A025174 -id:A025174 - cp's OEIS frontend

A001764 a(n) = binomial(3n,n)/(2n+1) (enumerates ternary trees and also noncrossing trees).

1, 1, 3, 12, 55, 273, 1428, 7752, 43263, 246675, 1430715, 8414640, 50067108, 300830572, 1822766520, 11124755664, 68328754959, 422030545335, 2619631042665, 16332922290300, 102240109897695, 642312451217745, 4048514844039120, 25594403741131680, 162250238001816900

Offset: 0

N. J. A. Sloane

Smallest number of straight line crossing-free spanning trees on n points in the plane.
Number of dissections of some convex polygon by nonintersecting diagonals into polygons with an odd number of sides and having a total number of 2n+1 edges (sides and diagonals). - Emeric Deutsch, Mar 06 2002
Number of lattice paths of n East steps and 2n North steps from (0,0) to (n,2n) and lying weakly below the line y=2x. - David Callan, Mar 14 2004
With interpolated zeros, this has g.f. 2*sqrt(3)*sin(arcsin(3*sqrt(3)*x/2)/3)/(3*x) and a(n) = C(n+floor(n/2),floor(n/2))*C(floor(n/2),n-floor(n/2))/(n+1). This is the first column of the inverse of the Riordan array (1-x^2,x(1-x^2)) (essentially reversion of y-y^3). - Paul Barry, Feb 02 2005
Number of 12312-avoiding matchings on [2n].
Number of complete ternary trees with n internal nodes, or 3n edges.
Number of rooted plane trees with 2n edges, where every vertex has even outdegree ("even trees").
a(n) is the number of noncrossing partitions of [2n] with all blocks of even size. E.g.: a(2)=3 counts 12-34, 14-23, 1234. - David Callan, Mar 30 2007
Pfaff-Fuss-Catalan sequence C^{m}_n for m=3, see the Graham et al. reference, p. 347. eq. 7.66.
Also 3-Raney sequence, see the Graham et al. reference, p. 346-7.
The number of lattice paths from (0,0) to (2n,0) using an Up-step=(1,1) and a Down-step=(0,-2) and staying above the x-axis. E.g., a(2) = 3; UUUUDD, UUUDUD, UUDUUD. - Charles Moore (chamoore(AT)howard.edu), Jan 09 2008
a(n) is (conjecturally) the number of permutations of [n+1] that avoid the patterns 4-2-3-1 and 4-2-5-1-3 and end with an ascent. For example, a(4)=55 counts all 60 permutations of [5] that end with an ascent except 42315, 52314, 52413, 53412, all of which contain a 4-2-3-1 pattern and 42513. - David Callan, Jul 22 2008
Central terms of pendular triangle A167763. - Philippe Deléham, Nov 12 2009
With B(x,t)=x+t*x^3, the comp. inverse in x about 0 is A(x,t) = Sum_{j>=0} a(j) (-t)^j x^(2j+1). Let U(x,t)=(x-A(x,t))/t. Then DU(x,t)/Dt=dU/dt+U*dU/dx=0 and U(x,0)=x^3, i.e., U is a solution of the inviscid Burgers's, or Hopf, equation. Also U(x,t)=U(x-t*U(x,t),0) and dB(x,t)/dt = U(B(x,t),t) = x^3 = U(x,0). The characteristics for the Hopf equation are x(t) = x(0) + t*U(x(t),t) = x(0) + t*U(x(0),0) = x(0) + t*x(0)^3 = B(x(0),t). These results apply to all the Fuss-Catalan sequences with 3 replaced by n>0 and 2 by n-1 (e.g., A000108 with n=2 and A002293 with n=4), see also A086810, which can be generalized to A133437, for associahedra. - Tom Copeland, Feb 15 2014
Number of intervals (i.e., ordered pairs (x,y) such that x<=y) in the Kreweras lattice (noncrossing partitions ordered by refinement) of size n, see the Bernardi & Bonichon (2009) and Kreweras (1972) references. - Noam Zeilberger, Jun 01 2016
Number of sum-indecomposable (4231,42513)-avoiding permutations. Conjecturally, number of sum-indecomposable (2431,45231)-avoiding permutations. - Alexander Burstein, Oct 19 2017
a(n) is the number of topologically distinct endstates for the game Planted Brussels Sprouts on n vertices, see Ji and Propp link. - Caleb Ji, May 14 2018
Number of complete quadrillages of 2n+2-gons. See Baryshnikov p. 12. See also Nov 10 2014 comments in A134264. - Tom Copeland, Jun 04 2018
a(n) is the number of 2-regular words on the alphabet [n] that avoid the patterns 231 and 221. Equivalently, this is the number of 2-regular tortoise-sortable words on the alphabet [n] (see the Defant and Kravitz link). - Colin Defant, Sep 26 2018
a(n) is the number of Motzkin paths of length 3n with n steps of each type, with the condition that (1, 0) and (1, 1) steps alternate (starting with (1, 0)). - Helmut Prodinger, Apr 08 2019
a(n) is the number of uniquely sorted permutations of length 2n+1 that avoid the patterns 312 and 1342. - Colin Defant, Jun 08 2019
The compositional inverse o.g.f. pair in Copeland's comment above are related to a pair of quantum fields in Balduf's thesis by Theorem 4.2 on p. 92. - Tom Copeland, Dec 13 2019
The sequences of Fuss-Catalan numbers, of which this is the first after the Catalan numbers A000108 (the next is A002293), appear in articles on random matrices and quantum physics. See Banica et al., Collins et al., and Mlotkowski et al. Interpretations of these sequences in terms of the cardinality of specific sets of noncrossing partitions are provided by A134264. - Tom Copeland, Dec 21 2019
Call C(p, [alpha], g) the number of partitions of a cyclically ordered set with p elements, of cyclic type [alpha], and of genus g (the genus g Faa di Bruno coefficients of type [alpha]). This sequence counts the genus 0 partitions (non-crossing, or planar, partitions) of p = 3n into n parts of length 3: a(n) = C(3n, [3^n], 0). For genus 1 see A371250, for genus 2 see A371251. - Robert Coquereaux, Mar 16 2024
a(n) is the total number of down steps before the first up step in all 2_1-Dyck paths of length 3*n for n > 0. A 2_1-Dyck path is a lattice path with steps (1,2), (1,-1) that starts and ends at y = 0 and does not go below the line y = -1. - Sarah Selkirk, May 10 2020
a(n) is the number of pairs (A<=B) of noncrossing partitions of [n]. - Francesca Aicardi, May 28 2022
a(n) is the number of parking functions of size n avoiding the patterns 231 and 321. - Lara Pudwell, Apr 10 2023
Number of rooted polyominoes composed of n square cells of the hyperbolic regular tiling with Schläfli symbol {4,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {4,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
This is instance k = 3 of the family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment in A130564. - _Wolfdieter Lang, Feb 05 2024
The number of Apollonian networks (planar 3-trees) with n+3 vertices with a given base triangle. - Allan Bickle, Feb 20 2024
Number of rooted polyominoes composed of n tetrahedral cells of the hyperbolic regular tiling with Schläfli symbol {3,3,oo}. A rooted polyomino has one external face identified, and chiral pairs are counted as two. a(n) = T(n) in the second Beineke and Pippert link. - Robert A. Russell, Mar 20 2024

			a(2) = 3 because the only dissections with 5 edges are given by a square dissected by any of the two diagonals and the pentagon with no dissecting diagonal.
G.f. = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 + 43263*x^8 + ...

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Jian Zhou, Fat and Thin Emergent Geometries of Hermitian One-Matrix Models, arXiv:1810.03883 [math-ph], 2018.
Index entries for "core" sequences
Index entries for sequences related to trees

Cf. A001762, A001763, A002294 - A002296, A006013, A025174, A063548, A064017, A072247, A072248, A134264, A143603, A258708, A256311, A188687 (binomial transform), A346628 (inverse binomial transform).
A column of triangle A102537.
Bisection of A047749 and A047761.
Row sums of triangles A108410 and A108767.
Second column of triangle A062993.
Mod 3 = A113047.
2D Polyominoes: A005034 (oriented), A005036 (unoriented), A369315 (chiral), A047749 (achiral), A000108 {3,oo}, A002293 {5,oo}.
3D Polyominoes: A007173 (oriented), A027610 (unoriented), A371350 (chiral), A371351 (achiral).
Cf. A130564 (for C(k, n) cases).

List([0..25],n->Binomial(3*n,n)/(2*n+1)); # Muniru A Asiru, Oct 31 2018

a001764 n = a001764_list !! n
a001764_list = 1 : [a258708 (2 * n) n | n <- [1..]]
-- Reinhard Zumkeller, Jun 23 2015

[Binomial(3*n,n)/(2*n+1): n in [0..30]]; // Vincenzo Librandi, Sep 04 2014

A001764 := n->binomial(3*n,n)/(2*n+1): seq(A001764(n), n=0..25);
with(combstruct): BB:=[T,{T=Prod(Z,F),F=Sequence(B),B=Prod(F,Z,F)}, unlabeled]:seq(count(BB,size=i),i=0..22); # Zerinvary Lajos, Apr 22 2007
with(combstruct):BB:=[S, {B = Prod(S,S,Z), S = Sequence(B)}, labelled]: seq(count(BB, size=n)/n!, n=0..21); # Zerinvary Lajos, Apr 25 2008
n:=30:G:=series(RootOf(g = 1+x*g^3, g),x=0,n+1):seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 03 2015
alias(PS=ListTools:-PartialSums): A001764List := proc(m) local A, P, n;
A := [1,1]; P := [1]; for n from 1 to m - 2 do P := PS(PS([op(P), P[-1]]));
A := [op(A), P[-1]] od; A end: A001764List(25); # Peter Luschny, Mar 26 2022

InverseSeries[Series[y-y^3, {y, 0, 24}], x] (* then a(n)=y(2n+1)=ways to place non-crossing diagonals in convex (2n+4)-gon so as to create only quadrilateral tiles *) (* Len Smiley, Apr 08 2000 *)
Table[Binomial[3n,n]/(2n+1),{n,0,25}] (* Harvey P. Dale, Jul 24 2011 *)

{a(n) = if( n<0, 0, (3*n)! / n! / (2*n + 1)!)};

{a(n) = if( n<0, 0, polcoeff( serreverse( x - x^3 + O(x^(2*n + 2))), 2*n + 1))};

{a(n) = my(A); if( n<0, 0, A = 1 + O(x); for( m=1, n, A = 1 + x * A^3); polcoeff(A, n))};

b=vector(22);b[1]=1;for(n=2,22,for(i=1,n-1,for(j=1,n-1,for(k=1,n-1,if((i-1)+(j-1)+(k-1)-(n-2),NULL,b[n]=b[n]+b[i]*b[j]*b[k])))));a(n)=b[n+1]; print1(a(0));for(n=1,21,print1(", ",a(n))) \\ Gerald McGarvey, Oct 08 2008

Vec(1 + serreverse(x / (1+x)^3 + O(x^30))) \\ Gheorghe Coserea, Aug 05 2015

from math import comb
def A001764(n): return comb(3*n,n)//(2*n+1) # Chai Wah Wu, Nov 10 2022

def A001764_list(n) :
    D = [0]*(n+1); D[1] = 1
    R = []; b = false; h = 1
    for i in range(2*n) :
        for k in (1..h) : D[k] += D[k-1]
        if not b : R.append(D[h])
        else : h += 1
        b = not b
    return R
A001764_list(22) # Peter Luschny, May 03 2012

From Karol A. Penson, Nov 08 2001: (Start)
G.f.: (2/sqrt(3*x))*sin((1/3)*arcsin(sqrt(27*x/4))).
E.g.f.: hypergeom([1/3, 2/3], [1, 3/2], 27/4*x).
Integral representation as n-th moment of a positive function on [0, 27/4]: a(n) = Integral_{x=0..27/4} (x^n*((1/12) * 3^(1/2) * 2^(1/3) * (2^(1/3)*(27 + 3 * sqrt(81 - 12*x))^(2/3) - 6 * x^(1/3))/(Pi * x^(2/3)*(27 + 3 * sqrt(81 - 12*x))^(1/3)))), n >= 0. This representation is unique. (End)
G.f. A(x) satisfies A(x) = 1+x*A(x)^3 = 1/(1-x*A(x)^2) [Cyvin (1998)]. - Ralf Stephan, Jun 30 2003
a(n) = n-th coefficient in expansion of power series P(n), where P(0) = 1, P(k+1) = 1/(1 - x*P(k)^2).
G.f. Rev(x/c(x))/x, where c(x) is the g.f. of A000108 (Rev=reversion of). - Paul Barry, Mar 26 2010
From Gary W. Adamson, Jul 07 2011: (Start)
Let M = the production matrix:
  1, 1
  2, 2, 1
  3, 3, 2, 1
  4, 4, 3, 2, 1
  5, 5, 4, 3, 2, 1
  ...
a(n) = upper left term in M^n. Top row terms of M^n = (n+1)-th row of triangle A143603, with top row sums generating A006013: (1, 2, 7, 30, 143, 728, ...). (End)
Recurrence: a(0)=1; a(n) = Sum_{i=0..n-1, j=0..n-1-i} a(i)a(j)a(n-1-i-j) for n >= 1 (counts ternary trees by subtrees of the root). - David Callan, Nov 21 2011
G.f.: 1 + 6*x/(Q(0) - 6*x); Q(k) = 3*x*(3*k + 1)*(3*k + 2) + 2*(2*(k^2) + 5*k +3) - 6*x*(2*(k^2) + 5*k + 3)*(3*k + 4)*(3*k + 5)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 27 2011
D-finite with recurrence: 2*n*(2n+1)*a(n) - 3*(3n-1)*(3n-2)*a(n-1) = 0. - R. J. Mathar, Dec 14 2011
REVERT transform of A115140. BINOMIAL transform is A188687. SUMADJ transform of A188678. HANKEL transform is A051255. INVERT transform of A023053. INVERT transform is A098746. - Michael Somos, Apr 07 2012
(n + 1) * a(n) = A174687(n).
G.f.: F([2/3,4/3], [3/2], 27/4*x) / F([2/3,1/3], [1/2], (27/4)*x) where F() is the hypergeometric function. - Joerg Arndt, Sep 01 2012
a(n) = binomial(3*n+1, n)/(3*n+1) = A062993(n+1,1). - Robert FERREOL, Apr 03 2015
a(n) = A258708(2*n,n) for n > 0. - Reinhard Zumkeller, Jun 23 2015
0 = a(n)*(-3188646*a(n+2) + 20312856*a(n+3) - 11379609*a(n+4) + 1437501*a(n+5)) + a(n+1)*(177147*a(n+2) - 2247831*a(n+3) + 1638648*a(n+4) - 238604*a(n+5)) + a(n+2)*(243*a(n+2) + 31497*a(n+3) - 43732*a(n+4) + 8288*a(n+5)) for all integer n. - Michael Somos, Jun 03 2016
a(n) ~ 3^(3*n + 1/2)/(sqrt(Pi)*4^(n+1)*n^(3/2)). - Ilya Gutkovskiy, Nov 21 2016
Given g.f. A(x), then A(1/8) = -1 + sqrt(5), A(2/27) = (-1 + sqrt(3))*3/2, A(4/27) = 3/2, A(3/64) = -2 + 2*sqrt(7/3), A(5/64) = (-1 + sqrt(5))*2/sqrt(5), etc. A(n^2/(n+1)^3) = (n+1)/n if n > 1. - Michael Somos, Jul 17 2018
From Peter Bala, Sep 14 2021: (Start)
A(x) = exp( Sum_{n >= 1} (1/3)*binomial(3*n,n)*x^n/n ).
The sequence defined by b(n) := [x^n] A(x)^n = A224274(n) for n >= 1 and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. Cf. A060941. (End)
G.f.: 1/sqrt(B(x)+(1-6*x)/(9*B(x))+1/3), with B(x):=((27*x^2-18*x+2)/54-(x*sqrt((-(4-27*x))*x))/(2*3^(3/2)))^(1/3). - Vladimir Kruchinin, Sep 28 2021
x*A'(x)/A(x) = (A(x) - 1)/(- 2*A(x) + 3) = x + 5*x^2 + 28*x^3 + 165*x^4 + ... is the o.g.f. of A025174. Cf. A002293 - A002296. - Peter Bala, Feb 04 2022
a(n) = hypergeom([1 - n, -2*n], [2], 1). Row sums of A108767. - Peter Bala, Aug 30 2023
G.f.: z*exp(3*z*hypergeom([1, 1, 4/3, 5/3], [3/2, 2, 2], (27*z)/4)) + 1.
 - Karol A. Penson, Dec 19 2023
G.f.: hypergeometric([1/3, 2/3], [3/2], (3^3/2^2)*x). See the e.g.f. above. - Wolfdieter Lang, Feb 04 2024
a(n) = (3*n)! / (n!*(2*n+1)!). - Allan Bickle, Feb 20 2024
Sum_{n >= 0} a(n)*x^n/(1 + x)^(3*n+1) = 1. See A316371 and A346627. - Peter Bala, Jun 02 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^5). - Seiichi Manyama, Jun 16 2025

A002293 Number of dissections of a polygon: binomial(4n, n)/(3n + 1).

Original entry on oeis.org

1, 1, 4, 22, 140, 969, 7084, 53820, 420732, 3362260, 27343888, 225568798, 1882933364, 15875338990, 134993766600, 1156393243320, 9969937491420, 86445222719724, 753310723010608, 6594154339031800, 57956002331347120, 511238042454541545

Offset: 0

N. J. A. Sloane

The number of rooted loopless n-edge maps in the plane (planar with a distinguished outside face). - Valery A. Liskovets, Mar 17 2005
Number of lattice paths from (1,0) to (3*n+1,n) which, starting from (1,0), only utilize the steps +(1,0) and +(0,1) and additionally, the paths lie completely below the line y = (1/3)*x (i.e., if (a,b) is in the path, then b < a/3). - Joseph Cooper (jecooper(AT)mit.edu), Feb 07 2006
Number of length-n restricted growth strings (RGS) [s(0), s(1), ..., s(n-1)] where s(0) = 0 and s(k) <= s(k-1) + 3, see fxtbook link below. - Joerg Arndt, Apr 08 2011
From Wolfdieter Lang, Sep 14 2007: (Start)
a(n), n >= 1, enumerates quartic trees (rooted, ordered, incomplete) with n vertices (including the root).
Pfaff-Fuss-Catalan sequence C^{m}_n for m = 4. See the Graham et al. reference, p. 347. eq. 7.66. (Second edition, p. 361, eq. 7.67.) See also the Pólya-Szegő reference.
Also 4-Raney sequence. See the Graham et al. reference, pp. 346-347.
(End)
Bacher: "We describe the statistics of checkerboard triangulations obtained by coloring black every other triangle in triangulations of convex polygons." The current sequence (A002293) occurs on p. 12 as one of two "extremal sequences" of an array of coefficients of polynomials, whose generating functions are given in terms of hypergeometric functions. - Jonathan Vos Post, Oct 05 2007
A generating function in terms of a (labyrinthine) solution to a depressed quartic equation is given in the Copeland link for signed A005810. With D(z,t) that g.f., a g.f. for signed A002293 is {[-1+1/D(z,t)]/(4t)}^(1/3). - Tom Copeland, Oct 10 2012
For a relation to the inviscid Burgers's equation, see A001764. - Tom Copeland, Feb 15 2014
For relations to compositional inversion, the Legendre transform, and convex geometry, see the Copeland, the Schuetz and Whieldon, and the Gross (p. 58) links. - Tom Copeland, Feb 21 2017 (See also Gross et al. in A062994. - Tom Copeland, Dec 24 2019)
This is the number of A'Campo bicolored forests of degree n and co-dimension 0. This can be shown using generating functions or a combinatorial approach. See Combe and Jugé link below. - Noemie Combe, Feb 28 2017
Conjecturally, a(n) is the number of 3-uniform words over the alphabet [n] that avoid the patterns 231 and 221 (see the Defant and Kravitz link). - Colin Defant, Sep 26 2018
The compositional inverse o.g.f. pair in Copeland's comment above are related to a pair of quantum fields in Balduf's thesis by Theorem 4.2 on p. 92. Cf. A001764. - Tom Copeland, Dec 13 2019
a(n) is the total number of down steps before the first up step in all 3_1-Dyck paths of length 4*n. A 3_1-Dyck path is a lattice path with steps (1, 3), (1, -1) that starts and ends at y = 0 and stays above the line y = -1. - Sarah Selkirk, May 10 2020
a(n) is the number of pairs (A<=B) of noncrossing partitions of [2n] such that every block of A has exactly two elements. In fact, it is proved that a(n) is the number of planar tied arc diagrams with n arcs (see Aicardi link below). A planar diagram with n arcs represents a noncrossing partition A of [2n] with n blocks, each block containing the endpoints of one arc; each tie connects two arcs, so that the ties define a partition B >= A: the endpoints of two arcs connected by a tie belong to the same block of B. Ties do not cross arcs nor other ties iff B has a planar diagram, i.e., B is a noncrossing partition. - Francesca Aicardi, Nov 07 2022
Dropping the initial 1 (starting 1, 4, 22 with offset 1) yields the REVERT transformation 1, -4 ,10, -20, 35.. essentially A000292 without leading 0. - R. J. Mathar, Aug 17 2023
Number of rooted polyominoes composed of n pentagonal cells of the hyperbolic regular tiling with Schläfli symbol {5,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {5,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
This is instance k = 4 of the generalized Catalan family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment of A130564. - _Wolfdieter Lang, Feb 05 2024
a(n) is the cardinality of the planar ramified Jones monoid PR(J_n). - Diego Arcis, Nov 21 2024

			There are a(2) = 4 quartic trees (vertex degree <= 4 and 4 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these four trees yields 4*4 + 6 = 22 = a(3) such trees.

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Column k=3 of triangle A062993 and A070914.
Cf. A000260, A002295, A002296, A027836, A062994, A346646 (binomial transform), A346664 (inverse binomial transform).
Cf. A006632, A006633, A006634, A025174, A069271, A196678, A224274, A233658, A233666, A233667, A277877, A283049, A283101, A283102, A283103.
Polyominoes: A005038 (oriented), A005040 (unoriented), A369471 (chiral), A369472 (achiral), A001764 {4,oo}, A002294 {6,oo}.
Cf. A130564 (for generalized Catalan C(k, n), for = 4).

List([0..22],n->Binomial(4*n,n)/(3*n+1)); # Muniru A Asiru, Nov 01 2018

[ Binomial(4*n,n)/(3*n+1): n in [0..50]]; // Vincenzo Librandi, Apr 19 2011

series(RootOf(g = 1+x*g^4, g),x=0,20); # Mark van Hoeij, Nov 10 2011
seq(binomial(4*n, n)/(3*n+1),n=0..20); # Robert FERREOL, Apr 02 2015
# Using the integral representation above:
Digits:=6;
R:=proc(x)((I + sqrt(3))*(4*sqrt(256 - 27*x) - 12*I*sqrt(3)*sqrt(x))^(1/3))/16 - ((I - sqrt(3))*(4*sqrt(256 - 27*x) + 12*I*sqrt(3)*sqrt(x))^(1/3))/16;end;
W:=proc(x) x^(-3/4) * sqrt(4*R(x) - 3^(3/4)*x^(1/4)/sqrt(R(x)))/(2*3^(1/4)*Pi);end;
# Attention: W(x) is singular at x = 0. Integration is done from  a very small positive x to x = 256/27.
# For a(8):  ... gives 420732
evalf(int(x^8*W(x),x=0.000001..256/27));
# Karol A. Penson, Jul 05 2024

CoefficientList[InverseSeries[ Series[ y - y^4, {y, 0, 60}], x], x][[Range[2, 60, 3]]]
Table[Binomial[4n,n]/(3n+1),{n,0,25}] (* Harvey P. Dale, Apr 18 2011 *)
CoefficientList[1 + InverseSeries[Series[x/(1 + x)^4, {x, 0, 60}]], x] (* Gheorghe Coserea, Aug 12 2015 *)
terms = 22; A[] = 0; Do[A[x] = 1 + x*A[x]^4 + O[x]^terms, terms];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 13 2018 *)

a(n)=binomial(4*n,n)/(3*n+1) /* Charles R Greathouse IV, Jun 16 2011 */

my(x='x+O('x^33)); Vec(1 + serreverse(x/(1+x)^4)) \\ Gheorghe Coserea, Aug 12 2015

A002293_list, x = [1], 1
for n in range(100):
    x = x*4*(4*n+3)*(4*n+2)*(4*n+1)//((3*n+2)*(3*n+3)*(3*n+4))
    A002293_list.append(x) # Chai Wah Wu, Feb 19 2016

O.g.f. satisfies: A(x) = 1 + x*A(x)^4 = 1/(1 - x*A(x)^3).
a(n) = binomial(4*n,n-1)/n, n >= 1, a(0) = 1. From the Lagrange series of the o.g.f. A(x) with its above given implicit equation.
From Karol A. Penson, Apr 02 2010: (Start)
Integral representation as n-th Hausdorff power moment of a positive function on the interval [0, 256/27]:
a(n) = Integral_{x=0..256/27}(x^n((3/256) * sqrt(2) * sqrt(3) * ((2/27) * 3^(3/4) * 27^(1/4) * 256^(/4) * hypergeom([-1/12, 1/4, 7/12], [1/2, 3/4], (27/256)*x)/(sqrt(Pi) * x^(3/4)) - (2/27) * sqrt(2) * sqrt(27) * sqrt(256) * hypergeom([1/6, 1/2, 5/6], [3/4, 5/4], (27/256)*x)/ (sqrt(Pi) * sqrt(x)) - (1/81) * 3^(1/4) * 27^(3/4) * 256^(1/4) * hypergeom([5/12, 3/4, 13/12], [5/4, 3/2], (27/256)*x/(sqrt(Pi)*x^(1/4)))/sqrt(Pi))).
This representation is unique as it represents the solution of the Hausdorff moment problem.
O.g.f.: hypergeom([1/4, 1/2, 3/4], [2/3, 4/3], (256/27)*x);
E.g.f.: hypergeom([1/4, 1/2, 3/4], [2/3, 1, 4/3], (256/27)*x). (End)
a(n) = upper left term in M^n, M = the production matrix:
  1, 1
  3, 3, 1
  6, 6, 3, 1
  ...
(where 1, 3, 6, 10, ...) is the triangular series. - Gary W. Adamson, Jul 08 2011
O.g.f. satisfies g = 1+x*g^4. If h is the series reversion of x*g, so h(x*g)=x, then (x-h(x))/x^2 is the o.g.f. of A006013. - Mark van Hoeij, Nov 10 2011
a(n) = binomial(4*n+1, n)/(4*n+1) = A062993(n+2,2). - Robert FERREOL, Apr 02 2015
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-1-i} Sum_{k=0..n-1-i-j} a(i)*a(j)*a(k)*a(n-1-i-j-k) for n>=1; and a(0) = 1. - Robert FERREOL, Apr 02 2015
a(n) ~ 2^(8*n+1/2) / (sqrt(Pi) * n^(3/2) * 3^(3*n+3/2)). - Vaclav Kotesovec, Jun 03 2015
From Peter Bala, Oct 16 2015: (Start)
A(x)^2 is o.g.f. for A069271; A(x)^3 is o.g.f. for A006632;
A(x)^5 is o.g.f. for A196678; A(x)^6 is o.g.f. for A006633;
A(x)^7 is o.g.f. for A233658; A(x)^8 is o.g.f. for A233666;
A(x)^9 is o.g.f. for A006634; A(x)^10 is o.g.f. for A233667. (End)
D-finite with recurrence: a(n+1) = a(n)*4*(4*n + 3)*(4*n + 2)*(4*n + 1)/((3*n + 2)*(3*n + 3)*(3*n + 4)). - Chai Wah Wu, Feb 19 2016
E.g.f.: F([1/4, 1/2, 3/4], [2/3, 1, 4/3], 256*x/27), where F is the generalized hypergeometric function. - Stefano Spezia, Dec 27 2019
x*A'(x)/A(x) = (A(x) - 1)/(- 3*A(x) + 4) = x + 7*x^2 + 55*x^3 + 455*x^4 + ... is the o.g.f. of A224274. Cf. A001764 and A002294 - A002296. - Peter Bala, Feb 04 2022
a(n) = hypergeom([1 - n, -3*n], [2], 1). Row sums of A173020. - Peter Bala, Aug 31 2023
G.f.: t*exp(4*t*hypergeom([1, 1, 5/4, 3/2, 7/4], [4/3, 5/3, 2, 2], (256*t)/27))+1. - Karol A. Penson, Dec 20 2023
From Karol A. Penson, Jul 03 2024: (Start)
a(n) = Integral_{x=0..256/27} x^(n)*W(x)dx, n>=0, where W(x) = x^(-3/4) * sqrt(4*R(x) - 3^(3/4)*x^(1/4)/sqrt(R(x)))/(2*3^(1/4)*Pi), with R(x) = ((i + sqrt(3))*(4*sqrt(256 - 27*x) -12*i*sqrt(3*x))^(1/3))/16 - ((i - sqrt(3))*(4*sqrt(256 - 27*x) + 12*i* sqrt(3*x))^(1/3))/16, where i is the imaginary unit.
The elementary function W(x) is positive on the interval x = (0, 256/27) and is equal to the combination of hypergeometric functions in my formula from 2010; see above.
(Pi*W(x))^6 satisfies an algebraic equation of order 6, with integer polynomials as coefficients. (End)
G.f.: (Sum_{n >= 0} binomial(4*n+1, n)*x^n) / (Sum_{n >= 0} binomial(4*n, n)*x^n). - Peter Bala, Dec 14 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^7). - Seiichi Manyama, Jun 16 2025

A045721 a(n) = binomial(3*n+1,n).

Original entry on oeis.org

1, 4, 21, 120, 715, 4368, 27132, 170544, 1081575, 6906900, 44352165, 286097760, 1852482996, 12033222880, 78378960360, 511738760544, 3348108992991, 21945588357420, 144079707346575, 947309492837400, 6236646703759395

Offset: 0

Emeric Deutsch

Number of leaves in all noncrossing rooted trees on n nodes on a circle.
Number of standard tableaux of shape (n-1,1^(2n-3)). - Emeric Deutsch, May 25 2004
a(n) = number of Dyck (2n-3)-paths with exactly one descent of odd length. For example, a(3) counts all 5 Dyck 3-paths except UDUDUD. - David Callan, Jul 25 2005
a(n+2) gives row sums of A119301. - Paul Barry, May 13 2006
a(n) is the number of paths avoiding UU from (0,0) to (3n,n) and taking steps from {U,H}. - Shanzhen Gao, Apr 15 2010
Central coefficients of triangle A078812. - Vladimir Kruchinin, May 10 2012
Row sums of A252501. - L. Edson Jeffery, Dec 18 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Milan Janjic, Two Enumerative Functions
D. Kruchinin and V. Kruchinin, A Method for Obtaining Generating Function for Central Coefficients of Triangles, Journal of Integer Sequence, Vol. 15 (2012), article 12.9.3.
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
Index entries for sequences related to rooted trees

Cf. A252501, A263134 (partial sums), A001764, A004319, A005809, A006013, A013698, A025174, A117671, A165817, A236194.

[Binomial(3*n+1, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

[seq( binomial(3*n+1,n),n=0..40)]; # N. J. A. Sloane, Jun 09 2007

Table[Binomial[3 n + 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

a(n)=binomial(3*n+1,n) \\ Charles R Greathouse IV, Mar 18 2014

a(n) is asymptotic to c/sqrt(n)*(27/4)^n with c=0.73... - Benoit Cloitre, Jan 27 2003
G.f.: gz^2/(1-3zg^2), where g=g(z) is given by g=1+zg^3, g(0)=1, i.e. (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n+2) = C(3n+1,n) = Sum_{k=0..n} C(3n-k,n-k). - Paul Barry, May 13 2006
a(n+2) = C(3n+1,2n+1) = A078812(2n,n). - Paul Barry, Nov 09 2006
G.f.: A(x)=(2*cos(asin((3^(3/2)*sqrt(x))/2)/3)* sin(asin((3^(3/2)* sqrt(x))/2)/3))/(sqrt(3)*sqrt(1-(27*x)/4)*sqrt(x)). - Vladimir Kruchinin, Jun 10 2012
From Peter Luschny, Sep 04 2012: (Start)
O.g.f.: hypergeometric2F1([2/3, 4/3], [3/2], x*27/4).
a(n) = (n+1)*hypergeometric2F1([-2*n, -n], [2], 1).  (End)
D-finite with recurrence 2*n*(2*n+1)*a(n) - 3*(3*n-1)*(3*n+1)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
a(n) = Sum_{r=0..n} C(n,r) * C(2*n+1,r). - J. M. Bergot, Mar 18 2014
From Peter Bala, Nov 04 2015: (Start)
a(n) = Sum_{k = 0..n} 1/(2*k + 1)*binomial(3*n - 3*k,n - k)*binomial(3*k, k).
O.g.f. equals f(x)*g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*(n+1)). - Ilya Gutkovskiy, Oct 10 2017
O.g.f.: (i/24)*((4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(2/3) - (4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(2/3))*sqrt(3)*8^(1/3)*sqrt(4 - 27*z)/(sqrt(z)*(-4 + 27*z)), where i = sqrt(-1). - Karol A. Penson, Dec 13 2023
a(n-1) = (1/(4*n))*binomial(2*n, n)^2 * (1 - 3*((n - 1)/(n + 1))^3 + 5*((n - 1)*(n - 2)/((n + 1)*(n + 2)))^3 - 7*((n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)))^3 + - ...) for n >= 1. Cf. A112029. - Peter Bala, Aug 08 2024

Simpler definition from Ira M. Gessel, May 26 2007. This change means that most of the offsets in the comments will now need to be changed too.

A092392 Triangle read by rows: T(n,k) = C(2*n - k,n), 0 <= k <= n.

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 20, 10, 4, 1, 70, 35, 15, 5, 1, 252, 126, 56, 21, 6, 1, 924, 462, 210, 84, 28, 7, 1, 3432, 1716, 792, 330, 120, 36, 8, 1, 12870, 6435, 3003, 1287, 495, 165, 45, 9, 1, 48620, 24310, 11440, 5005, 2002, 715, 220, 55, 10, 1, 184756, 92378, 43758, 19448, 8008, 3003, 1001, 286, 66, 11, 1

Offset: 0

Ralf Stephan, Mar 21 2004

First column is C(2*n,n) or A000984. Central coefficients are C(3*n,n) or A005809. - Paul Barry, Oct 14 2009
T(n,k) = A046899(n,n-k), k = 0..n-1. - Reinhard Zumkeller, Jul 27 2012
From  Peter Bala, Nov 03 2015: (Start)
Viewed as the square array [binomial (2*n + k, n + k)]n,k>=0  this is the generalized Riordan array ( 1/sqrt(1 - 4*x),c(x) ) in the sense of the Bala link, where c(x) is the o.g.f. for A000108.
The square array factorizes as ( 1/(2 - c(x)),x*c(x) ) * ( 1/(1 - x),1/(1 - x) ), which equals the matrix product of A100100 with the square Pascal matrix [binomial (n + k,k)]n,k>=0. See the example below. (End)

			From _Paul Barry_, Oct 14 2009: (Start)
Triangle begins
  1,
  2, 1,
  6, 3, 1,
  20, 10, 4, 1,
  70, 35, 15, 5, 1,
  252, 126, 56, 21, 6, 1,
  924, 462, 210, 84, 28, 7, 1,
  3432, 1716, 792, 330, 120, 36, 8, 1
Production array is
  2, 1,
  2, 1, 1,
  2, 1, 1, 1,
  2, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1, 1,
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1 (End)
As a square array = A100100 * square Pascal matrix:
  /1   1  1  1 ...\   / 1          \/1 1  1  1 ...\
  |2   3  4  5 ...|   | 1 1        ||1 2  3  4 ...|
  |6  10 15 21 ...| = | 3 2 1      ||1 3  6 10 ...|
  |20 35 56 84 ...|   |10 6 3 1    ||1 4 10 20 ...|
  |70 ...         |   |35 ...      ||1 ...        |
- _Peter Bala_, Nov 03 2015

Reinhard Zumkeller, Rows n=0..149 of triangle, flattened
P. Bala, Notes on generalized Riordan arrays
P. Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), #13.5.1.
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Ik-Pyo Kim, Michael J. Tsatsomeros, Inverse Relations in Shapiro's Open Questions, arXiv:1707.06590 [math.CO], 2017. See p. 7.

Rows 0-14 are A000984, A001700, A001791, A002054, A002694, A003516, A002696, A030053, A004310, A030054, A004311, A030055, A004312, A030056, A004313.
Columns are A000217, A000292, A000332, A000389, A000579.
Diagonals are A005809, A045721, A025174, A004319, A013698, A003408.
Cf. A100100.

a092392 n k = a092392_tabl !! (n-1) !! (k-1)
a092392_row n = a092392_tabl !! (n-1)
a092392_tabl = map reverse a046899_tabl
-- Reinhard Zumkeller, Jul 27 2012

/* As a triangle */ [[Binomial(2*n-k, n): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 22 2017

A092392 := proc(n,k)
    binomial(2*n-k,n-k) ;
end proc:
seq(seq(A092392(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Table[Binomial[2 n - k, n], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Mar 19 2016 *)

C(x):=(1-sqrt(1-4*x))/2;
A(x,y):=(1/sqrt(1-4*x))/(1-y*C(x));
taylor(A(x,y),y,0,10,x,0,10); /* Vladimir Kruchinin, Mar 19 2016 */

for(n=0,10, for(k=0,n, print1(binomial(2*n - k,n), ", "))) \\ G. C. Greubel, Nov 22 2017

As a number triangle, this is T(n, k) = if(k <= n, C(2*n - k, n), 0). Its row sums are C(2*n + 1, n + 1) = A001700. Its diagonal sums are A176287. - Paul Barry, Apr 23 2005
G.f. of column k: 2^k/[sqrt(1 - 4*x)*(1 + sqrt(1 - 4*x))^k].
As a number triangle, this is the Riordan array (1/sqrt(1 - 4*x), x*c(x)), c(x) the g.f. of A000108. - Paul Barry, Jun 24 2005
G.f.: A(x,y)=1/sqrt(1 - 4*x)/(1-y*x*C(x)), where C(x) is g.f. of Catalan numbers. - Vladimir Kruchinin, Mar 19 2016

Diagonal sums comment corrected by Paul Barry, Apr 14 2010

Offset corrected by R. J. Mathar, Feb 08 2013

A165817 Number of compositions (= ordered integer partitions) of n into 2n parts.

Original entry on oeis.org

1, 2, 10, 56, 330, 2002, 12376, 77520, 490314, 3124550, 20030010, 129024480, 834451800, 5414950296, 35240152720, 229911617056, 1503232609098, 9847379391150, 64617565719070, 424655979547800, 2794563003870330, 18412956934908690, 121455445321173600

Offset: 0

Thomas Wieder, Sep 29 2009

Number of ways to put n indistinguishable balls into 2*n distinguishable boxes.

Number of rankings of n unlabeled elements for 2*n levels.

			Let [1,1,1], [1,2] and [3] be the integer partitions of n=3.
Then [0,0,0,1,1,1], [0,0,0,0,1,2] and [0,0,0,0,0,3] are the corresponding partitions occupying 2*n = 6 positions.
We have to take into account the multiplicities of the parts including the multiplicities of the zeros.
Then
  [0,0,0,1,1,1] --> 6!/(3!*3!) = 20
  [0,0,0,0,1,2] --> 6!/(4!*1!*1!) = 30
  [0,0,0,0,0,3] --> 6!/(5!*1!) = 6
and thus a(3) = 20+30+6=56.
a(2)=10, since we have 10 ordered partitions of n=2 where the parts are distributed over 2*n=4 boxes:
  [0, 0, 0, 2]
  [0, 0, 1, 1]
  [0, 0, 2, 0]
  [0, 1, 0, 1]
  [0, 1, 1, 0]
  [0, 2, 0, 0]
  [1, 0, 0, 1]
  [1, 0, 1, 0]
  [1, 1, 0, 0]
  [2, 0, 0, 0].

Alois P. Heinz, Table of n, a(n) for n = 0..400
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.

Cf. A000079, A001700, A059481, A081204, A001764, A004319, A006013, A005809, A013698, A025174, A045721, A117671, A236194.

[Binomial(3*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014

for n from 0 to 16 do
a[n] := 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3))
end do;

Table[Binomial[3 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

vector(30, n, n--; binomial(3*n-1, n)) \\ Altug Alkan, Nov 04 2015

from math import comb
def A165817(n): return comb(3*n-1,n) if n else 1 # Chai Wah Wu, Oct 11 2023

def A165817(n):
    return rising_factorial(2*n,n)/falling_factorial(n,n)
[A165817(n) for n in (0..22)]   # Peter Luschny, Nov 21 2012

a(n) = 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3)).
a(n) = binomial(3*n-1, n);
Let denote P(n) = the number of integer partitions of n,
p(i) = the number of parts of the i-th partition of n,
d(i) = the number of different parts of the i-th partition of n,
m(i,j) = multiplicity of the j-th part of the i-th partition of n.
Then one has:
a(n) = Sum_{i=1..P(n)} (2*n)!/((2*n-p(i))!*(Prod_{j=1..d(i)} m(i,j)!)).
a(n) = rf(2*n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
G.f.: A(x) = x*B'(x)/B(x), where B(x) satisfies B(x)^3-2*B(x)^2+B(x)=x, B(x)=A006013(x). - Vladimir Kruchinin, Feb 06 2013
G.f.: A(x) = sqrt(3*x)*cot(asin((3^(3/2)*sqrt(x))/2)/3)/(sqrt(4-27*x)). - Vladimir Kruchinin, Mar 18 2015
a(n) = Sum_{k=0..n} binomial(n-1,n-k)*binomial(2*n,k). - Vladimir Kruchinin, Oct 06 2015
From Peter Bala, Nov 04 2015: (Start)
The o.g.f. equals f(x)/g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5) and A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*n). - Ilya Gutkovskiy, Oct 03 2017
a(n) = A059481(2n,n). - Alois P. Heinz, Oct 17 2022
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-2*n, n).
a(n) = hypergeom([1 - 2*n, -n], [1], 1).
The g.f. A(x) satisfies A(x/(1 + x)^3) = 1/(1 - 2*x).
Sum_{n >= 0} a(n)/9^n = (1 + 4*cos(Pi/9))/3.
Sum_{n >= 0} a(n)/27^n = (3 + 4*sqrt(3)*cos(Pi/18))/9.
Sum_{n >= 0} a(n)*(2/27)^n = (2 + sqrt(3))/3. (End)
From Peter Bala, Sep 16 2024: (Start)
a(n) = Sum_{k = 0..n} binomial(n+k-1, k)*binomial(2*n-k-1, n-k).
More generally, a(n) = Sum_{k = 0..n} (-1)^k*binomial(x*n, k)*binomial((x+3)*n-k-1, n-k) for arbitrary x.
a(n) = (2/3) * Sum_{k = 0..n} (-1)^k*binomial(x*n+k-1, k)*binomial((x+3)*n, n-k) for n >= 1 and arbitrary x. (End)
G.f.: 1/(3-2*g) where g = 1+x*g^3 is the g.f. of A001764. - Seiichi Manyama, Aug 17 2025

a(0) prepended and more terms from Alois P. Heinz, Apr 04 2012

A059481 Triangle read by rows. T(n, k) = binomial(n+k-1, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6435, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12870, 24310, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620, 92378

Offset: 0

Fabian Rothelius, Feb 04 2001

T(n,k) is the number of ways to distribute k identical objects in n distinct containers; containers may be left empty.
T(n,k) is the number of nondecreasing functions f from {1,...,k} to {1,...,n}. - Dennis P. Walsh, Apr 07 2011
Coefficients of Faber polynomials for function x^2/(x-1). - Michael Somos, Sep 09 2003
Consider k-fold Cartesian products CP(n,k) of the vector A(n)=[1,2,3,...,n].
An element of CP(n,k) is a n-tuple T_t of the form T_t=[i_1,i_2,i_3,...,i_k] with t=1,...,n^k.
We count members T of CP(n,k) which satisfy some condition delta(T_t), so delta(.) is an indicator function which attains values of 1 or 0 depending on whether T_t is to be counted or not; the summation sum_{CP(n,k)} delta(T_t) over all elements T_t of CP produces the count.
For the triangle here we have delta(T_t) = 0 if for any two i_j, i_(j+1) in T_t one has i_j > i_(j+1), T(n,k) = Sum_{CP(n,k)} delta(T_t) = Sum_{CP(n,k)} delta(i_j > i_(j+1)).
The indicator function which tests on i_j = i_(j+1) generates A158497, which contains further examples of this type of counting.
Triangle of the numbers of combinations of k elements with repetitions from n elements {1,2,...,n} (when every element i, i=1,...,n, appears in a k-combination either 0, or 1, or 2, ..., or k times). - Vladimir Shevelev, Jun 19 2012
G.f. for Faber polynomials is -log(-t*x-(1-sqrt(1-4*t))/2+1)=sum(n>0, T(n,k)*t^k/n). - Vladimir Kruchinin, Jul 04 2013
Values of complete homogeneous symmetric polynomials with all arguments equal to 1, or, equivalently, the number of monomials of degree k in n variables. - Tom Copeland, Apr 07 2014
Row k >= 0 of the infinite square array A[k,n] = C(n,k), n >= 0, would start with k zeros in front of the first nonzero element C(k,k) = 1; this here is the triangle obtained by taking the first k+1 nonzero terms C(k .. 2k, k) of rows k = 0, 1, 2, ... of that array. - M. F. Hasler, Mar 05 2017

			The triangle T(n,k), n >= 0, 0 <= k <= n, begins
  1      A000217
  1 1   /     A000292
  1 2  3    /    A000332
  1 3  6  10    /    A000389
  1 4 10  20  35    /     A000579
  1 5 15  35  70 126     /
  1 6 21  56 126 252  462
  1 7 28  84 210 462  924 1716
  1 8 36 120 330 792 1716 3432 6435
.
T(3,2)=6 considers the CP with the 3^2=9 elements (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3), and does not count the 3 of them which are (2,1),(3,1) and (3,2).
T(3,3) = 10 because the ways to distribute the 3 objects into the three containers are: (3,0,0) (0,3,0) (0,0,3) (2,1,0) (1,2,0) (2,0,1) (1,0,2) (0,1,2) (0,2,1) (1,1,1), for a total of 10 possibilities.
T(3,3)=10 since (x^2/(x-1))^3 = (x+1+1/x+O(1/x^2))^3 = x^3+3x^2+6x+10+O(x).
T(4,2)=10 since there are 10 nondecreasing functions f from {1,2} to {1,2,3,4}. Using <f(1),f(2)> to denote such a function, the ten functions are <1,1>, <1,2>, <1,3>, <1,4>, <2,2>, <2,3>, <2,4>, <3,3>, <3,4>, and <4,4>. - _Dennis P. Walsh_, Apr 07 2011
T(4,0) + T(4,1) + T(4,2) + T(4,3) = 1 + 4 + 10 + 20 = 35 = T(4,4). - _Jonathan Sondow_, Jun 28 2014
From _Paul Curtz_, Jun 18 2018: (Start)
Consider the array
2,    1,    1,    1,    1,    1,     ... = A054977(n)
1,    1/2,  1/3,  1/4,  1/5,  1/6,   ... = 1/(n+1) = 1/A000027(n)
1/3,  1/6,  1/10, 1/15, 1/21, 1/28,  ... = 2/((n+2)*(n+3)) = 1/A000217(n+2)
1/10, 1/20, 1/35, 1/56, 1/84, 1/120, ... = 6/((n+3)*(n+4)*(n+5)) =1/A000292(n+2) (see the triangle T(n,k)).
Every row is an autosequence of the second kind. (See OEIS Wiki, Autosequence.)
By decreasing antidiagonals the denominator of the array is a(n).
Successive vertical denominators: A088218(n), A000984(n), A001700(n), A001791(n+1), A002054(n), A002694(n).
Successive diagonal denominators: A165817(n), A005809(n), A045721(n), A025174(n+1), A004319(n). (End)
Without the first row (2, 1, 1, 1, ... ), the array leads to A165257(n) instead of a(n). - _Paul Curtz_, Jun 19 2018

R. Grimaldi, Discrete and Combinatorial Mathematics, Addison-Wesley, 4th edition, chapter 1.4.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6, (referring to A158498 before recycling)
M. A. A. Obaid, S. K. Nauman, W. M. Fakieh, and C. M. Ringel, The numbers of support-tilting modules for a Dynkin algebra, 2014 and J. Int. Seq. 18 (2015) 15.10.6.
C. M. Ringel, The Catalan combinatorics of the hereditary artin algebras, arXiv:1502.06553 [math.RT], 2015.
Dennis Walsh, Notes on finite monotonic and non-monotonic functions

Columns: T(n,1) = A000027(n), n >= 1. T(n,2) = A000217(n) = A161680(n+1), n >= 2. T(n,3) = A000292(n), n >= 3. T(n,4) = A000332(n+3), n >= 4. T(n,5) = A000389(n+4), n >= 5. T(n,6) = A000579(n+5), n >=6. T(n,k) = A001405(n+k-1) for k <= n <= k+2. [Corrected and extended by M. F. Hasler, Mar 05 2017]
Rows: T(5,k) = A000332(k+4). T(6,k) = A000389(k+5). T(7,k) = A000579(k+6).
Diagonals: T(n,n) = A001700(n-1). T(n,n-1) = A000984(n-1).
T(n,k) = A046899(n-1,k). - R. J. Mathar, Mar 26 2009
Take Pascal's triangle A007318, delete entries to the right of a vertical line just right of center, then scan the diagonals.
For a signed version of this triangle see A027555.
Row sums give A000984.
Cf. A007318, A158497, A100100 (mirrored), A009766.
A000984, A001700, A001791, A002054, A002694, A004319, A005809, A025174, A045721, A088218, A165817, A054977, A165257, A000027, A000217, A006134 (trace of the symmetric Pascal matrix).

Flat(List([0..10], n->List([0..n], k->Binomial(n+k-1, k)))); # Stefano Spezia, Oct 30 2018

a059481 n k = a059481_tabl !! n !! n
a059481_row n = a059481_tabl !! n
a059481_tabl = map reverse a100100_tabl
-- Reinhard Zumkeller, Jan 15 2014

&cat [[&*[ Binomial(n+k-1,k)]: k in [0..n]]: n in [0..30] ]; // Vincenzo Librandi, Apr 08 2011

for n from 0 to 10 do for k from 0 to n do print(binomial(n+k-1,k)) ; od: od: # R. J. Mathar, Mar 31 2009

t[n_, k_] := Binomial[n+k-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 09 2013 *)
(* The combinatorial objects defined in the first comment can, for n >= 1, be generated by: *) r[n_, k_] := FrobeniusSolve[ConstantArray[1,n],k]; (* Peter Luschny, Jan 24 2019 *)

sjoin(v, j) := apply(sconcat, rest(join(makelist(j, length(v)), v)))$ display_triangle(n) := for i from 0 thru n do disp(sjoin(makelist(binomial(i+j-1, j), j, 0, i), " ")); display_triangle(10); /* triangle output */ /* Stefano Spezia, Oct 30 2018 */

{T(n, k) = binomial( n+k-1, k)}; \\ Michael Somos, Sep 09 2003, edited by M. F. Hasler, Mar 05 2017

{T(n, k) = if( n<0, 0, polcoeff( Pol(((1 / (x - x^2) + x * O(x^n))^n + O(x)) * x^n), k))}; /* Michael Somos, Sep 09 2003 */

[[binomial(n+k-1,k) for k in range(n+1)] for n in range(11)] # G. C. Greubel, Nov 21 2018

T(n,0) + T(n,1) + . . . + T(n,n-1) = T(n,n). - Jonathan Sondow, Jun 28 2014
From Peter Bala, Jul 21 2015: (Start)
T(n, k) = Sum_{j = k..n} (-1)^(k+j)*binomial(2*n,n+j)*binomial(n+j-1,j)* binomial(j,k) (gives the correct value T(n,k) = 0 for k > n).
O.g.f.: 1/2*( x*(2*x - 1)/(sqrt(1 - 4*t*x)*(1 - x - t)) + (1 + 2*x)/sqrt(1 - 4*t*x) + (1 - t)/(1 - x - t) ) = 1 + (1 + t)*x + (1 + 2*t + 3*t^2)*x^2 + (1 + 3*t + 6*t^2 + 10*t^3)*x^3 + ....
n-th row polynomial R(n,t) = [x^n] ( (1 + x)^2/(1 + x(1 - t)) )^n.
exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + t)*x + (1 + 2*t + 2*t^2)*x^2 + (1 + 3*t + 5*t^2 + 5*t^3)*x^3 + ... is the o.g.f for A009766. (End)
a(n) = abs(A027555(n)). - M. F. Hasler, Mar 05 2017
For n >= k > 0, T(n, k) = Sum_{j=1..n} binomial(k + j - 2, k - 1) = Sum_{j=1..n} A007318(k + j - 2, k - 1). - Stefano Spezia, Oct 30 2018
T(n, k) = RisingFactorial(n, k) / k!. - Peter Luschny, Nov 24 2023

Offset changed from 1 to 0 by R. J. Mathar, Jan 15 2013

Edited by M. F. Hasler, Mar 05 2017

cp's OEIS Frontend

A370376 Number of compositions of n where there are A025174(k) sorts of part k.

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A001764 a(n) = binomial(3*n,n)/(2*n+1) (enumerates ternary trees and also noncrossing trees).

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A002293 Number of dissections of a polygon: binomial(4*n, n)/(3*n + 1).

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A045721 a(n) = binomial(3*n+1,n).

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A092392 Triangle read by rows: T(n,k) = C(2*n - k,n), 0 <= k <= n.

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A001764 a(n) = binomial(3n,n)/(2n+1) (enumerates ternary trees and also noncrossing trees).

A002293 Number of dissections of a polygon: binomial(4n, n)/(3n + 1).