A013937 a(n) = Sum_{k=1..n} floor(n/k^3).

0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 59, 60, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 75, 76, 77, 78, 79, 80, 81, 82

Offset: 0

Author

N. J. A. Sloane, Henri Lifchitz

Keywords

• nonn

Examples

			a(36) = [36/1]+[36/8]+[36/27]+[36/64]+... = 36+4+1+0+... = 41.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000
• Benoit Cloitre, Plot of (a(n)-zeta(3)*n)/n^(1/3)-zeta(1/3)

Crossrefs

Cf. A005187, A006218, A011371, A013936, A013939 for similar sequences.

Programs

Maple

A013937:=n->add(floor(n/k^3), k=1..n); seq(A013937(n), n=0..100); # Wesley Ivan Hurt, Feb 15 2014

Mathematica

Table[Sum[Floor[n/k^3], {k, n}], {n, 0, 100}] (* Wesley Ivan Hurt, Feb 15 2014 *)

PARI

a(n)=sum(k=1,ceil(n^(1/3)),n\k^3) /*Benoit Cloitre, Nov 05 2012 */

Formula

a(n) = a(n-1)+A061704(n). a(n) = Sum_{k=1..n} floor((n/k)^(1/3)) with asymptotic formula: a(n) = zeta(3)*n+zeta(1/3)*n^(1/3)+O(n^theta) where theta<1/3 and we conjecture that theta=1/4+epsilon is the best possible choice. - Benoit Cloitre, Nov 05 2012

G.f.: (1/(1 - x))*Sum_{k>=1} x^(k^3)/(1 - x^(k^3)). - Ilya Gutkovskiy, Feb 11 2017

Extensions

More terms from Henry Bottomley, Jul 03 2001