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 A182595 Number of prime factors of form cn+1 for numbers 2^n+1 1

%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

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Last modified December 3 02:05 EST 2023. Contains 367530 sequences. (Running on oeis4.)