A273464 The number of tilings of an equilateral triangle of side length n with k lozenges and n^2 - 2*k unit triangles. Triangle T(n, k) with n >= 1 and 0 <= k <= n*(n + 1)/2, read by rows.

1, 1, 3, 1, 9, 24, 18, 1, 18, 126, 434, 762, 630, 187, 1, 30, 387, 2814, 12699, 36894, 69242, 81936, 57672, 21432, 3135, 1, 45, 915, 11127, 90270, 515970, 2139120, 6523428, 14683401, 24256853, 28975770, 24383838, 13860321, 4966929, 989970, 81462, 1, 63

Offset: 1

Author

R. J. Mathar, May 23 2016

Keywords

• tabf
• nonn

Examples

			Triangle T(n,k) (with rows n >= 1 and columns k >= 0) begins as follows:
  1;
  1,  3;
  1,  9,  24,   18;
  1, 18, 126,  434,   762,   630,   187;
  1, 30, 387, 2814, 12699, 36894, 69242, 81936, 57672, 21432, 3135;
  ...
		

Links

• R. J. Mathar, Table of n, a(n) for n = 1..575
• J. A. De Loera, J. Rambau, F. Santos, Further topics, in: Triangulations, vol 25 of Algor. Computat. Math. (2010), 433-511.
• R. J. Mathar, Lozenge tilings of the equilateral triangle, arXiv:1909.06336 [math.CO], 2019.
• Francisco Santos, The Cayley trick and triangulations of products of simplices, arXiv:math/0312069 [math.CO], 2004.
• Francisco Santos, The Cayley trick and triangulations of products of simplices, Cont. Math. 374 (2005), 151-177.
• Wikipedia, Lozenge.

Crossrefs

Cf. A045943 (column k=1), A011555, A011556, A011781, A122722, A326367 (k=2), A326368 (k=3), A326369 (k=4), A000124 (row lengths).

Formula

T(n,2) = 3*(n-1)*(n-2)*(3*n^2+3*n-4)/8 . - R. J. Mathar, May 24 2016

T(n,3) = (n-2)*(9*n^5-9*n^4-81*n^3+81*n^2+160*n-192)/16. - Greg Dresden, Jul 03 2019

Conjecture: T(n,4) = 3*(n-2)*(n-3)*(9*n^6+9*n^5-135*n^4-81*n^3+670*n^2+104*n-1216)/128. - Greg Dresden, Jul 03 2019

Conjecture: T(n,5) = 3*(n-3)*(n+3)* (27*n^8 -135*n^7 -387*n^6 +2835*n^5 -168*n^4 -18732*n^3 +19568*n^2 +36992*n -56320)/1280. - R. J. Mathar, Jul 07 2019

From Petros Hadjicostas, Sep 13 2019: (Start)

Conjecture for rightmost terms: A122722(n) = n! * T(n, n*(n+1)/2) for n >= 1.

Conjectures for column k >= 0: Sum_{0 <= s <= 2*k + 1} (-1)^s * binomial(2*k+1, s) * T(n-s, k) = 0 for n >= 2*k+2.

Sum_{0 <= s <= 2*k} (-1)^s * binomial(2*k, s) * T(n-s, k) = A011781(k) for n >= 2*k+1. (End)