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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).
7
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
1,3
LINKS
Richard J. Mathar, Lozenge tilings of the equilateral triangle, arXiv:1909.06336 [math.CO], 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)
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 18 different tiling of the triangle of side length 3 with exactly three lozenges:
o
/ \
o o
/ \ / \
o - o o
/ / \ / \
o - o - o - o
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Greg Dresden, Jul 01 2019
STATUS
approved