A008481 If n = Product (p_j^k_j) then a(n) = Sum partition(k_j).
0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 2, 5, 1, 3, 1, 3, 2, 2, 1, 4, 2, 2, 3, 3, 1, 3, 1, 7, 2, 2, 2, 4, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 6, 2, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 4, 1, 2, 3, 11, 2, 3, 1, 3, 2, 3, 1, 5, 1, 2, 3, 3, 2, 3, 1, 6, 5, 2, 1, 4, 2, 2, 2
Offset: 1
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537 (first 10000 terms from Vincenzo Librandi)
- Index entries for sequences computed from exponents in factorization of n.
Crossrefs
Programs
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Maple
a:= n-> add(combinat[numbpart](i[2]), i=ifactors(n)[2]): seq(a(n), n=1..100); # Alois P. Heinz, Aug 30 2018
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Mathematica
Prepend[ Array[ Plus @@ (PartitionsP /@ Last[ Transpose[ FactorInteger[ # ] ] ])&, 100, 2 ], 0 ] (* Second program: *) Array[Total[PartitionsP /@ FactorInteger[#][[All, -1]] - Boole[# == 1]] &, 87] (* Michael De Vlieger, Sep 02 2018 *)
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PARI
A008481(n) = vecsum(apply(e -> numbpart(e),factor(n)[,2])); \\ Antti Karttunen, Aug 30 2018
Formula
From Antti Karttunen, Aug 30 2018: (Start)
Additive with a(p^e) = A000041(e).
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B + C), where B is Mertens's constant (A077761) and C = Sum_{p prime} f(1/p) = 1.03089282973521424158..., where f(x) = -1 + (1-x) * Product_{k>=1} (1 + x^k)/(1 - x^(2*k)). - Amiram Eldar, Sep 29 2023
Extensions
Term a(1) corrected from 1 to 0 (for an empty sum) by Antti Karttunen, Aug 30 2018
Comments