%I #36 Feb 16 2025 08:33:23

%S 1,1,1,3,3,1,1,6,12,8,1,10,36,55,33,9,1,15,87,248,378,339,187,63,12,1,

%T 1,21,180,820,2190,3606,3716,2340,825,125,1,28,333,2212,9110,24474,

%U 43928,53018,42774,22792,7945,1764,196,1,36,567,5163,30300,121077,339621

%N Number T(n, k) of ways to place k points on an n X n X n triangular grid so that no pair of them has distance sqrt(3). Triangle read by rows.

%C In the following triangular grid points x have Euclidean distance sqrt(3) from point o. It is the second closest distance possible among grid points.

%C x

%C . .

%C . o .

%C x . . x

%C Triangle T(n, k) is irregular: 0 <= k <= max(n), where max(n), the maximal number of points that can be placed on the grid, is:

%C for n = 3j-2: max(n) = A000326(j) = j(3j-1)/2;

%C for n = 3j-1 or n = 3j: max(n) = A045943(j) = 3j(j+1)/2; j = 1,2,3,...

%C Empirical: (1) The number of ways to place the maximal number of points for grid sizes n = 3j are cubes of Catalan numbers, i.e., for n = 3j: T(n, max(n)) = C(j+1)^3 = A033536(j+1). (2) For n = 3j-2: T(n, max(n)) = A244506(n) = A244507^2(n). (3) For n = 3j-1: T(n, max(n)) = A000012(n) = 1 and T(n, max(n)-1) = 3j^2.

%C Row n is also the coefficients of the independence polynomial of the n-triangular honeycomb acute knight graph. - _Eric W. Weisstein_, May 21 2017

%H Heinrich Ludwig, <a href="/A244500/b244500.txt">Table of n, a(n) for n = 1..153</a>

%H Stan Wagon, <a href="http://www.jstor.org/stable/10.4169/college.math.j.45.4.278">Graph Theory Problems from Hexagonal and Traditional Chess</a>, The College Mathematics Journal, Vol. 45, No. 4, September 2014, pp. 278-287.

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/IndependencePolynomial.html">Independence Polynomial</a>

%e On an 8 X 8 X 8 grid there is T(8, 18) = 1 way to place 18 points (x) so that no pair of points has the distance square root of 3.

%e x

%e x x

%e . . .

%e x . . x

%e x x . x x

%e . . . . . .

%e x . . x . . x

%e x x . x x . x x

%e Continuation of this pattern will give the unique maximal solution for all n = 3j-1.

%e Triangle T(n, k) begins:

%e 1, 1;

%e 1, 3, 3, 1;

%e 1, 6, 12, 8;

%e 1, 10, 36, 55, 33, 9;

%e 1, 15, 87, 248, 378, 339, 187, 63, 12, 1;

%e 1, 21, 180, 820, 2190, 3606, 3716, 2340, 825, 125;

%e First row refers to n = 1.

%Y Cf. A000108, A000326, A033536, A045943, A244506, A244507.

%Y Cf. A000217 (column 2), A086274 (1/3 * column 3), A244501 (column 4), A244502 (column 5), A244503 (column 6).

%Y Cf. A287195 (length of row n). - _Eric W. Weisstein_, May 21 2017

%Y Cf. A287204 (row sums). - _Eric W. Weisstein_, May 21 2017

%K nonn,tabf

%O 1,4

%A _Heinrich Ludwig_, Jun 29 2014