A008893 Number of equilateral triangles formed by triples of points taken from a hexagonal chunk of side n in the hexagonal lattice.

0, 8, 66, 258, 710, 1590, 3108, 5516, 9108, 14220, 21230, 30558, 42666, 58058, 77280, 100920, 129608, 164016, 204858, 252890, 308910, 373758, 448316, 533508, 630300, 739700, 862758, 1000566, 1154258, 1325010, 1514040, 1722608, 1952016, 2203608, 2478770

Offset: 0

Author

N. J. A. Sloane, R. K. Guy

Keywords

• nonn
• easy

Comments

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice. Here we consider a hexagonal chunk of the lattice in which each bounding edge contains n+1 points.

Links

• Nathaniel Johnston, Table of n, a(n) for n = 0..10000
• Gabriele Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2.
• N. J. A. Sloane, Illustration for a(1)=8. [The drawing was made for a different offset, so it says a(2)=8.]
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A045949, A152041.

Programs

Mathematica

A008893[n_] := n*(n + 1)*(7*n*(n + 1) + 2)/4; Array[A008893, 50, 0] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 66, 258, 710}, 50] (* Paolo Xausa, Aug 16 2025 *)

Maxima

A008893(n):=n*(n+1)*(7*n^2+7*n+2)/4$
makelist(A008893(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

Formula

a(n) = n*(n+1)*(7*n^2+7*n+2)/4.

G.f.: -2*x*(4*x^2+13*x+4)/(x-1)^5 [From Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009]

From Elmo R. Oliveira, Aug 15 2025: (Start)

E.g.f.: exp(x)*x*(2 + x)*(16 + 42*x + 7*x^2)/4.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

a(n) = 2*A152041(n). (End)

Extensions

Edited May 29 2012 by N. J. A. Sloane, May 29 2012