A326367 Number of tilings of an equilateral triangle of side length n with unit triangles (of side length 1) and exactly two unit "lozenges" or "diamonds" (also of side length 1).

0, 0, 24, 126, 387, 915, 1845, 3339, 5586, 8802, 13230, 19140, 26829, 36621, 48867, 63945, 82260, 104244, 130356, 161082, 196935, 238455, 286209, 340791, 402822, 472950, 551850, 640224, 738801, 848337, 969615, 1103445, 1250664, 1412136, 1588752, 1781430, 1991115

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 24 different tiling of the triangle of side length 3 with exactly two 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 (5,-10,10,-5,1).

Crossrefs

Cf. A273464, A326368, A326369.

Programs

Mathematica

Rest@ CoefficientList[Series[3 x^3*(4 - x) (2 + x)/(1 - x)^5, {x, 0, 37}], x] (* Michael De Vlieger, Jul 04 2019 *)

PARI

concat([0,0], Vec(3*x^3*(4 - x)*(2 + x) / (1 - x)^5 + O(x^40))) \\ Colin Barker, Jul 01 2019

Formula

a(n) = (3/8)*(n-2)*(n-1)*(3*n^2 + 3*n - 4) (conjectured by R. J. Mathar, proved by Greg Dresden and E. Sijaric).

From Colin Barker, Jul 01 2019: (Start)

G.f.: 3*x^3*(4 - x)*(2 + x) / (1 - x)^5.

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

(End)

E.g.f.: (3/8)*exp(x)*x^2*(32 + 24*x + 3*x^2). - Stefano Spezia, Jul 01 2019