A201206 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

Offset: 1

Author

Artur Jasinski, Jan 07 2012

Keywords

• nonn

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Programs

Maple

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

Mathematica

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

Formula

a(n) = floor((n+1/2)^(3/2)) - floor((n-1/2)^(3/2)). - Robert Israel, Jun 06 2014