A005810 -id:A005810 - cp's OEIS frontend

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

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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Olivier Danvy, Summa Summarum: Moessner's Theorem without Dynamic Programming, arXiv:2412.03127 [cs.DM], 2024. See p. 25.
C. Defant and N. Kravitz, Stack-sorting for words, arXiv:1809.09158 [math.CO], 2018.
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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

A005809 a(n) = binomial(3n,n).

Original entry on oeis.org

1, 3, 15, 84, 495, 3003, 18564, 116280, 735471, 4686825, 30045015, 193536720, 1251677700, 8122425444, 52860229080, 344867425584, 2254848913647, 14771069086725, 96926348578605, 636983969321700, 4191844505805495, 27619435402363035

Offset: 0

N. J. A. Sloane

Number of paths in Z X Z starting at (0,0) and ending at (3n,0) using steps in {(1,1),(1,-2)}.
Number of even trees with 2n edges and one distinguished vertex. Even trees are rooted plane trees where every vertex (including root) has even degree.
Hankel transform is 3^n*A051255(n), where A051255 is the Hankel transform of C(3n,n)/(2n+1). - Paul Barry, Jan 21 2007
a(n) is the number of stack polyominoes inscribed in an (n+1) X (n+1) box. Equivalently, a(n) is the number of unimodal compositions with n+1 parts in which the maximum value of the parts is n+1. For instance, for n = 2, we have the following compositions: (3,3,3), (2,3,3), (1,3,3), (3,3,1), (3,3,2), (2,2,3), (1,2,3), (2,3,1), (1,1,3), (1,3,1), (3,1,1), (2,3,2), (1,3,2), (3,2,1), (3,2,2). - Emanuele Munarini, Apr 07 2011
Conjecture: a(n)==3 (mod n^3) iff n is an odd prime. - Gary Detlefs, Mar 23 2013. The congruence a(p) = binomial(3*p,p) = 3 (mod p^3) for odd prime p is a known generalization of Wolstenholme's theorem. See Mestrovic, Section 6, equation 35. - Peter Bala, Dec 28 2014
In general, C(k*n,n) = C(k*n-1,n-1)*C((k*n)^2,2)/(3*n*C(k*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014

			G.f. = 1 + 3*x + 15*x^2 + 84*x^3 + 495*x^4 + 3003*x^5 + 18564*x^6 + ... - _Michael Somos_, Jan 30 2019

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.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000[Terms 0 to 100 computed by T. D. Noe; terms 101 to 1000 by G. C. Greubel, Jan 14 2017]
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Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Wojciech Młotkowski and Karol A. Penson, Probability distributions with binomial moments, Infinite Dimensional Analysis, Quantum Probability and Related Topics, Vol. 17, No. 2 (2014), 1450014; arXiv preprint, arXiv:1309.0595 [math.PR], 2013.
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, Journal of Integer Sequences, Vol. 18 (2015), Article 15.11.7.
Yash Puri and Thomas Ward, Arithmetic and growth of periodic orbits, Journal of Integer Sequences, Vol. 4 (2001), Article 01.2.1.

binomial(k*n,n): A000984 (k = 2), A005810 (k = 4), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).

Cf. A001764, A007318, A110608, A229705, A254157, A268196.

a005809 n = a007318 (3*n) n  -- Reinhard Zumkeller, May 06 2012

[ Binomial(3*n,n): n in [0..150] ]; // Vincenzo Librandi, Apr 21 2011

A005809:=n->binomial(3*n,n); seq(A005809(n), n=0..40); # Wesley Ivan Hurt, Mar 21 2014

R[ z_ ] := ((2-18*z + 27*z^2 + 3^(3/2)*z^(3/2)*(27*z-4)^(1/2))/2)^(1/3); f[ z_ ] := ( (R[ z ])^3 + (1-3*z)*(R[ z ])^2 + (1-6*z)*R[ z ] )/( (R[ z ])^4 + (1-6*z)*(R[ z ])^2 + (6*z-1)^2 )
Table[Binomial[3*n,n],{n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Mar 03 2011 *)

makelist(binomial(3*n,n),n,0,100); /* Emanuele Munarini, Apr 07 2011 */

B(x):=(2/sqrt(3*x))*sin((1/3)*asin(sqrt(27*x/4)))-1;
taylor(x*diff(B(x),x)/B(x),x,0,10); /* Vladimir Kruchinin, Oct 02 2015 */

a(n)=binomial(3*n,n) \\ Charles R Greathouse IV, Nov 20 2012

[binomial(3*n,n) for n in range(0, 22)] # Zerinvary Lajos, Dec 16 2009

The g.f. R[ z_ ] below (in the Mathematica field) was found by Kurt Persson (kurt(AT)math.chalmers.se) and communicated by Einar Steingrimsson (einar(AT)math.chalmers.se).
Using Stirling's formula in A000142, it is easy to get the asymptotic expression a(n) ~ (1/2) * (27/4)^n / sqrt(Pi*n / 3). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
a(n) = Sum_{k=0..n} C(n, k)*C(2n, k). - Paul Barry, May 15 2003
G.f.: 1/(1-3zg^2), where g=g(z) is given by g=1+zg^3, g(0)=1, i.e., (in Maple notation) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
G.f.: x*B'(x)/B(x), where B(x)+1 is the g.f. for A001764. -  Vladimir Kruchinin, Oct 02 2015
a(n) ~ (1/2)*3^(1/2)*Pi^(-1/2)*n^(-1/2)*2^(-2*n)*3^(3*n)*(1 - 7/72*n^-1 + 49/10368*n^-2 + 6425/2239488*n^-3 - ...). - Joe Keane (jgk(AT)jgk.org), Nov 07 2003
a(n) = A006480(n)/A000984(n). - Lior Manor, May 04 2004
a(n) = Sum_{i_1=0..n, i_2=0..n} binomial(n, i_1)*binomial(n, i_2)*binomial(n, i_1+i_2). - Benoit Cloitre, Oct 14 2004
a(n) = Sum_{k=0..n} A109971(k)*3^k; a(0)=1, a(n) = Sum_{k=0..n} 3^k*C(3n-k,n-k)2k/(3n-k), n>0. - Paul Barry, Jan 21 2007
a(n) = A085478(2n,n). - Philippe Deléham, Sep 17 2009
E.g.f.: 2F2(1/3,2/3;1/2,1;27*x/4), where F(a1,a2;b1,b2;z) is a hypergeometric series. - Emanuele Munarini, Apr 12 2011
a(n) = Sum_{k=0..n} binomial(2*n+k-1,k). - Arkadiusz Wesolowski, Apr 02 2012
G.f.: cos((1/3)*asin(sqrt(27x/4)))/sqrt(1-27x/4). - Tom Copeland, May 24 2012
G.f.: A(x) = 1 + 6*x/(G(0)-6*x) where G(k) = (2*k+2)*(2*k+1) + 3*x*(3*k+1)*(3*k+2) - 6*x*(k+1)*(2*k+1)*(3*k+4)*(3*k+5)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jun 30 2012
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
a(n) = (2n+1)*A001764(n). - Johannes W. Meijer, Aug 22 2013
a(n) = C(3*n-1,n-1)*C(9*n^2,2)/(3*n*C(3*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
a(n) = [x^n] 1/(1 - x)^(2*n+1). - Ilya Gutkovskiy, Oct 03 2017
a(n) = hypergeom([-2*n, -n], [1], 1). - Peter Luschny, Mar 19 2018
a(n) = Sum_{k=0..n} binomial(n, k) * binomial(2*n, n-k) = row sums of A110608. - Michael Somos, Jan 30 2019
0 = a(n)*(-3188646*a(n+2) +7322076*a(n+3) -2805111*a(n+4) +273585*a(n+5)) +a(n+1)*(+413343*a(n+2) -1252017*a(n+3) +538344*a(n+4) -55940*a(n+5)) +a(n+2)*(-4131*a(n+2) +38733*a(n+3) -21628*a(n+4) +2528*a(n+5)) for all n in Z. - Michael Somos, Jan 30 2019
Sum_{n>=1} 1/a(n) = A229705. - Amiram Eldar, Nov 14 2020
From Peter Bala, Feb 20 2022: (Start)
The o.g.f. A(x) satisfies the differential equation (4*x - 27*x^2)*A''(x) + (2 - 54*x)*A'(x) - 6*A(x) = 0, with A(0) = 1 and A'(0) = 3.
Algebraic equation: (1 - A(x))*(1 + 2*A(x))^2 +  27*x*A(x)^3 = 0.
Sum_{n >= 1} a(n)*( x*(2*x + 3)^2/(27*(1 + x)^3) )^n = x. (End)
From Vaclav Kotesovec, May 13 2022: (Start)
Sum_{n>=0} a(n) / 3^(2*n) = 2*cos(Pi/9).
Sum_{n>=0} a(n) / (27/2)^n = (1 + sqrt(3))/2.
Sum_{n>=0} a(n) / 3^(3*n) = 2*cos(Pi/18) / sqrt(3).
In general, for k > 27/4, Sum_{n>=0} a(n)/k^n = 2*cos(arccos(1 - 27/(2*k))/6) / sqrt(4 - 27/k). (End)
G.f.: hypergeom([1/3, 2/3], [1/2], 27*z/4), the Gauss hypergeometric function 2F1. - Karol A. Penson, Dec 12 2023
a(n) = 1/4^n * Sum_{k = n..3*n} binomial(k, n)*binomial(3*n, k). - Peter Bala, Jun 29 2025

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

A000897 a(n) = (4n)! / ((2n)!*n!^2).

Original entry on oeis.org

1, 12, 420, 18480, 900900, 46558512, 2498640144, 137680171200, 7735904619300, 441233078286000, 25467973278667920, 1484298740174927040, 87202550985276963600, 5157850293780050462400, 306839461354466267304000, 18344908596179023234548480

Offset: 0

N. J. A. Sloane

Appears in Ramanujan's theory of elliptic functions of signature 4.
H. A. Verrill proves that a(n) = Sum_{p + q + r = 3n} w^(p-q) * {(3n)!/(p! q! r!)}^2, with p, q, r >= 0 and w = primitive 3rd root of unity.
The family of elliptic curves "x=2*H1=p^2+q^2-(1/4)*q^4, 0sqrt(-1)*q" to H1 produces "x=2*H2=p^2-q^2-(1/4)*q^4, 0Bradley Klee, Feb 25 2018
Even-order terms in the diagonal of rational function 1/(1 - (x^2 + y^2 + z)). - Gheorghe Coserea, Aug 09 2018

			G.f.: 1 + 12*x + 420*x^2 + 18480*x^3 + 900900*x^4 + 46558512*x^5 + 2498640144*x^6 + ...

E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Alin Bostan, Armin Straub, and Sergey Yurkevich, On the representability of sequences as constant terms, arXiv:2212.10116 [math.NT], 2022.
H. J. Brothers, Pascal's Prism: Supplementary Material
B. Klee, Geometric G.F. for Ramanujan Periods, seqfan mailing list, 2017.
S. Ramanujan, Modular Equations and Approximations to Pi, Quarterly Journal of Mathematics, XLV (1914), 350-372.
L. C. Shen, A note on Ramanujan's identities involving the hypergeometric function 2F1(1/6,5/6;1;z), The Ramanujan Journal, 30.2 (2013), 211-222.
H. A. Verrill, Sums of squares of binomial coefficients, with applications to Picard-Fuchs equations, arXiv:math/0407327v1 [math.CO], 2004.

Cf. A002897, A008977, A186420, A188662. Elliptic Integrals: A002894, A113424, A006480. Factors: A005809, A005810, A000984, A001448.

a:=n->Sum([0..3*n],k->(-1)^k*Binomial(3*n,k)*Binomial(6*n-k,3*n)*
Binomial(2*k,k));;
A000897:=List([0..14],n->a(n)); # Muniru A Asiru, Feb 11 2018

seq((4*n)!/(n!)^4/binomial(2*n,n), n=0..14); # Zerinvary Lajos, Jun 28 2007

Table[(4n)!/((2n)! n!^2), {n, 0, 30}] (* Stefan Steinerberger, Apr 14 2006 *)
a[ n_] := Binomial[ 4 n, 2 n] Binomial[ 2 n, n]; (* Michael Somos, Mar 24 2013 *)
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/4, 3/4, 1, 64 x], {x, 0, n}]; (* Michael Somos, Mar 24 2013 *)
a[ n_] := If[ n < 0, 0, With[{m = 4 n}, (-1)^n m! SeriesCoefficient[ BesselI[ 0, 2 x] BesselJ[ 0, 2 x], {x, 0, m}]]]; (* Michael Somos, Aug 12 2014 *)
a[ n_] := 64^n Pochhammer[1/4, n] Pochhammer[3/4, n] / n!^2; (* Michael Somos, Aug 12 2014 *)

{a(n) = if( n<0, 0, (4*n)! / ((2*n)! * n!^2))}; /* Michael Somos, Oct 31 2005 */

E.g.f.: Sum_{k>=0} (-1)^k * a(k) * x^(4*k) / (4*k)! = BesselI(0, 2x) * BesselJ(0, 2x).
G.f.: F(1/4, 3/4; 1; 64*x). - Michael Somos, Oct 31 2005
a(n) = A008977(n)/A000984(n) - Zerinvary Lajos, Jun 28 2007
Sum_{k>=0} a(k) * x^(3k)/(3k)!^2 = f(x)*f(x*w)*f(x/w) where f(x) = BesselI(0, 2*sqrt(x)) and w = primitive 3rd root of unity. - Michael Somos, Jul 25 2007
In general, for (BesselI(b, 2x))*(BesselJ(b, 2x))=((x^(2*b))/((GAMMA(b+1))^2)*(1-(x^4)/(Q(0)+(x^4))); Q(k)=(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)-(x^4)+(x^4)*(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)/Q(k+1)) ; (continued fraction). - Sergei N. Gladkovskii, Nov 24 2011
D-finite with recurrence 0 = a(n)*4*(4*n + 1)*(4*n + 3) - a(n+1)*(n + 1)^2 for all n in Z. - Michael Somos, Aug 12 2014
0 = a(n)*(-4026531840*a(n+2) +2005401600*a(n+3) -103896576*a(n+4) +1251948*a(n+5)) + a(n+1)*(+41418752*a(n+2) -30435328*a(n+3) +1863228*a(n+4) -24604*a(n+5)) + a(n+2)*(-16896*a(n+2) +75608*a(n+3) -6740*a(n+4) +105*a(n+5)) for all n in Z. - Michael Somos, Aug 12 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(3*n,n)*binomial(4*n,n) = A005809(n)*A005810(n) = ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(12*n)), where F(x) = 1 + x + 6*x^2 + 105*x^3 + 2448*x^4 + 67043*x^5 + 2028307*x^6 + ... appears to have integer coefficients. Cf. A002894, A002897, A006480, A008977, A186420 and A188662. (End)
a(n) ~ 2^(6*n-1/2)/(Pi*n). - Ilya Gutkovskiy, Jul 12 2016
G.f.: 2*EllipticK(sqrt((sqrt(1-64*x)-1)/(2*sqrt(1-64*x))))/(Pi*(1-64*x)^(1/4)) where EllipticK is the complete elliptic integral of the first kind (in Maple's notation). - Robert Israel, Jul 12 2016
a(n) = Sum_{k = 0..3*n} (-1)^k*C(3*n,k)*C(6*n-k,3*n)*C(2*k,k). - Peter Bala, Feb 10 2018
From Bradley Klee, Feb 27 2018: (Start)
a(n) = A000984(n)*A001448(n).
G.f.: (1/(sqrt(2)*Pi))*Integral_{q=-oo..oo} 1/sqrt(q^2+(1/4)*q^4+(1-64*x)) dq.
G.f.: (1/(2*Pi))*Integral_{phi=0..2*Pi} 1/sqrt(1-64*x*sin^4(phi)) dphi. (End)
From Peter Bala, Mar 20 2022: (Start)
Right-hand side of the following identities valid for n >= 1:
Sum_{k = 0..2*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (4*n)!/((2*n)!*n!^2);
(3/2)*Sum_{k = 0..n} 2*n*(3*n+k-1)!/(k!*n!*(2*n)!) = (4*n)!/((2*n)!*n!^2).
Cf. A001451. (End)
a(n) = (4^n/n!^2)*Product_{k = 0..2*n-1} (2*k + 1). - Peter Bala, Feb 26 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(3*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - Peter Bala, Oct 15 2024

A001449 Binomial coefficients binomial(5n,n).

Original entry on oeis.org

1, 5, 45, 455, 4845, 53130, 593775, 6724520, 76904685, 886163135, 10272278170, 119653565850, 1399358844975, 16421073515280, 193253756909160, 2280012686716080, 26958221130508525

Offset: 0

N. J. A. Sloane

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

T. D. Noe, Table of n, a(n) for n = 0..100

binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A005810 (k = 4), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).

[ Binomial(5*n,n): n in [0..100] ]; // Vincenzo Librandi, Apr 13 2011

Maple
```
f := n->(5*n)!/((4*n)!*(n)!);
```
Mathematica
```
Table[ Binomial[5n, n], {n, 0, 18} ]
```

B(x):=sum(binomial(5*n,n-1)/n*x^n,n,1,30);
taylor(x*diff(B(x),x)/B(x),x,0,10); /* Vladimir Kruchinin, Oct 05 2015 */

a(n) = binomial(5*n, n) \\ Altug Alkan, Oct 05 2015

a(n) = (5*n)!/((4*n)!*(n)!).
a(n) is asymptotic to c*(3125/256)^n/sqrt(n), with c = sqrt(5/(8*Pi)) = 0.44603102903819277863474159... - Benoit Cloitre, Jan 23 2008
a(n) = C(5*n-1,n-1)*C(25*n^2,2)/(3*n*C(5*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
G.f.: A(x) = x*B'(x)/B(x), where B(x)+1 is g.f. of A002294. - Vladimir Kruchinin, Oct 05 2015
From Ilya Gutkovskiy, Jan 16 2017: (Start)
O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1/4,1/2,3/4; 3125*x/256).
E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1/4,1/2,3/4,1; 3125*x/256). (End)
a(n) = hypergeom([-4*n, -n], [1], 1). - Peter Luschny, Mar 19 2018
From Peter Bala, Feb 20 2022: (Start)
4*n(4*n-1)*(4*n-2)*(4*n-3)*a(n) = 5*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*a(n-1).
The o.g.f. A(x) is algebraic: (1 - A(x))*(1 + 4*A(x))^4 +  3125*x*A(x)^5 = 0.
Sum_{n >= 1} a(n)*( x*(4*x + 5)^4/(3125*(1 + x)^5) )^n = x. (End)
From Peter Bala, Oct 17 2024: (Start)
Let G******(x) denote the o.g.f. of sequence A******.
For n >= 1 , a(n) = (5/2) * [x^n] G006013(x)^n.
For n >= 1, a(n) = [x^n] (1 + x)^(5*n) = (5/4) * [x^n] (1/(1 - x))^(4*n) = (5/3) * [x^n] G000108(x)^(3*n) = (5/2) * [x^n] G001764(x)^(2*n) = 5 * [x^n] G002293(x)^n.
a(n) = 5 * [x^n] (1 - G006632(x))^(-n) = (5/2) * [x^n] (1 - x*G006013(x))^(-2*n) =  (5/3) * [x^n] (1 - x*G000108(x))^(-3*n) (apply Concrete Mathematics, equation 5.60, p. 201). (End)

A004355 Binomial coefficient C(6n,n).

Original entry on oeis.org

1, 6, 66, 816, 10626, 142506, 1947792, 26978328, 377348994, 5317936260, 75394027566, 1074082795968, 15363284301456, 220495674290430, 3173734438530120, 45795673964460816, 662252084388541314

Offset: 0

N. J. A. Sloane

a(n) is asymptotic to c*(46656/3125)^n/sqrt(n), with c = sqrt(3/(5*Pi)) = 0.437019372236831628217354... - Benoit Cloitre, Jan 23 2008

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.

T. D. Noe, Table of n, a(n) for n = 0..100
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].

binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A005810 (k = 4), A001449 (k = 5), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).

Cf. A002295

[Binomial(6*n,n): n in [0..100]]; // Vincenzo Librandi, Apr 13 2011

Table[Binomial[6 n, n], {n, 0, 16}] (* Michael De Vlieger, Oct 05 2015 *)

B(x):=sum(binomial(6*n,n-1)/n*x^n,n,1,30);
taylor(x*diff(B(x),x)/B(x),x,0,10); /* Vladimir Kruchinin, Oct 05 2015 */

a(n) = binomial(6*n,n) \\ Altug Alkan, Oct 05 2015

a(n) = C(6*n-1,n-1)*C(36*n^2,2)/(3*n*C(6*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
G.f.: A(x) = x*B'(x)/B(x), where B(x)+1 is g.f. of A002295. - Vladimir Kruchinin, Oct 05 2015
a(n) = GegenbauerC(n, -3*n, -1). - Peter Luschny, May 07 2016
From Ilya Gutkovskiy, Jan 16 2017: (Start)
O.g.f.: 5F4(1/6,1/3,1/2,2/3,5/6; 1/5,2/5,3/5,4/5; 46656*x/3125).
E.g.f.: 5F5(1/6,1/3,1/2,2/3,5/6; 1/5,2/5,3/5,4/5,1; 46656*x/3125). (End)
RHS of identities Sum_{k = 0..n} binomial(3*n, k)*binomial(3*n, n-k) =
Sum_{k = 0..2*n} (-1)^(n+k)*binomial(6*n, k)*binomial(6*n, 2*n-k) = binomial(6*n,n). - Peter Bala, Oct 07 2021
From Peter Bala, Feb 20 2022: (Start)
5*n*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*a(n) = 6*(6*n-1)*(6*n-2)*(6*n-3)(6*n-4)*(6*n-5)*a(n-1).
The o.g.f. A(x) is algebraic: (1 - A(x))*(1 + 5*A(x))^5 +  (6^6)*x*A(x)^6 = 0.
Sum_{n >= 1} a(n)*( x*(5*x + 6)^5/(6^6*(1 + x)^6) )^n = x. (End)

A078995 a(n) = Sum_{k=0..n} C(4k,k)C(4*(n-k),n-k).

Original entry on oeis.org

1, 8, 72, 664, 6184, 57888, 543544, 5113872, 48180456, 454396000, 4288773152, 40503496536, 382701222296, 3617396099936, 34203591636048, 323492394385824, 3060238763412072, 28955508198895584, 274018698082833760, 2593539713410178528, 24550565251665845664

Offset: 0

N. J. A. Sloane, Jan 19 2003

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (Y_n for s=4).
Rui Duarte and António Guedes de Oliveira, Short note on the convolution of binomial coefficients, arXiv:1302.2100 [math.CO], 2013 and J. Int. Seq. 16 (2013) #13.7.6.

See A049235 for more information.
Cf. A006256, A079563, A079678, A079679.
Cf. A005810, A147855, A226733, A226761, A385605, A386811.

series(eval(g/(3*g-4), g=RootOf(g = 1+x*g^4,g))^2, x=0, 30); # Mark van Hoeij, May 06 2013

Table[Sum[Binomial[4*k, k]*Binomial[4*(n - k), n - k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 06 2012 *)

a(n) = sum(k=0, n, binomial(4*k, k)*binomial(4*(n-k), n-k)); \\ Michel Marcus, May 09 2020

a(n) = 2/3*(256/27)^n*(1+c/sqrt(n)+o(n^-1/2)) where c = 2/3*sqrt(2/(3*Pi)) = 0.307105910641187... More generally, a(n, m)=sum(k=0, n, binomial(m*k, k)*binomial(m*(n-k), n-k)) is asymptotic to 1/2*m/(m-1)*(m^m/(m-1)^(m-1))^n. See A000302, A006256 for cases m=2 and 3. - Benoit Cloitre, Jan 26 2003, corrected and extended by Vaclav Kotesovec, Nov 06 2012
243*n*(8*n - 17)*(3*n - 1)*(3*n - 4)*(3*n - 2)*(3*n - 5)*a(n) = 72*(3*n - 5)*(3*n - 4)*(6912*n^4 - 33120*n^3 + 58256*n^2 - 47798*n + 15309)*a(n - 1) - 3072*(2*n - 3)*(6912*n^5 - 55008*n^4 + 175696*n^3 - 282180*n^2 + 227825*n - 73710)*a(n - 2) + 262144*(n - 2)*(4*n - 7)*(2*n - 3)*(2*n - 5)*(4*n - 9)*(8*n - 9)*a(n - 3). - Vladeta Jovovic, Jul 16 2004
Shorter recurrence: 81*n*(3*n-2)*(3*n-1)*(8*n-11)*a(n) = 24*(4608*n^4-14400*n^3+15776*n^2-7346*n+1215)*a(n-1) - 2048*(2*n-3)*(4*n-5)*(4*n-3)*(8*n-3)*a(n-2). - Vaclav Kotesovec, Nov 06 2012
a(n) = Sum_{k=0..n} binomial(4*k+l,k) * binomial(4*(n-k)-l,n-k) for every real number l. - Rui Duarte and António Guedes de Oliveira, Feb 16 2013
From Rui Duarte and António Guedes de Oliveira, Feb 17 2013: (Start)
a(n) = Sum_{k=0..n} 3^(n-k) * binomial(4*n+1,k).
a(n) = Sum_{k=0..n} 4^(n-k) * binomial(3*n+k,k). (End)
G.f.: g^2/(3*g-4)^2 where g=ogf(A002293) satisfies g = 1+x*g^4. - Mark van Hoeij, May 06 2013
a(n) = [x^n] 1/((1-4*x) * (1-x)^(3*n+1)). - Seiichi Manyama, Aug 03 2025
a(n) = Sum_{k=0..n} 4^k * (-3)^(n-k) * binomial(4*n+1,k) * binomial(4*n-k,n-k). - Seiichi Manyama, Aug 15 2025

A224274 a(n) = binomial(4*n,n)/4.

Original entry on oeis.org

1, 7, 55, 455, 3876, 33649, 296010, 2629575, 23535820, 211915132, 1917334783, 17417133617, 158753389900, 1451182990950, 13298522298180, 122131734269895, 1123787895356412, 10358022441395860, 95615237915961100, 883829035553043580, 8179808679272664720, 75788358475481302185

Offset: 1

Gary Detlefs, Apr 02 2013

In general, binomial(k*n,n)/k = binomial(k*n-1,n-1).
Sequences in the OEIS related to this identity are:
. C(2n,n) = A000984, C(2n,n)/2 = A001700;
. C(3n,n) = A005809, C(3n,n)/3 = A025174;
. C(4n,n) = A005810, C(4n,n)/4 = a(n);
. C(5n,n) = A001449, C(5n,n)/5 = A163456;
. C(6n,n) = A004355, C(6n,n)/6 is not in the OEIS.
Conjecture: a(n) == 1 (mod n^3) iff n is an odd prime.
It is known that a(p) == 1(mod p^3) for prime p >= 3. See Mestrovic, Section 3. - Peter Bala, Oct 09 2015

			For n=2, binomial(4*n,n) = binomial(8,2) = 8*7/2 = 28, so a(2) = 28/4 = 7. - _Michael B. Porter_, Jul 12 2016

G. C. Greubel, Table of n, a(n) for n = 1..1000
Dmitry Kruchinin and Vladimir Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.
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. A001700, A001764, A025174, A163456, A002293, A069271, A004331, A005810, A052203, A257633, A262977, A227726.

[Binomial(4*n,n) div 4: n in [1..25]]; // Vincenzo Librandi, Jun 03 2015

Maple
```
seq(binomial(4*n,n)/4, n=1..17);
```

Table[Binomial[4 n, n]/4, {n, 30}] (* Vincenzo Librandi, Jun 03 2015 *)

a(n) = binomial(4*n,n)/4; /* Joerg Arndt, Apr 02 2013 */

a(n) = binomial(4*n,n)/4 = A005810(n)/4.
a(n) = binomial(4*n-1,n-1).
G.f.: A(x) = B'(x)/B(x), where B(x) = 1 + x*B(x)^4 is g.f. of A002293. - Vladimir Kruchinin, Aug 13 2015
From Peter Bala, Oct 08 2015: (Start)
a(n) = 1/2*[x^n] (C(x)^2)^n, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A163456.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 4*x^2 + 22*x^3 + ... is the o.g.f. for A002293.
exp( 2*Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 9*x^2 + 52*x^3 + ... is the o.g.f. for A069271. (End)
From Peter Bala, Nov 04 2015: (Start)
With an offset of 1, the o.g.f. equals f(x)*g(x)^3, 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 A004331 (k = 4). (End)
a(n) = 1/5*[x^n] (1 + x)/(1 - x)^(3*n + 1) = 1/5*[x^n]( 1/C(-x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A227726. - Peter Bala, Jul 12 2016
a(n) ~ 2^(8*n-3/2)*3^(-3*n-1/2)*n^(-1/2)/sqrt(Pi). - Ilya Gutkovskiy, Jul 12 2016
O.g.f.: A(x) = f(x)/(1 - 3*f(x)), where f(x) = series reversion (x/(1 + x)^4) = x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + ... is the o.g.f. of A002293 with the initial term omitted. Cf. A025174. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/3)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+3)*n+k-1,k) = C(4*n,n)/4 and (1/4)*Sum_{k = 0..n} (-1)^k*C(x*n,n-k)*C((x-4)*n+k-1,k) = C(4*n,n)/4, both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
Right-hand side of the identity (1/3)*Sum_{k = 0..2*n} (-1)^k*binomial(5*n-k-1,2*n-k)*binomial(3*n+k-1,k) = binomial(4*n,n)/4. - Peter Bala, Mar 09 2022
a(n) = [x^n] G(x)^n, where G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the g.f. of A001764. - Peter Bala, Oct 17 2024

cp's OEIS Frontend

A208977 Self-convolution square-root of A005810, where A005810(n) = binomial(4*n,n).

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

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A005809 a(n) = binomial(3n,n).

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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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A000897 a(n) = (4*n)! / ((2*n)!*n!^2).

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

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

A000897 a(n) = (4n)! / ((2n)!*n!^2).

A078995 a(n) = Sum_{k=0..n} C(4k,k)C(4*(n-k),n-k).