A368883 The number of infinitary divisors of n that are cubefree.
1, 2, 2, 2, 2, 4, 2, 3, 2, 4, 2, 4, 2, 4, 4, 1, 2, 4, 2, 4, 4, 4, 2, 6, 2, 4, 3, 4, 2, 8, 2, 2, 4, 4, 4, 4, 2, 4, 4, 6, 2, 8, 2, 4, 4, 4, 2, 2, 2, 4, 4, 4, 2, 6, 4, 6, 4, 4, 2, 8, 2, 4, 4, 2, 4, 8, 2, 4, 4, 8, 2, 6, 2, 4, 4, 4, 4, 8, 2, 2, 1, 4, 2, 8, 4, 4, 4
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Vaclav Kotesovec, Graph - the asymptotic ratio (100000 terms)
Programs
-
Mathematica
f[p_, e_] := Switch[Mod[e, 4], 1, 2, 2, 2, 3, 3, 0, 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
-
PARI
a(n) = vecprod(apply(x -> [1,2,2,3][x%4+1], factor(n)[, 2]));
Formula
Multiplicative with a(p^e) = 2 if e == 1 or 2 (mod 4), 3 if e == 3 (mod 4), and 1 if e == 0 (mod 4).
a(n) >= 1, with equality if and only if n is a 4th power (A000583).
Dirichlet g.f.: zeta(4*s) * Product_{p prime} (1 + 2/p^s + 2/p^(2*s) + 3/p^(3*s)).
From Vaclav Kotesovec, Jan 09 2024: (Start)
Dirichlet g.f.: zeta(4*s) * zeta(s)^2 * Product_{p prime} (1 - 1/p^(2*s) + 1/p^(3*s) - 4/p^(4*s) + 3/p^(5*s)).
Let f(s) = Product_{p prime} (1 - 1/p^(2*s) + 1/p^(3*s) - 4/p^(4*s) + 3/p^(5*s)).
Sum_{k=1..n} a(k) ~ f(1) * zeta(4) * n * (log(n) + 2*gamma - 1 + f'(1)/f(1) + 4*zeta'(4)/zeta(4)), where
f(1) = Product_{p prime} (1 - 1/p^2 + 1/p^3 - 4/p^4 + 3/p^5) = 0.5857770602270641007515615375942370402509903724261557972367075945186871...,
f'(1) = f(1) * Sum_{p prime} (2*p^2 - p + 15) * log(p) / (p^4 + p^3 + p - 3) = f(1) * 1.319786264712492218167871116508220489817987315752197198819256094...,
Comments