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Count of the 3-full divisors of n.
4

%I #43 Sep 21 2023 01:45:47

%S 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,

%T 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,

%U 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

%N Count of the 3-full divisors of n.

%C a(n) is the number of divisors d of n with d an element of A036966.

%C This is the 3-full analog of the 2-full case A005361.

%H Antti Karttunen, <a href="/A190867/b190867.txt">Table of n, a(n) for n = 1..10000</a>

%H Aleksandar Ivić, <a href="https://eudml.org/doc/257502">On the asymptotic formulas for powerful numbers</a>, Publications de l'Institut Mathématique, Vol. 23, No. 37 (1978), pp. 85-94; <a href="http://elib.mi.sanu.ac.rs/files/journals/publ/43/12.pdf">alternative link</a>.

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.

%F a(n) = Sum_{d|n, d in A036966} 1.

%F a(n) <= A005361(n).

%F Multiplicative with a(p^e) = max(1,e-1).

%F 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

%F Dirichlet g.f.: zeta(s)^2 * Product_{p prime} (1 - 1/p^s + 1/p^(3*s)). - _Amiram Eldar_, Sep 21 2023

%e 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.

%p f:= n -> convert(map(t -> max(1,t[2]-1), ifactors(n)[2]),`*`):

%p map(f, [$1..200]); # _Robert Israel_, Jul 19 2017

%t Table[Product[Max[{1, i - 1}], {i, FactorInteger[n][[All, 2]]}], {n, 1, 200}] (* _Geoffrey Critzer_, Feb 12 2015 *)

%t Table[1 + DivisorSum[n, 1 &, AllTrue[FactorInteger[#][[All, -1]], # > 2 &] &], {n, 120}] (* _Michael De Vlieger_, Jul 19 2017 *)

%o (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

%o (Python)

%o from sympy import factorint

%o from operator import mul

%o def a(n): return 1 if n==1 else reduce(mul, [max(1, e - 1) for e in factorint(n).values()])

%o print([a(n) for n in range(1, 101)]) # _Indranil Ghosh_, Jul 19 2017

%Y Cf. A005361, A036966, A065483.

%K nonn,mult,easy

%O 1,8

%A _R. J. Mathar_, May 27 2011