A140074 Excess over the asymptote of the number of perfect squares between cubes.

1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1

Offset

0,1

Comments

There are always at least two squares between positive consecutive cubes, starting with the perfect squares 1 and 4 between the perfect cubes 1 (included) and 8 (excluded).

The number of squares between the cube of n (included) and the cube of n+1 (excluded) is always one of the two integers bracketing 3*sqrt(n)/2.

The number a(n) in the sequence is 0 if the correct count is the lower number or 1 if the actual count is the higher number.

Links

Table of n, a(n) for n=0..100.

G. P. Michon, A Sequence of Bits with Strange Statistics.

Formula

a(n) = floor(sqrt((n+1)^3-1)) - ceiling(sqrt(n^3)) + 1 - floor(1.5 sqrt(n))

Example

The sequence starts with a(0)=1 for n=0 because there is just one perfect square (0) between the cube of 0 (included) and the cube of 1 (excluded).
This exceeds by a(0)=1 the asymptotic expression floor(1.5*sqrt(n)) for the value n=0.

Crossrefs

Sequence in context: A189166 A092079 A342877 * A351957 A342004 A284881

Adjacent sequences: A140071 A140072 A140073 * A140075 A140076 A140077

Keyword

easy,nonn

Author

Gerard P. Michon, May 06 2008

Status

approved