A379141 If n = Product (p_j^k_j) then a(n) = numerator of Sum 1/k_j.
0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 1, 3, 1, 1, 2, 2, 2, 1, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 5, 1, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 5, 1, 2, 3, 1, 2, 3, 1, 3, 2, 3, 1, 5, 1, 2, 3, 3, 2, 3, 1, 5, 1, 2, 1, 5, 2, 2, 2, 4, 1, 5, 2, 3, 2, 2, 2, 6, 1, 3, 3, 1, 1, 3, 1, 4, 3, 2, 1, 5, 1, 3
Offset: 1
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Maple
a:= n-> numer(add(1/i[2], i=ifactors(n)[2])): seq(a(n), n=1..110); # Alois P. Heinz, Dec 16 2024
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Mathematica
Join[{0}, Table[Plus @@ (1/#[[2]] & /@ FactorInteger[n]), {n, 2, 110}]] // Numerator
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PARI
a(n) = my(f=factor(n)); numerator(sum(k=1, #f~, 1/f[k,2])); \\ Michel Marcus, Dec 16 2024