%I #15 Mar 13 2020 20:37:20
%S 1,0,1,1,1,1,1,1,1,1,1,1,2,1,1,1,2,1,1,2,2,1,2,2,3,1,1,2,2,1,2,2,3,2,
%T 2,2,3,2,1,2,2,1,2,1,4,2,2,1,3,3,2,2,3,2,2,3,2,3,3,1,4,2,2,4,4,2,2,2,
%U 2,2,2,2,4,2,3,3,5,1,2,3,4,5,3,2,4,2
%N Number of prime factors of form cn+1 for numbers 2^n+1
%C Repeated prime factors are counted.
%H Seppo Mustonen, <a href="/A182595/b182595.txt">Table of n, a(n) for n = 2..250</a>
%H S. Mustonen, <a href="http://www.survo.fi/papers/MustonenPrimes.pdf">On prime factors of numbers m^n+1</a>
%H Seppo Mustonen, <a href="/A182590/a182590.pdf">On prime factors of numbers m^n+1</a> [Local copy]
%e For n=14, 2^n+1=16385=5*29*113 has two prime factors of form, namely 29=2n+1, 113=8n+1. Thus a(14)=2.
%t m = 2; n = 2; nmax = 250;
%t While[n <= nmax, {l = FactorInteger[m^n + 1]; s = 0;
%t For[i = 1, i <= Length[l],
%t i++, {p = l[[i, 1]];
%t If[IntegerQ[(p  1)/n] == True, s = s + l[[i, 2]]];}];
%t a[n] = s;} n++;];
%t Table[a[n], {n, 2, nmax}]
%t Table[{p, e}=Transpose[FactorInteger[2^n+1]]; Sum[If[Mod[p[[i]], n] == 1, e[[i]], 0], {i, Length[p]}], {n, 2, 50}]
%K nonn
%O 2,13
%A _Seppo Mustonen_, Nov 24 2010
