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%I #10 Mar 13 2020 20:37:39
%S 1,1,1,1,1,1,2,2,1,2,1,1,2,3,1,3,2,2,2,3,2,1,4,2,2,2,2,3,1,2,1,4,2,1,
%T 2,3,2,5,2,2,2,1,3,3,2,3,2,2,3,6,2,2,2,3,3,5,2,2,5,2,3,5,1,2,1,2,3,6,
%U 3,5,3,2,3,6,4,1,2,3,4,7,3,4,5,4,5,8,3,3,3,6,2,6,2,4,4,3,5,6,3,2,5,3,4,6,4,3,7,4,4,7,7,3,4,3,3,6,1,2,5,4,4,6,2,3,4,4,5,6,3
%N Number of prime factors of form cn+1 for numbers 3^n+1
%C Repeated prime factors are counted.
%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=8, 3^n+1=6562=2*17*193 has two prime factors of form, namely 17=2n+1, 193=24n+1. Thus a(8)=2.
%t m = 3; n = 2; nmax = 130;
%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[3^n+1]]; Sum[If[Mod[p[[i]], n] == 1, e[[i]], 0], {i, Length[p]}], {n, 2, 50}]
%K nonn
%O 2,7
%A _Seppo Mustonen_, Nov 24 2010
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