A045721 -id:A045721 - cp's OEIS frontend

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

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

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

Original entry on oeis.org

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

Offset: 0

Ralf Stephan, Mar 21 2004

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

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

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

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

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

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

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

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

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

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

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

Diagonal sums comment corrected by Paul Barry, Apr 14 2010

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

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

Original entry on oeis.org

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

Offset: 0

Thomas Wieder, Sep 29 2009

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Original entry on oeis.org

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

Offset: 0

Fabian Rothelius, Feb 04 2001

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Edited by M. F. Hasler, Mar 05 2017

A004319 a(n) = binomial(3*n, n - 1).

Original entry on oeis.org

1, 6, 36, 220, 1365, 8568, 54264, 346104, 2220075, 14307150, 92561040, 600805296, 3910797436, 25518731280, 166871334960, 1093260079344, 7174519270695, 47153358767970, 310325523515700, 2044802197953900, 13488561475572645, 89067326568860640, 588671286046028640

Offset: 1

N. J. A. Sloane

Milton Abramowitz and Irene 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
Milan Janjic, Two Enumerative Functions.
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Emanuele Munarini, Shifting Property for Riordan, Sheffer and Connection Constants Matrices, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.2.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.

Cf. A045721 (k=1), A025174 (k=2), A236194 (k=4), A013698 (k=5), A165817 (k=-1), A117671 (k=-2).

Cf. A000139, A001764, A005809, A006013, A236194, A001791, A004331, A004343, A004356, A004369, A004382.

A004319 := proc(n)
binomial(3*n,n-1);
end proc: # R. J. Mathar, Aug 10 2015

Table[Binomial[3n, n - 1], {n, 20}] (* Harvey P. Dale, Sep 21 2011 *)

a(n):=sum((binomial(3*i-1,2*i-1)*binomial(3*n-3*i-3,2*n-2*i-2))/(2*n-2*i-1),i,1,n-1)/2; /* Vladimir Kruchinin, May 15 2013 */

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

G.f.: (g-1)/(1-3*z*g^2), where g = g(z) is given by g = 1 + z*g^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
a(n) = Sum_{i=0..n-1} binomial(i+2*n, i). - Ralf Stephan, Jun 03 2005
D-finite with recurrence -2*(2*n+1)*(n-1)*a(n) + 3*(3*n-1)*(3*n-2)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
a(n) = (1/2) * Sum_{i=1..n-1} binomial(3*i - 1, 2*i - 1)*binomial(3*n - 3*i - 3, 2*n - 2*i - 2)/(2*n - 2*i - 1). - Vladimir Kruchinin, May 15 2013
G.f.: x*hypergeom2F1(5/3, 4/3; 5/2; 27x/4). - R. J. Mathar, Aug 10 2015
a(n) = n*A001764(n). - R. J. Mathar, Aug 10 2015
From Peter Bala, Nov 04 2015: (Start)
With offset 0, the o.g.f. equals f(x)*g(x)^3, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k, n). See the cross-references. (End)
G.f.: cos(t)/(2*sqrt(1 - (27*x)/4)) - sin(t)/(sqrt(3)*sqrt(x)), where t = arcsin((sqrt(27*x))/2)/3. - Vladimir Kruchinin, May 13 2016
a(n) = [x^(2*n+1)] 1/(1 - x)^n. - Ilya Gutkovskiy, Oct 10 2017
a(n) = binomial(n+1, 2) * A000139(n). - F. Chapoton, Feb 23 2024

A013698 a(n) = binomial(3*n+2, n-1).

Original entry on oeis.org

1, 8, 55, 364, 2380, 15504, 100947, 657800, 4292145, 28048800, 183579396, 1203322288, 7898654920, 51915526432, 341643774795, 2250829575120, 14844575908435, 97997533741800, 647520696018735, 4282083008118300

Offset: 1

Joachim.Rosenthal(AT)nd.edu (Joachim Rosenthal), Emeric Deutsch

Degree of variety K_{2,n}^1. Also number of double-rises (or odd-level peaks) in all generalized {(1,2),(1,-1)}-Dyck paths of length 3(n+1).
Number of dissections of a convex (2n+2)-gon by n-2 noncrossing diagonals into (2j+2)-gons, 1<=j<=n-1.
a(n) is the number of lattice paths from (0,0) to (3n+1,n-1) avoiding two consecutive up-steps. - Shanzhen Gao, Apr 20 2010

Robert Israel, Table of n, a(n) for n = 1..1000
M. S. Ravi et al., Dynamic Pole Assignment and Schubert Calculus, SIAM J. Control Optimization, 34 (1996), 813-832, esp. p. 825.
Daniel W. Stasiuk, An Enumeration Problem for Sequences of n-ary Trees Arising from Algebraic Operads, Master's Thesis, University of Saskatchewan-Saskatoon (2018).

Cf. A013699 (q=2), A013700 (q=3), A013701 (q=4), A013702 (q=5).

A column of triangle A102537. Cf. A001764, A004319, A005809, A006013, A025174, A045721, A117671, A165817, A236194.

List([1..25], n-> Binomial(3*n+2, n-1)) # G. C. Greubel, Mar 21 2019

[Binomial(3*n+2, n-1): n in [1..25]]; // Vincenzo Librandi, Aug 10 2015

seq(binomial(3*n+2,n-1), n=0..30); # Robert Israel, Aug 09 2015

Table[Binomial[3*n+2, n-1], {n, 25}] (* Arkadiusz Wesolowski, Apr 02 2012 *)

first(m)=vector(m,n,binomial(3*n+2, n-1)); /* Anders Hellström, Aug 09 2015 */

[binomial(3*n+2, n-1) for n in (1..25)] # G. C. Greubel, Mar 21 2019

G.f.: g/((g-1)^3*(3*g-1)) where g*(1-g)^2 = x. - Mark van Hoeij, Nov 09 2011
a(n) = Sum_{k=0..n-1} binomial(2*n+k+2,k). - Arkadiusz Wesolowski, Apr 02 2012
D-finite with recurrence 2*(2*n+3)*(n+1)*a(n) -n*(67*n+34)*a(n-1) +30*(3*n-1)*(3*n-2)*a(n-2)=0. - R. J. Mathar, Feb 05 2013
a(n+1) = (3*n+5)*(3*n+4)*(3*n+3)*a(n)/((2*n+5)*(2*n+4)*n). - Robert Israel, Aug 09 2015
With offset 0, the o.g.f. equals f(x)*g(x)^5, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A165817 (k = -1), A117671 (k = -2). - Peter Bala, Nov 04 2015

A117671 a(n) = binomial(3*n+1, n+1).

Original entry on oeis.org

1, 6, 35, 210, 1287, 8008, 50388, 319770, 2042975, 13123110, 84672315, 548354040, 3562467300, 23206929840, 151532656696, 991493848554, 6499270398159, 42671977361650, 280576272201225

Offset: 0

Zerinvary Lajos, Apr 12 2006

a(n) = A258993(2*n+1, n). - Reinhard Zumkeller, Jun 22 2015

			if n=0 then C(3*0+1,0+1) = C(1,1) = 1.
if n=10 then C(3*10+1,10+1) = C(31,11) = 84672315.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions

Cf. A025174: binomial(3n-1,n-1), A006013.

Cf. A258993, A001764, A004319, A005809, A013698, A045721, A165817, A236194.

a117671 n = a258993 (2 * n + 1) n  -- Reinhard Zumkeller, Jun 22 2015

seq(binomial(3*n+1,n+1),n=0..30); # Robert Israel, Oct 10 2017

Table[Binomial[3n+1,n+1],{n,0,20}] (* Harvey P. Dale, Jul 19 2011 *)

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

G.f.: (2*(-1+Hypergeometric2F1[-(1/3),1/3,-(1/2),(27*x)/4]))/(3*x). - Harvey P. Dale, Jul 19 2011
G.f.: A(x) = B'(x)/B(x)-B'(x)-1/x, where B(x) = 4/3*sin(1/3*asin(sqrt((27*x)/4)))^2. - Vladimir Kruchinin, Nov 26 2014
From Peter Bala, Nov 04 2015: (Start)
With an extra initial term equal to 1, 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.
More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1). (End)
a(n) = [x^(2*n)] 1/(1 - x)^(n+2). - Ilya Gutkovskiy, Oct 10 2017
a(n+1) = 3*(3*n+2)*(3*n+4)*a(n)/(2*(n+2)*(2*n+1)). - Robert Israel, Oct 10 2017

A264772 Triangle T(n,k) = binomial(3n - 2k, 2*n - k), 0 <= k <= n.

Original entry on oeis.org

1, 3, 1, 15, 4, 1, 84, 21, 5, 1, 495, 120, 28, 6, 1, 3003, 715, 165, 36, 7, 1, 18564, 4368, 1001, 220, 45, 8, 1, 116280, 27132, 6188, 1365, 286, 55, 9, 1, 735471, 170544, 38760, 8568, 1820, 364, 66, 10, 1, 4686825, 1081575, 245157, 54264, 11628, 2380, 455, 78, 11, 1

Offset: 0

Peter Bala, Nov 24 2015

Riordan array (f(x), x*g(x)), where g(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + ... is the o.g.f. for A001764 and f(x) = g(x)/(3 - 2*g(x)) = 1 + 3*x + 15*x^2 + 84*x^3 + 495*x^4 + ... is the o.g.f. for A005809.
The even numbered columns give the Riordan array A119301, the odd numbered columns give the Riordan array A144484. A159841 is the array formed from columns 1,4,7,10,....
More generally, if R = (R(n,k))n,k>=0 is a proper Riordan array, m is a nonnegative integer and a > b are integers then the array with (n,k)-th element R((m + 1)*n - a*k, m*n - b*k) is also a Riordan array (not necessarily proper). Here we take R as Pascal's triangle and m = a = 2, b = 1. See A092392, A264773, A264774 and A113139 for further examples.

			Triangle begins
.n\k.|......0.....1....2....3...4..5...6..7...
----------------------------------------------
..0..|      1
..1..|      3     1
..2..|     15     4    1
..3..|     84    21    5    1
..4..|    495   120   28    6   1
..5..|   3003   715  165   36   7  1
..6..|  18564  4368 1001  220  45  8  1
..7..| 116280 27132 6188 1365 286 55  9  1
...

Michael De Vlieger, Table of n, a(n) for n = 0..11475
Peter Bala, A 4-parameter family of embedded Riordan arrays
Paul Barry, On the halves of a Riordan array and their antecedents, arXiv:1906.06373 [math.CO], 2019.
E. Lebensztayn, On the asymptotic enumeration of accessible automata, Discrete Mathematics and Theoretical Computer Science, Vol. 12, No.3, 2010, 75-80, Section 2.
R. Sprugnoli, An Introduction to Mathematical Methods in Combinatorics, CreateSpace Independent Publishing Platform 2006, Section 5.6, ISBN-13: 978-1502925244.

Cf. A005809 (column 0), A045721 (column 1), A025174 (column 2), A004319 (column 3), A236194 (column 4), A013698 (column 5). Cf. A001764, A007318, A092392, A119301 (C(3n-k,2n)), A144484 (C(3n+1-k,2n+1)), A159841 (C(3n+1,2n+k+1)), A264773, A264774.

/* As triangle */ [[Binomial(3*n-2*k, n-k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Dec 02 2015

A264772:= proc(n,k) binomial(3*n - 2*k, 2*n - k); end proc:
seq(seq(A264772(n,k), k = 0..n), n = 0..10);

Table[Binomial[3 n - 2 k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 01 2015 *)

T(n,k) = binomial(3*n - 2*k, n - k).

O.g.f.: f(x)/(1 - t*x*g(x)), where f(x) = Sum_{n >= 0} binomial(3*n,n)*x^n and g(x) = Sum_{n >= 0} 1/(2*n + 1)*binomial(3*n,n)*x^n.

A236194 a(n) = binomial(3n+1, n-1).

Original entry on oeis.org

1, 7, 45, 286, 1820, 11628, 74613, 480700, 3108105, 20160075, 131128140, 854992152, 5586853480, 36576848168, 239877544005, 1575580702584, 10363194502115, 68248282427325, 449972009097765, 2969831763694950, 19619725782651120, 129728497393775280

Offset: 1

Bruno Berselli, Jan 20 2014

This sequence is related to A006013 by a(n)/n = A006013(n)/2.

Bruno Berselli, Table of n, a(n) for n = 1..100
Emanuele Munarini, Shifting Property for Riordan, Sheffer and Connection Constants Matrices, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.2.

Cf. A006013; A025174: C(3n-1, n-1); A117671: C(3n+1, n+1).
Second column of the triangle A159841.
Third column of the triangle A119301.
Cf. A001764, A004319, A005809, A013698, A045721, A165817.
Cf. A045721, A117671.

Magma
```
[Binomial(3*n+1,n-1): n in [1..30]];
```
Mathematica
```
Table[Binomial[3n+1, n-1], {n, 30}]
```

makelist(binomial(3*n+4,n),n,0,40); /* Emanuele Munarini, Oct 14 2014 */

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

[binomial(3*n+1,n-1) for n in range(1,31)] # G. C. Greubel, Nov 09 2022

G.f.: (sqrt(4-27*x)*cos((2/3)*arcsin((3/2)*sqrt(3*x))) + sqrt(3*x)*sin((2/3)*arcsin((3/2)*sqrt(3*x))) - sqrt(4-27*x))/(3*sqrt(4-27*x)*x^2). - Emanuele Munarini, Oct 14 2014
From Peter Bala, Nov 04 2015: (Start)
With offset 0, the o.g.f. equals f(x)*g(x)^4, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764.
More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
a(n) = [x^n] x/(1 - x)^(2*n+3). - Ilya Gutkovskiy, Oct 10 2017
From Karol A. Penson, Mar 02 2024: (Start)
G.f.: ((sqrt(3)*sqrt(x)*i + sqrt(4 - 27*x))*(4*sqrt(4 - 27*x) - 12*i*sqrt(3)*sqrt(x))^(2/3) + (-sqrt(3)*sqrt(x)*i + sqrt(4 - 27*x))*(4*sqrt(4 - 27*x) + 12*i*sqrt(3)*sqrt(x))^(2/3) - 8*sqrt(4 - 27*x))/(24*sqrt(4 - 27*x)*x^2), where i is the imaginary unit, i=sqrt(-1).
G.f.: hypergeometric3F2([5/3,2,7/3],[5/2,3],27*x/4).
G.f. = G satisfies the algebraic equation: 1 + (7*z-1)*G + (27*z-4)*z^2*G^2 + (27*z-4)*z^4*G^3 = 0. (End)

A343832 a(n) = Sum_{k=0..n} k! * binomial(n,k) * binomial(2*n+1,k).

Original entry on oeis.org

1, 4, 31, 358, 5509, 106096, 2456299, 66471826, 2059640713, 71920704124, 2794938616471, 119653108240414, 5595650767265101, 283841520215780008, 15523069639558351459, 910529206043204428426, 57023540590242398853649, 3797750659849704886903156, 268025698704886063968108943

Offset: 0

Seiichi Manyama, May 01 2021

nonn

Let A(x) be the e.g.f. of this sequence, and B(x) be the e.g.f. of A082545, then A(x)/B(x) = C(x) where C(x) = 1 + x*C(x)^2 is the Catalan function (A000108). This follows from the fact that this sequence and A082545 form adjacent semi-diagonals of table A088699. - Paul D. Hanna, Aug 16 2022

Seiichi Manyama, Table of n, a(n) for n = 0..365
Wikipedia, Laguerre polynomials
Index entries for sequences related to Laguerre polynomials

Cf. A000522, A045721, A082545, A152059, A248668, A343830, A343831, A000108.

[Factorial(n)*Evaluate(LaguerrePolynomial(n, n+1), -1): n in [0..40]]; // G. C. Greubel, Aug 11 2022

a := n -> add(k!*binomial(n, k)*binomial(2*n+1, k), k=0..n):
a := n -> n!*add(binomial(2*n+1, k)/(n-k)!, k=0..n):
a := n -> (-1)^n*KummerU(-n, n+2, -1):
a := n -> n!*LaguerreL(n, n+1, -1): # Peter Luschny, May 02 2021

a[n_] := Sum[k! * Binomial[n, k] * Binomial[2*n+1, k], {k, 0, n}]; Array[a, 20, 0] (* Amiram Eldar, May 01 2021 *)
Table[(-1)^n * HypergeometricU[-n, 2 + n, -1], {n, 0, 20}] (* Vaclav Kotesovec, May 02 2021 *)

a(n) = sum(k=0, n, k!*binomial(n, k)*binomial(2*n+1, k));

a(n) = (2*n+1)!*sum(k=0, n, binomial(n, k)/(k+n+1)!);

a(n) = n!*sum(k=0, n, binomial(2*n+1, k)/(n-k)!);

PARI
```
a(n) = n!*pollaguerre(n, n+1, -1);
```

[factorial(n)*gen_laguerre(n, n+1, -1) for n in (0..40)] # G. C. Greubel, Aug 11 2022

a(n) = (2*n+1)! * Sum_{k=0..n} binomial(n,k)/(k+n+1)!.
a(n) = n! * Sum_{k=0..n} binomial(2*n+1,k)/(n-k)!.
a(n) = n! * LaguerreL(n, n+1, -1).
a(n) = n! * [x^n] exp(x/(1 - x))/(1 - x)^(n+2).
a(n) == 1 (mod 3).
a(n) ~ 2^(2*n + 3/2) * n^n / exp(n-1). - Vaclav Kotesovec, May 02 2021
From Paul D. Hanna, Aug 16 2022: (Start)
E.g.f.: exp( (1-2*x - sqrt(1-4*x))/(2*x) ) * (1 - sqrt(1-4*x)) / (2*x*sqrt(1-4*x)), derived from the e.g.f for A082545 given by Mark van Hoeij.
E.g.f.: exp(C(x) - 1) * C(x) / sqrt(1-4*x), where C(x) = (1 - sqrt(1-4*x))/(2*x) is the Catalan function (A000108). (End)

cp's OEIS Frontend

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

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

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

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

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A004319 a(n) = binomial(3*n, n - 1).

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A264772 Triangle T(n,k) = binomial(3n - 2k, 2*n - k), 0 <= k <= n.