A005809 -id:A005809 - cp's OEIS frontend

A005810 a(n) = binomial(4n,n).

1, 4, 28, 220, 1820, 15504, 134596, 1184040, 10518300, 94143280, 847660528, 7669339132, 69668534468, 635013559600, 5804731963800, 53194089192720, 488526937079580, 4495151581425648, 41432089765583440, 382460951663844400

Offset: 0

N. J. A. Sloane

Start off with 0 balls in a box. Find the number of ways you can throw 3 balls back out. Then continue to throw 4 balls into the box after each stage. (I.e., the first stage is 0. Then at the next stage there are 4 ways to throw 3 balls back out.) - Ruppi Rana (ruppirana007(AT)hotmail.com), Mar 03 2004
Central coefficients of A094527. - Paul Barry, Mar 08 2011
This is the case m = 2n in Catalan's formula (2m)!*(2n)!/(m!*(m+n)!*n!) - see Umberto Scarpis in References. - Bruno Berselli, Apr 27 2012
A generating function in terms of a (labyrinthine) solution to a depressed quartic equation is given in the Copeland link for signed A005810. - Tom Copeland, Oct 10 2012
Conjecture: a(n) == 4 (mod n^3) iff n is prime. - Gary Detlefs, Apr 03 2013
For prime p, the congruence a(p) = binomial(4*p,p) = 4 (mod p^3)  is a known generalization of Wolstenholme's theorem. See Mestrovic, Section 6, equation 35. - Peter Bala, Dec 28 2014

			G.f. = 1 + 4*x + 28*x^2 + 220*x^3 + 1820*x^4 + 15504*x^5 + 134596*x^6 + ...

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.
Umberto Scarpis, Sui numeri primi e sui problemi dell'analisi indeterminata in Questioni riguardanti le matematiche elementari, Nicola Zanichelli Editore (1924-1927, third Edition), page 11.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000 (terms 0..100 from T. D. Noe, terms 101..213 from Muniru A Asiru)
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].
Tom Copeland, Discriminating Deltas, Depressed Equations, and Generalized Catalan Numbers, 2012, pp. 5-6.
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Ruppi Rana, Title? [Broken link]

Row 4 of A060539.
binomial(k*n,n): A000984 (k = 2), A005809 (k = 3), A001449 (k = 5), A004355 (k = 6), A004368 (k = 7), A004381 (k = 8), A169958 - A169961 (k = 9 thru 12).
Cf. A007318, A182400, A262261, A378806, A378807.

List([0..20],n->Binomial(4*n,n)); # Muniru A Asiru, Mar 19 2018

a005810 n = a007318 (4*n) n  -- Reinhard Zumkeller, Mar 04 2012

[ Binomial(4*n,n): n in [0..100] ]; // Vincenzo Librandi, Apr 13 2011

seq(binomial(4*n,n),n=0..20); # Muniru A Asiru, Mar 19 2018

Table[Binomial[4n,n],{n,0,19}] (* Geoffrey Critzer, Sep 15 2013 *)

a(n) = binomial(4*n, n); \\ Altug Alkan, Mar 19 2018

from math import comb
def A005810(n): return comb(n<<2,n) # Chai Wah Wu, Aug 01 2023

a(n) is asymptotic to c*(256/27)^n/sqrt(n) with c = sqrt(2 / (3 Pi)) = 0.460658865961780639... - Benoit Cloitre, Jan 26 2003; corrected by Charles R Greathouse IV, Dec 14 2006
a(n) = Sum_{k=0..2n} binomial(2n,k) * binomial(2n,k-n). - Paul Barry, Mar 08 2011
G.f.: g/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
D-finite with recurrence: 3*n*(3*n-1)*(3*n-2)*a(n) - 8*(4*n-3)*(2*n-1)*(4*n-1)*a(n-1) = 0. - R. J. Mathar, Dec 02 2012
a(n) = binomial(4*n,n-1)*(3*n+1)/n. - Gary Detlefs, Apr 03 2013
a(n) = C(4*n-1,n-1)*C(16*n^2,2)/(3*n*C(4*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
a(n) = Sum_{i,j,k = 0..n} binomial(n,i)*binomial(n,j)*binomial(n,k)* binomial(n,i+j+k). - Peter Bala, Dec 28 2014
a(n) = GegenbauerC(n, -2*n, -1). - Peter Luschny, May 07 2016
From Ilya Gutkovskiy, Nov 22 2016: (Start)
O.g.f.: 3F2(1/4,1/2,3/4; 1/3,2/3; 256*x/27).
E.g.f.: 3F3(1/4,1/2,3/4; 1/3,2/3,1; 256*x/27). (End)
a(n) = hypergeom([-3*n, -1*n], [1], 1). - Peter Luschny, Mar 19 2018
RHS of the identity Sum_{k = 0..2*n} (-1)^(n+k)*binomial(4*n, k)* binomial(4*n, 2*n-k) = binomial(4*n,n). - Peter Bala, Oct 07 2021
From Peter Bala, Feb 20 2022: (Start)
The o.g.f. A(x) satisfies the differential equation
(-256*x^3 + 27*x^2)*A(x)''' + (-1152*x^2 + 54*x)*A(x)'' + (-816*x + 6)*A(x)' - 24*A(x) = 0 with A(0) = 1, A'(0) = 4 and A''(0) = 56.
Algebraic equation: (1 - A(x))*(1 + 3*A(x))^3 +  256*x*A(x)^4 = 0.
Sum_{n >= 1} a(n)*( x*(3*x + 4)^3/(256*(1 + x)^4) )^n = x. (End)
From Amiram Eldar, Dec 07 2024: (Start)
Sum_{n>=1} 1/a(n) = A378806.
Sum_{n>=1} (-1)^n/a(n) = A378807. (End)
From Peter Bala, Jun 29 2025: (Start)
a(n) = (1/8)^n * Sum_{k = n..4*n} binomial(k, n) * binomial(4*n, k).
Sum_{n >= 0 } a(n)*(1/128)^n = (1/5)*(sqrt(2) + sqrt(7 + 5*sqrt(2))). (End)
From Seiichi Manyama, Aug 16 2025: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(4*n+1,k).
G.f.: 1/(1 - 4*x*g^3) where g = 1+x*g^4 is the g.f. of A002293. (End)

More terms from Henry Bottomley, Oct 06 2000

Corrected by T. D. Noe, Jan 16 2007

A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 15, 7, 1, 1, 15, 35, 28, 9, 1, 1, 21, 70, 84, 45, 11, 1, 1, 28, 126, 210, 165, 66, 13, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1

Offset: 0

Philippe Deléham, Aug 14 2003

Coefficient array for Morgan-Voyce polynomial b(n,x). A053122 (unsigned) is the coefficient array for B(n,x). Reversal of A054142. - Paul Barry, Jan 19 2004
This triangle is formed from even-numbered rows of triangle A011973 read in reverse order. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k+1 peaks. T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k peaks at height >= 2. T(n,k) is the number of directed column-convex polyominoes of area n+1, having k+1 columns. - Emeric Deutsch, May 31 2004
Riordan array (1/(1-x), x/(1-x)^2). - Paul Barry, May 09 2005
The triangular matrix a(n,k) = (-1)^(n+k)*T(n,k) is the matrix inverse of A039599. - Philippe Deléham, May 26 2005
The n-th row gives absolute values of coefficients of reciprocal of g.f. of bottom-line of n-wave sequence. - Floor van Lamoen (fvlamoen(AT)planet.nl), Sep 24 2006
Unsigned version of A129818. - Philippe Deléham, Oct 25 2007
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k >=1 (height(alpha) = |Im(alpha)|) and of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
A085478 is jointly generated with A078812 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x)+x*v(n-1)x and v(n,x) = u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Per Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle given by (0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 26 2012
T(n,k) is also the number of compositions (ordered partitions) of 2*n+1 into 2*k+1 parts which are all odd. Proof: The o.g.f. of column k, x^k/(1-x)^(2*k+1) for k >= 0, is the o.g.f. of the odd-indexed members of the sequence with o.g.f. (x/(1-x^2))^(2*k+1) (bisection, odd part). Thus T(n,k) is obtained from the sum of the multinomial numbers A048996 for the partitions of 2*n+1 into 2*k+1 parts, all of which are odd. E.g., T(3,1) = 3 + 3 from the numbers for the partitions [1,1,5] and [1,3,3], namely 3!/(2!*1!) and 3!/(1!*2!), respectively. The number triangle with the number of these partitions as entries is A152157. - Wolfdieter Lang, Jul 09 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*A039599(n,k). - R. J. Mathar, Mar 12 2013
T(n,k) = A258993(n+1,k) for k = 0..n-1. - Reinhard Zumkeller, Jun 22 2015
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the algebraic function F(x)*G(x)^n about 0, where F(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) and G(x) = ((1 + sqrt(1 + 4*x))/2)^2. For example, for n = 4, (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) * ((1 + sqrt(1 + 4*x))/2)^8 = (x^4  + 10*x^3 + 15*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 23 2018
Row n also gives the coefficients of the characteristc polynomial of the tridiagonal n X n matrix M_n given in A332602: Phi(n, x) := Det(M_n - x*1_n) = Sum_{k=0..n} T(n, k)*(-x)^k, for n >= 0, with Phi(0, x) := 1. - Wolfdieter Lang, Mar 25 2020
It appears that the largest root of the n-th degree polynomial is equal to the sum of the distinct diagonals of a (2*n+1)-gon including the edge, 1. The largest root of x^3 - 6*x^2 + 5*x - 1 is 5.048917... = the sum of (1 + 1.80193... + 2.24697...). Alternatively, the largest root of the n-th degree polynomial is equal to the square of sigma(2*n+1). Check: 5.048917... is the square of sigma(7), 2.24697.... Given N = 2*n+1, sigma(N) (N odd) can be defined as 1/(2*sin(Pi/(2*N))). Relating to the 9-gon, the largest root of x^4 - 10*x^3 + 15*x^2 - 7*x + 1 is 8.290859..., = the sum of (1 + 1.879385... + 2.532088... + 2.879385...), and is the square of sigma(9), 2.879385... Refer to A231187 for a further clarification of sigma(7). - Gary W. Adamson, Jun 28 2022
For n >=1, the n-th row is given by the coefficients of the minimal polynomial of -4*sin(Pi/(4*n + 2))^2. - Eric W. Weisstein, Jul 12 2023
Denoting this lower triangular array by L, then L * diag(binomial(2*k,k)^2) * transpose(L) is the LDU factorization of A143007, the square array of crystal ball sequences for the A_n X A_n lattices. - Peter Bala, Feb 06 2024
T(n, k) is the number of occurrences of the periodic substring (01)^k in the periodic string (01)^n (see Proposition 4.7 at page 7 in Fang). - Stefano Spezia, Jun 09 2024

			Triangle begins as:
  1;
  1    1;
  1    3    1;
  1    6    5    1;
  1   10   15    7    1;
  1   15   35   28    9    1;
  1   21   70   84   45   11    1;
  1   28  126  210  165   66   13    1;
  1   36  210  462  495  286   91   15    1;
  1   45  330  924 1287 1001  455  120   17    1;
  1   55  495 1716 3003 3003 1820  680  153   19    1;
...
From _Philippe Deléham_, Mar 26 2012: (Start)
(0, 1, 0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) begins:
  1
  0, 1
  0, 1,  1
  0, 1,  3,   1
  0, 1,  6,   5,   1
  0, 1, 10,  15,   7,   1
  0, 1, 15,  35,  28,   9,  1
  0, 1, 21,  70,  84,  45, 11,  1
  0, 1, 28, 126, 210, 165, 66, 13, 1. (End)

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - _N. J. A. Sloane_, May 10 2012
Peter Bala, A 4-parameter family of embedded Riordan arrays
E. Barcucci, A. Del Lungo, S. Fezzi, and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, J. Integer Sequ., Vol. 8 (2005), Article 05.4.5.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
Paul Barry, The second production matrix of a Riordan array, arXiv:2011.13985 [math.CO], 2020.
Paul Barry and Aoife Hennessy, Notes on a Family of Riordan Arrays and Associated Integer Hankel Transforms , JIS 12 (2009) 09.5.3.
Eduardo H. M. Brietzke, Recurrence Relations for Sums of Binomial Coefficients and Some Generalizations, Journal of Integer Sequences, Vol. 27 (2024), Article 24.3.4.
E. Czabarka et al, Enumerations of peaks and valleys on non-decreasing Dyck paths, Disc. Math. 341 (2018) 2789-2807, Theorem 3.
Emeric Deutsch and H. Prodinger, A bijection between directed column-convex polyominoes and ordered trees of height at most three, Theoretical Comp. Science, 307, 2003, 319-325.
James East and Nicholas Ham, Lattice paths and submonoids of Z^2, arXiv:1811.05735 [math.CO], 2018.
Wenjie Fang, Maximal number of subword occurrences in a word, arXiv:2406.02971 [math.CO], 2024.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62. - _Abdullahi Umar_, Oct 02 2008
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
Yidong Sun, Numerical Triangles and Several Classical Sequences, Fib. Quart. 43, no. 4, (2005) 359-370.
Eric Weisstein's World of Mathematics, Morgan-Voyce Polynomials

Row sums: A001519. Signed versions: A123970, A129818.
Cf. A000108, A001006, A007318, A098158, A183160, A078812, A091042.
Cf. A258993, A025581, A051162, A054142, A332602.
Cf. A231187, A143007.

Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k, 2*k) ))); # G. C. Greubel, Aug 01 2019

a085478 n k = a085478_tabl !! n !! k
a085478_row n = a085478_tabl !! n
a085478_tabl = zipWith (zipWith a007318) a051162_tabl a025581_tabl
-- Reinhard Zumkeller, Jun 22 2015

[Binomial(n+k, 2*k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019

T := (n,k) -> binomial(n+k,2*k): seq(seq(T(n,k), k=0..n), n=0..11);

(* First program *)
u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A085478 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A078812 *) (*Clark Kimberling, Feb 25 2012 *)
(* Second program *)
Table[Binomial[n + k, 2 k], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 01 2019 *)
CoefficientList[Table[Fibonacci[2 n + 1, Sqrt[x]], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Jul 03 2023 *)
Join[{{1}}, CoefficientList[Table[MinimalPolynomial[-4 Sin[Pi/(4 n + 2)]^2, x], {n, 20}], x]] (* Eric W. Weisstein, Jul 12 2023 *)

PARI
```
T(n,k) = binomial(n+k,n-k)
```

[[binomial(n+k,2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

T(n, k) = (n+k)!/((n-k)!*(2*k)!).
G.f.: (1-z)/((1-z)^2-tz). - Emeric Deutsch, May 31 2004
Row sums are A001519 (Fibonacci(2n+1)). Diagonal sums are A011782. Binomial transform of A026729 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
T(n, 0) = 1, T(n, k) = 0 if n=0} T(n-1-j, k-1)*(j+1). T(0, 0) = 1, T(0, k) = 0 if k>0; T(n, k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j). For the column k, g.f.: Sum_{n>=0} T(n, k)*x^n = (x^k) / (1-x)^(2*k+1). - Philippe Deléham, Feb 15 2004
Sum_{k=0..n} T(n,k)*x^(2*k) = A000012(n), A001519(n+1), A001653(n), A078922(n+1), A007805(n), A097835(n), A097315(n), A097838(n), A078988(n), A097841(n), A097727(n), A097843(n), A097730(n), A098244(n), A097733(n), A098247(n), A097736(n), A098250(n), A097739(n), A098253(n), A097742(n), A098256(n), A097767(n), A098259(n), A097770(n), A098262(n), A097773(n), A098292(n), A097776(n) for x=0,1,2,...,27,28 respectively. - Philippe Deléham, Dec 31 2007
T(2*n,n) = A005809(n). - Philippe Deléham, Sep 17 2009
A183160(n) = Sum_{k=0..n} T(n,k)*T(n,n-k). - Paul D. Hanna, Dec 27 2010
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Feb 06 2012
O.g.f. for column k: x^k/(1-x)^(2*k+1), k >= 0. [See the o.g.f. of the triangle above, and a comment on compositions. - Wolfdieter Lang, Jul 09 2012]
E.g.f.: (2/sqrt(x + 4))*sinh((1/2)*t*sqrt(x + 4))*cosh((1/2)*t*sqrt(x)) = t + (1 + x)*t^3/3! + (1 + 3*x + x^2)*t^5/5! + (1 + 6*x + 5*x^2 + x^3)*t^7/7! + .... Cf. A091042. - Peter Bala, Jul 29 2013
T(n, k) = A065941(n+3*k, 4*k) = A108299(n+3*k, 4*k) = A194005(n+3*k, 4*k). - Johannes W. Meijer, Sep 05 2013
Sum_{k=0..n} (-1)^k*T(n,k)*A000108(k) = A000007(n) for n >= 0. - Werner Schulte, Jul 12 2017
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A001006(n) for n >= 0. - Werner Schulte, Jul 12 2017
From  Peter Bala, Jun 26 2025: (Start)
The n-th row polynomial b(n, x) = (-1)^n * U(2*n, (i/2)*sqrt(x)), where U(n,x) is the n-th Chebyshev polynomial of the second kind.
b(n, x) = (-1)^n * Dir(n, -1 - x/2), where Dir(n, x) is the n-th row polynomial of the triangle A244419.
b(n, -1 - x) is the n-th row polynomial of A098493. (End)

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

Original entry on oeis.org

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

Offset: 0

Wouter Meeussen, Emeric Deutsch

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

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

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

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

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

Cf. A000108, A001045, A005156, A224274.

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

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

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

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

G.f.: z*g^2/(1-3*z*g^2), where g=g(z) is given by g=1+z*g^3, g(0)=1, that is, (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n) = Sum_{k=0..n} ((3k+1)/(2n+k+1))C(3n, 2n+k)*A001045(k). - Paul Barry, Oct 07 2005
Hankel transform of a(n+1) is A005156(n+1). - Paul Barry, Apr 14 2008
G.f.: x*B'(x)/B(x) where B(x) is the g.f. of A001764. - Vladimir Kruchinin Feb 03 2013
D-finite with recurrence: 2*n*(2*n-1)*a(n) -3*(3*n-1)*(3*n-2)*a(n-1)=0. - R. J. Mathar, Feb 05 2013
Logarithmic derivative of A001764; g.f. of A001764 satisfies G(x) = 1 + x*G(x)^3. - Paul D. Hanna, Jul 14 2013
G.f.: (2*cos((1/3)*arcsin((3/2)*sqrt(3*x)))-sqrt(4-27*x))/(3*sqrt(4-27*x)). - Emanuele Munarini, Oct 14 2014
a(n) = Sum_{k=1..n} binomial(n-1,n-k)*binomial(2*n,n-k). - Vladimir Kruchinin, Nov 12 2014
a(n) = [x^n] C(x)^n for n >= 1, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function for A000108 (Ramanujan). - Peter Bala, Jun 24 2015
From Peter Bala, Nov 04 2015: (Start)
Without the initial term 0, the o.g.f. equals f(x)*g(x)^2, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. g(x)^2 is the o.g..f for A006013. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
G.f.: ( 2F1(1/3,2/3;1/2;27*x/4)-1)/3. - R. J. Mathar, Jan 27 2020
O.g.f. without the initial term 0, in the form g(x)=(2*cos(arcsin((3*sqrt(3)*sqrt(x))/2)/3)/sqrt(4-27*x)-1)/(3*x), satisfies the following algebraic equation: 1+(9*x-1)*g(x)+x*(27*x-4)*g(x)^2+x^2*(27*x-4)*g(x)^3=0. - Karol A. Penson, Oct 11 2021
O.g.f. equals f(x)/(1 - 2*f(x)), where f(x) = series reversion (x/(1 + x)^3) = x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the o.g.f. of A001764 with the initial term omitted. Cf. A224274. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/2)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+2)*n+k-1,k) = C(3*n-1,n-1) and (1/3)*Sum_{k = 0..n} (-1)^k* C(x*n,n-k)*C((x-3)*n+k-1,k) = C(3*n-1,n-1), both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
a(n) ~ 2^(-2*n)*3^(3*n)/(2*sqrt(3*n*Pi)). - Stefano Spezia, Apr 25 2024
a(n) = Sum_{k = 0..n-1} binomial(2*n+k-1, k) = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(3*n, k). - Peter Bala, Jul 21 2024
E.g.f.: (2 + hypergeom([1/3, 2/3], [1/2, 1], 27*x/4))/3 - 1. - Stefano Spezia, Dec 31 2024
a(n+1) = Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n+3, k)*binomial(2*n-k, n-k). - Peter Bala, Sep 04 2025

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

Original entry on oeis.org

1, 4, 21, 120, 715, 4368, 27132, 170544, 1081575, 6906900, 44352165, 286097760, 1852482996, 12033222880, 78378960360, 511738760544, 3348108992991, 21945588357420, 144079707346575, 947309492837400, 6236646703759395, 41107996877935680, 271250494550621040, 1791608261879217600

Offset: 0

Emeric Deutsch

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

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

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

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

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

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

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

a(n) is asymptotic to c/sqrt(n)*(27/4)^n with c=0.73... - Benoit Cloitre, Jan 27 2003
G.f.: gz^2/(1-3zg^2), where g=g(z) is given by g=1+zg^3, g(0)=1, i.e. (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n+2) = C(3n+1,n) = Sum_{k=0..n} C(3n-k,n-k). - Paul Barry, May 13 2006
a(n+2) = C(3n+1,2n+1) = A078812(2n,n). - Paul Barry, Nov 09 2006
G.f.: A(x)=(2*cos(asin((3^(3/2)*sqrt(x))/2)/3)* sin(asin((3^(3/2)* sqrt(x))/2)/3))/(sqrt(3)*sqrt(1-(27*x)/4)*sqrt(x)). - Vladimir Kruchinin, Jun 10 2012
From Peter Luschny, Sep 04 2012: (Start)
O.g.f.: hypergeometric2F1([2/3, 4/3], [3/2], x*27/4).
a(n) = (n+1)*hypergeometric2F1([-2*n, -n], [2], 1).  (End)
D-finite with recurrence 2*n*(2*n+1)*a(n) - 3*(3*n-1)*(3*n+1)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
a(n) = Sum_{r=0..n} C(n,r) * C(2*n+1,r). - J. M. Bergot, Mar 18 2014
From Peter Bala, Nov 04 2015: (Start)
a(n) = Sum_{k = 0..n} 1/(2*k + 1)*binomial(3*n - 3*k,n - k)*binomial(3*k, k).
O.g.f. equals f(x)*g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*(n+1)). - Ilya Gutkovskiy, Oct 10 2017
O.g.f.: (i/24)*((4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(2/3) - (4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(2/3))*sqrt(3)*8^(1/3)*sqrt(4 - 27*z)/(sqrt(z)*(-4 + 27*z)), where i = sqrt(-1). - Karol A. Penson, Dec 13 2023
a(n-1) = (1/(4*n))*binomial(2*n, n)^2 * (1 - 3*((n - 1)/(n + 1))^3 + 5*((n - 1)*(n - 2)/((n + 1)*(n + 2)))^3 - 7*((n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)))^3 + - ...) for n >= 1. Cf. A112029. - Peter Bala, Aug 08 2024
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n+2, k)*binomial(2*n-k, n-k). - Peter Bala, Sep 04 2025
a(n) ~ 3^(3*n+3/2) / (4^(n+1) * sqrt(Pi*n)). - Amiram Eldar, Sep 05 2025

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

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

Original entry on oeis.org

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

Offset: 0

Ralf Stephan, Mar 21 2004

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

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

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

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

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

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

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

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

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

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

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

Diagonal sums comment corrected by Paul Barry, Apr 14 2010

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

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

Original entry on oeis.org

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

Offset: 0

Thomas Wieder, Sep 29 2009

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Original entry on oeis.org

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

Offset: 0

Fabian Rothelius, Feb 04 2001

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Edited by M. F. Hasler, Mar 05 2017

A188687 Partial binomial sums of binomial(3n,n)/(2n+1) = A001764(n).

Original entry on oeis.org

1, 2, 6, 25, 126, 704, 4183, 25897, 165166, 1077520, 7156352, 48222354, 328859011, 2265428728, 15740837575, 110187356134, 776336572878, 5501042194580, 39177463572112, 280277949384146, 2013277273220064, 14514764553512488, 104993261648226446

Offset: 0

Emanuele Munarini, Apr 08 2011

Nathaniel Johnston, Table of n, a(n) for n = 0..400
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.

Cf. A005809, A001764, A188675, A188676, A104859, A188678, A188679, A188680, A188681, A188682, A188683, A188684, A188685, A188686.

Table[Sum[Binomial[n,k]Binomial[3k,k]/(2k+1),{k,0,n}],{n,0,22}]

makelist(sum(binomial(n,k)*binomial(3*k,k)/(2*k+1),k,0,n),n,0,20);

a(n) = Sum_{k=0..n} binomial(n,k)*binomial(3k,k)/(2k+1).
G.f.: (2/sqrt(3x*(1-x)))*sin((1/3)*arcsin(3/2*sqrt(3*x/(1-x)))).
Recurrence: 2*n*(2*n+1)*a(n) = (39*n^2-35*n+8)*a(n-1) - 2*(n-1)*(33*n-32)*a(n-2) + 31*(n-2)*(n-1)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 31^(n+3/2)/(3^4*2^(2*n+2)*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x) * A(x)^3. - Ilya Gutkovskiy, Jul 25 2021

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)

cp's OEIS Frontend

A005810 a(n) = binomial(4n,n).

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A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

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

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

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

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