A370256 The number of ways in which n can be expressed as b^2 * c^3, with b and c >= 1.

1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0

Offset: 1

Author

Amiram Eldar, Feb 23 2024

Keywords

• nonn
• easy
• mult

Comments

First differs from A075802 and A112526 at n = 64.

The least number k such that a(k) = n is A005179(n)^6.

The indices of records are the sixth powers of the highly composite numbers, A002182(n)^6.

Examples

			1 = 1^2 * 1^3, so a(1) = 1.
64 = 1^2 * 4^3 = 8^2 * 1^3, so a(64) = 2.
4096 = 64^2 * 1^3 = 8^2 * 4^3 = 1^2 * 16^3, so a(4096)= 3.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A001694, A002182, A005179, A057523, A075802, A078434, A103221, A112526.

Programs

Mathematica

f[p_, e_] := Floor[(e + 2)/2] - Floor[(e + 2)/3]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

PARI

a(n) = vecprod(apply(x -> (x+2)\2 - (x+2)\3, factor(n)[, 2]));

PARI

for(n=1, 100, print1(direuler(p=2, n, 1/((1 - X^2)*(1 - X^3)))[n], ", ")) \\ Vaclav Kotesovec, Feb 23 2024

Python

from math import prod
from sympy import factorint
def A370256(n): return prod((e>>1)+1-(e+2)//3 for e in factorint(n).values()) # Chai Wah Wu, Apr 15 2025

Formula

Multiplicative with a(p^e) = A103221(e).

a(n) > 0 if and only if n is a powerful number (A001694).

a(A001694(n)) = A057523(n).

a(n^6) = A000005(n).

Sum_{k=1..n} a(k) ~ zeta(3/2) * sqrt(n) + zeta(2/3) * n^(1/3).

Dirichlet generating function: zeta(2*s)*zeta(3*s). - Vaclav Kotesovec, Feb 23 2024