A286680 Smallest nonnegative m such that (1 + n)^(2^m) + n is not prime.
0, 5, 4, 2, 0, 3, 1, 0, 3, 3, 0, 1, 0, 0, 2, 4, 0, 0, 2, 0, 2, 1, 0, 2, 0, 0, 1, 0, 0, 2, 3, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 1, 1, 0, 3, 2, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 1, 2, 0, 0, 0, 0, 1, 2, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 2
Offset: 0
Keywords
Examples
a(0) = 0 because (1 + 0)^(2^0) + 0 = 1 is not prime.
Links
- Robert G. Wilson v, Table of n, a(n) for n = 0..10000
- Wikipedia, Bunyakovsky conjecture.
Programs
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Maple
f:= proc(n) local k; for k from 0 while isprime((1+n)^(2^k)+n) do od: k; end proc: map(f, [$0..100]); # Robert Israel, May 17 2017
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Mathematica
f[n_] := Block[{k = 0}, While[ PrimeQ[(1 + n)^(2^k) + n], k++]; k]; Array[f, 105, 0] (* Robert G. Wilson v, May 14 2017 *)
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PARI
a(n) = {my(m = 0); while (isprime((1 + n)^(2^m) + n), m++); m;} \\ Michel Marcus, May 19 2017
Comments