A000260 -id:A000260 - cp's OEIS frontend

A103942 Number of unrooted n-edge isthmusless maps in the plane (planar with a distinguished outside face).

1, 1, 3, 9, 38, 187, 1120, 7083, 47990, 337676, 2455517, 18310155, 139447034, 1080773098, 8502896424, 67763884363, 546147639926, 4445389286380, 36501274080076, 302060508150976, 2517213486505592, 21110062391001119, 178052027949519768, 1509631210682469661, 12860805940582898474

Offset: 0

Valery A. Liskovets, Mar 17 2005

V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005.

Andrew Howroyd, Table of n, a(n) for n = 0..500
V. A. Liskovets and T. R. Walsh, Counting unrooted maps on the plane, Advances in Applied Math., 36, No.4 (2006), 364-387.

Cf. A027836, A006390, A103941, A000260.

a[n_] := (1/(2n)) ((5n^2 + 13n + 2) Binomial[4n, n]/((n+1)(3n+1)(3n+2)) + Sum[Boole[0 < k < n] EulerPhi[n/k] Binomial[4k, k], {k, Divisors[n]}] + q[n]);
q[n_] := If[EvenQ[n], 0, (n-1) Binomial[2n, (n-1)/2]]/(n+1);
Array[a, 20] (* Jean-François Alcover, Sep 01 2019 *)

a(n) = {if(n==0, 1, (sumdiv(n, d, if(dAndrew Howroyd, Mar 28 2021

For n > 0, a(n) = (1/(2n))*[(5n^2+13n+2)*binomial(4n, n)/((n+1)(3n+1)(3n+2)) + Sum_{0A000010), q(n)=0 if n is even and q(n)=(n-1)*binomial(2n, (n-1)/2)/(n+1) if n is odd.

a(0)=1 prepended and terms a(21) and beyond from Andrew Howroyd, Mar 28 2021

A027432 Related to sorting procedure studied by West: number of permutations that are both sorted (i.e., obtainable as output of the sorting procedure) and one-stack sortable.

Original entry on oeis.org

1, 1, 1, 2, 4, 10, 25, 69, 192, 562, 1663, 5065, 15592, 48874, 154651, 495418, 1599816, 5212650, 17098590, 56473664, 187572584, 626430568, 2101977231, 7084963950, 23976649328, 81447876258, 277627821135, 949393445553, 3256266981128, 11199653726786, 38620292110925

Offset: 0

Mireille Bousquet-Mélou

Series reversion of g.f. A(x) is -A(-x) (if offset 1).

Vaclav Kotesovec, Table of n, a(n) for n = 0..500
M. Bousquet-Mélou, Sorted or sortable permutations, Discrete Math., 225 (2000), 25-50.
J. West, Sorting twice through a stack, Theoret. Comput. Sci. 117 (1993) 303-313.
Index entries for sequences related to sorting

Cf. A027361.

max = 29; f[x_] = Sum[a[n]*x^n, {n, 0, max}]; a[0] = a[1] = 1; y = x*f[x]; coes = CoefficientList[ Series[(x^4-3x^3+3x^2-x) + y*(4x^3+29x^2-7x+1) + y^2*(6x^2-29x+3) + y^3*(4x+3) + y^4, {x, 0, max}], x]; Table[a[n], {n, 0, max-1}] /. First[ Solve[ Thread[coes == 0]]] (* Jean-François Alcover, Nov 14 2011 *)
a[0] = a[1] = a[2] = 1; a[3] = 2; a[n_] := a[n] = (1/(3*(n - 1)*(n + 1)* (3*n + 1)*(3*n + 2)*(n*(24*n - 1) - 90)))*(2*(n*(n*(n*(n*(1680*n^2 - 1294*n - 10977) + 16676) - 3843) - 2602) + 720)*a[n-1] + 4*(n*(n*(n* (768*n^3 - 3872*n^2 + 3560*n + 6195) - 11498) + 6887) - 2520)*a[n-2] - 32*n*(n*(2*n*(16*n*(n*(24*n - 133) + 29) + 16153) - 63331) + 33165)* a[n-3]); Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Apr 20 2018, after Vaclav Kotesovec *)

{a(n) = local(A); if( n<0, 0, n++; A = O(x); for( k=1, n, A = subst( x - 3*(x^2 + y^2) + 7*x*y + 3*(x^3 - y^3) - 29*x*y*(x - y) - (x^4 + y^4) - 4*x*y*(x^2 + y^2) - 6*x^2*y^2, y, A)); polcoeff(A, n))}

G.f. is algebraic of degree 4.
If g.f. is A(x), y = x*A(x) satisfies (x^4 - 3*x^3 + 3*x^2 - x) + y * (4*x^3 + 29*x^2 - 7*x + 1) + y^2 * (6*x^2 - 29*x + 3) + y^3 * (4*x + 3) + y^4 = 0.
G.f. A(x) satisfies A(x) = x + B(x*A(x)) where B(x) is g.f. for A000260 (offset 1). - Michael Somos, Sep 07 2005
Recurrence: 3*(n-1)*(n+1)*(3*n+1)*(3*n+2)*(24*n^2 - n - 90)*a(n) = 2*(1680*n^6 - 1294*n^5 - 10977*n^4 + 16676*n^3 - 3843*n^2 - 2602*n + 720)*a(n-1) + 4*(768*n^6 - 3872*n^5 + 3560*n^4 + 6195*n^3 - 11498*n^2 + 6887*n - 2520)*a(n-2) - 32*n*(2*n-5)*(4*n-11)*(4*n-9)*(24*n^2 + 47*n - 67)*a(n-3). - Vaclav Kotesovec, Mar 18 2014
a(n) ~ c * (1/r)^n / (sqrt(Pi) * n^(5/2)), where r = (2*sqrt(7)-1)/16 = 0.268218913883... and c = sqrt((9653 + 3619*sqrt(7))/1944) = 3.14498539342675985580088726043277778... - Vaclav Kotesovec, Jul 01 2014

A185113 Number of dissections of a convex (3n+3)-sided polygon into n pentagons and one triangle (up to equivalence).

Original entry on oeis.org

1, 3, 18, 130, 1020, 8379, 70840, 610740, 5340060, 47187580, 420412278, 3770221338, 33991902308, 307826695050, 2798052616800, 25514463687720, 233296537299228, 2138295980859588, 19639886707062280, 180724535020583400, 1665767679910654320, 15376467276901980315

Offset: 0

F. Chapoton, Feb 03 2011

nonn

This sequence counts dissections of a convex 3n+3-sided polygon into one triangle and n pentagons, modulo a simple equivalence relation. This equivalence relation is defined by moving the triangle according to a simple rule (not detailed here).
(The equivalence relation is not defined by a group, but by local moves. Consider the hexagon formed by a pentagon adjacent to the triangle. The local move is half-rotation of such hexagons.)
The terms seem to be odd exactly for indices in A002450. - F. Chapoton Mar 08 2020

			For n=0, there is just one triangle, so that a(0)=1. For n=1, one can dissect an hexagon in 6 ways into a pentagon and a triangle. In this case, the equivalence relation just relates every such dissection to its half rotated image, so that a(1)=3.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A001700, A002293, A174687, A069271, A000217, A000260.

f := RootOf(81*x - 8 - (75*x - 8)*Z + (288*x^2 - 30*x)*_Z^2 + (256*x^3 - 27*x^2)*_Z^4): seq(coeff(series(f, x, 20), x, n), n = 0..19); # _Peter Luschny, Apr 06 2023

Table[Binomial[4*n + 1, n]*(n + 2)/(3*n + 2), {n, 0, 50}] (* G. C. Greubel, Jun 23 2017 *)

for(n=0,25, print1(binomial(4*n+1,n)*(n+2)/(3*n+2), ", ")) \\ G. C. Greubel, Jun 23 2017

def A185113(n):
    return binomial(4*n+1, n) * (n+2) / (3*n+2)

a(n) = binomial(4*n+1,n-1)*(n+2)/n = binomial(4*n+1,n)*(n+2)/(3*n+2).

a(n) = binomial(n+2,2) * A000260(n). - F. Chapoton Feb 22 2024

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).

A253882 Number of 3-connected planar triangulations of the sphere with n vertices up to orientation preserving isomorphisms.

Original entry on oeis.org

1, 1, 2, 6, 17, 73, 389, 2274, 14502, 97033, 672781, 4792530, 34911786, 259106122, 1954315346, 14949368524, 115784496932, 906736988527, 7171613842488, 57231089062625, 460428456484557, 3731572377382341, 30447133566946517, 249968326771680542, 2063931874299323140

Offset: 4

Danny Rorabaugh, Feb 27 2015

nonn

Andrew Howroyd, Table of n, a(n) for n = 4..500
CombOS - Combinatorial Object Server, generate planar graphs
Pascal Honvault, Equivalent classes of degree sequences for triangulated polyhedra and their convex realization, Contributions to Disc. Math. (2021) Vol. 16, No. 1, 128-137.
Pascal Honvault, Local geometry of polyhedra, hal-03744217 [math], 2022.
The House of Graphs, Planar graphs

Cf. A000109 (full automorphism group), A000260 (rooted at an edge), A000944, A002709 (with a distinguished face).

a(n)={if(n<3, 0, (2*binomial(4*(n-3)+1, n-3)/((n-2)*(3*n-7))
  + 3*sumdiv(n-2, d, if(d>=2, my(s=(n-2)/d); eulerphi(d)*binomial(4*s,s))/4)
  + if(n%2==1, my(s=(n-3)/2); 3*binomial(4*s,s)*(2*s+1)/(3*s+1))
  + if(n%3==1, my(s=(n-4)/3); 8*binomial(4*s,s)*(4*s+1)/(3*s+1))
  + if(n%3==0, my(s=(n-3)/3); 2*binomial(4*s,s)) )/(6*(n-2)))} \\ Andrew Howroyd, Mar 02 2021

Name clarified and terms a(24) and beyond from Andrew Howroyd, Mar 02 2021

A263191 Triangle read by rows: T(n>=0, 1<=k<=A000108(n)) is the number of Dyck paths of length 2n having k smaller elements in Tamari order.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 0, 1, 1, 3, 2, 2, 2, 0, 2, 0, 1, 0, 0, 0, 0, 1, 1, 4, 3, 5, 4, 2, 4, 0, 5, 2, 0, 2, 0, 3, 0, 1, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 4, 9, 6, 7, 6, 3, 10, 6, 4, 4, 0, 9, 5, 2, 0, 4, 4, 4, 0, 0, 4, 3, 1, 0, 2, 4, 0, 4, 0, 0, 0, 3, 0, 0, 2

Offset: 0

Christian Stump, Oct 19 2015

Row sums give A000108.

			Triangle begins:
1;
1;
1,1;
1,2,1,0,1;
1,3,2,2,2,0,2,0,1,0,0,0,0,1;
1,4,3,5,4,2,4,0,5,2,0,2,0,3,0,1,0,0,2,0,0,0,2,0,0,0,0,1,0,0,0,0,0,0,0, 0,0,0,0,0,0,1;
...

Alois P. Heinz, Rows n = 0..10, flattened
FindStat - Combinatorial Statistic Finder, The number of elements smaller than the given Dyck path in the Tamari Order.
Wikipedia, Tamari lattice.

Cf. A000108, A000260.

Sum_{k=1..A000108(n)} k * T(n,k) = A000260(n). - Alois P. Heinz, Nov 15 2015

Two terms (for rows 0 and 1) prepended by Alois P. Heinz, Nov 15 2015

A268315 Decimal expansion of 256/27.

Original entry on oeis.org

9, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8, 1, 4, 8

Offset: 1

Gheorghe Coserea, Feb 01 2016

			9.481481481481481481481481481481...

Exponential growth rate for A000260, A002293, A005810, A006633, A006634, A052203, A066381, A069271, A078516, A078995, A153396, A196678, A268316.

[9] cat &cat[[4, 8, 1]^^45]; // Vincenzo Librandi, Feb 04 2016

Join[{9}, PadRight[{}, 120, {4, 8, 1}]] (* Vincenzo Librandi, Feb 04 2016 *)

PARI
```
1.0 * 256/27
```

More digits from Jon E. Schoenfield, Mar 15 2018

A255918 Array a(n,m) read by descending antidiagonals giving the number of intervals in a generalized Tamari lattice of m-ballot paths of size n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 13, 1, 1, 10, 58, 68, 1, 1, 15, 170, 703, 399, 1, 1, 21, 395, 3685, 9729, 2530, 1, 1, 28, 791, 13390, 91881, 146916, 16965, 1, 1, 36, 1428, 38591, 524256, 2509584, 2359968, 118668, 1, 1, 45, 2388, 94738, 2180262, 22533126

Offset: 1

Jean-François Alcover, Mar 11 2015

This array occurs in counting the degeneracies in the supersymmetric ground states of the Kronecker model of quiver quantum mechanics. See Cordova and Shao, 1.4. - Peter Bala, Oct 29 2017

In mathematical terms, this corresponds to the homology of some moduli spaces of semi-stable representations of the Kronecker quiver K_m with dimension vector (n,n+1). F. Chapoton, Jun 09 2021

			Array begins:
1,   1,    1,     1,      1,       1,       1,        1,        1, ...
1,   3,    6,    10,     15,      21,      28,       36,       45, ...
1,  13,   58,   170,    395,     791,    1428,     2388,     3765, ...
1,  68,  703,  3685,  13390,   38591,   94738,   206718,   412095, ...
1, 399, 9729, 91881, 524256, 2180262, 7291550, 20787390, 52450587, ...
...
2nd row is A000217 (triangular numbers);
3rd row is A103220;
4th row is not in the OEIS;
2nd column is A000260 (number of intervals in the usual Tamari lattice of size n);
3rd column is not in the OEIS.

Olivier Bernardi and Nicolas Bonichon, Intervals in Catalan lattices and realizers of triangulations, Journal of Combinatorial Theory, Series A 116:1 (2009), pp. 55-75.
M. Bousquet-Mélou, E. Fusy, and L.-F. Préville Ratelle, The number of intervals in the m-Tamari lattices, arXiv:1106.1498 [math.CO], The Electronic Journal of Combinatorics 18, 2 (2011) P31.
Clay Cordova and Shu-Heng Shao, Counting Trees in Supersymmetric Quantum Mechanics arXiv:1502.08050v2 [hep-th], 2015.
Thorsten Weist, Localization in quiver moduli spaces, arXiv:0903.5442 [math.RT], 2009.

Cf. A000217, A000260, A070914 (generalized Catalan numbers giving the number of paths), A103220.

a[n_, m_] := ((m + 1)/(n*(m*n + 1)))*Binomial[(m + 1)^2*n + m, n - 1]; Table[a[n - m, m], {n, 1, 12}, {m, n - 1, 0, -1}] // Flatten

def a(n, m):
    return (m + 1) * binomial((m + 1)**2 * n + m, n - 1) // (n*(m*n + 1)) # F. Chapoton, Mar 24 2021

a(n,m) = ((m + 1)/(n*(m*n + 1)))*binomial((m + 1)^2*n + m, n - 1).

A294084 Number of indecomposable intervals in the Tamari lattices.

Original entry on oeis.org

0, 1, 2, 8, 41, 240, 1528, 10312, 72647, 528992, 3954488, 30201504, 234798627, 1853076528, 14814453896, 119763949936, 977709717091, 8050816106176, 66803956281592, 558146870481760, 4692269111973668, 39669049950811328, 337082395954643168, 2877697636252004168, 24672447821197834553

Offset: 0

F. Chapoton, Feb 26 2018

nonn

This is also the number of interval-posets with connected Hasse diagram.

			Among the 3 interval-posets of size 2 :
1 --> 2 ; 1 <-- 2 ; 1  2,
only the third (which is an antichain) is not a connected poset.

Alois P. Heinz, Table of n, a(n) for n = 0..1031
F. Chapoton, Sur le nombre d'intervalles dans les treillis de Tamari, arXiv:math/0602368 [math.CO], 2006.
F. Chapoton, Sur le nombre d'intervalles dans les treillis de Tamari, Séminaire Lotharingien de combinatoire, vol. 55 (2006).
Alois Panholzer, Parking function varieties for combinatorial tree models, arXiv:2007.14676 [math.CO], 2020.

Cf. A000260.

using Nemo
s(n) = div(Nemo.binom(4*n + 2, n + 1), (2*n + 1) * (3*n + 2))
R, z = PowerSeriesRing(ZZ, 25, "z")
F = sum(s(n)z^n for n in 0:25)
G = 1 - inv(F)
println([coeff(G,n) for n in 0:24]) # Peter Luschny, Feb 26 2018

h:= proc(n) h(n):= 2*(4*n+1)!/((n+1)!*(3*n+2)!) end:
a:= proc(n) a(n):= `if`(n=0, 0, h(n)-add(a(n-i)*h(i), i=1..n-1)) end:
seq(a(n), n=0..25);  # Alois P. Heinz, Feb 28 2019

terms = 25;
G[] = 0; Do[G[x] = 1 + x G[x]^4 + O[x]^terms, terms];
F[x_] = 1 - 1/((2 - G[x]) G[x]^2);
CoefficientList[F[x], x] (* Jean-François Alcover, Feb 14 2019 *)

F = PowerSeriesRing(ZZ,'t')([1] + [(2 * binomial(4 * n + 1, n - 1)) // (n * (n + 1)) for n in range(1, 20)])
1 - F.inverse()

The generating series can be obtained by inverting the generating series of A000260.

A336110 Irregular triangle of Catalan-based numbers, read by rows.

Original entry on oeis.org

1, 1, 2, -2, 5, -14, 5, 14, -74, 74, -14, 42, -352, 668, -352, 42, 132, -1588, 4808, -4808, 1588, -132, 429, -6946, 30371, -48540, 30371, -6946, 429, 1430, -29786, 176270, -407810, 407810, -176270, 29786, -1430

Offset: 1

Sergii Voloshyn, Jul 08 2020

Calculation of the sum over the partitions of r of products of dimensions of two different representations of a symmetric group S_r gives
Sum_{L |- S_r} f(L)*f(l+q^N) = (r+q^N)! * G[N+1] * G[q+1]/(G[N+q+1]) * B_r(1!c_1, ..., r!c_r) where f(L) is the dimension of the symmetric group S_r, G[x] is Barnes function, and B_r() is the complete exponential Bell polynomial.
In the limit N -> infinity the coefficients [are?]
  c_1 = 1/(1+x), c_i = 1/(i*N^(2*(i-1)))*P(i-1), for i >= 2.
Coefficient of x^n in the numerator of P(i) is T(s, i).
This triangle of coefficients was discovered by Borisenko et al. In mathematical physics these coefficients appear as an important ingredient of series that define the free energy of the SU(N) standard lattice model in the large N limit.
They are easily obtained from the g.f.
Some special cases are given by A000108 (first column of the triangle), A138156 (second column of the triangle).
The sum of the numbers in row 2*k+1 is (-1)^k * A000260(k) * 2^(2*k).

			     1;
     1;
     2,     -2;
     5,    -14,      5;
    14,    -74,     74,     -14;
    42,   -352,    668,    -352,     42;
   132,  -1588,   4808,   -4808,   1588,    -132;
   429,  -6946,  30371,  -48540,  30371,   -6946,   429;
  1430, -29786, 176270, -407810, 407810, -176270, 29786, -1430;
  ...

Sergii Voloshyn, Table of n, a(n) for n = 1..120
O. Borisenko, V. Chelnokov, and Sergii Voloshyn, The large N limit of SU(N) integrals in lattice, arXiv:2008.00773 [hep-lat], 2020. See formula (48).
F. Green and S. Samuel, Calculating the large-N phase transition in gauge and matrix models, Nuclear Physics B 194, Issue 1, 11 January 1982, Pages 107-156, See Appendix A.

Cf. A000108, A000309, A138156, A000260.

T[L_, n_] := CatalanNumber[L] Sum[u[L, a]/u[0, a] M[n - 1 - 2*a, L], {a, 0, (n - 1)/2}]-4^L Sum[v[L, a]/v[0, a] M[n - 2 - 2*a, L], {a,0,(n-2)/2}];
M[n_, l_] := Sum[k!/n! BellY[n,k,Table[(-1)^(j-1) j!Binomial[3l+j,j], {j,n}]], {k, 0, n}];
u[k_, n_] := Product[Binomial[2 k + 2 l + 1, 3], {l, 1, n}];
v[k_, n_] := Product[ Binomial[2 k + 2 l + 2, 3], {l, 1, n}];
(* alternate program using coefficients in numerator *)
P[s_] := x CatalanNumber[s] HypergeometricPFQ[{s + 1/2, s + 1, s + 3/2}, {1/2, 3/2}, x^2] - x^2*4^s HypergeometricPFQ[{s + 1, s + 3/2, s + 2}, {3/2, 2}, x^2];
Table[CoefficientList[P[s] // Together // Numerator, x] // Rest, {s, 0, 10}] // Flatten (* amended by Jean-François Alcover, Sep 25 2020 *)
(* another program using coefficients in numerator *)
Needs["Combinatorica`"];
OA[p_,x_]:= (2^p(-(1/(x+1)))^(2p+1))/((2p+1)!(p+1)!) Sum[(-x/(1+x))^(p-r+1)Product[Pochhammer[1+Plus@@Table[3*k[[i]]-1,{i,1,j-1}],3*k[[j]]],{j,1,r}],{r, 1, p},{k,Compositions[p-r,r]+1}]; Table[CoefficientList[OA[s, x] // Together // Numerator, x] //
   Rest, {s, 0, 10}] // Flatten (* Sergii Voloshyn, Sep 03 2021 *)

Coefficients of x^n in the numerator of P(s) = (x * C[s]* 3F2[ s+ 1/2, s+1, s+3/2; 1/2,3/2; x^2] - x^2 * 4^s * 3F2[ s+1,s+3/2, s+2; 3/2, 2; x^2]), where C(s) are Catalan numbers.
or in a more explicit way (only for k >= 1)
T(s, n) = C(s) * U(s, n) - 4^s * V(s, n), where
U(s, n) = Sum_{a=0..(n-1)/2} (u(s, a)/u(0,a)) * M(s, n-1- 2a),
V(s, n) = Sum_{a=0..(n-2)/2} (v(s, a)/v(0,a)) * M(s, n-2- 2a), and
u(s, n) = Product_{L=1..n} binomial(2s+2L+1, 3),
v(s, n) = Product_{L=1..n} binomial(2s+2L+2, 3), and
M(m, n) = Sum_{L=1..n} L!/n! B_{n,l} ( x_1, ..., x_{n-L+1}), and
x_i = (-1)^{1+i} * (3s+i)_i = (-1)^{1+i} * i! * binomial(3s + i, i), where
B_{n,l} (x_1, ..., x_{n-L+1}) is the n-th partial or incomplete exponential Bell polynomial with monomials sorted into graded lexicographic order.
Sum of numbers in the particular row:
Sum_{n=1..2*k+1} T(2*k+1, n) = 2*(4*k+1)!/((k+1)!*(3*k+2)!) *2^(2*k) (odd s);
Sum_{n=1..2*k} T(2*k, n) = 0 (even s).
From Sergii Voloshyn, Oct 22 2020: (Start)
Formulae for particular columns:
T(s, 1) = C(s);
T(s, 2) = C(s)*(3*s+1) - 4^s;
T(s, 3) = C(s)*(binomial(2*s+3,3) + (3*s+1)^2 - binomial(3*s +2,2)) - 4^s*(3*s+1);
T(s, 4) = C(s)*((2*s+1)binomial(2*s+3,3) +(3*s+1)^3 - 2(3*s+1)* binomial(3*s+2,2)+ binomial(3*s+3, 3)) - 4^s*(binomial(3*s+4, 3)/4 + (3*s+1)^2 - binomial(3*s+2,2));
...
T(s, s) = (-1)^(s+1)*C(s). (End)
From Sergii Voloshyn, Mar 17 2021: (Start)
Recursion ( P[0] = x/(1+x) ):
x*(d^3/dx^3)*P[s] = (1/4)*(2*k+2)*(2*k+3)*(2*k+4)*P[s+1]
for P[s_] := x CatalanNumber[s] HypergeometricPFQ[{s + 1/2, s + 1, s + 3/2}, {1/2, 3/2}, x^2] - x^2*4^s HypergeometricPFQ[{s + 1, s + 3/2, s + 2}, {3/2, 2}, x^2].
(End)
Recursion for array ( T(1,1) = 1 ): T(k, p) = (1/(k*(k+1)*(2k+1)))*[p*(p+1)*(p+2)*T(k-1,p+2) - 3*p*(p+1)*(3*k-p-1)*T(k-1,p+1) + 3*p*(3*k-p)*(3*k-p-1)*T(k-1,p) - (3*k-p+1)*(3*k-p)*(3*k-p-1)*T(k-1,p-1)]; p =[1,...,k]. - Sergii Voloshyn, Apr 14 2021
From Sergii Voloshyn, Apr 17 2021: (Start)
G.f.: Sum_{m>=1} x^m Om(k, m) = (1/(1+x)^(3*k+1))*Sum_{n=1..k} x^k * T(k,n);
Om(k, m) = (12^k/((1+k)!*(2k+1)!)) * Product_{L=1..k} binomial(m+2L, 3).
(End)
From Sergii Voloshyn, Apr 25 2021: (Start)
Differential equation for P[k_]:
x*(x^2-1)*(d^3/dx^3)P[s] + (2*k+2)*3*x^2*(d^2/dx^2)P[s] +(2*k+2)*(2*k+1)*3*x (d/dx)P[s] + (2*k+2)(2*k+1)*2*k*P[s] = 0.
Discrete set equation for T(k,n) (n=-1..k-2) at fixed k:
(k-n-1)*(k-n)*(k-n+1)*T(k,n) - (k-n-1)*(k-n)*(8*k+n+5)*T(k,n+1) - (n+1)*(n+2)*(9-n+3)*T(k,n+2) + (n+1)*(n+2)*(n+3)*T(k,n+3) = 0
and
Sum_{m=1..k} (k*(5+7*k) + 12*n*(n-1-k))*T(k,n) = 0. (End)
P[k_]:= (2^k)/((2*k+1)!*(k+1)!)*(-1/(1+x))^{2k+1}[Sum_{r(1)+...+ r(L)=k} (-x/(1+x))^{k-L+1}* 1^{(3*r(1))}*(1+3*r(1)-1)^{(3*r(2))}*... *(1+Sum_{i=1..L-1} 3 r(i) -L+1)^{(3*r(L))}] where Sum_ r_i = k runs over all integer compositions of k, L is a number of parts of this composition and 1^{(3 r(1))} is a rising factorial. - Sergii Voloshyn, Sep 03 2021
Sum_{k} abs(T(n,k)) = A000309(n-1). - Sergii Voloshyn, Nov 20 2024

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