A110608 -id:A110608 - cp's OEIS frontend

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

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]
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].
Paul Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), Article 13.5.1.
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Naiomi Tuere Cameron, Random walks, trees and extensions of Riordan group techniques, Dissertation, Howard University, 2002.
Tom C. Copeland, Discriminating Deltas, Depressed Equations, and Generalized Catalan Numbers
Maciej Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv:1410.5747 [math.CO], 2014.
Milan Janjic, Two Enumerative Functions.
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 436.
Yaping Liu, On the Recursiveness of Pascal Sequences, Global J. of Pure and Appl. Math. (2022) Vol. 18, No. 1, 71-80.
Mathematics Stack Exchange, Ordinary generating function for binomial(3n,n), Nov 2013.
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
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n+1, k)*binomial(2*n-k, n-k). - Peter Bala, Sep 04 2025

A110609 a(n) = n * binomial(2*n, n-1).

Original entry on oeis.org

0, 1, 8, 45, 224, 1050, 4752, 21021, 91520, 393822, 1679600, 7113106, 29953728, 125550100, 524190240, 2181340125, 9051563520, 37467344310, 154754938800, 637982011590, 2625648168000, 10789623755820, 44277560801760, 181478535620850, 742984788858624, 3038716500907500

Offset: 0

Paul Barry, Jul 30 2005

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

Cf. A000108, A002390, A002736, A253487.

Column k=1 of A110608.

[0] cat [((4*n+4)*(2*n+1)*Binomial(2*n, n)/(n+2))/2: n in [0..25]]; // Vincenzo Librandi, Jan 09 2015

with(combinat):with(combstruct):a[0]:=0:for n from 1 to 30 do a[n]:=sum((count(Composition(n*2+1),size=n)),j=1..n) od: seq(a[n], n=0..22); # Zerinvary Lajos, May 09 2007
a:=n->sum(sum(binomial(2*n,n)/(n+1), j=1..n),k=1..n): seq(a(n), n=0..22); # Zerinvary Lajos, May 09 2007
series(simplify(x*diff(x*diff((1-sqrt(1-4*x))/(2*x), x), x)), x, 20):
seq(coeff(%, x, k), k=0..18); # Karol A. Penson, Apr 25 2025

Table[CatalanNumber[n]*n^2, {n, 0, 22}] (* Zerinvary Lajos, Jul 08 2009 *)
CoefficientList[Series[x (1 / x^2 - (1 - 6 x + 4 x^2) / ((1 - 4 x)^(3/2) x^2)) / 2, {x, 0, 30}], x] (* Vincenzo Librandi, Jan 09 2015 *)

for(n=0,25, print1(n*binomial(2*n,n-1), ", ")) \\ G. C. Greubel, Sep 01 2017

a(n) = n^2*binomial(2*n, n)/(n+1) = n^2*A000108(n) = A002736(n)/(n+1).
G.f.: -(2*x*(2*x+2*sqrt(1-4*x)-3) - sqrt(1-4*x) + 1)/(2*sqrt((1 - 4*x)^3)*x). - Marco A. Cisneros Guevara, Jul 23 2011; amended by Georg Fischer, Apr 09 2020
(n+1)*(10*n-7)*a(n)+2*n*(5*n-88)*a(n-1) -4*(25*n-22)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 07 2012
From Ilya Gutkovskiy, Jan 20 2017: (Start)
E.g.f.: x*(BesselI(0,2*x) + 2*BesselI(1,2*x) + BesselI(2,2*x))*exp(2*x).
a(n) ~ 4^n*sqrt(n)/sqrt(Pi).
Sum_{n>=1} 1/a(n) = Pi*(2*sqrt(3) + Pi)/18 = 1.152911143694148... (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (2/sqrt(5))*log(phi) + 2*log(phi)^2, where log(phi) = A002390. - Amiram Eldar, Feb 20 2021
G.f.: (x*(d/dx))^2 [g.f. of A000108]. - Karol A. Penson, Apr 25 2025

A120986 Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k middle edges (n >= 0, k >= 0).

Original entry on oeis.org

1, 2, 1, 5, 6, 1, 14, 28, 12, 1, 42, 120, 90, 20, 1, 132, 495, 550, 220, 30, 1, 429, 2002, 3003, 1820, 455, 42, 1, 1430, 8008, 15288, 12740, 4900, 840, 56, 1, 4862, 31824, 74256, 79968, 42840, 11424, 1428, 72, 1, 16796, 125970, 348840, 465120, 325584

Offset: 0

Emeric Deutsch, Jul 21 2006

A ternary tree is a rooted tree in which each vertex has at most three children and each child of a vertex is designated as its left or middle or right child.
T(n,k) is the number of Dyck paths of semilength 2n+2 with all descent runs of even length and n+1-k peaks. - Alexander Burstein, May 23 2020
T(n,k) is the number of Dyck paths of semilength 2n+2 with all descent runs of even length and k+1 peaks at even height. - Alexander Burstein, Jun 03 2020
T(n,k) is the number of Dyck paths of semilength 2n+2 with all descent runs of even length and k peaks at odd height. - Alexander Burstein, Jun 18 2020
An apparent refinement is A338135. - Tom Copeland, Oct 12 2020

			Triangle starts:
    1;
    2,   1;
    5,   6,   1;
   14,  28,  12,   1;
   42, 120,  90,  20,  1;
  132, 495, 550, 220, 30, 1;
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1274
Alexander Burstein, Distribution of peak heights modulo k and double descents on k-Dyck paths, arXiv:2009.00760 [math.CO], 2020.
Alexander Burstein and Megan Martinez, Pattern classes equinumerous to the class of ternary forests, Permutation Patterns Virtual Workshop, Howard University (2020).
Helmut Prodinger, Counting ternary trees according to the number of middle edges and factorizing into (3/2)-ary trees, arXiv:2009.06793 [math.CO], 2020.
Helmut Prodinger, Counting edges according to edge-type in t-ary trees, arXiv:2205.13374 [math.CO], 2022.
Chao-Jen Wang, Applications of the Goulden-Jackson cluster method to counting Dyck paths by occurrences of subwords, Dissertation, Brandeis University, 2011.

Cf. A001764 (row sums), A000108, A108767, A013698, A110608.

T:=(n,k)->binomial(n+1,k)*binomial(2*(n+1),n-k)/(n+1): for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_, k_] := Binomial[n+1, k]*Binomial[2*(n+1), n-k]/(n+1);
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 06 2018 *)

T(n,k) = binomial(n+1,k)*binomial(2*(n+1),n-k)/(n+1); \\ Andrew Howroyd, Nov 06 2017

from sympy import binomial
def T(n, k): return binomial(n + 1, k)*binomial(2*(n + 1), n - k)//(n + 1)
for n in range(21): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Nov 07 2017

T(n,k) = (1/(n+1))*binomial(n+1,k)*binomial(2*(n+1),n-k).
T(n,0) = A000108(n+1) (the Catalan numbers).
T(n,k) = A108767(n+1,n+1-k).
Sum_{k>=1} k*T(n,k) = binomial(3*n+2,n-1) = A013698(n).
G.f.: G = G(t,z) satisfies G = (1+t*z*G)(1+z*G)^2.
O.g.f.: A(x,t) = 1 + (2 + t)*x + (5 + 6*t + t^2)*x^2 + ... satisfies 1 + x*d/dx(A(x,t))/A(x,t) = 1 + (2 + t)*x + (6 + 8*t + t^2)*x^2 + ..., which is the o.g.f. for A110608. - Peter Bala, Oct 13 2015

A357613 Triangle read by rows T(n, k) = binomial(2n, k) binomial(3n - k, 2n).

Original entry on oeis.org

1, 3, 2, 15, 20, 6, 84, 168, 105, 20, 495, 1320, 1260, 504, 70, 3003, 10010, 12870, 7920, 2310, 252, 18564, 74256, 120120, 100100, 45045, 10296, 924, 116280, 542640, 1058148, 1113840, 680680, 240240, 45045, 3432

Offset: 0

F. Chapoton, Oct 06 2022

Each line should be the f-vector of a cellular complex. The sequence seems to give the coefficients in a binomial basis of the integer-valued polynomials (x+1)*(x+2)*...*(x+2*n)*(x+1)*(x+2)*...*(x+n)/(n!*(2n)!).

The precise expansion is (x+1)*(x+2)*...*(x+2*n)*(x+1)*(x+2)*...*(x+n)/(n!*(2*n)!) = Sum_{k = 0..n} (-1)^k*T(n,k)*binomial(x+3*n-k, 3*n-k), as can be verified using the WZ algorithm. For example, n = 3 gives (x+1)^2*(x+2)^2*(x+3)^2*(x+4)*(x+5)*(x+6)/(3!*6!) = 84*binomial(x+9, 9) - 168*binomial(x+8, 8) + 105*binomial(x+7, 7) - 20*binomial(x+6, 6). - Peter Bala, Jun 25 2023

			As a triangle of numbers, this starts with
  1;
  3, 2;
  15, 20, 6;
  84, 168, 105, 20;
  495, 1320, 1260, 504, 70.
Here is an example for n=1 as coefficients (up to sign) in the binomial basis of integer-valued polynomials:
(x+1)*(x+2)*(x+1)/2 = 3*binomial(x+3,3)-2*binomial(x+2,2).

Row sums A026000. Cf. A000984, A005809 (k=0), A144485 (k=1), A033282, A110608, A243660.

A357613 := proc(n,k)
    binomial(2*n,k)*binomial(3*n-k,2*n) ;
end proc:
seq(seq(A357613(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Jul 06 2023

Table[Binomial[2n,k]Binomial[3n-k,2n],{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Oct 11 2023 *)

def a(n):
    return [binomial(2 * n, k) * binomial(3 * n - k, 2 * n)
            for k in range(n + 1)]

T(n,k) = binomial(2*n, k) * binomial(3*n - k, 2*n) for 0 <= k <= n

A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 10, 15, 1, 21, 84, 84, 1, 36, 270, 660, 495, 1, 55, 660, 2860, 5005, 3003, 1, 78, 1365, 9100, 27300, 37128, 18564, 1, 105, 2520, 23800, 107100, 244188, 271320, 116280, 1, 136, 4284, 54264, 339150, 1139544, 2089164, 1961256, 735471, 1, 171, 6840, 111720, 921690, 4239774, 11306064

Offset: 0

Peter Bala, Feb 13 2024

Compare with A063007(n, k) = binomial(n, k)*binomial(n+k, k), the table of coefficients of the shifted Legendre polynomials P(n, 2*x + 1).

			Triangle begins
n\k| 0    1     2      3       4       5       6       7
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -
 0 | 1
 1 | 1    3
 2 | 1   10    15
 3 | 1   21    84     84
 4 | 1   36   270    660     495
 5 | 1   55   660   2860    5005    3003
 6 | 1   78  1365   9100   27300   37128   18564
 7 | 1  105  2520  23800  107100  244188  271320  116280
 ...

A114496 (row sums), A000984 (alt. row sums unsigned), A005809 (main diagonal), A090763 (first subdiagonal), A014105 (column 1).

Cf. A026000, A063007, A102537, A110608, A339710.

seq(print(seq(binomial(n, k)*binomial(2*n+k, k), k = 0..n)), n = 0..10);

n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(2*n+k, k)*x^k = (1 + x)^n * Sum_{k = 0..n} binomial(n, k)*binomial(2*n, k)*(x/(1 + x))^k = Sum_{k = 0..n} A110608(n, n-k)*x^k*(1 + x)^(n-k).
(x - 1)^n * R(n, 1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)*binomial(2*n, n-k)*x^k = the n-th row polynomial of A110608.
R(n, x) = hypergeom([-n, 2*n + 1], [1], -x).
Second-order differential equation: ( (1 + x)^n * (x + x^2)*R(n, x)' )' = n*(2*n + 1)*(1 + x)^n * R(n, x), where the prime indicates differentiation w.r.t. x.
Equivalently, x*(1 + x)*R(n, x)'' + ((n + 2)*x + 1)*R(n, x)' - n*(2*n + 1)*R(n, x)' = 0.
Analog of Rodrigues' formula for the shifted Legendre polynomials:
R(n, x) = 1/(1 + x)^n * 1/n! * (d/dx)^n (x*(1 + x)^2)^n.
Analog of Rodrigues' formula for the Legendre polynomials:
R(n, (x-1)/2) = 1/(n!*2^n) * 1/(1 + x)^n *(d/dx)^n ((x - 1)*(x + 1)^2)^n.
Orthogonality properties:
Integral_{x = -1..0} (1 + x)^n * R(n, x) * R(m, x) dx = 0 for n > m.
Integral_{x = -1..0} (1 + x)^n * R(n, x)^2 dx = 1/(3*n + 1).
Integral_{x = -1..0} (1 + x)^(n+m) * R(n, x) * R(m, x) dx = 0 for m >= 2*n + 1 or m <= (n - 1)/2.
Integral_{x = -1..0} (1 + x)^k * R(n, x) dx = 0 for n <= k <= 2*n - 1;
Integral_{x = -1..0} (1 + x)^(2*n) * R(n, x) dx = (2*n)!*n!/(3*n+1)! = 1/A090816(n).
Recurrence for row polynomials:
2*n*(2*n - 1)*((9*n - 12)*x + 8*n - 11)*(1 + x)*R(n, x) = (9*(3*n - 1)*(3*n - 2)*(3*n - 4)*x^3 + 3*(3*n - 1)*(3*n - 2)*(20*n - 27)*x^2 + 6*(3*n - 2)*(20*n^2 - 34*n + 9)*x + 2*(32*n^3 - 76*n^2 + 50*n - 9))*R(n-1, x) - 2*(n - 1)*(2*n - 3)*((9*n - 3)*x + 8*n - 3)*R(n-2, x), with R(0, x) = 1, R(1, x) = 1 + 3*x.
Conjecture: exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + 3*t)*x + (1 + 8*t + 12*t^2)*x^2 + ... is the o.g.f. for A102537. If true, then it would follows that, for each integer t, the sequence u = {R(n,t) : n >= 0} satisfies the Gauss congruences u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all primes p and positive integers m and r.
R(n, 1) = A114496(n); R(n, -1) = (-1)^n * A000984(n).
R(n, 2) = A339710(n); R(n, -2) = (-1)^n * A026000(n).
(2^n)*R(n, -1/2) = A234839(n).

cp's OEIS Frontend

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

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A110609 a(n) = n * binomial(2*n, n-1).

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A120986 Triangle read by rows: T(n,k) is the number of ternary trees with n edges and having k middle edges (n >= 0, k >= 0).

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A357613 Triangle read by rows T(n, k) = binomial(2*n, k) * binomial(3*n - k, 2*n).

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A370258 Triangle read by rows: T(n, k) = binomial(n, k)*binomial(2*n+k, k), 0 <= k <= n.

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A357613 Triangle read by rows T(n, k) = binomial(2n, k) binomial(3n - k, 2n).

A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.