OFFSET
1,7
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.
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.
EXAMPLE
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.
MATHEMATICA
a[n_] := Floor[Surd[2^n-1, 3]] - Floor[Surd[2^(n-1)-1, 3]]; Array[a, 56] (* Amiram Eldar, Oct 30 2023 *)
PROG
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
Karl-Heinz Hofmann, Oct 05 2023
STATUS
approved