A242136 Number of strong triangulations of a fixed square with n interior vertices.

0, 1, 6, 36, 228, 1518, 10530, 75516, 556512, 4194801, 32224114, 251565996, 1991331720, 15953808780, 129171585690, 1055640440268, 8698890336576, 72215877581844, 603532770013080, 5074488683389840

Offset: 0

Author

David Callan, Aug 15 2014

Keywords

• nonn

Comments

A strong triangulation is one in which no interior edge joins two vertices of the square (see W. G. Brown reference).

If the restriction "strong" is dropped, the counting sequence is A197271 (shifted left).

Examples

			The 6 triangulations for n=2 are as follows. Four have a central vertex joined to all 4 vertices of the square creating 4 triangular regions, one of which contains the second interior vertex. In these 4 cases, the central vertex has degree 5, the other interior  vertex has degree 3. In the other 2 triangulations, both interior vertices have degree 4, an opposite pair a, c of vertices of the square both have degree 3 (so 1 interior edge), and the other 2 opposite vertices have degree 4.
		

Links

• William G. Brown, Enumeration of Triangulations of the Disk, Proc. Lond. Math. Soc. s3-14, Issue 4, (1964) 746-768.
• William T. Tutte, A census of planar triangulations (Eq. 5.12), Canad. J. Math. 14 (1962), 21-38.

Crossrefs

Column k=1 of A341856.

Cf. A000260 for triangulations of a triangle.

Programs

Maple

A242136:=n->24*binomial(4*n+3,n-1)/((3*n+5)*(n+2)): seq(A242136(n), n=0..30); # Wesley Ivan Hurt, Aug 16 2014

Mathematica

Table[24 Binomial[4n+3,n-1]/((3n+5)(n+2)), {n, 0, 15}]

Formula

a(n) = 72 * (4*n+3)!/((3*n+6)!*(n-1)!) = 24 * binomial(4*n+3,n-1)/((3*n+5)*(n+2)) = binomial(4*n+3,n-1) - 5 * binomial(4*n+3,n-2) + 6 * binomial(4*n+3,n-3).