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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).
7
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
1,3
LINKS
Richard J. Mathar, Lozenge tilings of the equilateral triangle, arXiv:1909.06336 [math.CO], 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
EXAMPLE
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
MATHEMATICA
Rest@ CoefficientList[Series[3 x^3*(4 - x) (2 + x)/(1 - x)^5, {x, 0, 37}], x] (* Michael De Vlieger, Jul 04 2019 *)
PROG
(PARI) concat([0, 0], Vec(3*x^3*(4 - x)*(2 + x) / (1 - x)^5 + O(x^40))) \\ Colin Barker, Jul 01 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Greg Dresden, Jul 01 2019
STATUS
approved