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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
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.
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
KEYWORD
easy,nonn
AUTHOR
Gerard P. Michon, May 06 2008
STATUS
approved