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A140074 Excess over the asymptote of the number of perfect squares between cubes. 0
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 (list; graph; refs; listen; history; text; internal format)
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: A132194 A174897 A092079 * A090174 A165556 A127243

Adjacent sequences:  A140071 A140072 A140073 * A140075 A140076 A140077

KEYWORD

easy,nonn

AUTHOR

Gerard P. Michon, May 06 2008

STATUS

approved

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Last modified August 23 22:23 EDT 2017. Contains 291021 sequences.