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a(n) = floor((n+1/2)^(3/2)) - floor((n-1/2)^(3/2)). _- _Robert Israel_, Jun 06 2014
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Robert Israel, <a href="/A201206/b201206.txt">Table of n, a(n) for n = 1..10000</a>
a(n) = floor((n+1/2)^(3/2)) - floor((n-1/2)^(3/2)). Robert Israel, Jun 06 2014
A:= n -> floor((n+1/2)^(3/2)) - floor((n-1/2)^(3/2)):
seq(A(n), n=1..100); # Robert Israel, Jun 06 2014
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_Artur Jasinski (grafix(AT)csl.pl), _, Jan 07 2012
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Number of successive decreasing values of round(n^(2/3))^3 - n^2.
1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 4, 6, 5, 6, 6, 6, 6, 6, 7, 6, 7, 7, 7, 8, 7, 8, 8, 8, 8, 8, 8, 9, 8, 9, 9, 9, 9, 9, 10, 9, 10, 10, 9, 10, 10, 11, 10, 10, 11, 10, 11, 11, 11, 11, 11, 11, 12, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 13, 14
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allocated for Artur JasinskiNumber of successive decreasing values of round(n^(2/3))^3 - n^2
1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 4, 6, 5, 6, 6, 6, 6, 6, 7, 6, 7, 7, 7, 8, 7, 8, 8, 8, 8, 8, 8, 9, 8, 9, 9, 9, 9, 9, 10, 9, 10, 10, 9, 10, 10, 11, 10, 10, 11, 10, 11, 11, 11, 11, 11, 11, 12, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 13, 14
1,2
ile = 0; aa = {}; d1 = -1; Do[ile = ile + 1; xx = Round[y^(2/3)]; dd = xx^3 - y^2; If[dd > 0 && d1 < 0, AppendTo[aa, ile]; ile = 0]; d1 = dd, {y, 2, 1000}]; aa
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Artur Jasinski (grafix(AT)csl.pl), Jan 07 2012
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