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A140074
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Excess over the asymptote of the number of perfect squares between cubes.
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0
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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; internal format)
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OFFSET
| 0,1
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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.
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LINKS
| G. P. Michon, A Sequence of Bits with Strange Statistics.
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FORMULA
| a(n) = floor(sqrt((n+1)^3-1)) - ceiling(sqrt(n^3)) + 1 - floor(1.5 sqrt(n))
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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.
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CROSSREFS
| Sequence in context: A132194 A174897 A092079 * A090174 A165556 A127243
Adjacent sequences: A140071 A140072 A140073 * A140075 A140076 A140077
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KEYWORD
| easy,nonn
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AUTHOR
| Gerard P. Michon (g.michon(AT)att.net), May 06 2008
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