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Number of equilateral triangles formed by triples of points taken from a hexagonal chunk of side n in the hexagonal lattice.
6

%I #35 Apr 25 2024 15:26:08

%S 0,8,66,258,710,1590,3108,5516,9108,14220,21230,30558,42666,58058,

%T 77280,100920,129608,164016,204858,252890,308910,373758,448316,533508,

%U 630300,739700,862758,1000566,1154258,1325010,1514040,1722608,1952016,2203608,2478770

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

%C 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.

%H Nathaniel Johnston, <a href="/A008893/b008893.txt">Table of n, a(n) for n = 0..10000</a>

%H G. Nebe and N. J. A. Sloane, <a href="http://www.math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/A2.html">Home page for hexagonal (or triangular) lattice A2</a>

%H N. J. A. Sloane, <a href="/A008893/a008893.jpg">Illustration for a(1)=8.</a> [The drawing was made for a different offset, so it says a(2)=8.]

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5, -10, 10, -5, 1).

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

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

%o (Maxima) A008893(n):=n*(n+1)*(7*n^2+7*n+2)/4$

%o makelist(A008893(n),n,0,30); /* _Martin Ettl_, Nov 03 2012 */

%Y Cf. A045949, A152041.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, _R. K. Guy_

%E Edited May 29 2012 by _N. J. A. Sloane_, May 29 2012