A326369 Number of tilings of an equilateral triangle of side length n with unit triangles (of side length 1) and exactly four unit "lozenges" or "diamonds" (also of side length 1).
0, 0, 0, 762, 12699, 90270, 417435, 1478160, 4354497, 11203269, 25970895, 55414395, 110505120, 208300257, 374375664, 645922095, 1075615380, 1736379630, 2727171042, 4179918384, 6267764745, 9214763640, 13307191065, 18906643602, 26465101179, 36542141595, 49824502425
Offset: 1
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=4, here is one of the 762 different tiling of the triangle of side length 4 with exactly four lozenges:
o
/ \
o - o
/ \ / \
o - o 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 (9,-36,84,-126,126,-84,36,-9,1).
Programs
-
Mathematica
Rest@ CoefficientList[Series[3 x^4*(254 + 1947 x + 1137 x^2 - 613 x^3 + 87 x^4 + 33 x^5 - 10 x^6)/(1 - x)^9, {x, 0, 27}], x] (* Michael De Vlieger, Jul 07 2019 *) -
PARI
concat([0,0,0], Vec(3*x^4*(254 + 1947*x + 1137*x^2 - 613*x^3 + 87*x^4 + 33*x^5 - 10*x^6) / (1 - x)^9 + O(x^40))) \\ Colin Barker, Jul 01 2019
Formula
a(n) = (3/128)*(n-3)*(n-2)*(9*n^6 + 9*n^5 - 135*n^4 - 81*n^3 + 670*n^2 + 104*n - 1216) for n >= 2 (proved by Greg Dresden and Eldin Sijaric).
From Colin Barker, Jul 01 2019: (Start)
G.f.: 3*x^4*(254 + 1947*x + 1137*x^2 - 613*x^3 + 87*x^4 + 33*x^5 - 10*x^6) / (1 - x)^9.
a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n>10.
a(n) = (3/128)*(-7296 + 6704*n + 2284*n^2 - 3732*n^3 + 265*n^4 + 648*n^5 - 126*n^6 - 36*n^7 + 9*n^8) for n>1.
(End)