A326368 Number of tilings of an equilateral triangle of side length n with unit triangles (of side length 1) and exactly three unit "lozenges" or "diamonds" (also of side length 1).

0, 0, 18, 434, 2814, 11127, 33365, 83568, 184254, 369254, 686952, 1203930, 2009018, 3217749, 4977219, 7471352, 10926570, 15617868, 21875294, 30090834, 40725702, 54318035, 71490993, 92961264, 119547974, 152182002, 191915700, 239933018, 297560034, 366275889

Offset: 1

Author

Greg Dresden, Jul 01 2019

Keywords

• nonn
• easy

Examples

			We can represent a unit triangle this way:
       o
      / \
     o - o
and a unit "lozenge" or "diamond" has these three orientations:
     o
    / \          o - o            o - o
   o   o  and   /   /   and also   \   \
    \ /        o - o                o - o
     o
and for n=3, here is one of the 18 different tiling of the triangle of side length 3 with exactly three lozenges:
          o
         / \
        o   o
       / \ / \
      o - o   o
     /   / \ / \
    o - o - o - o
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Richard J. Mathar, Lozenge tilings of the equilateral triangle, arXiv:1909.06336 [math.CO], 2019.
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Crossrefs

Cf. A273464, A326367, A326369.

Programs

Mathematica

Rest@ CoefficientList[Series[x^3*(18 + 308 x + 154 x^2 - 87 x^3 + 10 x^4 + 2 x^5)/(1 - x)^7, {x, 0, 30}], x] (* Michael De Vlieger, Jul 07 2019 *)

PARI

concat([0,0], Vec(x^3*(18 + 308*x + 154*x^2 - 87*x^3 + 10*x^4 + 2*x^5) / (1 - x)^7 + O(x^40))) \\ Colin Barker, Jul 02 2019

Formula

a(n) = (1/16)*(n-2)*(9*n^5 - 9*n^4 - 81*n^3 + 81*n^2 + 160*n - 192) for n >= 2 (proved by Greg Dresden and E. Sijaric).

From Colin Barker, Jul 02 2019: (Start)

G.f.: x^3*(18 + 308*x + 154*x^2 - 87*x^3 + 10*x^4 + 2*x^5) / (1 - x)^7.

a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>8.

(End)