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

0,5

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

Also, expected number of intersection points, when n cords are randomly chosen on a circle (i.e., each endpoint is chosen with uniform random distribution on the circle), rounded to the nearest integer. The rounding (down) occurs only for n==2 (mod 3), where the expectation value is 1/3 above the nearest integer. For n=2, there are 3 possible configurations of the two cords with equal probability, in one of which there is an intersection point, so the exact expectation value is 1/3. Applying this to each of the C(n,2) = n(n-1)/2 possible pairs yields the result also for other n. - M. F. Hasler, Jun 16 2026

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).

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

a(n) = floor(n*(n-1)/6), where "floor" could also be replaced with "round". - M. F. Hasler, Jun 16 2026

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 *)

Prog

(SageMath) [floor(binomial(n, 2)/3) for n in range(0, 60)] # Zerinvary Lajos, Dec 01 2009
(Magma) [Round(n*(n-1)/6): n in [0..60]]; // Vincenzo Librandi, Jun 25 2011
(PARI) a(n)=n*(n-1)\/6 \\ Charles R Greathouse IV, Jun 05 2013
(GAP) List([0..60], n-> Int(n*(n-1)/6)); # G. C. Greubel, Aug 31 2019
(Python)
from math import comb
def A130518(n): return comb(n, 2)//3 # Chai Wah Wu, Jun 17 2026

Crossrefs

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

Sequence in context: A145919 A058937 A062781 * A001840 A022794 A025693

Adjacent sequences: A130515 A130516 A130517 * A130519 A130520 A130521

Keyword

nonn,easy

Author

Hieronymus Fischer, Jun 01 2007

Status

approved