A362973 The number of cubefull numbers (A036966) not exceeding 10^n.

1, 2, 7, 20, 51, 129, 307, 713, 1645, 3721, 8348, 18589, 41136, 90619, 198767, 434572, 947753, 2062437, 4480253, 9718457, 21055958, 45575049, 98566055, 213028539, 460160083, 993533517, 2144335391, 4626664451, 9980028172, 21523027285, 46408635232, 100053270534

Offset: 0

Author

Amiram Eldar, May 11 2023

Keywords

• nonn

Comments

The number of cubefull numbers not exceeding x is N(x) = c_0 * x^(1/3) + c_1 * x^(1/4) + c_2 * x^(1/5) + o(x^(1/8)), where c_0 (A362974), c_1 (A362975) and c_2 (A362976) are constants (Bateman and Grosswald, 1958; Finch, 2003).

The digits of a(3k) converge to A362974 as k -> oo. - Chai Wah Wu, May 13 2023

Examples

			There are 2 cubefull numbers not exceeding 10, 1 and 8, therefore a(1) = 2.
		

References

• Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, section 2.6.1, pp. 113-115.

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..36
• Paul T. Bateman and Emil Grosswald, On a theorem of Erdős and Szekeres, Illinois Journal of Mathematics, Vol. 2, No. 1 (1958), pp. 88-98.
• A. Ivić and P. Shiu, The distribution of powerful integers, Illinois Journal of Mathematics, Vol. 26, No. 4 (1982), pp. 576-590.
• Ekkehard Krätzel, On the distribution of square-full and cube-full numbers, Monatshefte für Mathematik, Vol. 120, No. 2 (1995), pp. 105-119.
• P. Shiu, The distribution of cube-full numbers, Glasgow Mathematical Journal, Vol. 33, No. 3 (1991), pp. 287-295.
• P. Shiu, Cube-full numbers in short intervals, Mathematical Proceedings of the Cambridge Philosophical Society, Vol. 112, No. 1 (1992), pp. 1-5.

Crossrefs

Cf. A036966, A362974, A362975, A362976.

Similar sequences: A070428, A118896.

Programs

Mathematica

a[n_] := Module[{max = 10^n}, CountDistinct@ Flatten@ Table[i^5 * j^4 * k^3, {i, Surd[max, 5]}, {j, Surd[max/i^5, 4]}, {k, CubeRoot[max/(i^5*j^4)]}]]; Array[a, 15, 0]

Python

from math import gcd
from sympy import factorint, integer_nthroot
def A362973(n):
    m, c = 10**n, 0
    for x in range(1,integer_nthroot(m,5)[0]+1):
        if all(d<=1 for d in factorint(x).values()):
            for y in range(1,integer_nthroot(z:=m//x**5,4)[0]+1):
                if gcd(x,y)==1 and all(d<=1 for d in factorint(y).values()):
                    c += integer_nthroot(z//y**4,3)[0]
    return c # Chai Wah Wu, May 11-13 2023

Extensions

a(23)-a(31) from Chai Wah Wu, May 11 2023