|
%I #12 Mar 13 2020 20:37:53
%S 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 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 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 Number of prime factors of form cn+1 for numbers 5^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=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 m = 5; n = 2; nmax = 107;
%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[5^n+1]]; Sum[If[Mod[p[[i]], n] == 1, e[[i]], 0], {i, Length[p]}], {n, 2, 50}]
%K nonn
%O 2,5
%A _Seppo Mustonen_, Nov 24 2010
|
