A130518 a(n) = Sum_{k=0..n} floor(k/3). (Partial sums of A002264.)

0, 0, 0, 1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 26, 30, 35, 40, 45, 51, 57, 63, 70, 77, 84, 92, 100, 108, 117, 126, 135, 145, 155, 165, 176, 187, 198, 210, 222, 234, 247, 260, 273, 287, 301, 315, 330, 345, 360, 376, 392, 408, 425, 442, 459, 477, 495, 513, 532, 551, 570

Offset: 0

Author

Hieronymus Fischer, Jun 01 2007

Keywords

• nonn
• easy

Comments

Complementary with A130481 regarding triangular numbers, in that A130481(n) + 3*a(n) = n(n+1)/2 = A000217(n).

Apart from offset, the same as A062781. - R. J. Mathar, Jun 13 2008

Apart from offset, the same as A001840. - Michael Somos, Sep 18 2010

The sum of any three consecutive terms is a triangular number. - J. M. Bergot, Nov 27 2014

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Crossrefs

Cf. A002265, A002266, A004526, A010872, A010873, A010874, A062781, A130482, A130483.

Programs

GAP

List([0..60], n-> Int(n*(n-1)/6)); # G. C. Greubel, Aug 31 2019

Magma

[Round(n*(n-1)/6): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011

Maple

seq(floor(n*(n-1)/6),n=0..60); # Robert Israel, Nov 27 2014

Mathematica

Table[n, {n, 0, 19}, {3}] // Flatten // Accumulate (* Jean-François Alcover, Jun 05 2013 *)

PARI

a(n)=n*(n-1)\/6 \\ Charles R Greathouse IV, Jun 05 2013

Sage

[floor(binomial(n,2)/3) for n in range(0,60)] # Zerinvary Lajos, Dec 01 2009

Formula

G.f.: x^3 / ((1-x^3)*(1-x)^2).

a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).

a(n) = (1/2)*floor(n/3)*(2*n - 1 - 3*floor(n/3)) = A002264(n)*(2n - 1 - 3*A002264(n))/2.

a(n) = (1/2)*A002264(n)*(n - 1 + A010872(n)).

a(n) = round(n*(n-1)/6) = round((n^2-n-1)/6) = floor(n*(n-1)/6) = ceiling((n+1)*(n-2)/6). - Mircea Merca, Nov 28 2010

a(n) = a(n-3) + n - 2, n > 2. - Mircea Merca, Nov 28 2010

a(n) = A214734(n, 1, 3). - Renzo Benedetti, Aug 27 2012

a(3n) = A000326(n), a(3n+1) = A005449(n), a(3n+2) = 3*A000217(n) = A045943(n). - Philippe Deléham, Mar 26 2013

a(n) = (3*n*(n-1) - (-1)^n*((1+i*sqrt(3))^(n-2) + (1-i*sqrt(3))^(n-2))/2^(n-3) - 2)/18, where i=sqrt(-1). - Bruno Berselli, Nov 30 2014

Sum_{n>=3} 1/a(n) = 20/3 - 2*Pi/sqrt(3). - Amiram Eldar, Sep 17 2022