A363195 Number of divisors of the n-th cubefull number A036966(n).

1, 4, 5, 4, 6, 7, 5, 4, 8, 16, 6, 9, 4, 20, 10, 5, 20, 7, 24, 16, 11, 25, 4, 28, 24, 20, 12, 8, 4, 5, 30, 16, 6, 16, 32, 30, 24, 13, 4, 20, 35, 20, 28, 9, 4, 36, 36, 28, 14, 16, 25, 20, 40, 16, 24, 35, 4, 40, 5, 42, 7, 32, 15, 6, 20, 32, 16, 20, 10, 30, 45, 20

Offset: 1

Author

Amiram Eldar, May 21 2023

Keywords

• nonn
• easy

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Rafael Jakimczuk and Matilde Lalín, Asymptotics of sums of divisor functions over sequences with restricted factorization structure, Notes on Number Theory and Discrete Mathematics, Vol. 28, No. 4 (2022), pp. 617-634, eq. (8).

Crossrefs

Cf. A000005, A036966, A362986.

Similar sequences: A072048, A076400, A363194.

Programs

Mathematica

DivisorSigma[0, Select[Range[25000], # == 1 || Min[FactorInteger[#][[;; , 2]]] > 2 &]]

PARI

lista(kmax) = for(k = 1, kmax, if(k==1 || vecmin(factor(k)[, 2]) > 2, print1(numdiv(k), ", ")));

Python

from itertools import count, islice
from math import prod
from sympy import factorint
def A363195_gen(): # generator of terms
    for n in count(1):
        f = factorint(n).values()
        if all(e>2 for e in f):
            yield prod(e+1 for e in f)
A363195_list = list(islice(A363195_gen(),20)) # Chai Wah Wu, May 21 2023

Formula

a(n) = A000005(A036966(n)).

Sum_{A036966(k) < x} a(k) = c_1 * x^(1/3) * log(x)^3 + c_2 * x^(1/3) * log(x)^2 + c_3 * x^(1/3) * log(x) + c_4 * x^(1/3) + O(x^(7/24 + eps)), where c_1, c_2, c_3 and c_4 are constants. c_1 = Product_{p prime} ((1-1/p)^4 * (1 + 1/((p^(1/3) - 1)^2*p^(1/3)) + 3/(p-p^(2/3))))/162 = 0.1346652397135839416... . [corrected Sep 21 2024]