A197271 -id:A197271 - cp's OEIS frontend

A146305 Array T(n,m) = 2(2m+3)!(4n+2m+1)!/(m!(m+2)!n!*(3n+2m+3)!) read by antidiagonals.

1, 1, 2, 3, 5, 5, 13, 20, 21, 14, 68, 100, 105, 84, 42, 399, 570, 595, 504, 330, 132, 2530, 3542, 3675, 3192, 2310, 1287, 429, 16965, 23400, 24150, 21252, 16170, 10296, 5005, 1430, 118668, 161820, 166257, 147420, 115500, 78936, 45045, 19448, 4862, 857956

Offset: 0

R. J. Mathar, Oct 29 2008

T(n,m) is the number of rooted nonseparable (2-connected) triangulations of the disk with n internal nodes and 3 + m nodes on the external face. The triangulation has 2*n + m + 1 triangles and 3*(n+1) + 2*m edges. - Andrew Howroyd, Feb 21 2021

			The array starts at row n=0 and column m=0 as
.....1......2.......5......14.......42.......132
.....1......5......21......84......330......1287
.....3.....20.....105.....504.....2310.....10296
....13....100.....595....3192....16170.....78936
....68....570....3675...21252...115500....602316
...399...3542...24150..147420...844074...4628052
..2530..23400..166257.1057224..6301680..35939904
.16965.161820.1186680.7791168.47948670.282285432

Andrew Howroyd, Table of n, a(n) for n = 0..1325
Alin Bostan, Frédéric Chyzak, and Vincent Pilaud, Refined product formulas for Tamari intervals, arXiv:2303.10986 [math.CO], 2023.
William G. Brown, Enumeration of Triangulations of the Disk, Proc. Lond. Math. Soc. s3-14 (1964) 746-768.

Columns m=0..3 are A000260, A197271(n+1), A341853, A341854.
Rows n=0..2 are A000108(n+1), A002054(n+1) and A000917.
Antidiagonal sums are A000260(n+1).
Cf. A169808 (unrooted), A169809 (achiral), A262586 (oriented).

A146305 := proc(n,m)
    2*(2*m+3)!*(4*n+2*m+1)!/m!/(m+2)!/n!/(3*n+2*m+3)! ;
end proc:
for d from 0 to 13 do
    for m from 0 to d do
        printf("%d,", A146305(d-m,m)) ;
    end do:
end do:

T[n_, m_] := 2*(2*m+3)!*(4*n+2*m+1)!/m!/(m+2)!/n!/(3*n+2*m+3)!; Table[T[n-m, m], {n, 0, 13}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jan 06 2014, after Maple *)

T(n,m)={2*(2*m+3)!*(4*n+2*m+1)!/(m!*(m+2)!*n!*(3*n+2*m+3)!)} \\ Andrew Howroyd, Feb 21 2021

A197272 a(n) = 6/((4n+1)(4n+2))binomial(5*n,n).

Original entry on oeis.org

3, 1, 3, 15, 95, 690, 5481, 46376, 411255, 3781635, 35791910, 346821930, 3427001253, 34425730640, 350732771160, 3617153918640, 37703805776935, 396716804816265, 4209161209968825, 44993046668984145, 484176486362971710

Offset: 0

Peter Bala, Oct 12 2011

A combinatorial interpretation for this sequence in terms of a family of plane trees is given in [Schaeffer, Corollary 2 with k = 5].

A combinatorial interpretation for this sequence in terms of a family of four-dimensional stacked spheres is given in [Thorlieffson, Table 3 in Appendix B]. - Robert A. Russell, Mar 15 2012

G. Schaeffer, A combinatorial interpretation of super-Catalan numbers of order two (2001).
G. Thorlieffson, P. Bialis, B. Petersson, The weak-coupling limit of simplicial quantum gravity, Nuclear Physics B, Volume 550, Issues 1-2, 14 June 1999, Pages 465-491.

Cf. A000139, A197271.

A197272 := proc(n)
    6/((4*n+1)*(4*n+2))*binomial(5*n,n)
end proc:
seq(A197272(n),n=0..40) ; # R. J. Mathar, Mar 29 2023

Table[6/((4n+1)(4n+2)) Binomial[5n,n],{n,0,20}] (* Harvey P. Dale, Aug 08 2013 *)

a(n) = 6/((4*n+1)*(4*n+2))*binomial(5*n,n).

D-finite with recurrence 8*n*(4*n+1)*(2*n+1)*(4*n-1)*a(n) -5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Mar 29 2023

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

Original entry on oeis.org

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

Offset: 0

David Callan, Aug 15 2014

nonn

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).

			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.

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.

Column k=1 of A341856.

Cf. A000260 for triangulations of a triangle.

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

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

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).

A362103 a(n) = K(4,n), where K(M,n) = 2(2M+3)!(4n+2M+1)!/((M+2)!M!n!(3n+2M+3)!).

Original entry on oeis.org

42, 330, 2310, 16170, 115500, 844074, 6301680, 47948670, 370952010, 2911858950, 23150207388, 186127769100, 1511405695800, 12382019142570, 102244420000800, 850316530400304, 7117336900424520, 59922942071869800, 507204902536897950

Offset: 0

N. J. A. Sloane, Apr 13 2023

nonn

K. A. Penson, K. Górska, A. Horzela, and G. H. E. Duchamp, Hausdorff moment problem for combinatorial numbers of Brown and Tutte: exact solution, arXiv:2209.06574 [math.CO], 2022.

Cf. A000260, A197271, A341853, A341854, A362104.

A362104 a(n) = K(5,n), where K(M,n) = 2(2M+3)!(4n+2M+1)!/((M+2)!M!n!(3n+2M+3)!).

Original entry on oeis.org

132, 1287, 10296, 78936, 602316, 4628052, 35939904, 282285432, 2241753228, 17986968714, 145691385840, 1190292863760, 9801223583580, 81284313900636, 678513446252928, 5697583720939968, 48104529963993360, 408179183470170255, 3479470856108521560

Offset: 0

N. J. A. Sloane, Apr 13 2023

nonn

K. A. Penson, K. Górska, A. Horzela, and G. H. E. Duchamp, Hausdorff moment problem for combinatorial numbers of Brown and Tutte: exact solution, arXiv:2209.06574 [math.CO], 2022.

Cf. A000260, A197271, A341853, A341854, A362103.

cp's OEIS Frontend

A146305 Array T(n,m) = 2*(2m+3)!*(4n+2m+1)!/(m!*(m+2)!*n!*(3n+2m+3)!) read by antidiagonals.

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A197272 a(n) = 6/((4*n+1)*(4*n+2))*binomial(5*n,n).

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

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Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A362103 a(n) = K(4,n), where K(M,n) = 2*(2*M+3)!*(4*n+2*M+1)!/((M+2)!*M!*n!*(3*n+2*M+3)!).

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Author

Keywords

Links

Crossrefs

A362104 a(n) = K(5,n), where K(M,n) = 2*(2*M+3)!*(4*n+2*M+1)!/((M+2)!*M!*n!*(3*n+2*M+3)!).

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Author

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Crossrefs

A146305 Array T(n,m) = 2(2m+3)!(4n+2m+1)!/(m!(m+2)!n!*(3n+2m+3)!) read by antidiagonals.

A197272 a(n) = 6/((4n+1)(4n+2))binomial(5*n,n).

A362103 a(n) = K(4,n), where K(M,n) = 2(2M+3)!(4n+2M+1)!/((M+2)!M!n!(3n+2M+3)!).

A362104 a(n) = K(5,n), where K(M,n) = 2(2M+3)!(4n+2M+1)!/((M+2)!M!n!(3n+2M+3)!).