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A182595
Number of prime factors of form cn+1 for numbers 2^n+1
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
2,13
COMMENTS
Repeated prime factors are counted.
LINKS
EXAMPLE
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.
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}]
CROSSREFS
Sequence in context: A077479 A335225 A070106 * A109706 A174541 A029444
KEYWORD
nonn
AUTHOR
Seppo Mustonen, Nov 24 2010
STATUS
approved