A224274 -id:A224274 - 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

A025174 a(n) = binomial(3n-1, n-1).

Original entry on oeis.org

0, 1, 5, 28, 165, 1001, 6188, 38760, 245157, 1562275, 10015005, 64512240, 417225900, 2707475148, 17620076360, 114955808528, 751616304549, 4923689695575, 32308782859535, 212327989773900, 1397281501935165, 9206478467454345, 60727722660586800, 400978991944396320

Offset: 0

Wouter Meeussen, Emeric Deutsch

Number of standard tableaux of shape (2n-1,n). Example: a(2)=5 because in the top row we can have 123, 124, 125, 134, or 135. - Emeric Deutsch, May 23 2004
Number of peaks in all generalized {(1,2),(1,-1)}-Dyck paths of length 3n.
Positive terms in this sequence are the numbers k such that k and 2k are consecutive terms in a row of Pascal's triangle. 1001 is the only k such that k, 2k, and 3k are consecutive terms in a row of Pascal's triangle. - J. Lowell, Mar 11 2023

			L.g.f.: L(x) = x + 5*x^2/2 + 28*x^3/3 + 165*x^4/4 + 1001*x^5/5 + 6188*x^6/6 + ...
where G(x) = exp(L(x)) satisfies G(x) = 1 + x*G(x)^3, and begins:
exp(L(x)) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... + A001764(n)*x^n + ...

B. C. Berndt, Ramanujan's Notebooks Part I, Springer-Verlag, see Entry 14, Corollary 1, p. 71.

Robert Israel, Table of n, a(n) for n = 0..1190
Paul Barry, On the Central Antecedents of Integer (and Other) Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.3.
D. Kruchinin and V. Kruchinin, A Generating Function for the Diagonal T2n,n in Triangles, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
Emanuele Munarini, Shifting Property for Riordan, Sheffer and Connection Constants Matrices, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.2.
Stepan Orevkov, Asymptotics of the number of lattice triangulations of rectangles of width 4 and 5, arXiv:2412.17065 [math.CO], 2024. See p. 16.
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A001764 (binomial(3n,n)/(2n+1)), A117671 (C(3n+1,n+1)), A004319, A005809, A006013, A013698, A045721, A117671, A165817, A224274, A236194.

Cf. A000108, A001045, A005156, A224274.

[Binomial(3*n-1,n-1): n in [0..30]]; // Vincenzo Librandi, Nov 12 2014

with(combinat):seq(numbcomp(3*i,i), i=0..20); # Zerinvary Lajos, Jun 16 2007

Table[ GegenbauerC[ n, n, 1 ]/2, {n, 0, 24} ]
Join[{0},Table[Binomial[3n-1,n-1],{n,20}]] (* Harvey P. Dale, Oct 19 2022 *)
nmax=23; CoefficientList[Series[(2+HypergeometricPFQ[{1/3,2/3},{1/2,1},27x/4])/3-1,{x,0,nmax}],x]Range[0,nmax]! (* Stefano Spezia, Dec 31 2024 *)

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

G.f.: z*g^2/(1-3*z*g^2), where g=g(z) is given by g=1+z*g^3, g(0)=1, that is, (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n) = Sum_{k=0..n} ((3k+1)/(2n+k+1))C(3n, 2n+k)*A001045(k). - Paul Barry, Oct 07 2005
Hankel transform of a(n+1) is A005156(n+1). - Paul Barry, Apr 14 2008
G.f.: x*B'(x)/B(x) where B(x) is the g.f. of A001764. - Vladimir Kruchinin Feb 03 2013
D-finite with recurrence: 2*n*(2*n-1)*a(n) -3*(3*n-1)*(3*n-2)*a(n-1)=0. - R. J. Mathar, Feb 05 2013
Logarithmic derivative of A001764; g.f. of A001764 satisfies G(x) = 1 + x*G(x)^3. - Paul D. Hanna, Jul 14 2013
G.f.: (2*cos((1/3)*arcsin((3/2)*sqrt(3*x)))-sqrt(4-27*x))/(3*sqrt(4-27*x)). - Emanuele Munarini, Oct 14 2014
a(n) = Sum_{k=1..n} binomial(n-1,n-k)*binomial(2*n,n-k). - Vladimir Kruchinin, Nov 12 2014
a(n) = [x^n] C(x)^n for n >= 1, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function for A000108 (Ramanujan). - Peter Bala, Jun 24 2015
From Peter Bala, Nov 04 2015: (Start)
Without the initial term 0, the o.g.f. equals f(x)*g(x)^2, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. g(x)^2 is the o.g..f for A006013. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
G.f.: ( 2F1(1/3,2/3;1/2;27*x/4)-1)/3. - R. J. Mathar, Jan 27 2020
O.g.f. without the initial term 0, in the form g(x)=(2*cos(arcsin((3*sqrt(3)*sqrt(x))/2)/3)/sqrt(4-27*x)-1)/(3*x), satisfies the following algebraic equation: 1+(9*x-1)*g(x)+x*(27*x-4)*g(x)^2+x^2*(27*x-4)*g(x)^3=0. - Karol A. Penson, Oct 11 2021
O.g.f. equals f(x)/(1 - 2*f(x)), where f(x) = series reversion (x/(1 + x)^3) = x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the o.g.f. of A001764 with the initial term omitted. Cf. A224274. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/2)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+2)*n+k-1,k) = C(3*n-1,n-1) and (1/3)*Sum_{k = 0..n} (-1)^k* C(x*n,n-k)*C((x-3)*n+k-1,k) = C(3*n-1,n-1), both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
a(n) ~ 2^(-2*n)*3^(3*n)/(2*sqrt(3*n*Pi)). - Stefano Spezia, Apr 25 2024
a(n) = Sum_{k = 0..n-1} binomial(2*n+k-1, k) = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(3*n, k). - Peter Bala, Jul 21 2024
E.g.f.: (2 + hypergeom([1/3, 2/3], [1/2, 1], 27*x/4))/3 - 1. - Stefano Spezia, Dec 31 2024

A069271 a(n) = binomial(4n+1,n)2/(3*n+2).

Original entry on oeis.org

1, 2, 9, 52, 340, 2394, 17710, 135720, 1068012, 8579560, 70068713, 580034052, 4855986044, 41043559340, 349756577100, 3001701610320, 25921837477692, 225083787458904, 1963988670706228, 17211860478150800, 151433425446423120

Offset: 0

Henry Bottomley, Mar 12 2002

nonn

This sequence counts the set B_n of plane trees defined in the Poulalhon and Schaeffer link (Definition 2.2 and Section 4.2, Proposition 4). - David Callan, Aug 20 2014
a(n) is the number of lattice paths of length 4n starting and ending on the x-axis consisting of steps {(1, 1), (1, -3)} that remain on or above the line y=-1. - Sarah Selkirk, Mar 31 2020
a(n) is the number of ordered pairs of 4-ary trees with a (summed) total of n internal nodes. - Sarah Selkirk, Mar 31 2020

			a(3) = C(4*3+1,3)*2/(3*3+2) = C(13,3)*2/11 = 286*2/11 = 52.
a(3) = 52 since the top row of M^3 = (22, 22, 7, 1).
1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + 2394*x^5 + 17710*x^6 + 135720*x^7 + ...
q + 2*q^3 + 9*q^5 + 52*q^7 + 340*q^9 + 2394*q^11 + 17710*q^13 + 135720*q^15 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Thomas Anderson, T. Bruce McLean, Homeira Pajoohesh, and Chasen Smith , The combinatorics of all regular flexagons, Eu. J. Combinat. 31 (2010) 72-80, Theorem 2.
Roland Bacher, Fair Triangulations, arXiv:0710.0960 [math.CO], 2007.
Esther Banaian, Elise Catania, Christian Gaetz, Miranda Moore, Gregg Musiker, and Kayla Wright, Twists, Higher Dimer Covers, and Web Duality for Grassmannian Cluster Algebras, arXiv:2507.15211 [math.CO], 2025. See p. 28.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
Gi-Sang Cheon, S.-T. Jin and L. W. Shapiro, A combinatorial equivalence relation for formal power series, Linear Algebra and its Applications, Volume 491, 15 February 2016, Pages 123-137.
Emmanuel Guitter, The distance-dependent two-point function of triangulations: a new derivation from old results, Ann. Inst. Henri Poincaré Comb. Phys. Interact. Vol. 4 (2017), 177-211. DOI: 10.4171/AIHPD/38.
Clemens Heuberger, Sarah J. Selkirk, and Stephan Wagner, Enumeration of Generalized Dyck Paths Based on the Height of Down-Steps Modulo k, arXiv:2204.14023 [math.CO], 2022.
Ionut E. Iacob, T. Bruce McLean and Hua Wang, The V-flex, Triangle Orientation, and Catalan Numbers in Hexaflexagons, The College Mathematics Journal, Vol. 43, No. 1 (January 2012), pp. 6-10.
Jean-Christophe Novelli and Jean-Yves Thibon, Noncommutative Symmetric Functions and Lagrange Inversion, arXiv:math/0512570 [math.CO], 2005-2006.
C. O. Oakley and R. J. Wisner, Flexagons, Am. Math. Monthly 64 (3) (1957) 143-154, u_{3k+2}.
Karol A. Penson and Karol Zyczkowski, Product of Ginibre matrices : Fuss-Catalan and Raney distribution, arXiv:1103.3453 [math-ph], 2011.
Karol A. Penson and Karol Zyczkowski, Product of Ginibre matrices : Fuss-Catalan and Raney distribution, Phys. Rev. E 83, 061118, 2011.
Dominique Poulalhon and Gilles Schaeffer, Optimal Coding and Sampling of Triangulations, in Automata, Languages and Programming, Lecture Notes in Computer Science, Volume 2719, 2003, pp 1080-1094.
Sarah J. Selkirk, On a generalisation of k-Dyck paths, MSc Thesis, 2019.

Cf. A002293, A006013, A006632, A069270 for similar generalized Catalan sequences.

Cf. A000260, A115141, A163456, A224274.

[2*Binomial(4*n+1, n)/(3*n+2): n in [0..20]];  // Bruno Berselli, Mar 04 2011

BB:=[T,{T=Prod(Z,Z,Z,F,F),F=Sequence(B),B=Prod(F,F,F,Z)}, unlabeled]: seq(count(BB,size=i),i=3..23); # Zerinvary Lajos, Apr 22 2007

f[n_] := 2 Binomial[4 n + 1, n]/(3 n + 2); Array[f, 21, 0] (* Robert G. Wilson v *)

a(n)=if(n<0,0,polcoeff(serreverse(x/(1+x^2)^2+O(x^(2*n+2))),2*n+1)) /* Ralf Stephan */

{a(n) =  binomial(4*n + 2, n)*2 / (2*n + 1)} /* Michael Somos, Mar 28 2012 */

{a(n) =  local(A); if( n<0, 0, A = 1 + O(x); for( k=1, n, A = (1 + x * A^2)^2); polcoeff( A, n))} /* Michael Somos, Mar 28 2012 */

a(n) = A069270(n+1, n) = A005810(n)*A016813(n)/A060544(n+1)
O.g.f. A(x) satisfies 2*x^2*A(x)^3 = 1-2*x*A(x)-sqrt(1-4*x*A(x)). - Vladimir Kruchinin, Feb 23 2011
a(n) is the sum of top row terms in M^n, where M is the infinite square production matrix with the triangular series in each column as follows, with the rest zeros:
   1,  1,  0, 0, 0, 0, ...
   3,  3,  1, 0, 0, 0, ...
   6,  6,  3, 1, 0, 0, ...
  10, 10,  6, 3, 1, 0, ...
  15, 15, 10, 6, 3, 1, ...
  ... - Gary W. Adamson, Aug 11 2011
Given g.f. A(x) then B(x) = x * A(x^2) satisfies x = B(x) / (1 + B(x)^2)^2. - Michael Somos, Mar 28 2012
Given g.f. A(x) then A(x) = (1 + x * A(x)^2)^2. - Michael Somos, Mar 28 2012
a(n) / (n+1) = A000260(n). - Michael Somos, Mar 28 2012
REVERT transform is A115141. - Michael Somos, Mar 28 2012
D-finite with recurrence 3*n*(3*n+2)*(3*n+1)*a(n) - 8*(4*n+1)*(2*n-1)*(4*n-1)*a(n-1) = 0. - R. J. Mathar, Jun 07 2013
a(n) = 2*binomial(4n+1,n-1)/n for n>0, a(0)=1. - Bruno Berselli, Jan 19 2014
G.f.: hypergeom([1/2, 3/4, 5/4], [4/3, 5/3], (256/27)*x). - Robert Israel, Aug 24 2014
From Peter Bala, Oct 08 2015: (Start)
O.g.f. A(x) = (1/x) * series reversion (x/C(x)^2), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A163456.
(1/2)*x*A'(x)/A(x) is the o.g.f. for A224274. (End)
E.g.f.: hypergeom([1/2, 3/4, 5/4], [1, 4/3, 5/3], (256/27)*x). - Karol A. Penson, Jun 26 2017
a(n) = binomial(4*n+2,n)/(2*n+1). - Alexander Burstein, Nov 08 2021

A262977 a(n) = binomial(4*n-1,n).

Original entry on oeis.org

1, 3, 21, 165, 1365, 11628, 100947, 888030, 7888725, 70607460, 635745396, 5752004349, 52251400851, 476260169700, 4353548972850, 39895566894540, 366395202809685, 3371363686069236, 31074067324187580, 286845713747883300, 2651487106659130740, 24539426037817994160

Offset: 0

Vladimir Kruchinin, Oct 06 2015

From Gus Wiseman, Sep 28 2022: (Start)
Also the number of integer compositions of 4n with alternating sum 2n, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A348614. The a(12) = 21 compositions are:
  (6,2)  (1,2,5)  (1,1,5,1)  (1,1,1,1,4)
         (2,2,4)  (2,1,4,1)  (1,1,2,1,3)
         (3,2,3)  (3,1,3,1)  (1,1,3,1,2)
         (4,2,2)  (4,1,2,1)  (1,1,4,1,1)
         (5,2,1)  (5,1,1,1)  (2,1,1,1,3)
                             (2,1,2,1,2)
                             (2,1,3,1,1)
                             (3,1,1,1,2)
                             (3,1,2,1,1)
                             (4,1,1,1,1)
The following pertain to this interpretation:
- The case of partitions is A000712, reverse A006330.
- Allowing any alternating sum gives A013777 (compositions of 4n).
- A011782 counts compositions of n.
- A034871 counts compositions of 2n with alternating sum 2k.
- A097805 counts compositions by alternating (or reverse-alternating) sum.
- A103919 counts partitions by sum and alternating sum (reverse: A344612).
- A345197 counts compositions by length and alternating sum.
(End)

V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A006632, A002293, A004331, A005810, A052203, A224274, A257633.
Cf. A000346, A000984, A001700, A002458, A025047, A058622, A081294, A294175.
Cf. A088218, A163455, A165817.

[Binomial(4*n-1,n): n in [0..20]]; // Vincenzo Librandi, Oct 06 2015

Table[Binomial[4 n - 1, n], {n, 0, 40}] (* Vincenzo Librandi, Oct 06 2015 *)

B(x):=sum(binomial(4*n-1,n-1)*3/(4*n-1)*x^n,n,1,30);
taylor(x*diff(B(x),x,1)/B(x),x,0,20);

a(n) = binomial(4*n-1,n); \\ Michel Marcus, Oct 06 2015

G.f.: A(x)=x*B'(x)/B(x), where B(x) if g.f. of A006632.
a(n) = Sum_{k=0..n}(binomial(n-1,n-k)*binomial(3*n,k)).
a(n) = 3*A224274(n), for n > 0. - Michel Marcus, Oct 12 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 A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A005810 (k = 0), A052203 (k = 1), A257633 (k = 2), A224274 (k = 3) and A004331 (k = 4). (End)
a(n) = [x^n] 1/(1 - x)^(3*n). - Ilya Gutkovskiy, Oct 03 2017
a(n) = A071919(3n-1,n+1) = A097805(4n,n+1). - Gus Wiseman, Sep 28 2022
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-3*n, n).
a(n) = hypergeom([1 - 3*n, -n], [1], 1).
The g.f. A(x) satisfies A(x/(1 + x)^4) = 1/(1 - 3*x). (End)
a(n) = Sum_{k = 0..n} binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). - Peter Bala, Sep 16 2024
G.f.: 1/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 17 2025

A052203 a(n) = (4n+1)*binomial(4n,n)/(3n+1).

Original entry on oeis.org

1, 5, 36, 286, 2380, 20349, 177100, 1560780, 13884156, 124403620, 1121099408, 10150595910, 92263734836, 841392966470, 7694644696200, 70539987842520, 648045936942300, 5964720367660956, 54991682779774384, 507749884105448600, 4694436188839116720

Offset: 0

Barry E. Williams, Jan 28 2000

Central terms of the triangles in A122366 and A111418. - Reinhard Zumkeller, Aug 30 2006 and Mar 14 2014

a(n) is the number of paths from (0,0) to (4n,n), taking north and east steps while avoiding exactly 2 consecutive north steps. - Shanzhen Gao, Apr 15 2010

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Cf. A002293, A051944, A004331, A005810, A224274, A257633, A262977.

a052203 n = a122366 (2 * n) n  -- Reinhard Zumkeller, Mar 14 2014

[Binomial(4*n+1, n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014

Table[Binomial[4 n + 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)

vector(30, n, n--; (4*n+1)*binomial(4*n,n)/(3*n+1)) \\ Altug Alkan, Nov 05 2015

a(n) = C(4n+1, n); a(n) is asymptotic to c/sqrt(n)*(256/27)^n with c=0.614... - Benoit Cloitre, Jan 27 2003 [c = 2^(5/2)/(3^(3/2)*sqrt(Pi)) = 0.61421182128... - Vaclav Kotesovec, Feb 14 2019]
G.f.: g^2/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
G.f.: hypergeom([1/2, 3/4, 5/4], [2/3, 4/3], (256/27)*x). - Robert Israel, Aug 07 2014
D-finite with recurrence 3*n*(3*n-1)*(3*n+1)*a(n) - 8*(4*n+1)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Nov 26 2012
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 A005810 and g(x) is the o.g.f. for A002293.
More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A257633 (k = 2), A224274 (k = 3) and A004331 (k = 4). (End)
a(n) = [x^n] 1/(1 - x)^(3*n+2). - Ilya Gutkovskiy, Oct 03 2017
a(n) = Sum_{k = 0..n} binomial(2*n+k+1, k)*binomial(2*n-k-1, n-k). - Peter Bala, Sep 17 2024

More terms from James Sellers, Jan 31 2000

A004331 Binomial coefficient C(4n,n-1).

Original entry on oeis.org

1, 8, 66, 560, 4845, 42504, 376740, 3365856, 30260340, 273438880, 2481256778, 22595200368, 206379406870, 1889912732400, 17345898649800, 159518999862720, 1469568786235308, 13559593014190944, 125288932441604200

Offset: 1

N. J. A. Sloane

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.

Seiichi Manyama, Table of n, a(n) for n = 1..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A002293, A005810, A052203, A224274, A257633, A262977.

Cf. A001791, A004319, A004343, A004356, A004369, A004382.

#A004331
seq(binomial(4*n - 1,n), n = 0..20);

a[n_] := Binomial[4*n, n - 1]; Array[a, 19] (* Amiram Eldar, May 09 2020 *)

vector(30, n, binomial(4*n, n-1)) \\ Altug Alkan, Nov 05 2015

G.f.: (g^2-g)/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
With an offset of 0, the o.g.f. equals f(x)*g(x)^4, where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A052203 (k = 1), A257633 (k = 2) and A224274 (k = 3). - Peter Bala, Nov 04 2015
D-finite with recurrence 3*(n-1)*(3*n-1)*(3*n+1)*a(n) -8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1)=0. - R. J. Mathar, Mar 19 2025

A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.

Original entry on oeis.org

0, 9, 11, 14, 130, 133, 135, 138, 141, 143, 148, 153, 155, 158, 168, 177, 179, 182, 188, 208, 225, 227, 230, 236, 248, 2052, 2057, 2059, 2062, 2066, 2069, 2071, 2074, 2077, 2079, 2084, 2089, 2091, 2094, 2098, 2101, 2103, 2106, 2109, 2111, 2120, 2129, 2131

Offset: 1

Gus Wiseman, Oct 29 2021

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The terms together with their binary indices begin:
    0: ()
    9: (3,1)
   11: (2,1,1)
   14: (1,1,2)
  130: (6,2)
  133: (5,2,1)
  135: (5,1,1,1)
  138: (4,2,2)
  141: (4,1,2,1)
  143: (4,1,1,1,1)
  148: (3,2,3)
  153: (3,1,3,1)
  155: (3,1,2,1,1)
  158: (3,1,1,1,2)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The unordered case (partitions) is counted by A000712, reverse A006330.
These compositions are counted by A262977.
Except for 0, a subset of A345917 (which is itself a subset of A345913).
A000346 = even-length compositions with alt sum != 0, complement A001700.
A011782 counts compositions.
A025047 counts wiggly compositions, ranked by A345167.
A034871 counts compositions of 2n with alternating sum 2k.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
A345197 counts compositions by length and alternating sum.
Cf. A000984, A002458, A004331, A005810, A224274.
Cf. A008549, A013777, A027306, A058622, A088218, A114121, A120452, A126869, A163493, A294175, A344604.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,1000],Total[stc[#]]==2*ats[stc[#]]&]

A163456 a(n) = binomial(5*n,n)/5.

Original entry on oeis.org

1, 9, 91, 969, 10626, 118755, 1344904, 15380937, 177232627, 2054455634, 23930713170, 279871768995, 3284214703056, 38650751381832, 456002537343216, 5391644226101705, 63871405575418665, 757929628541719755

Offset: 1

Zak Seidov, Jul 28 2009

For prime p, a(p) == 1 (mod p). - Gary Detlefs, Aug 03 2013
In fact, a(p) == 1 (mod p^3) for prime p >= 5. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015
From Robert Israel, Jul 12 2016: (Start)
a(p+1) == 5 (mod p) for primes p >= 5.
a(p^(k+1)) == a(p^k) mod p^(3(k+1)) for primes p >= 5. (End)

Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, 2nd ed. 1994.

Robert Israel, Table of n, a(n) for n = 1..920
Romeo Meštrović, Lucas’ theorem: its generalizations, extensions and applications (1878-2014), arXiv:1409.3820v1 [math.NT], 2014.
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A000108, A000245, A001764, A002293, A025174, A118970, A163455, A224274, A227726.

seq(binomial(5*n,n)/5, n=1..20); # Robert Israel, Jul 12 2016

Array[Binomial[5 #, #]/5 &, {18}] (* Michael De Vlieger, Oct 09 2015 *)

a(n) = binomial(5*n,n)/5 \\ Altug Alkan, Oct 09 2015

a(n) = (5*n-1)!/(4*n!*(4*n-1)!) = A001449(n)/5 = A163455(n)/4.
a(n) = binomial(5*n,n)/5. - Gary Detlefs, Aug 03 2013
From Peter Bala, Oct 08 2015: (Start)
a(n) = (1/3)*[x^n] (C(x)^3)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A224274.
exp( 3*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 18*x^2 + 136*x^3 + ... is the o.g.f. for A118970. (End)
From Peter Bala,Jul 12 2016: (Start)
a(n) = 1/6*[x^n] (1 + x)/(1 - x)^(4*n + 1).
a(n) = 1/6*[x^n] ( 1/C(-x)^6 )^n. Cf. A227726. (End)
a(n) ~ 2^(-8*n-3/2)*5^(5*n-1/2)*n^(-1/2)/sqrt(Pi). - Ilya Gutkovskiy, Jul 12 2016
From Robert Israel, Jul 12 2016: (Start)
G.f.: x*hypergeom([1, 6/5, 7/5, 8/5, 9/5], [5/4, 3/2, 7/4, 2], (3125/256)*x).
a(n) = 5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)/(8*n*(4*n-3)*(2*n-1)*(4*n-1)). (End)
O.g.f.: f(x)/(1 - 4*f(x)), where f(x) = series reversion (x/(1 + x)^5) = x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + ... is the o.g.f. of A002294 with the initial term omitted. Cf. A025174. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/4)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+4)*n+k-1,k) = C(5*n,n)/5 and (1/5)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-5)*n+k-1,k) = C(5*n,n)/5, both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
Right-hand side of the identity (1/4)*Sum_{k = 0..2*n} (-1)^k*binomial(6*n-k-1,2*n-k)*binomial(4*n+k-1,k) = binomial(5*n,n)/5, for n >= 1. - Peter Bala, Mar 09 2022
a(n) = (1/2)* [x*n] F(x)^(2*n) = [x^n] G(x)^n for n >= 1, where F(x) = Sum_{k >= 0} 1/(2*k + 1)*binomial(3*k,k)*x^k is the o.g.f. of A001764 and G(x) = Sum_{k >= 0} 1/(3*k + 1)*binomial(4*k,k)*x^k is the o.g.f. of A002293 (apply Concrete Mathematics, equation 5.60, p. 201). - Peter Bala, Apr 26 2023

Renamed by Peter Bala, Oct 08 2015

A257633 a(n) = binomial(4*n + 2,n).

Original entry on oeis.org

1, 6, 45, 364, 3060, 26334, 230230, 2035800, 18156204, 163011640, 1471442973, 13340783196, 121399651100, 1108176102180, 10142940735900, 93052749919920, 855420636763836, 7877932561061640, 72667580816130436, 671262558647881200, 6208770443303347920

Offset: 0

Peter Bala, Nov 04 2015

Michael De Vlieger, Table of n, a(n) for n = 0..1025
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A002293, A004331, A005810, A052203, A224274, A262977.

#A257633
seq(binomial(4*n + 2,n), n = 0..20);

Table[Binomial[4*n + 2, n], {n, 0, 120}] (* Michael De Vlieger, Apr 11 2025 *)

vector(30, n, n--; binomial(4*n+2, n)) \\ Altug Alkan, Nov 05 2015

The o.g.f. equals f(x)*g(x)^2, where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A052203 (k = 1), A224274 (k = 3) and A004331 (k = 4).

cp's OEIS Frontend

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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A025174 a(n) = binomial(3n-1, n-1).

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A069271 a(n) = binomial(4*n+1,n)*2/(3*n+2).

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A262977 a(n) = binomial(4*n-1,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).

A069271 a(n) = binomial(4n+1,n)2/(3*n+2).