A119624 Least k>0 such that, for n>1, 2*n^k + 1 is prime; or 0 if no such prime possible as 2*n^k + 1 is 0 mod(3).
1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 3, 0, 1, 1, 0, 47, 1, 0, 1, 1, 0, 1, 2, 0, 1, 2, 0, 1, 1, 0, 3, 1, 0, 1, 1, 0, 2729, 1, 0, 1, 2, 0, 1, 2, 0, 175, 1, 0, 1, 1, 0, 1, 1, 0, 1, 3, 0, 3, 3, 0, 43, 1, 0, 1, 2, 0, 1, 1, 0, 3, 2, 0, 1, 1, 0, 3, 1, 0, 11, 1, 0, 1, 4, 0, 1, 2, 0, 1, 1, 0, 3, 2, 0, 1, 1, 0, 1, 1, 0
Offset: 1
Keywords
Links
- Robert Israel, Table of n, a(n) for n = 1..217
Programs
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Maple
f:= proc(n) local k; if n mod 3 = 1 then return 0 fi; if n mod 3 = 2 then r:= 2 else r:= 1 fi; for k from 1 by r do if isprime(2*n^k+1) then return k fi od end proc: f(1):= 1: map(f, [$1..100]); # Robert Israel, Apr 02 2018
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Mathematica
f[n_] := Block[{k = 0}, If[Mod[n, 3] != 1, k = 1; While[ ! PrimeQ[2*n^k + 1], k++ ]; ]; k ]; Table[f[n], {n, 2, 100}] (* Ray Chandler, Jun 08 2006 *) Table[If[n>1 && Mod[n,3]==1, 0, k=1; While[ !PrimeQ[2n^k+1], k++ ]; k], {n,100}] (* T. D. Noe, Jun 08 2006 *) -
PARI
a(n) = if(n%3==1, 0, for(k=1, 2^24, if(ispseudoprime(2*n^k+1),return(k)))) \\ Eric Chen, Mar 20 2015
Extensions
Extended by Ray Chandler and T. D. Noe, Jun 08 2006
Comments