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A182596
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Number of prime factors of form cn+1 for numbers 3^n+1
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0
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1, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 3, 1, 3, 2, 2, 2, 3, 2, 1, 4, 2, 2, 2, 2, 3, 1, 2, 1, 4, 2, 1, 2, 3, 2, 5, 2, 2, 2, 1, 3, 3, 2, 3, 2, 2, 3, 6, 2, 2, 2, 3, 3, 5, 2, 2, 5, 2, 3, 5, 1, 2, 1, 2, 3, 6, 3, 5, 3, 2, 3, 6, 4, 1, 2, 3, 4, 7, 3, 4, 5, 4, 5, 8, 3, 3, 3, 6, 2, 6, 2, 4, 4, 3, 5, 6, 3, 2, 5, 3, 4, 6, 4, 3, 7, 4, 4, 7, 7, 3, 4, 3, 3, 6, 1, 2, 5, 4, 4, 6, 2, 3, 4, 4, 5, 6, 3
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OFFSET
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2,7
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COMMENTS
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Repeated prime factors are counted.
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LINKS
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EXAMPLE
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For n=8, 3^n+1=6562=2*17*193 has two prime factors of form, namely 17=2n+1, 193=24n+1. Thus a(8)=2.
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MATHEMATICA
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m = 3; n = 2; nmax = 130;
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[3^n+1]]; Sum[If[Mod[p[[i]], n] == 1, e[[i]], 0], {i, Length[p]}], {n, 2, 50}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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