A189240 Least number k such that 2*k*n + 1 is a prime dividing 3^n + 1.
1, 1, 5, 6, 6, 39, 1, 1, 59, 3, 270, 15330, 1, 1, 672605, 3, 2, 75, 1, 1, 125, 511647711, 2, 3, 1, 360, 7691, 9, 796056, 111, 14476720225405, 1, 14064, 5355114024, 90, 249, 69757, 1, 180
Offset: 2
Keywords
Examples
a(4) = 5 because 3^4+1 = 2*41 => the smallest prime divisor of the form 2k*n+1 is 41 = 2*5*4+1.
Links
- Amiram Eldar, Table of n, a(n) for n = 2..658
Programs
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Mathematica
Table[p=First/@FactorInteger[3^n+1]; (Select[p, Mod[#1, n] == 1 &, 1][[1]] - 1)/(2n), {n, 2, 40}] -
PARI
a(n)=forstep(K=2*n+1,3^n+1,2*n,if(Mod(3,K)^n==0,return((k-1)/2/n))) \\ Charles R Greathouse IV, May 15 2013
Comments