0,3

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

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

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Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007.

Gilles Clement and Johannes Bader, Tighter Upper Bound for the Number of Kobon Triangles, Unpublished, 2007.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).

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

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

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

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

Cf. A001082, A032766, A000217, A001399.

Sequence in context: A130258 A380545 A186496 * A375298 A340931 A261416

Adjacent sequences: A032762 A032763 A032764 * A032766 A032767 A032768

nonn,easy

Patrick De Geest, May 15 1998

Name change suggested by Wesley Ivan Hurt, Dec 20 2013

approved