A194480 T(n,k) = number of ways to arrange k indistinguishable points on an n X n X n triangular grid so that no three points are in the same row or diagonal.
1, 0, 3, 0, 3, 6, 0, 1, 15, 10, 0, 0, 17, 45, 15, 0, 0, 6, 105, 105, 21, 0, 0, 0, 114, 410, 210, 28, 0, 0, 0, 39, 879, 1225, 378, 36, 0, 0, 0, 1, 909, 4284, 3066, 630, 45, 0, 0, 0, 0, 337, 8568, 15729, 6762, 990, 55, 0, 0, 0, 0, 15, 8733, 50526, 47565, 13560, 1485, 66, 0, 0, 0
Offset: 1
Examples
Some solutions for n=4, k=4: .....1........0........0........0........0........0........1........1 ....1.0......1.0......0.1......0.1......1.0......1.1......0.1......0.1 ...0.1.0....1.0.1....0.1.0....0.0.1....0.1.1....0.1.0....1.1.0....0.1.0 ..0.0.0.1..0.0.0.1..1.0.1.0..0.1.1.0..0.0.0.1..1.0.0.0..0.0.0.0..0.0.1.0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..290
- Manuel Kauers and Christoph Koutschan, Some D-finite and some Possibly D-finite Sequences in the OEIS, arXiv:2303.02793 [cs.SC], 2023, pp. 31-33.
Formula
Empirical: T(n,k) is a polynomial of degree 2k in n with lead coefficient 1/(2^k*k!) for k <= 5.
T(n,1) = (1/2)*n^2 + (1/2)*n
T(n,2) = (1/8)*n^4 + (1/4)*n^3 - (1/8)*n^2 - (1/4)*n
T(n,3) = (1/48)*n^6 + (1/16)*n^5 - (3/16)*n^4 + (1/48)*n^3 + (1/6)*n^2 - (1/12)*n
T(n,4) = (1/384)*n^8 + (1/96)*n^7 - (5/64)*n^6 + (13/240)*n^5 + (27/128)*n^4 - (23/96)*n^3 - (13/96)*n^2 + (7/40)*n
T(n,5) = (1/3840)*n^10 + (1/768)*n^9 - (7/384)*n^8 + (37/1920)*n^7 + (737/3840)*n^6 - (2347/3840)*n^5 + (101/192)*n^4 + (93/320)*n^3 - (7/10)*n^2 + (3/10)*n
Comments