A337736 The number of cubefull numbers (A036966) between the consecutive cubes n^3 and (n+1)^3.

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 0, 4, 2, 1, 3, 0, 3, 1, 2, 1, 3, 2, 0, 2, 5, 1, 3, 1, 1, 3, 3, 2, 1, 3, 1, 2, 2, 2, 1, 2, 2, 3, 6, 1, 1, 1, 4, 1, 1, 3, 3, 1, 3, 4, 1, 2, 3, 1, 2, 3, 2, 3, 2, 3, 3, 2, 1, 4, 2, 1, 1, 0, 7, 1, 1, 4, 3, 2, 2, 2, 3, 3, 2, 0, 4, 2, 4

Offset: 1

Author

Amiram Eldar, Sep 17 2020

Keywords

• nonn

Comments

For each k >= 0 the sequence of solutions to a(x) = k has a positive asymptotic density (Shiu, 1991).

Examples

			a(2) = 1 since there is one cubefull number, 16 = 2^4, between 2^3 = 8 and 3^3 = 27.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• P. Shiu, The distribution of cube-full numbers, Glasgow Mathematical Journal, Vol. 33, No. 3 (1991), pp. 287-295. See section 3, p. 291.

Crossrefs

Cf. A000578, A036966, A119241, A337737, A362974.

Programs

Mathematica

cubQ[n_] := Min[FactorInteger[n][[;; , 2]]] > 2; a[n_] := Count[Range[n^3 + 1, (n + 1)^3 - 1], _?cubQ]; Array[a, 100]

Python

from math import gcd
from sympy import integer_nthroot, factorint
def A337736(n):
    def f(x):
        c = 0
        for w in range(1,integer_nthroot(x,5)[0]+1):
            if all(d<=1 for d in factorint(w).values()):
                for y in range(1,integer_nthroot(z:=x//w**5,4)[0]+1):
                    if gcd(w,y)==1 and all(d<=1 for d in factorint(y).values()):
                        c += integer_nthroot(z//y**4,3)[0]
        return c
    return f((n+1)**3-1)-f(n**3) # Chai Wah Wu, Sep 10 2024

Formula

Asymptotic mean: lim_{m->oo} (1/m) Sum_{k=1..m} a(k) = A362974 - 1 = 3.659266... . - Amiram Eldar, May 11 2023