A069470 a(n) = Sum_{k>=1} floor(n/(k*(k+1)/2)).

0, 1, 2, 4, 5, 6, 9, 10, 11, 13, 15, 16, 19, 20, 21, 24, 25, 26, 29, 30, 32, 35, 36, 37, 40, 41, 42, 44, 46, 47, 52, 53, 54, 56, 57, 58, 62, 63, 64, 66, 68, 69, 73, 74, 75, 79, 80, 81, 84, 85, 87, 89, 90, 91, 94, 96, 98, 100, 101, 102, 107, 108, 109, 112, 113, 114, 118

Offset: 0

Author

Henry Bottomley, Mar 25 2002

Keywords

• nonn

Comments

The summation has floor(1/2 + sqrt(2*n)) = A002024(n) nonzero terms. - Enrique Pérez Herrero, Apr 05 2010

Examples

			a(11) = floor(11/1) + floor(11/3) + floor(11/6) + floor(11/10) + floor(11/15) + ... = 11 + 3 + 1 + 1 + 0 + ... = 16.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000217, A006218, A007862, A013936, A013937, A013938, A013939.

Cf. A002024, A004275. - Enrique Pérez Herrero, Apr 05 2010

Programs

Magma

[(&+[Floor(n/(k*(k+1)/2)): k in [1..100]]): n in [0..30]]; // G. C. Greubel, May 23 2018

Mathematica

A069470[n_]:=Sum[Floor[(2*n)/(k*(1 + k))], {k, 1, Floor[1/2 + Sqrt[2*n]]}] (* Enrique Pérez Herrero, Apr 05 2010 *)

PARI

for(n=0, 30, print1(sum(k=1, 100, floor(n/(k*(k+1)/2))), ", ")) \\ G. C. Greubel, May 23 2018

Formula

a(n) = a(n-1) + A007862(n).

It appears that limit((sum(floor((1/2)*n/(k*(k+1))), k=1..n))/n, n=infinity) = 1/2. - Stephen Crowley, Aug 12 2009

From Enrique Pérez Herrero, Apr 05 2010: (Start)

a(n) <= floor((2*n^2)/(1 + n)) = A004275(n).

a(n) <= floor((2*n*floor((1 + 2*sqrt(2*n))/2))/(1+floor((1+2*sqrt(2*n))/2))). (End)

G.f.: (1/(1 - x)) * Sum_{k>=1} x^(k*(k+1)/2)/(1 - x^(k*(k+1)/2)). - Ilya Gutkovskiy, Jul 11 2019