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

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, 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, 2, 2, 2, 2, 4, 2, 3, 3, 5, 1, 2, 3, 4, 5, 3, 2, 4, 2

Offset: 2

Author

Seppo Mustonen, Nov 24 2010

Keywords

• nonn

Comments

Repeated prime factors are counted.

Examples

			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.
		

Links

• Seppo Mustonen, Table of n, a(n) for n = 2..250
• S. Mustonen, On prime factors of numbers m^n+-1
• Seppo Mustonen, On prime factors of numbers m^n+-1 [Local copy]

Programs

Mathematica

m = 2; n = 2; nmax = 250;
While[n <= nmax, {l = FactorInteger[m^n + 1]; s = 0;
     For[i = 1, i <= Length[l],
      i++, {p = l[[i, 1]];
       If[IntegerQ[(p - 1)/n] == True, s = s + l[[i, 2]]];}];
     a[n] = s;} n++;];
Table[a[n], {n, 2, nmax}]
Table[{p, e}=Transpose[FactorInteger[2^n+1]]; Sum[If[Mod[p[[i]], n] == 1, e[[i]], 0], {i, Length[p]}], {n, 2, 50}]