A190867 Count of the 3-full divisors of n.
1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1
Offset: 1
Examples
a(16)=3 because the divisors of 16 are {1,2,4,8,16}, and three of these are 3-full: 1, 8=2^3 and 16=2^4.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
- Aleksandar Ivić, On the asymptotic formulas for powerful numbers, Publications de l'Institut Mathématique, Vol. 23, No. 37 (1978), pp. 85-94; alternative link.
- Index entries for sequences computed from exponents in factorization of n.
Programs
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Maple
f:= n -> convert(map(t -> max(1,t[2]-1), ifactors(n)[2]),`*`): map(f, [$1..200]); # Robert Israel, Jul 19 2017
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Mathematica
Table[Product[Max[{1, i - 1}], {i, FactorInteger[n][[All, 2]]}], {n, 1, 200}] (* Geoffrey Critzer, Feb 12 2015 *) Table[1 + DivisorSum[n, 1 &, AllTrue[FactorInteger[#][[All, -1]], # > 2 &] &], {n, 120}] (* Michael De Vlieger, Jul 19 2017 *)
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PARI
A190867(n) = { my(f = factor(n), m = 1); for (k=1, #f~, m *= max(1,f[k, 2]-1); ); m; } \\ Antti Karttunen, Jul 19 2017
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Python
from functools import reduce from sympy import factorint from operator import mul def a(n): return 1 if n==1 else reduce(mul, [max(1, e - 1) for e in factorint(n).values()]) print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 19 2017
Formula
a(n) = Sum_{d|n, d in A036966} 1.
a(n) <= A005361(n).
Multiplicative with a(p^e) = max(1,e-1).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Product_{p prime} (1 + 1/(p^2*(p-1))) (A065483). (Ivić, 1978). - Amiram Eldar, Jul 23 2022
Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/p^s + 1/p^(3*s)). - Amiram Eldar, Sep 21 2023
Comments