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A182596 Number of prime factors of form cn+1 for numbers 3^n+1.

%I A182596 #12 Jun 02 2025 03:14:34
%S A182596 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 A182596 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 A182596 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 A182596 Number of prime factors of form cn+1 for numbers 3^n+1.
%C A182596 Repeated prime factors are counted.
%H A182596 S. Mustonen, <a href="http://www.survo.fi/papers/MustonenPrimes.pdf">On prime factors of numbers m^n+-1</a>
%H A182596 Seppo Mustonen, <a href="/A182590/a182590.pdf">On prime factors of numbers m^n+-1</a> [Local copy]
%e A182596 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 A182596 m = 3; n = 2; nmax = 130;
%t A182596 While[n <= nmax, {l = FactorInteger[m^n + 1]; s = 0;
%t A182596      For[i = 1, i <= Length[l],
%t A182596       i++, {p = l[[i, 1]];
%t A182596        If[IntegerQ[(p - 1)/n] == True, s = s + l[[i, 2]]];}];
%t A182596      a[n] = s;} n++;];
%t A182596 Table[a[n], {n, 2, nmax}]
%t A182596 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 A182596 nonn
%O A182596 2,7
%A A182596 _Seppo Mustonen_, Nov 24 2010