cp's OEIS Frontend

A182597 Number of prime factors of form cn+1 for numbers 5^n+1.

%I A182597 #14 Jun 02 2025 03:14:40
%S A182597 1,1,1,1,2,2,2,1,2,3,2,2,1,2,2,2,2,3,2,3,2,2,1,2,2,3,1,2,3,3,3,4,4,2,
%T A182597 3,3,2,4,2,4,3,4,1,1,1,3,4,3,3,5,4,3,1,2,4,3,1,4,4,4,2,6,3,4,2,1,5,4,
%U A182597 3,3,2,3,3,5,3,2,4,4,4,5,4,3,4,6,3,4,4,3,3,2,2,4,4,4,4,5,4,1,4,1,7,1,5,5,2,2
%N A182597 Number of prime factors of form cn+1 for numbers 5^n+1.
%C A182597 Repeated prime factors are counted.
%H A182597 S. Mustonen, <a href="http://www.survo.fi/papers/MustonenPrimes.pdf">On prime factors of numbers m^n+-1</a>
%H A182597 Seppo Mustonen, <a href="/A182590/a182590.pdf">On prime factors of numbers m^n+-1</a> [Local copy]
%e A182597 For n=11, 5^n+1=48828126=2*3*23*67*5281 has three prime factors of form, namely 23=2n+1, 67=6n+1, 5281=480n+1. Thus a(11)=3.
%t A182597 m = 5; n = 2; nmax = 107;
%t A182597 While[n <= nmax, {l = FactorInteger[m^n + 1]; s = 0;
%t A182597      For[i = 1, i <= Length[l],
%t A182597       i++, {p = l[[i, 1]];
%t A182597        If[IntegerQ[(p - 1)/n] == True, s = s + l[[i, 2]]]; }];
%t A182597      a[n] = s; } n++; ];
%t A182597 Table[a[n], {n, 2, nmax}]
%t A182597 Table[{p,e}=Transpose[FactorInteger[5^n+1]]; Sum[If[Mod[p[[i]], n] == 1, e[[i]], 0], {i, Length[p]}], {n, 2, 50}]
%K A182597 nonn
%O A182597 2,5
%A A182597 _Seppo Mustonen_, Nov 24 2010