,1,1,2,2,1,2,2,2,1,1,1,1,1,1,1,2,1,2,1,2,4,2,2,1,2,1,4,1,1,1,
%U 1,5,2,1,1,1,2,2,1,1,2,1,1,1,2,2,1,5,2,1,1,3,2,2,3,1,1,1,1
%N a(n) = b(n)/c(n) where b(n) = smallest positive k such that (2*k)^2 + 2*n - 1 is prime and c(n) = gcd(n,3) = A109007(n).
%e a(1) = b(1)/c(1) = 1/1 = 1 because b(1) = (2*1)^2 + 2*1 - 1 = 5 and 5 is prime, c(1) = gcd(1,3) + A109007(1) = 1,
%e a(2) = b(2)/c(2) = 1/1 = 1 because b(2) = (2*1)^2 + 2*2 - 1 = 7 and 7 is prime, c(2) = gcd(2,3) + A109007(2) = 1,
%e a(3) = b(3)/c(3) = 3/3 = 1 because b(2) = (2*3)^2 + 2*3 - 1 = 41 and 41 is prime, c(3) = gcd(3,3) + A109007(3) = 3.
%t Table[k = 1; While[! PrimeQ[(2 k)^2 + 2 n - 1], k++]; k/GCD[n, 3], {n, 97}] (* _Michael De Vlieger_, Aug 31 2016 *)
%Y Cf. A109007, A023204.
%K nonn
%O 1,11
%A _Juri-Stepan Gerasimov_, Aug 29 2016
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