A032765 a(n) = floor(n*(n+1)*(n+2) / (n + n+1 + n+2)) = floor(n*(n+2)/3).

0, 1, 2, 5, 8, 11, 16, 21, 26, 33, 40, 47, 56, 65, 74, 85, 96, 107, 120, 133, 146, 161, 176, 191, 208, 225, 242, 261, 280, 299, 320, 341, 362, 385, 408, 431, 456, 481, 506, 533, 560, 587, 616, 645, 674, 705, 736, 767, 800, 833, 866, 901, 936, 971, 1008

Offset: 0

Author

Patrick De Geest, May 15 1998

Keywords

• nonn
• easy

Comments

Satisfies a(n+1) - 2*a(n) + a(n-1) = (2/3)*(1 + w^(n+1) + w^(2*n+2)), a(0)=0 & a(1)=1 where w is the imaginary cubic root of unity. - Robert G. Wilson v, Jun 24 2002

First differences have this pattern: (+1) +1 +1 +3 +3 +3 +5 +5 +5 +7 +7 +7 +9 +9 +9. - Alexandre Wajnberg, Dec 19 2005

In Duistermaat (2010) section 11.3 The Planar Four-Bar Link on page 516: "It follows from (4.3.2) that the number of k-periodic fibers of the QRT automorphism, counted with multiplicities, is equal to nu((tau^S)^k) = 3*n^2 - 1 when k = 3*n, 3*n^2 + 2*n when k = 3*n + 1, 3*n^2 + 4*n + 1 when k = 3*n + 2, for every integer n." - Michael Somos, Mar 14 2023

a(n-1) is the maximum number of edges in a graph with vertices labeled from 1 to n, such that there is no triangle whose sum of vertex labels is divisible by n. - Hoang Xuan Thanh, Jun 11 2025

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 8*x^4 + 11*x^5 + 16*x^6 + 21*x^7 + 26*x^8 + ... - _Michael Somos_, Mar 14 2023
Let n = 5: a(5-1) = 8. Consider the graph G(5) with vertex set {1, 2, 3, 4, 5} and the edge set: E = {12, 23, 34, 45, 51, 13, 24, 35}, which contains 8 edges. The triangles in this graph are: {123}, {234}, {345}, {135}. Their vertex sums are: 1+2+3 = 6, 2+3+4 = 9, 3+4+5 = 12, 1+3+5=9. Since none of these sums are divisible by 5, the graph satisfies the required condition. If we add the edge 14, a new triangle {145} would appear, whose vertex sum is 1+4+5 = 10, divisible by 5. Similarly, if we add the edge 25, the triangle {235} would form, with vertex sum 2+3+5 = 10, also divisible by 5. - _Hoang Xuan Thanh_, Jun 11 2025
		

References

• Johannes J. Duistermaat, Discrete Integrable Systems, Springer Science+Business Media, 2010.

Links

• Johannes Bader, Kobon Triangles.
• Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007.
• Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007. [Cached copy, with permission]
• Taylor Short, The saturation number of carbon nanocones and nanotubes, arXiv:1807.11355 [math.CO], 2018.
• Eric Weisstein's World of Mathematics, Kobon Triangle.
• Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Crossrefs

Cf. A001082, A032766, A000217, A001399.

Programs

Maple

A032765:=n->floor(n*(n+2)/3); seq(A032765(n), n=0..100); # Wesley Ivan Hurt, Dec 20 2013

Mathematica

Table[Floor[n (n + 1)(n + 2)/(n + (n + 1) + (n + 2))], {n, 0, 55}]
Table[Floor[n (n + 2)/3], {n, 0, 100}] (* Wesley Ivan Hurt, Dec 20 2013 *)
LinearRecurrence[{2, -1, 1, -2, 1}, {0, 1, 2, 5, 8}, 60] (* Harvey P. Dale, Jun 06 2016 *)
Table[(3 n (n + 2) + 2 Cos[2 n Pi/3] + 2 Sqrt[3] Sin[2 n Pi/3] - 2)/9, {n, 0, 20}] (* Eric W. Weisstein, Jul 27 2025 *)
CoefficientList[Series[x (-1 - 2 x^2 + x^3)/((-1 + x)^3 (1 + x + x^2)), {x, 0, 20}], x] (* Eric W. Weisstein, Jul 27 2025 *)

PARI

a(n)=n*(n+2)\3 \\ Charles R Greathouse IV, Jun 11 2015

Formula

From Ralf Stephan, May 05 2004: (Start)

a(n) = n^2 - ceiling(n*(n-1)/3).

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

a(n) = floor(n*(n+2)/3). - Saburo Tamura, sent by Alexandre Wajnberg, Dec 19 2005

a(3*n - 1) = 3*n^2 - 1, a(3*n) = 3*n^2 + 2*n, a(3*n + 1) = 3*n^2 + 4*n + 1. - Michael Somos, Mar 14 2023

a(n+5) = A000217(n+5) - A001399(n+3) - A001399(n). - Hoang Xuan Thanh, Jun 11 2025

Sum_{n>=1} (-1)^(n+1)/a(n) = -1/4 + (Pi/(2*sqrt(3)))*(2-cosec(Pi/sqrt(3))). - Amiram Eldar, Jun 18 2025

Extensions

Name change suggested by Wesley Ivan Hurt, Dec 20 2013