A361936 Indices of the squares in the sequence of powerful numbers (A001694).

1, 2, 4, 5, 6, 9, 10, 11, 13, 14, 16, 19, 20, 21, 24, 26, 28, 29, 31, 33, 35, 36, 39, 40, 41, 44, 45, 46, 48, 50, 51, 55, 56, 59, 60, 61, 65, 67, 68, 70, 71, 73, 75, 76, 79, 81, 84, 85, 87, 88, 90, 92, 94, 96, 97, 100, 102, 104, 107, 109, 110, 111, 114, 116, 117, 119, 120

Offset: 1

Author

Amiram Eldar, Mar 31 2023

Keywords

• nonn

Comments

Equivalently, the number of powerful numbers that do not exceed n^2.

The asymptotic density of this sequence is zeta(3)/zeta(3/2) = 1/A090699 = 0.460139... .

If k is a term of A336175 then a(k) and a(k+1) are consecutive integers, i.e., a(k+1) = a(k) + 1.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Paul Erdős and George Szekeres, Über die Anzahl der Abelschen Gruppen gegebener Ordnung und über ein verwandtes zahlentheoretisches Problem, Acta Sci. Math. (Szeged), Vol. 7, No. 2 (1935), pp. 95-102.
• Solomon W. Golomb, Powerful numbers, Amer. Math. Monthly, Vol. 77, No. 8 (1970), pp. 848-852.

Crossrefs

Cf. A000290, A001694, A090699, A119241, A217038, A336175.

Programs

Mathematica

Position[Select[Range[5000], # == 1 || Min[FactorInteger[#][[;; , 2]]] > 1 &], _?(IntegerQ[Sqrt[#]] &)] // Flatten

PARI

lista(kmax) = {my(c = 0); for(k = 1, kmax, if(ispowerful(k), c++); if(issquare(k), print1(c, ", "))); }

Python

from math import isqrt
from sympy import integer_nthroot, factorint
def A361936(n):
    m = n**2
    return int(sum(isqrt(m//k**3) for k in range(1, integer_nthroot(m, 3)[0]+1) if all(d<=1 for d in factorint(k).values()))) # Chai Wah Wu, Sep 10 2024

Formula

a(n) = A217038(n^2).

a(n+1) - a(n) = A119241(n) + 1.

a(n) = (zeta(3/2)/zeta(3)) * n + O(n^(2/3)).