A365932 a(n) = the number of cubes (of integers > 0) that have bit length n.

1, 0, 0, 1, 1, 0, 2, 1, 1, 3, 2, 3, 5, 5, 6, 9, 10, 13, 17, 21, 26, 34, 42, 52, 67, 84, 105, 134, 167, 211, 267, 335, 422, 533, 670, 845, 1065, 1341, 1690, 2130, 2682, 3380, 4259, 5365, 6760, 8518, 10730, 13520, 17035, 21461, 27040, 34069, 42923, 54080, 68137, 85847

Offset: 1

Author

Karl-Heinz Hofmann, Oct 05 2023

Keywords

• nonn
• easy
• base

Comments

Number of cubes in the range: 2^(n-1) <= k^3 < 2^n-1.

There is no need to include 2^n-1 because it is a Mersenne number and it cannot be a power anyway.

Examples

			For n = 13; a(n) = 5; following 5 cubes have a bit length of 13: 16^3, 17^3, 18^3, 19^3 and 20^3.
		

Crossrefs

Cf. A004632.

Cf. A017981 (partial sums).

Cf. A002580, A126726, A365930.

Programs

Mathematica

a[n_] := Floor[Surd[2^n-1, 3]] - Floor[Surd[2^(n-1)-1, 3]]; Array[a, 56] (* Amiram Eldar, Oct 30 2023 *)

Python

from sympy import integer_nthroot
def A365932(n):
    return integer_nthroot(2**n-1, 3)[0] - integer_nthroot(2**(n-1)-1, 3)[0]
print([A365932(n) for n in range(1,57)])

Formula

a(n) = floor((2^n-1)^(1/3)) - floor((2^(n-1)-1)^(1/3)) for n > 0.

Limit_{n->oo} a(n)/a(n-1) = 2^(1/3) = A002580.