A382489 The number of unitary 5-smooth divisors of n.
1, 2, 2, 2, 2, 4, 1, 2, 2, 4, 1, 4, 1, 2, 4, 2, 1, 4, 1, 4, 2, 2, 1, 4, 2, 2, 2, 2, 1, 8, 1, 2, 2, 2, 2, 4, 1, 2, 2, 4, 1, 4, 1, 2, 4, 2, 1, 4, 1, 4, 2, 2, 1, 4, 2, 2, 2, 2, 1, 8, 1, 2, 2, 2, 2, 4, 1, 2, 2, 4, 1, 4, 1, 2, 4, 2, 1, 4, 1, 4, 2, 2, 1, 4, 2, 2, 2
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).
Crossrefs
Programs
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Mathematica
a[n_] := Product[If[Divisible[n, p], 2, 1], {p, {2, 3, 5}}]; Array[a, 100]
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PARI
a(n) = vecprod(apply(x -> !((n % 30) % x) + 1, [2, 3, 5]))
Formula
Multiplicative with a(p^e) = 2 if p <= 5, and 1 otherwise.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 12/5.
In general, the asymptotic mean of the number of unitary prime(k)-smooth divisors of n is A054640(k)/A002110(k) = A236435(k)/A236436(k).
Dirichlet g.f.: (1 + 1/2^s) * (1 + 1/3^s) * (1 + 1/5^s) * zeta(s).
In general, Dirichlet g.f. of the number of unitary prime(k)-smooth divisors of n is zeta(s) * Product_{p prime <= prime(k)} (1 + 1/p^s).
Comments