A001448 -id:A001448 - cp's OEIS frontend

A000984 Central binomial coefficients: binomial(2n,n) = (2n)!/(n!)^2.

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720, 8233430727600, 32247603683100, 126410606437752, 495918532948104, 1946939425648112

Offset: 0

N. J. A. Sloane

Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).
Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.
Number of possible interleavings of a program with n atomic instructions when executed by two processes. - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001
Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002
Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of directed, convex polyominoes having semiperimeter n+2.
Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002
The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its g.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003
Number of possible values of a 2n-bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003
Ordered partitions of n with zeros to n+1, e.g., for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003
Number of nondecreasing sequences of n integers from 0 to n: a(n) = Sum_{i_1=0..n} Sum_{i_2=i_1..n}...Sum_{i_n=i_{n-1}..n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003
Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003
a(n-1) = number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. E.g., n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004
The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.
Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004
Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2) * (sqrt(n)) * zeta(1/2). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. - Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009. - N. J. A. Sloane, Oct 25 2015]
The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 70-80.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73-107) who showed that it holds for all n>4. - Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]
p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006
The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006
Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007
So this is the 2-dimensional analog of A008793. - William Entriken, Aug 06 2013
The number of walks of length 2n on an infinite linear lattice that begins and ends at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Integral representation: C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e., C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008
Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 2008
Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11." - Jonathan Vos Post, Nov 14 2008
Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333, ...). - Gary W. Adamson, May 15 2009
Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009
Number of n X n binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.
Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011
Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012
From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1 + a2*b2 + ... + a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1) + 4*(5) + 6*(10) + 4*(10) + 1*(5)) = 252, the sixth term in this sequence.
Take from Pascal's triangle row(n) with terms b1, b2, ..., b(n+1) and row(n+2) with terms c1, c2, ..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1).  Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012
a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p. 4.) - Gary Detlefs, Feb 16 2013
Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = Sum_{k=0..n} (m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
This comment generalizes the comment dated Oct 31 2012 and the second of the sequence's original comments.  For j = 1 to n, a(n) = Sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*Sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). - Charlie Marion, Jun 07 2013
The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013
Number of distinct strings of length 2n using n letters A and n letters B. - Hans Havermann, May 07 2014
From Fung Lam, May 19 2014: (Start)
Expansion of  G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=-1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=-1 recovers the g.f. of A000108 as xA^2-A+1=0. The present sequence A000984 refers to q=-2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q-4*q-4)*n + 2*(q*q-q-1))*a(n+1) - 2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=-q. Asymptotics: a(n) ~ ((q+2)/(q+1))*(q^2/(-q-1))^n, q<=-3, a(n) ~ (-1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ -Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=-3), A076035 (q=-4), A076036 (q=-5), A127628 (q=-6), A126694 (q=-7), A115970 (q=-8). (End)
a(n)*(2^n)^(j-2) equals S(n), where S(n) is the n-th number in the self-convolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(4-2) = 252*1024 = 258048.  The self-convolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing). - Bob Selcoe, Jul 16 2014
The variance of the n-th difference of a sequence of pairwise uncorrelated random variables each with variance 1. - Liam Patrick Roche, Jun 04 2015
Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(1-4z). - Emeric Deutsch, Jun 17 2015
Number of monomials of degree at most n in n variables. - Ran Pan, Sep 26 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r, then V(n, 2^n) / Pi = V(n-1, 2^n) * a(n/2) for all even n. - Peter Luschny, Oct 12 2015
a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0). - Anton Zakharov, Jul 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)/(-2)^k = 1/sqrt(3).
Sum_{k>=0} a(k)/(-1)^k = 1/sqrt(5).
Sum_{k>=0} a(k)/(-1/2)^k = 1/3.
Sum_{k>=0} a(k)/(1/2)^k = -1/sqrt(7)i.
Sum_{k>=0} a(k)/(1)^k = -1/sqrt(3)i.
Sum_{k>=0} a(k)/2^k  = -i. (End)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18] - Eric M. Schmidt, Jul 17 2017
The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1 - (x + y)), 1/(1 - (x + y*z)), 1/(1 - (x + x*y + y*z)) or 1/(1 - (x + y + y*z)). - Peter Bala, Jan 30 2018
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.
a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)
All binary self-dual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n.  More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together. - Nathan J. Russell, Nov 25 2018
From Isaac Saffold, Dec 28 2018: (Start)
Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,
[(m+n)/p] == [n/p]*Sum_{k=0..(p-1)/2} (-m/(4*n))^k * a(k) (mod p).
Evaluating this identity for m = -1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p-1)/2} (1/4)^k * a(k) is divisible by p. (End)
Number of vertices of the subgraph of the (2n-1)-dimensional hypercube induced by all bitstrings with n-1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle. - Torsten Muetze, Feb 11 2019
a(n) is the number of walks of length 2n from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {3>1, 1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element but smaller than the third elements. - Sergey Kitaev, Dec 08 2020
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 1 through a(2) = 6 compositions are:
  (1)  (2,1)    (3,2)
       (1,1,1)  (1,2,2)
                (2,2,1)
                (1,1,2,1)
                (2,1,1,1)
                (1,1,1,1,1)
The following relate to these compositions:
- The unordered version is A000070.
- The alternating sum -1 version is counted by A001791, ranked by A345910/A345912.
- The alternating sum 0 version is counted by A088218, ranked by A344619.
- Including even indices gives A126869.
- The complement is counted by A202736.
- Ranked by A345909 (reverse: A345911).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 1 than 0's. For example, the a(2) = 6 binary numbers are: 10011, 10101, 10110, 11001, 11010, 11100.
(End)
From Michael Wallner, Jan 25 2022: (Start)
a(n) is the number of nx2 Young tableaux with a single horizontal wall between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
  Example for a(2)=6:
   3 4  2 4  3 4  3|4  4|3  2|4
   1|2, 1|3, 2|1, 1 2, 1 2, 1 3
a(n) is also the number of nx2 Young tableaux with n "walls" between the first and second column.
  Example for a(2)=6:
   3|4  2|4  4|3  3|4  4|3  4|2
   1|2, 1|3, 1|2, 2|1, 2|1, 3|1 (End)
From Shel Kaphan, Jan 12 2023: (Start)
a(n)/4^n is the probability that a fair coin tossed 2n times will come up heads exactly n times and tails exactly n times, or that a random walk with steps of +-1 will return to the starting point after 2n steps (not necessarily for the first time). As n becomes large, this number asymptotically approaches 1/sqrt(n*Pi), using Stirling's approximation for n!.
a(n)/(4^n*(2n-1)) is the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. The absolute value of the n-th term of A144704 is denominator of this fraction.
Considering all possible random walks of exactly 2n steps with steps of +-1, a(n)/(2n-1) is the number of such walks that return to the starting point for the first time after 2n steps. See the absolute values of A002420 or A284016 for these numbers. For comparison, as mentioned by Stefan Hollos, Dec 10 2007, a(n) is the number of such walks that return to the starting point after 2n steps, but not necessarily for the first time. (End)
p divides a((p-1)/2) + 1 for primes p of the form 4*k+3 (A002145). - Jules Beauchamp, Feb 11 2023
Also the size of the shuffle product of two words of length n, such that the union of the two words consist of 2n distinct elements. - Robert C. Lyons, Mar 15 2023
a(n) is the number of vertices of the n-dimensional cyclohedron W_{n+1}. - Jose Bastidas, Mar 25 2025
Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of increasing sizes, followed by a series of reversals of decreasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025

			G.f.: 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
For n=2, a(2) = 4!/(2!)^2 = 24/4 = 6, and this is the middle coefficient of the binomial expansion (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. - _Michael B. Porter_, Jul 06 2016

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.
Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 160.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line -3, with a=b=n.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 101.
Emeric Deutsch and Louis W. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31 (2001), 31-38.
Henry W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
Ronald. L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
Martin Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
Leonard Lipshitz and A. van der Poorten, "Rational functions, diagonals, automata and arithmetic", in Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), 339-358.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe and Edward Jiang, Table of n, a(n) for n = 0..500 (Previously 0..200 by T. D. Noe)
Joseph Abate and Ward Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011), Article 11.2.6, example section 3.
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].
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Fixed Sequences for a Generalization of the Binomial Interpolated Operator and for some Other Operators, J. Int. Seq. 14 (2011), Article 11.8.1.
B. Adamczewski, Jason P. Bell, and E. Delaygue, Algebraic independence of G-functions and congruences "a la Lucas", arXiv preprint arXiv:1603.04187 [math.NT], 2016.
Martin Aigner, Enumeration via ballot numbers, Discrete Math., 308 (2008), 2544-2563.
Michael Anshelevich, Product formulas on posets, Wick products, and a correction for the q-Poisson process, arXiv:1708.08034 [math.OA], 2017, See Proposition 34 p. 25.
David H. Bailey, Jonathan M. Borwein, and David M. Bradley, Experimental determination of Apéry-like identities for zeta(4n+2), arXiv:math/0505124 [math.CA], 2005.
Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
Paul Barry, Riordan-Bernstein Polynomials, Hankel Transforms and Somos Sequences, Journal of Integer Sequences, Vol. 15 (2012), Article 12.8.2.
Paul Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), Article 13.5.1.
Paul Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), Article 13.5.4.
Paul Barry, The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.
Paul Barry, Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.
Paul Barry, On the Central Antecedents of Integer (and Other) Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.3.
Paul Barry and Aoife Hennessy, Generalized Narayana Polynomials, Riordan Arrays, and Lattice Paths, Journal of Integer Sequences, Vol. 15, (2012), Article 12.4.8.
Paul Barry, On the Connection Coefficients of the Chebyshev-Boubaker polynomials, The Scientific World Journal, Volume 2013 (2013), Article ID 657806, 10 pages.
Antonio Bernini, Filippo Disanto, Renzo Pinzani, and Simone Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007), Article 07.9.7.
Robert J. Betts, Lack of Divisibility of {2N choose N} by three fixed odd primes infinitely often, through the Extension of a Result by P. Erdős, et al., arXiv:1010.3070 [math.NT], 2010. [It is not clear if the results in this paper have been confirmed. There appears to be no mention of this work in MathSciNet, for example. - _N. J. A. Sloane_, Oct 29 2015]
Jonathan M. Borwein and David M. Bradley, Empirically determined Apéry-like formulas for zeta(4n+3), arXiv:math/0505124 [math.CA], 2005.
Jonathan M. Borwein, Dirk Nuyens, Armin Straub and James Wan, Random Walk Integrals, 2010.
Jonathan M. Borwein and Armin Straub, Mahler measures, short walks and log-sine integrals.
Harlan J. Brothers, Pascal's Prism: Supplementary Material.
Marie-Louise Bruner, Central binomial coefficients also count (2431,4231,1432,4132)-avoiders, arXiv:1505.04929 [math.CO], 2015.
Kevin Buchin, Man-Kwun Chiu, Stefan Felsner, Günter Rote, and André Schulz, The Number of Convex Polyominoes with Given Height and Width, arXiv:1903.01095 [math.CO], 2019.
Naiomi Tuere Cameron, Random walks, trees and extensions of Riordan group techniques, Dissertation, Howard University, 2002.
Grégory Chatel and Vincent Pilaud, The Cambrian and Baxter-Cambrian Hopf Algebras, arXiv preprint arXiv:1411.3704 [math.CO], 2014-2015.
Hongwei Chen, Evaluations of Some Variant Euler Sums, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.3.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014.
Johann Cigler, Some nice Hankel determinants, arXiv:1109.1449 [math.CO], 2011.
Johann Cigler and Christian Krattenthaler, Hankel determinants of linear combinations of moments of orthogonal polynomials, arXiv:2003.01676 [math.CO], 2020.
CombOS - Combinatorial Object Server, Generate middle levels Gray codes
B. N. Cooperstein and E. E. Shult, A note on embedding and generating dual polar spaces. Adv. Geom. 1 (2001), 37-48. See Theorem 5.4.
Kristina Crona, Mengming Luo, and Devin Greene, An Uncertainty Law for Microbial Evolution, Journal of Theoretical Biology (2020) Vol. 489, Article No. 110155.
Dan Daly and Lara Pudwell, Pattern avoidance in rook monoids, 2013.
Colin Defant, Stack-sorting preimages of permutation classes, arXiv:1809.03123 [math.CO], 2018.
Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, and Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.8.
Thierry Dana-Picard, Sequences of Definite Integrals, Factorials and Double Factorials, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.6.
Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
E. Delaygue, Arithmetic properties of Apéry-like numbers, arXiv preprint arXiv:1310.4131 [math.NT], 2013.
Nachum Dershowitz, Touchard's Drunkard, Journal of Integer Sequences, Vol. 20 (2017), #17.1.5.
Emeric Deutsch, Enumerating symmetric directed convex polyominoes, Discrete Math., 280 (2004), 225-231.
Satyan L. Devadoss, A realization of graph associahedra, Discrete Math. 309 (2009), no. 1, 271-276.
J. C. F. de Winter, Using the Student's t-test with extremely small sample sizes, Practical Assessment, Research & Evaluation, 18(10), 2013.
R. M. Dickau, Shortest-path diagrams
Phan Thuan Do, Thi Thu Huong Tran, and Vincent Vajnovszki, Exhaustive generation for permutations avoiding a (colored) regular sets of patterns, arXiv:1809.00742 [cs.DM], 2018.
R. Duarte and A. G. de Oliveira, Short note on the convolution of binomial coefficients, arXiv preprint arXiv:1302.2100 [math.CO], 2013 and J. Int. Seq. 16 (2013) #13.7.6.
Bryan Ek, Unimodal Polynomials and Lattice Walk Enumeration with Experimental Mathematics, arXiv:1804.05933 [math.CO], 2018.
P. Erdős, R. L. Graham, I. Z. Russa and E. G. Straus, On the prime factors of C(2n,n), Math. Comp. 29 (1975), 83-92.
Gennady Eremin, Factoring Middle Binomial Coefficients, arXiv:2003.01494 [math.CO], 2020.
A. Erickson and F. Ruskey, Enumerating maximal tatami mat coverings of square grids with v vertical dominoes, arXiv preprint arXiv:1304.0070 [math.CO], 2013.
Luca Ferrari and Emanuele Munarini, Enumeration of edges in some lattices of paths, arXiv preprint arXiv:1203.6792 [math.CO], 2012 and J. Int. Seq. 17 (2014) #14.1.5.
Francesc Fité and Andrew V. Sutherland, Sato-Tate distributions of twists of y^2= x^5-x and y^2= x^6+1, arXiv preprint arXiv:1203.1476 [math.NT], 2012.
Francesc Fité, Kiran S. Kedlaya, Victor Rotger and Andrew V. Sutherland, Sato-Tate distributions and Galois endomorphism modules in genus 2, arXiv:1110.6638 [math.NT], 2011.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 77.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.
Joël Gay and Vincent Pilaud, The weak order on Weyl posets, arXiv:1804.06572 [math.CO], 2018.
Solomon W. Golomb, Facts about C(2n, n), Golomb's Puzzle Column, IEEE Information Theory Society Newsletter, Vol. 53, No. 1 (March 2003), p. 28; Facts about C(2n, n) Solutions, ibid., Vol. 53, No. 2 (June 2003), pp. 17-18.
H. W. Gould, Tables of Combinatorial Identities, Vol. 7, Edited by J. Quaintance.
A. Granville and O. Ramaré, Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients, Mathematika 43 (1996), 73-107, [DOI].
T. Halverson and M. Reeks, Gelfand Models for Diagram Algebras, arXiv preprint arXiv:1302.6150 [math.RT], 2013.
Oktay Haracci (timetunnel3(AT)hotmail.com), Regular Polygons
R. H. Hardin, Binary arrays with both rows and cols sorted, symmetries
P.-Y. Huang, S.-C. Liu, and Y.-N. Yeh, Congruences of Finite Summations of the Coefficients in certain Generating Functions, The Electronic Journal of Combinatorics, 21 (2014), #P2.45.
W. Cary Huffman and Vera Pless, Fundamentals of Error Correcting Codes, Cambridge University Press, 2003, Pages 7,252-282,338-393.
Anders Hyllengren, Four integer sequences, Oct 04 1985. Observes essentially that A000984 and A002426 are inverse binomial transforms of each other, as are A000108 and A001006.
Milan Janjic, Two Enumerative Functions
I. Jensen, Series expansions for self-avoiding polygons
Pakawut Jiradilok and Elchanan Mossel, Gaussian Broadcast on Grids, arXiv:2402.11990 [cs.IT], 2024. See p. 27.
C. Kimberling, Matrix Transformations of Integer Sequences, J. Integer Seqs., Vol. 6, 2003.
Sergey Kitaev and Jeffrey Remmel, Simple marked mesh patterns, arXiv preprint arXiv:1201.1323 [math.CO], 2012.
V. V. Kruchinin and D. V. Kruchinin, A Method for Obtaining Generating Function for Central Coefficients of Triangles, arXiv:1206.0877 [math.CO], 2012.
D. Kruchinin and V. Kruchinin, A Method for Obtaining Generating Function for Central Coefficients of Triangles, Journal of Integer Sequence, Vol. 15 (2012), article 12.9.3.
Jean-Philippe Labbé and Carsten Lange, Cambrian acyclic domains: counting c-singletons, arXiv:1802.07978 [math.CO], 2018.
Marie-Louise Lackner and M. Wallner, An invitation to analytic combinatorics and lattice path counting; Preprint, Dec 2015.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
D. H. Lehmer, Interesting series involving the Central Binomial Coefficient, Am. Math. Monthly 92, no 7 (1985) 449-457.
Huyile Liang, Jeffrey Remmel, and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 19.
L. Lipshitz and A. J. van der Poorten, Rational functions, diagonals, automata and arithmetic
T. Manneville and V. Pilaud, Compatibility fans for graphical nested complexes, arXiv preprint arXiv:1501.07152 [math.CO], 2015.
Megan A. Martinez and Carla D. Savage, Patterns in Inversion Sequences II: Inversion Sequences Avoiding Triples of Relations, arXiv:1609.08106 [math.CO], 2016.
MathOverflow, Divisibility of a binomial coefficient by p^2 — current status
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv preprint arXiv:1111.3057 [math.NT], 2011.
R. Mestrovic, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv preprint arXiv:1409.3820 [math.NT], 2014.
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.
T. Motzkin, The hypersurface cross ratio, Bull. Amer. Math. Soc., 51 (1945), 976-984.
T. S. Motzkin, Relations between hypersurface cross ratios and a combinatorial formula for partitions of a polygon, for permanent preponderance and for non-associative products, Bull. Amer. Math. Soc., 54 (1948), 352-360.
Torsten Mütze, Proof of the middle levels conjecture, arXiv preprint arXiv:1404.4442 [math.CO], 2014.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Article 12.3.3, 2012. - From _N. J. A. Sloane_, Sep 16 2012
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
Valentin Ovsienko, Shadow sequences of integers, from Fibonacci to Markov and back, arXiv:2111.02553 [math.CO], 2021.
Ran Pan, Exercise I, Project P.
P. Peart and W.-J. Woan, Generating Functions via Hankel and Stieltjes Matrices, J. Integer Seqs., Vol. 3 (2000), #00.2.1.
M. A. Perlstadt, Some Recurrences for Sums of Powers of Binomial Coefficients, Journal of Number Theory 27 (1987), pp. 304-309.
A. Petojevic and N. Dapic, The vAm(a,b,c;z) function, Preprint 2013.
C. Pomerance, Divisors of the middle binomial coefficient, Amer. Math. Monthly, 112 (2015), 636-644.
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
T. M. Richardson, The Reciprocal Pascal Matrix, arXiv preprint arXiv:1405.6315 [math.CO], 2014.
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their N-numbers, not their A-numbers.
D. P. Roberts and A. Venkatesh, Hurwitz monodromy and full number fields, 2014. Also arXiv:1401:7379, 2014.
H. P. Robinson, Letter to N. J. A. Sloane, Oct 1981
A. Sárközy, On Divisors of Binomial Coefficients. I., J. Number Th. 20, 70-80, 1985.
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)
L. W. Shapiro, S. Getu, Wen-Jin Woan and L. C. Woodson, The Riordan Group, Discrete Appl. Maths. 34 (1991) 229-239.
N. J. A. Sloane, Notes on A984 and A2420-A2424
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
Armin Straub, Arithmetic aspects of random walks and methods in definite integration, Ph. D. Dissertation, School Of Science And Engineering, Tulane University, 2012.
Armin Straub, Tewodros Amdeberhan and Victor H. Moll, The p-adic valuation of k-central binomial coefficients, arXiv:0811.2028 [math.NT], 2008, pp. 10-11.
V. Strehl, Recurrences and Legendre transform, Séminaire Lotharingien de Combinatoire, B29b (1992), 22 pp.
R. A. Sulanke, Moments of generalized Motzkin paths, J. Integer Sequences, Vol. 3 (2000), #00.1.
Hua Sun and Yi Wang, A Combinatorial Proof of the Log-Convexity of Catalan-Like Numbers, J. Int. Seq. 17 (2014) # 14.5.2.
Michael Torpey, Semigroup congruences: computational techniques and theoretical applications, Ph.D. Thesis, University of St. Andrews (Scotland, 2019).
H. A. Verrill, Sums of squares of binomial coefficients, ..., arXiv:math/0407327 [math.CO], 2004.
M. Wallner, Lattice Path Combinatorics, Diplomarbeit, Institut für Diskrete Mathematik und Geometrie der Technischen Universität Wien, 2013.
Eric Weisstein's World of Mathematics, Binomial Sums.
Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
Eric Weisstein's World of Mathematics, Staircase Walk.
Eric Weisstein's World of Mathematics, Circle Line Picking.
Wikipedia, Shuffle product.
Herbert S. Wilf, Generatingfunctionology, Academic Press, NY, 1990. See p. 50.
Index entries for "core" sequences.

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A081696, A182400. Differs from A071976 at 10th term.
Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.
Cf. A226078, A051924 (first differences).
Row sums of A059481, A008459, A152229, A158815, A205946.
Cf. A258290 (arithmetic derivative). Cf. A098616, A214377.
See A261009 for a conjecture about this sequence.
Cf. A046521 (first column).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Cf. A000346, A001700, A001791, A008549, A032443, A073016, A097805, A126869.
Cf. A003418, A056040, A001316, A261130, A356637.

List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018

a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011

a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];

A000984 := n-> binomial(2*n,n); seq(A000984(n), n=0..30);
with(combstruct); [seq(count([S,{S=Prod(Set(Z,card=i),Set(Z,card=i))}, labeled], size=(2*i)), i=0..20)];
with(combstruct); [seq(count([S,{S=Sequence(Union(Arch,Arch)), Arch=Prod(Epsilon, Sequence(Arch),Z)},unlabeled],size=i), i=0..25)];
with(combstruct):bin := {B=Union(Z,Prod(B,B))}: seq (count([B,bin,unlabeled],size=n)*n, n=1..25); # Zerinvary Lajos, Dec 05 2007
A000984List := proc(m) local A, P, n; A := [1,2]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
A := [op(A), 2*P[-1]] od; A end: A000984List(28); # Peter Luschny, Mar 24 2022

Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)
CoefficientList[Series[1/Sqrt[1-4x],{x,0,25}],x]  (* Harvey P. Dale, Mar 14 2011 *)

A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */

A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014

fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=prodeuler(p=2,2*n,p^(fv(2*n,p)-2*fv(n,p))) \\ Charles R Greathouse IV, Aug 21 2013

fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=my(s=1);forprime(p=2,2*n,s*=p^(fv(2*n,p)-2*fv(n,p)));s \\ Charles R Greathouse IV, Aug 21 2013

from _future_ import division
A000984_list, b = [1], 1
for n in range(10**3):
    b = b*(4*n+2)//(n+1)
    A000984_list.append(b) # Chai Wah Wu, Mar 04 2016

a(n)/(n+1) = A000108(n), the Catalan numbers.
G.f.: A(x) = (1 - 4*x)^(-1/2) = 1F0(1/2;;4x).
a(n+1) = 2*A001700(n) = A030662(n) + 1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n) =A099976.
D-finite with recurrence: n*a(n) + 2*(1-2*n)*a(n-1)=0.
a(n) = 2^n/n! * Product_{k=0..n-1} (2*k+1).
a(n) = a(n-1)*(4-2/n) = Product_{k=1..n} (4-2/k) = 4*a(n-1) + A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = -A005408(n-1)*A002420(n). - Henry Bottomley, Nov 10 2000
Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 4^n / sqrt(Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
Integral representation as n-th moment of a positive function on the interval [0, 4]: a(n) =  Integral_{x=0..4}(x^n*((x*(4-x))^(-1/2))/Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Sep 17 2001
Sum_{n>=1} 1/a(n) = (2*Pi*sqrt(3) + 9)/27. [Lehmer 1985, eq. (15)] - Benoit Cloitre, May 01 2002 (= A073016. - Bernard Schott, Jul 20 2022)
a(n) = Max_{ (i+j)!/(i!j!) | 0<=i,j<=n }. - Benoit Cloitre, May 30 2002
a(n) = Sum_{k=0..n} binomial(n+k-1,k), row sums of A059481. - Vladeta Jovovic, Aug 28 2002
E.g.f.: exp(2*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 08 2002
E.g.f.: I_0(2*x) = Sum a(n)*x^(2*n)/(2*n)!, where I_0 is Bessel function. - Michael Somos, Sep 09 2002
a(n) = Sum_{k=0..n} binomial(n, k)^2. - Benoit Cloitre, Jan 31 2003
Determinant of n X n matrix M(i, j) = binomial(n+i, j). - Benoit Cloitre, Aug 28 2003
Given m = C(2*n, n), let f be the inverse function, so that f(m) = n. Letting q denote -log(log(16)/(m^2*Pi)), we have f(m) = ceiling( (q + log(q)) / log(16) ). - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Oct 30 2003
a(n) = 2*Sum_{k=0..(n-1)} a(k)*a(n-k+1)/(k+1). - Philippe Deléham, Jan 01 2004
a(n+1) = Sum_{j=n..n*2+1} binomial(j, n). E.g., a(4) = C(7,3) + C(6,3) + C(5,3) + C(4,3) + C(3,3) = 35 + 20 + 10 + 4 + 1 = 70. - Jon Perry, Jan 20 2004
a(n) = (-1)^(n)*Sum_{j=0..(2*n)} (-1)^j*binomial(2*n, j)^2. - Helena Verrill (verrill(AT)math.lsu.edu), Jul 12 2004
a(n) = Sum_{k=0..n} binomial(2n+1, k)*sin((2n-2k+1)*Pi/2). - Paul Barry, Nov 02 2004
a(n-1) = (1/2)*(-1)^n*Sum_{0<=i, j<=n}(-1)^(i+j)*binomial(2n, i+j). - Benoit Cloitre, Jun 18 2005
a(n) = C(2n, n-1) + C(n) = A001791(n) + A000108(n). - Lekraj Beedassy, Aug 02 2005
G.f.: c(x)^2/(2*c(x)-c(x)^2) where c(x) is the g.f. of A000108. - Paul Barry, Feb 03 2006
a(n) = A006480(n) / A005809(n). - Zerinvary Lajos, Jun 28 2007
a(n) = Sum_{k=0..n} A106566(n,k)*2^k. - Philippe Deléham, Aug 25 2007
a(n) = Sum_{k>=0} A039599(n, k). a(n) = Sum_{k>=0} A050165(n, k). a(n) = Sum_{k>=0} A059365(n, k)*2^k, n>0. a(n+1) = Sum_{k>=0} A009766(n, k)*2^(n-k+1). - Philippe Deléham, Jan 01 2004
a(n) = 4^n*Sum_{k=0..n} C(n,k)(-4)^(-k)*A000108(n+k). - Paul Barry, Oct 18 2007
a(n) = Sum_{k=0..n} A039598(n,k)*A059841(k). - Philippe Deléham, Nov 12 2008
A007814(a(n)) = A000120(n). - Vladimir Shevelev, Jul 20 2009
From Paul Barry, Aug 05 2009: (Start)
G.f.: 1/(1-2x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction);
G.f.: 1/(1-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
If n>=3 is prime, then a(n) == 2 (mod 2*n). - Vladimir Shevelev, Sep 05 2010
Let A(x) be the g.f. and B(x) = A(-x), then B(x) = sqrt(1-4*x*B(x)^2). - Vladimir Kruchinin, Jan 16 2011
a(n) = (-4)^n*sqrt(Pi)/(gamma((1/2-n))*gamma(1+n)). - Gerry Martens, May 03 2011
a(n) = upper left term in M^n, M = the infinite square production matrix:
  2, 2, 0, 0, 0, 0, ...
  1, 1, 1, 0, 0, 0, ...
  1, 1, 1, 1, 0, 0, ...
  1, 1, 1, 1, 1, 0, ...
  1, 1, 1, 1, 1, 1, .... - Gary W. Adamson, Jul 14 2011
a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011
E.g.f.: hypergeometric([1/2],[1],4*x). - Wolfdieter Lang, Jan 13 2012
a(n) = 2*Sum_{k=0..n-1} a(k)*A000108(n-k-1). - Alzhekeyev Ascar M, Mar 09 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 28 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2*H(k)/(2*H(n)-H(2*n)), n>0, where H(n) is the n-th harmonic number. - Gary Detlefs, Mar 19 2013
G.f.: Q(0)*(1-4*x), where Q(k) = 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,0,-1/2-n,-1). - Karol A. Penson, Jul 27 2013
a(n) = 2^(4*n)/((2*n+1)*Sum_{k=0..n} (-1)^k*C(2*n+1,n-k)/(2*k+1)). - Mircea Merca, Nov 12 2013
a(n) = C(2*n-1,n-1)*C(4*n^2,2)/(3*n*C(2*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
Sum_{n>=0} a(n)/n! = A234846. - Richard R. Forberg, Feb 10 2014
0 =  a(n)*(16*a(n+1) - 6*a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 17 2014
a(n+1) = 4*a(n) - 2*A000108(n). Also a(n) = 4^n*Product_{k=1..n}(1-1/(2*k)). - Stanislav Sykora, Aug 09 2014
G.f.: Sum_{n>=0} x^n/(1-x)^(2*n+1) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n) = (-4)^n*binomial(-1/2,n). - Jean-François Alcover, Feb 10 2015
a(n) = 4^n*hypergeom([-n,1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
a(n) ~ 4^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)/sqrt((4*n+1) *Pi). - Peter Luschny, Oct 14 2015
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). Compare with the o.g.f. B(x) of A098616, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
a(n) = GegenbauerC(n,-n,-1). - Peter Luschny, May 07 2016
a(n) = gamma(1+2*n)/gamma(1+n)^2. - Andres Cicuttin, May 30 2016
Sum_{n>=0} (-1)^n/a(n) = 4*(5 - sqrt(5)*log(phi))/25 = 0.6278364236143983844442267..., where phi is the golden ratio. - Ilya Gutkovskiy, Jul 04 2016
From Peter Bala, Jul 22 2016: (Start)
This sequence occurs as the closed-form expression for several binomial sums:
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(2*n + 1,k).
a(n) = 2*Sum_{k = 0..2*n-1} (-1)^(n+k)*binomial(2*n - 1,k)*binomial(2*n,k) for n >= 1.
a(n) = 2*Sum_{k = 0..n-1} binomial(n - 1,k)*binomial(n,k) for n >= 1.
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x - k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... both Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y + k,n) and Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x - k,n)*binomial(y - k,n) appear to equal Kronecker's delta(n,0).
a(n) = (-1)^n*Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y - k,n) appears to equal Kronecker's delta(n,0).
a(n) = Sum_{k = 0..2n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)* binomial(n + k,n)^2. (Gould, Vol. 7, 5.23).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n,n + k)*binomial(n + k,n)^2. (End)
From Ralf Steiner, Apr 07 2017: (Start)
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for q in N, p in Z/{-4q< (some p) <-2}.
...
Sum_{k>=0} a(k)/(-4)^k = 1/sqrt(2).
Sum_{k>=0} a(k)/(17/4)^k = sqrt(17).
Sum_{k>=0} a(k)/(18/4)^k = 3.
Sum_{k>=0} a(k)/5^k = sqrt(5).
Sum_{k>=0} a(k)/6^k = sqrt(3).
Sum_{k>=0} a(k)/8^k = sqrt(2).
...
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for p>4q.(End)
Boas-Buck recurrence: a(n) = (2/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n, 0). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(2*n+1, k) for n in N. - Rene Adad, Sep 30 2017
a(n) = A034870(n,n). - Franck Maminirina Ramaharo, Nov 26 2018
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*(sqrt(4-x) + sqrt(x)*arcsin(sqrt(x)/2)) / (4-x)^(3/2).
E.g.f. for {1/a(n)}: 1 + exp(x/4)*sqrt(Pi*x)*erf(sqrt(x)/2)/2.
Sum_{n>=0} (-1)^n/a(n) = 4*(1/5 - arcsinh(1/2)/(5*sqrt(5))). (End)
From Peter Luschny, Sep 08 2022: (Start)
a(n) = 2^(2*n)*Product_{k=1..2*n} k^((-1)^(k+1)) = A056040(2*n).
a(n) = A001316(n) * A356637(n) * A261130(n) for n >= 2. (End)
a(n) = 4^n*binomial(n-1/2,-1/2) = 4^n*GegenbauerC(n,1/4,1). - Gerry Martens, Oct 19 2022
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^2 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^2 * binomial(2*n, n+k)^3 = x*(x + 2*n)*a(n) (x arbitrary). Compare with the identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^2 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Mar 31 2024: (Start)
4^n*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
16^n = Sum_{k = 0..2*n} a(k)*a(2*n-k). (End)
From Gary Detlefs, May 28 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k)*binomial(2*k,k)*2^(n-2*k). (H. W. Gould) - Gary Detlefs, May 28 2024
a(n) = Sum_{k=0..2*n} (-1)^k*binomial(2n,k)*binomial(2*n+2*k,n+k)*3^(2*n-k). (H. W. Gould) (End)
a(n) = Product_{k>=n+1} k^2/(k^2 - n^2). - Antonio Graciá Llorente, Sep 08 2024
a(n) = Product_{k=1..n} A003418(floor(2*n/k))^((-1)^(k+1)) (Golomb, 2003). - Amiram Eldar, Aug 08 2025

A001700 a(n) = binomial(2*n+1, n+1): number of ways to put n+1 indistinguishable balls into n+1 distinguishable boxes = number of (n+1)-st degree monomials in n+1 variables = number of monotone maps from 1..n+1 to 1..n+1.

Original entry on oeis.org

1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052

Offset: 0

N. J. A. Sloane

To show for example that C(2n+1, n+1) is the number of monotone maps from 1..n + 1 to 1..n + 1, notice that we can describe such a map by a nondecreasing sequence of length n + 1 with entries from 1 to n + 1. The number k of increases in this sequence is anywhere from 0 to n. We can specify these increases by throwing k balls into n+1 boxes, so the total is Sum_{k = 0..n} C((n+1) + k - 1, k) = C(2*n+1, n+1).
Also number of ordered partitions (or compositions) of n + 1 into n + 1 parts. E.g., a(2) = 10: 003, 030, 300, 012, 021, 102, 120, 210, 201, 111. - Mambetov Bektur (bektur1987(AT)mail.ru), Apr 17 2003
Also number of walks of length n on square lattice, starting at origin, staying in first and second quadrants. - David W. Wilson, May 05 2001. (E.g., for n = 2 there are 10 walks, all starting at 0, 0: 0, 1 -> 0, 0; 0, 1 -> 1, 1; 0, 1 -> 0, 2; 1, 0 -> 0, 0; 1, 0 -> 1, 1; 1, 0 -> 2, 0; 1, 0 -> 1, -1; -1, 0 -> 0, 0; -1, 0 -> -1, 1; -1, 0-> -2, 0.)
Also total number of leaves in all ordered trees with n + 1 edges.
Also number of digitally balanced numbers [A031443] from 2^(2*n+1) to 2^(2*n+2). - Naohiro Nomoto, Apr 07 2001
Also number of ordered trees with 2*n + 2 edges having root of even degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of paths of length 2*d(G) connecting two neighboring nodes in optimal chordal graph of degree 4, G(2*d(G)^2 + 2*d(G) + 1, 2d(G) + 1), where d(G) = diameter of graph G. - S. Bujnowski (slawb(AT)atr.bydgoszcz.pl), Feb 11 2002
Define an array by m(1, j) = 1, m(i, 1) = i, m(i, j) = m(i, j-1) + m(i-1, j); then a(n) = m(n, n), diagonal of A165257 - Benoit Cloitre, May 07 2002
Also the numerator of the constant term in the expansion of cos^(2*n)(x) or sin^(2*n)(x) when the denominator is 2^(2*n-1). - Robert G. Wilson v
Consider the expansion of cos^n(x) as a linear combination of cosines of multiple angles. If n is odd, then the expansion is a combination of a*cos((2*k-1)*x)/2^(n-1) for all 2*k - 1 <= n. If n is even, then the expansion is a combination of a*cos(2k*x)/2^(n-1) terms plus a constant. "The constant term, [a(n)/2^(2n-1)], is due to the fact that [cos^2n(x)] is never negative, i.e., electrical engineers would say the average or 'dc value' of [cos^(2*n)(x)] is [a(n)/2^(2*n-1)]. The dc value of [cos^(2*n-1)(x)] on the other hand, is zero because it is symmetrical about the horizontal axis, i.e., it is negative and positive equally." Nahin[62] - Robert G. Wilson v, Aug 01 2002
Also number of times a fixed Dyck word of length 2*k occurs in all Dyck words of length 2*n + 2*k. Example: if the fixed Dyck word is xyxy (k = 2), then it occurs a(1) = 3 times in the 5 Dyck words of length 6 (n = 1): (xy[xy)xy], xyxxyy, xxyyxy, x(xyxy)y, xxxyyy (placed between parentheses). - Emeric Deutsch, Jan 02 2003
a(n+1) is the determinant of the n X n matrix m(i, j) = binomial(2*n-i, j). - Benoit Cloitre, Aug 26 2003
a(n-1) = (2*n)!/(2*n!*n!), formula in [Davenport] used by Gauss for the special case prime p = 4*n + 1: x = a(n-1) mod p and y = x*(2n)! mod p are solutions of p = x^2 + y^2. - Frank Ellermann. Example: For prime 29 = 4*7 + 1 use a(7-1) = 1716 = (2*7)!/(2*7!*7!), 5 = 1716 mod 29 and 2 = 5*(2*7)! mod 29, then 29 = 5*5 + 2*2.
The number of compositions of 2*n, say c_1 + c_2 + ... + c_k = 2n, satisfy that Sum_{i = 1..j} c_i < 2*j for all j = 1..k, or equivalently, the number of subsets, say S, of [2*n-1] = {1, 2, ..., 2*n-1} with at least n elements such that if 2k is in S, then there must be at least k elements in S smaller than 2k. E.g., a(2) = 3 because we can write 4 = 1 + 1 + 1 + 1 = 1 + 1 + 2 = 1 + 2 + 1. - Ricky X. F. Chen (ricky_chen(AT)mail.nankai.edu.cn), Jul 30 2006
The number of walks of length 2*n + 1 on an infinite linear lattice that begin at the origin and end at node (1). Also the number of paths on a square lattice from the origin to (n+1, n) that use steps (1,0) and (0,1). Also number of binary numbers of length 2*n + 1 with n + 1 ones and n zeros. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
If Y is a 3-subset of an 2*n-set X then, for n >= 3, a(n-1) is the number of n-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
Also the number of rankings (preferential arrangements) of n unlabeled elements onto n levels when empty levels are allowed. - Thomas Wieder, May 24 2008
Also the Catalan transform of A000225 shifted one index, i.e., dropping A000225(0). - R. J. Mathar, Nov 11 2008
With offset 1. The number of solutions in nonnegative integers to X1 + X2 + ... + Xn = n. The number of terms in the expansion of (X1 + X2 + ... + Xn)^n. The coefficient of x^n in the expansion of (1 + x + x^2 + ...)^n. The number of distinct image sets of all functions taking [n] into [n]. - Geoffrey Critzer, Feb 22 2009
The Hankel transform of the aerated sequence 1, 0, 3, 0, 10, 0, ... is 1, 3, 3, 5, 5, 7, 7, ... (A109613(n+1)). - Paul Barry, Apr 21 2009
Also the number of distinct network topologies for a network of n items with 1 to n - 1 unidirectional connections to other objects in the network. - Anthony Bachler, May 05 2010
Equals INVERT transform of the Catalan numbers starting with offset 1. E.g.: a(3) = 35 = (1, 2, 5) dot (10, 3, 1) + 14 = 21 + 14 = 35. - Gary W. Adamson, May 15 2009
The integral of 1/(1+x^2)^(n+1) is given by a(n)/2^(2*n - 1) * (x/(1 + x^2)^n*P(x) + arctan(x)), where P(x) is a monic polynomial of degree 2*n - 2 with rational coefficients. - Christiaan van de Woestijne, Jan 25 2011
a(n) is the number of Schroder paths of semilength n in which the (2,0)-steps at level 0 come in 2 colors and there are no (2,0)-steps at a higher level. Example: a(2) = 10 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 2^2 = 4 paths of shape HH, 2 paths of shape HUD, 2 paths of shape UDH, and 1 path of each of the shapes UDUD and UUDD. - Emeric Deutsch, May 02 2011
a(n) is the number of Motzkin paths of length n in which the (1,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=35 because, denoting  U = (1,1), H = (1,0), and D = (1,-1), we have 3^3 = 27 paths of shape HHH, 3 paths of shape HUD, 3 paths of shape UDH, and 2 paths of shape UHD. - Emeric Deutsch, May 02 2011
Also number of digitally balanced numbers having length 2*(n + 1) in binary representation: a(n) = #{m: A070939(A031443(m)) = 2*(n + 1)}. - Reinhard Zumkeller, Jun 08 2011
a(n) equals 2^(2*n + 3) times the coefficient of Pi in 2F1([1/2, n+2]; [3/2]; -1). - John M. Campbell, Jul 17 2011
For positive n, a(n) equals 4^(n+2) times the coefficient of Pi^2 in Integral_{x = 0..Pi/2} x sin^(2*n + 2)x. - John M. Campbell, Jul 19 2011 [Apparently, the contributor means Integral_{x = 0..Pi/2} x * (sin(x))^(2*n + 2).]
a(n-1) = C(2*n, n)/2 is the number of ways to assign 2*n people into 2 (unlabeled) groups of size n. - Dennis P. Walsh, Nov 09 2011
Equals row sums of triangle A205945. - Gary W. Adamson, Feb 01 2012
a(n-1) gives the number of n-regular sequences defined by Erdős and Gallai in 1960 in connection with the degree sequences of simple graphs. - Matuszka Tamás, Mar 06 2013
a(n) is the sum of falling diagonals of squares in the comment in A085812 (equivalent to the Cloitre formula of Aug 2002). - John Molokach, Sep 26 2013
For n > 0: largest terms of Zigzag matrices as defined in A088961. - Reinhard Zumkeller, Oct 25 2013
Also the number of different possible win/loss round sequences (from the perspective of the eventual winner) in a "best of 2*n + 1" two-player game. For example, a(2) = 10 means there are 10 different win/loss sequences in a "best of 5" game (like a tennis match in which the first player to win 3 sets, out of a maximum of 5, wins the match); the 10 sequences are WWW, WWLW, WWLLW, WLWW, WLWLW, WLLWW, LWWW, LWWLW, LWLWW, LLWWW. See also A072600. - Philippe Beaudoin, May 14 2014; corrected by Jon E. Schoenfield, Nov 23 2014
When adding 1 to the beginning of the sequence: Convolving a(n)/2^n with itself equals 2^(n+1).  For example, when n = 4: convolving {1, 1/1, 3/2, 10/4, 35/8, 126/16} with itself is 32 = 2^5. - Bob Selcoe, Jul 16 2014
From Tom Copeland, Nov 09 2014: (Start)
The shifted array belongs to a family of arrays associated to the Catalan A000108 (t = 1), and Riordan, or Motzkin sums A005043 (t = 0), with the o.g.f. [1 - sqrt(1 - 4x/(1 + (1 - t)x))]/2 and inverse x*(1 - x)/[1 + (t - 1)*x*(1 - x)]. See A091867 for more info on this family. Here is t = -3 (mod signs in the results).
Let C(x) = [1 - sqrt(1-4x)]/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1 + t*x) with inverse P(x, -t).
O.g.f: G(x) = [-1 + sqrt(1 + 4*x/(1 - 4*x))]/2 = -C[P(-x, 4)].
Inverse o.g.f: Ginv(x) = x*(1 + x)/(1 + 4*x*(1 + x)) = -P(Cinv(-x), -4) (shifted signed A001792). A088218(x) = 1 + G(x).
Equals A001813/2 omitting the leading 1 there. (End)
Placing n distinguishable balls into n indistinguishable boxes gives A000110(n) (the number of set partitions). - N. J. A. Sloane, Jun 19 2015
The sequence is the INVERTi transform of A049027: (1, 4, 17, 74, 326, ...). - Gary W. Adamson, Jun 23 2015
a(n) is the number of compositions of 2*n + 2 such that the sum of the elements at odd positions is equal to the sum of the elements at even positions. a(2) = 10 because there are 10 such compositions of 6: (3, 3), (1, 3, 2), (2, 3, 1), (1, 1, 2, 2), (1, 2, 2, 1), (2, 2, 1, 1), (2, 1, 1, 2), (1, 2, 1, 1, 1), (1, 1, 1, 2, 1), (1, 1, 1, 1, 1, 1). - Ran Pan, Oct 08 2015
a(n-1) is also the Schur function of the partition (n) of n evaluated at x_1 = x_2 = ... = x_n = 1, i.e., the number of semistandard Young tableaux of shape (n) (weakly increasing rows with n boxes with numbers from {1, 2, ..., n}). - Wolfdieter Lang, Oct 11 2015
Also the number of ordered (rooted planar) forests with a total of n+1 edges and no trivial trees. - Nachum Dershowitz, Mar 30 2016
a(n) is the number of sets (i1,...in) of length n so that n >= i1 >= i2 >= ...>= in >= 1. For instance, n=3 as there are only 10 such sets (3,3,3) (3,3,2) (3,3,1) (3,2,2) (3,2,1) (3,1,1) (2,2,2) (2,2,1) (2,1,1) (1,1,1,) 3,2,1 is each used 10 times respectively. - Anton Zakharov, Jul 04 2016
The repeated middle term in the odd rows of Pascal's triangle, or half the central binomial coefficient in the even rows of Pascal's triangle, n >= 2. - Enrique Navarrete, Feb 12 2018
a(n) is the number of walks of length 2n+1 from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n+1 from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Total number of nodes summed over all Dyck paths of semilength n. - Alois P. Heinz, Mar 08 2020
a(n-1) is the determinant of the n X n matrix m(i, j) = binomial(n+i-1, j). - Fabio Visonà, May 21 2022
Let X_i be iid standard Gaussian random variable N(0,1), and S_n be the partial sum S_n = X_1+...+X_n. Then P(S_1>0,S_2>0,...,S_n>0) = a(n+1)/2^(2n-1) = a(n+1) / A004171(n+1). For example,  P(S_1>0) = 1/2, P(S_1>0,S_2>0) = 3/8, P(S_1>0,S_2>0,S_3>0) = 5/16, etc. This probability is also equal to the volume of the region x_1 > 0, x_2 > -x_1, x_3 > -(x_1+x_2), ..., x_n > -(x_1+x_2+...+x_(n-1)) in the hypercube [-1/2, 1/2]^n. This also holds for the Cauchy distribution and other stable distributions with mean 0, skew 0 and scale 1. - Xiaohan Zhang, Nov 01 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 132, 213, and 321. - Lara Pudwell, Apr 10 2023
Number of vectors in (Z_>=0)^(n+1) such that the sum of the components is n+1. binomial(2*n-1, n) provides this property for n. - Michael Richard, Jun 12 2023
Also number of discrete negations on the finite chain L_n={0,1,...,n-1,n}, i.e., monotone decreasing unary operators such that N(0)=n and N(n)=0. - Marc Munar, Oct 10 2023
a(n) is the number of Dyck paths of semilength n+1 having one of its peaks marked. - Juan B. Gil, Jan 03 2024
a(n) is the dimension of the (n+1)-st symmetric power of an (n+1)-dimensional vector space. - Mehmet A. Ates, Feb 15 2024
a(n) is the independence number of the twisted odd graph O^(sigma)(n+2). - _Miquel A. Fiol, Aug 26 2024
a(n) is the number of non-descending sequences with length n and the last number is less or equal to n. a(n) is also the number of integer partitions (of any positive integer) with length n and largest part is less or equal to n. - Zlatko Damijanic, Dec 06 2024
a(n) is the number of triangulations of a once-punctured (n+1)-gon [from Fontaine & Plamondon's Theorem 3.6]. - Esther Banaian, May 06 2025

			There are a(2)=10 ways to put 3 indistinguishable balls into 3 distinguishable boxes, namely, (OOO)()(), ()(OOO)(), ()()(OOO), (OO)(O)(), (OO)()(O), (O)(OO)(), ()(OO)(O), (O)()(OO), ()(O)(OO), and (O)(O)(O). - _Dennis P. Walsh_, Apr 11 2012
a(2) = 10: Semistandard Young tableaux for partition (3) of 3 (the indeterminates x_i, i = 1, 2, 3 are omitted and only their indices are given): 111, 112, 113, 122, 123, 133, 222, 223, 233, 333. - _Wolfdieter Lang_, Oct 11 2015

H. Davenport, The Higher Arithmetic. Cambridge Univ. Press, 7th ed., 1999, ch. V.3 (p. 122).
A. Frosini, R. Pinzani, and S. Rinaldi, About half the middle binomial coefficient, Pure Math. Appl., 11 (2000), 497-508.
Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 449.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
Paul J. Nahin, "An Imaginary Tale, The Story of [Sqrt(-1)]," Princeton University Press, Princeton, NJ 1998, p. 62.
L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Matuszka Tamás, Table of n, a(n) for n = 0..1200 (first 100 terms from T. D. Noe)
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011), #11.2.6, theorem 4.
Ayomikun Adeniran and Lara Pudwell, Pattern avoidance in parking functions, Enumer. Comb. Appl. 3:3 (2023), Article S2R17.
José Agapito, Ângela Mestre, Maria M. Torres, and Pasquale Petrullo, On One-Parameter Catalan Arrays, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.1.
Martin Aigner, Enumeration via ballot numbers, Discrete Math., 308 (2008), 2544-2563.
Elena Barcucci, Andrea Frosini, and Simone Rinaldi, On directed-convex polyominoes in a rectangle, Discr. Math., 298 (2005), 62-78.
Elena Barcucci, Antonio Bernini, and Renzo Pinzani, Exhaustive generation of positive lattice paths, Semantic Sensor Networks Workshop 2018, CEUR Workshop Proceedings (2018), Vol. 2113.
Jean-Luc Baril, David Bevan, and Sergey Kirgizov, Bijections between directed animals, multisets and Grand-Dyck paths, arXiv:1906.11870 [math.CO], 2019.
Jean-Luc Baril, Pamela E. Harris, Kimberly J. Harry, Matt McClinton, and José L. Ramírez, Enumerating runs, valleys, and peaks in Catalan words, arXiv:2404.05672 [math.CO], 2024. See p. 5.
Jean-Luc Baril and Nathanaël Hassler, Intervals in a family of Fibonacci lattices, Univ. de Bourgogne (France, 2024). See p. 17.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, 9 (2006), Article 06.2.4.
Paul Barry, The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths, J. Int. Seq., 22 (2019), Article 19.1.3.
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Antonio Bernini, Filippo Disanto, Renzo Pinzani, and Simone Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007), #07.9.7.
Ciprian Borcea and Ileana Streinu, On the number of embeddings of minimally rigid graphs, arXiv:math/0207126 [math.MG], 2002.
Mireille Bousquet-Mélou, New enumerative results on two-dimensional directed animals, Discr. Math., 180 (1998), 73-106. See Eq. (1).
M. Bousquet-Mélou and A. Rechnitzer, Lattice animals and heaps of dimers.
Jean-Paul Bultel and Samuele Giraudo, Combinatorial Hopf algebras from PROs, arXiv:1406.6903 [math.CO], 2014. [DOI]
Libor Caha and Daniel Nagaj, The pair-flip model: a very entangled translationally invariant spin chain, arXiv:1805.07168 [quant-ph], 2018.
David Callan, A Combinatorial Interpretation for a Super-Catalan Recurrence, Journal of Integer Sequences, 8 (2005), Article 05.1.8.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs., 3 (2000), #00.1.5.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, Mutation effects in ordered trees, arXiv:1410.1249 [math.CO], 2014.
Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, and Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, 18 (2015), Article 15.5.8.
Nachum Dershowitz, 1700 Forests, arXiv:1608.08740 [cs.DM], 2016.
Leonard E. Dickson, On the Inscription of Regular Polygons, Annals of Mathematics, Vol. 9, No. 1/6 (1894 - 1895), pp. 73-84.
Leonard E. Dickson, Problem 44, The American Mathematical Monthly, Vol. 2, No. 7/8 (Jul. - Aug., 1895), pp. 229-230.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
Richard Ehrenborg, Gábor Hetyei, and Margaret Readdy, Catalan-Spitzer permutations, arXiv:2310.06288 [math.CO], 2023. See p. 20.
Miquel A. Fiol, E. Garriga, and J. L. A. Yebra, On twisted odd graphs, Combin. Probab. Comput. 9 (2000), 227-240.
Rigoberto Flórez, Leandro Junes, and José L. Ramírez, Further Results on Paths in an n-Dimensional Cubic Lattice, Journal of Integer Sequences, 21 (2018), Article 18.1.2.
B. Fontaine and P.-G. Plamondon, Counting friezes in type D_n, Journal of Algebraic Combinatorics, 44.2 (2016), 433-445; arXiv:1409.3698 [math.CO], 2014-2016.
Shishuo Fu and Yaling Wang, Bijective recurrences concerning two Schröder triangles, arXiv:1908.03912 [math.CO], 2019.
Juan B. Gil, Emma G. Hoover, and Jessica A. Shearer, Bijections between colored compositions, Dyck paths, and polygon partitions, arXiv:2403.04575 [math.CO], 2024. See p. 4.
N. Gromov and P. Vieira, Tailoring Three-Point Functions and Integrability IV. Theta-morphism, arXiv:1205.5288 [hep-th], 2012. - From _N. J. A. Sloane_, Oct 23 2012
R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, 3 (2000), Article #00.1.6.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 145.
A. Ivanyi, L. Lucz, T. Matuszka, and S. Pirzada, Parallel enumeration of degree sequences of simple graphs, Acta Univ. Sapientiae, Informatica, 4(2) (2012) 260-288.
Milan Janjic, Two Enumerative Functions.
Christian Krattenthaler and Daniel Yaqubi, Some determinants of path generating functions, II, Adv. Appl. Math. 101 (2018), 232-265.
Dmitry Kruchinin, Superposition's properties of logarithmic generating functions, arXiv:1109.1683 [math.CO], 2011-2015.
Markus Kuba and Alois Panholzer, Stirling permutations containing a single pattern of length three, Australasian Journal of Combinatorics 74(2) (2019), 216-239.
Wolfdieter Lang, On generalizations of Stirling number triangles, J. Integer Seqs., 3 (2000), #00.2.4.
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
H. Li and T. MacHenry, Permanents and Determinants, Weighted Isobaric Polynomials, and Integer Sequences, J. Int. Seq. 16 (2013) #13.3.5, example 40.
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
T. Mansour and M. Shattuck, A statistic on n-color compositions and related sequences, Proc. Indian Acad. Sci. (Math. Sci.) 124(2) (2014), pp. 127-140.
M. D. McIlroy, Letter to N. J. A. Sloane (no date).
Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, 21 (2018), Article 18.1.3.
Hanna Mularczyk, Lattice Paths and Pattern-Avoiding Uniquely Sorted Permutations, arXiv:1908.04025 [math.CO], 2019.
M. Munar, S. Massanet, and D. Ruiz-Aguilera, On the cardinality of some families of discrete connectives, Information Sciences, Volume 621, 2023, 708-728.
A. Nkwanta, Riordan matrices and higher-dimensional lattice walks, J. Stat. Plann. Infer. 140 (2010), 2321-2334.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, 15 (2012), Article 12.3.3. - From _N. J. A. Sloane_, Sep 16 2012
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., 4 (2001), #01.2.1.
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their N-numbers, not their A-numbers.
Louis Shapiro, Problem 10753 Amer. Math. Monthly, 106(8) (1999), p. 777.
Louis Shapiro et al., Leaves of Ordered Trees: 10753, Amer. Math. Monthly, 108(9) (Nov., 2001), 873-874.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See pp. 96, 98-99.
Eric Weisstein's World of Mathematics, Binomial Coefficient.
Eric Weisstein's World of Mathematics, Odd Graph.
Index entries for "core" sequences.

Cf. A000110, A007318, A030662, A046097, A060897-A060900, A049027, A076025, A076026, A060150, A001263, A005773, A001405, A132813, A134285.
Equals A000984(n+1)/2.
a(n) = (2*n+1)*Catalan(n) [A000108] = A035324(n+1, 1) (first column of triangle).
Row sums of triangles A028364, A050166, A039598.
Bisections: a(2*k) = A002458(k), a(2*k+1) = A001448(k+1)/2, k >= 0.
Other versions of the same sequence: A088218, A110556, A138364.
Diagonals 1 and 2 of triangle A100257.
Second row of array A102539.
Column of array A073165.
Row sums of A103371. - Susanne Wienand, Oct 22 2011
Cf. A002054: C(2*n+1, n-1). - Bruno Berselli, Jan 20 2014
Cf. A005043, A091867, A001792, A001813, A166454.

List([0..30],n->Binomial(2*n+1,n+1)); # Muniru A Asiru, Feb 26 2019

a001700 n = a007318 (2*n+1) (n+1)  -- Reinhard Zumkeller, Oct 25 2013

[Binomial(2*n, n)/2: n in [1..40]]; // Vincenzo Librandi, Nov 10 2014

A001700 := n -> binomial(2*n+1,n+1); seq(A001700(n), n=0..20);
A001700List := proc(m) local A, P, n; A := [1]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
A := [op(A), P[-1]] od; A end: A001700List(27); # Peter Luschny, Mar 24 2022

Table[ Binomial[2n + 1, n + 1], {n, 0, 23}]
CoefficientList[ Series[2/((Sqrt[1 - 4 x] + 1)*Sqrt[1 - 4 x]), {x, 0, 22}], x] (* Robert G. Wilson v, Aug 08 2011 *)

B(n,a,x):=coeff(taylor(exp(x*t)*(t/(exp(t)-1))^a,t,0,20),t,n)*n!;
makelist((-1)^(n)*B(n,n+1,-n-1)/n!,n,0,10); /* Vladimir Kruchinin, Apr 06 2016 */

PARI
```
a(n)=binomial(2*n+1,n+1)
```

z='z+O('z^50); Vec((1/sqrt(1-4*z)-1)/(2*z)) \\ Altug Alkan, Oct 11 2015

from _future_ import division
A001700_list, b = [], 1
for n in range(10**3):
    A001700_list.append(b)
    b = b*(4*n+6)//(n+2) # Chai Wah Wu, Jan 26 2016

[rising_factorial(n+1,n+1)/factorial(n+1) for n in (0..22)] # Peter Luschny, Nov 07 2011

a(n-1) = binomial(2*n, n)/2 = A000984(n)/2 = (2*n)!/(2*n!*n!).
D-finite with recurrence: a(0) = 1, a(n) = 2*(2*n+1)*a(n-1)/(n+1) for n > 0.
G.f.: (1/sqrt(1 - 4*x) - 1)/(2*x).
L.g.f.: log((1 - sqrt(1 - 4*x))/(2*x)) = Sum_{n >= 0} a(n)*x^(n+1)/(n+1). - Vladimir Kruchinin, Aug 10 2010
G.f.: 2F1([1, 3/2]; [2]; 4*x). - Paul Barry, Jan 23 2009
G.f.: 1/(1 - 2*x - x/(1 - x/(1 - x/(1 - x/(1 - ... (continued fraction). - Paul Barry, May 06 2009
G.f.: c(x)^2/(1 - x*c(x)^2), c(x) the g.f. of A000108. - Paul Barry, Sep 07 2009
O.g.f.: c(x)/sqrt(1 - 4*x) = (2 - c(x))/(1 - 4*x), with c(x) the o.g.f. of A000108. Added second formula. - Wolfdieter Lang, Sep 02 2012
Convolution of A000108 (Catalan) and A000984 (central binomial): Sum_{k=0..n} C(k)*binomial(2*(n-k), n-k), C(k) Catalan. - Wolfdieter Lang, Dec 11 1999
a(n) = Sum_{k=0..n} C(n+k, k). - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n+1, k+1). - Benoit Cloitre, Oct 19 2002
a(n) = Sum_{k = 0..n+1} binomial(2*n+2, k)*cos((n - k + 1)*Pi). - Paul Barry, Nov 02 2004
a(n) = 4^n*binomial(n+1/2, n)/(n+1). - Paul Barry, May 10 2005
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n + 1)/(2*n + 1)! = BesselI(1, 2*x). - Michael Somos, Jun 22 2005
E.g.f. in Maple notation: exp(2*x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). Integral representation as n-th moment of a positive function on [0, 4]: a(n) = Integral_{x = 0..4} x^n * (x/(4 - x))^(1/2)/(2*Pi) dx, n >= 0. This representation is unique. - Karol A. Penson, Oct 11 2001
Narayana transform of [1, 2, 3, ...]. Let M = the Narayana triangle of A001263 as an infinite lower triangular matrix and V = the Vector [1, 2, 3, ...]. Then A001700 = M * V. - Gary W. Adamson, Apr 25 2006
a(n) = A122366(n,n). - Reinhard Zumkeller, Aug 30 2006
a(n) = C(2*n, n) + C(2*n, n-1) = A000984(n) + A001791(n). - Zerinvary Lajos, Jan 23 2007
a(n-1) = (n+1)*(n+2)*...*(2*n-1)/(n-1)! (product of n-1 consecutive integers, divided by (n-1)!). - Jonathan Vos Post, Apr 09 2007; [Corrected and shortened by Giovanni Ciriani, Mar 26 2019]
a(n-1) = (2*n - 1)!/(n!*(n - 1)!). - William A. Tedeschi, Feb 27 2008
a(n) = (2*n + 1)*A000108(n). - Paul Barry, Aug 21 2007
Binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double binomial transform of A001405. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A132813. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A134285. - Gary W. Adamson, Nov 19 2007
a(n) = 2*A000984(n) - A000108(n), that is, a(n) = 2*C(2*n, n) - n-th Catalan number. - Joseph Abate, Jun 11 2010
Conjectured: 4^n GaussHypergeometric(1/2,-n; 2; 1) -- Solution for the path which stays in the first and second quadrant. - Benjamin Phillabaum, Feb 20 2011
a(n)= Sum_{k=0..n} A038231(n,k) * (-1)^k * A000108(k). - Philippe Deléham, Nov 27 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1] = -1, A[i,j] = Catalan(j-i), (i <= j), and A[i,j] = 0, otherwise. Then, for n >= 1, a(n) = (-1)^n * charpoly(A,-2). - Milan Janjic, Jul 08 2010
a(n) is the upper left term of M^(n+1), where M is the infinite matrix in which a column of (1,2,3,...) is prepended to an infinite lower triangular matrix of all 1's and the rest zeros, as follows:
   1, 1, 0, 0, 0, ...
   2, 1, 1, 0, 0, ...
   3, 1, 1, 1, 0, ...
   4, 1, 1, 1, 1, ...
   ...
Alternatively, a(n) is the upper left term of M^n where M is the infinite matrix:
   3, 1, 0, 0, 0, ...
   1, 1, 1, 0, 0, ...
   1, 1, 1, 1, 0, ...
   1, 1, 1, 1, 1, ...
   ...
- Gary W. Adamson, Jul 14 2011
a(n) = (n + 1)*hypergeom([-n, -n], [2], 1). - Peter Luschny, Oct 24 2011
a(n) = Pochhammer(n+1, n+1)/(n+1)!. - Peter Luschny, Nov 07 2011
E.g.f.: 1 + 6*x/(U(0) - 6*x); U(k) = k^2 + (4*x + 3)*k + 6*x + 2 - 2*x*(k + 1)*(k + 2)*(2*k + 5)/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2011
a(n) = 2*A000984(n) - A000108(n). [Abate & Whitt]
a(n) = 2^(2*n+1)*binomial(n+1/2, -1/2). - Peter Luschny, May 06 2014
For n > 1: a(n-1) = A166454(2*n, n), central terms in A166454. - Reinhard Zumkeller, Mar 04 2015
a(n) = 2*4^n*Gamma(3/2 + n)/(sqrt(Pi)*Gamma(2+n)). - Peter Luschny, Dec 14 2015
a(n) ~ 2*4^n*(1 - (5/8)/n + (73/128)/n^2 - (575/1024)/n^3 + (18459/32768)/n^4)/sqrt(n*Pi). - Peter Luschny, Dec 16 2015
a(n) = (-1)^(n)*B(n, n+1, -n-1)/n!, where B(n,a,x) is a generalized Bernoulli polynomial. - Vladimir Kruchinin, Apr 06 2016
a(n) = Gamma(2 + 2*n)/(n!*Gamma(2 + n)). Andres Cicuttin, Apr 06 2016
a(n) = (n + (n + 1))!/(Gamma(n)*Gamma(1 + n)*A002378(n)), for n > 0. Andres Cicuttin, Apr 07 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
Sum_{n >= 0} 1/a(n) = 2*(9 + 2*sqrt(3)*Pi)/27 = A248179.
Sum_{n >= 0} (-1)^n/a(n) = 2*(5 + 4*sqrt(5)*arcsinh(1/2))/25 = 2*(5*A145433 - 1).
Sum_{n >= 0} (-1)^n*a(n)/n! = BesselI(2,2)*exp(-2) = A229020*A092553. (End)
Conjecture: a(n) = Sum_{k=2^n..2^(n+1)-1} A178244(k). - Mikhail Kurkov, Feb 20 2021
a(n-1) = 1 + (1/n)*Sum_{t=1..n/2} (2*cos((2*t-1)*Pi/(2*n)))^(2*n). - Greg Dresden, Oct 11 2022
a(n) = Product_{1 <= i <= j <= n} (i + j + 1)/(i + j - 1). Cf. A006013. - Peter Bala, Feb 21 2023
Sum_{n >= 0} a(n)*x^(n+1)/(n+1) = x + 3*x^2/2 + 10*x^3/3 + 35*x^4/4 + ... =  the series reversion of exp(-x)*(1 - exp(-x)). - Peter Bala, Sep 06 2023

Name corrected by Paul S. Coombes, Jan 11 2012

Name corrected by Robert Tanniru, Feb 01 2014

A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

Original entry on oeis.org

1, 2, 1, 6, 6, 1, 20, 30, 10, 1, 70, 140, 70, 14, 1, 252, 630, 420, 126, 18, 1, 924, 2772, 2310, 924, 198, 22, 1, 3432, 12012, 12012, 6006, 1716, 286, 26, 1, 12870, 51480, 60060, 36036, 12870, 2860, 390, 30, 1, 48620, 218790, 291720, 204204, 87516, 24310

Offset: 0

Wolfdieter Lang

Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).
This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306.
As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005
The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007
As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - Paul Barry, Apr 14 2010
From Wolfdieter Lang, Aug 10 2017: (Start)
The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):
  (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)).
This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m).
For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End)
From Peter Bala, Mar 04 2018: (Start)
The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + ....
2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5).
Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End)

			Array begins:
  1,  2,   6,  20,   70, ...
  1,  6,  30, 140,  630, ...
  1, 10,  70, 420, 2310, ...
  1, 14, 126, 924, 6006, ...
Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30.
Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1.
From _Paul Barry_, Apr 14 2010: (Start)
As a number triangle, T(n, m) begins:
n\k       0      1       2       3      4      5     6    7   8  9 10 ...
0:        1
1:        2      1
2:        6      6       1
3:       20     30      10       1
4:       70    140      70      14      1
5:      252    630     420     126     18      1
6:      924   2772    2310     924    198     22     1
7:     3432  12012   12012    6006   1716    286    26    1
8:    12870  51480   60060   36036  12870   2860   390   30   1
9:    48620 218790  291720  204204  87516  24310  4420  510  34  1
10:  184756 923780 1385670 1108536 554268 184756 41990 6460 646 38  1
... [Reformatted and extended by _Wolfdieter Lang_, Aug 10 2017]
Production matrix begins
      2, 1,
      2, 4, 1,
     -4, 0, 4, 1,
     10, 0, 0, 4, 1,
    -28, 0, 0, 0, 4, 1,
     84, 0, 0, 0, 0, 4, 1,
   -264, 0, 0, 0, 0, 0, 4, 1,
    858, 0, 0, 0, 0, 0, 0, 4, 1,
  -2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End)
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (2*(2*2+1)/2) * Sum_{k=2..3} 4^(3-k)*T(k, 2) = 5*(4*1 + 1*10) = 70. - _Wolfdieter Lang_, Aug 10 2017
From _Peter Bala_, Feb 15 2018: (Start)
With C(x) = (1 - sqrt( 1 - 4*x))/(2*x),
-x^3/3! * (d/dx)^3(C(x)) = 1/(2*x)*( 1 - (1 - 10*x + 30*x^2 - 20*x^3)/(1 - 4*x)^(5/2) ).
x^4/4! * (d/dx)^4(C(x)) = 1/(2*x)*( 1 - (1 - 14*x + 70*x^2 - 140*x^3 + 70*x^4 )/(1 - 4*x)^(7/2) ). (End)

Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Muniru A Asiru, Antidiagonals n = 0..50, flattened
Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
Wolfdieter Lang, First 10 rows.
Wolfdieter Lang, On polynomials related to derivatives of the generating function of Catalan numbers, Fib. Quart. 40,4 (2002) 299-313; T(n,m) is called B(n,m) there.
Benjamin Lovitz and Nathaniel Johnston, A hierarchy of eigencomputations for polynomial optimization on the sphere, arXiv:2310.17827 [math.OC], 2023. See pp. 35, 39. See also Math. Prog. (2025) Series A.
Helmut Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.

Columns include: A000984 (m=0), A002457 (m=1), A002802 (m=2), A020918 (m=3), A020920 (m=4), A020922 (m=5), A020924 (m=6), A020926 (m=7), A020928 (m=8), A020930 (m=9), A020932 (m=10).
Row sums: A046748.
Cf. A001147, A006882, A007318, A068555, A111959, A283150, A283151.

Flat(List([0..9],n->List([0..n],m->Binomial(2*n,n)*Binomial(n,m)/Binomial(2*m,m)))); # Muniru A Asiru, Jul 19 2018

[Binomial(n+1,k+1)*Catalan(n)/Catalan(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 28 2024

t[i_, j_] := If[i < 0 || j < 0, 0, (2*i + 2*j)!*i!/(2*i)!/(i + j)!/j!]; Flatten[Reverse /@ Table[t[n, k - n] , {k, 0, 9}, {n, k, 0, -1}]][[1 ;; 51]] (* Jean-François Alcover, Jun 01 2011, after PARI prog. *)

T(i,j)=if(i<0 || j<0,0,(2*i+2*j)!*i!/(2*i)!/(i+j)!/j!)

def A046521(n,k): return binomial(n+1, k+1)*catalan_number(n)/catalan_number(k)
flatten([[A046521(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 28 2024

T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0.
G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x).
Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1.
Recurrence from the Z-sequence given above: a(n,0) = Sum_{j=0..n-1} Z(j)*a(n-1,j), n >= 1; a(0,0)=1.
As a number triangle, T(n,k) = C(2*n,n)*C(n,k)/C(2*k,k) = C(n-1/2,n-k)*4^(n-k). - Paul Barry, Apr 14 2010
From Peter Bala, Apr 11 2012: (Start):
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A068555.
The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere.
Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). (End)
T(n,k) = 4^(n-k)*A006882(2*n - 1)/(A006882(2*n - 2*k)*A006882(2*k - 1)) = 4^(n-k)*(2*n - 1)!!/((2*n - 2*k)!*(2*k - 1)!!). - Peter Bala, Nov 07 2016
Boas-Buck recurrence for column m, m > n >= 0: T(n, m) = (2*(2*m+1)/(n-m))*Sum_{k=m..n-1} 4^(n-1-k)*T(k, m), with input T(n, n) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
From Peter Bala, Aug 13 2021: (Start)
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} (-1)^(n-k)*T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 4*b, c = 1 and d = 1/2.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of  formal power series then
G(x) = 1/sqrt(1 - 4*b*x) * F(x/(1 - 4*b*x)) iff F(x) = 1/sqrt(1 + 4*b*x) * G(x/(1 + 4*b*x)).
The m-th power of this array has entries m^(n-k)*T(n,k). (End)

A065097 a(n) = ((2n+1) + (2n-1) - 1)!/((2n+1)!*(2n-1)!).

Original entry on oeis.org

1, 1, 7, 66, 715, 8398, 104006, 1337220, 17678835, 238819350, 3282060210, 45741281820, 644952073662, 9183676536076, 131873975875180, 1907493251046152, 27767032438524099, 406472021074865382, 5979899192930226746, 88366931393503350700, 1311063521138246054410

Offset: 0

Len Smiley, Nov 11 2001

A Catalan-like formula using consecutive odd numbers. Recall that Catalan numbers (A000108) are given by ((n+1)+(n)-1)!/((n+1)!(n)!).
From David Callan, Jun 01 2006: (Start)
a(n) = number of Dyck (2n)-paths (i.e., semilength = 2n) all of whose interior returns to ground level (if any) occur at or before the (2n-2)-nd step, that is, they occur strictly before the midpoint of the path.
For example, a(2)=7 counts UUUUDDDD, UUUDUDDD, UUDUUDDD, UUDUDUDD, UUUDDUDD, UD.UUUDDD, UD.UUDUDD ("." denotes an interior return to ground level).
This result follows immediately from an involution on Dyck paths, due to Emeric Deutsch, defined by E->E, UPDQ -> UQDP (where E is the empty Dyck path; U=upstep, D=downstep and P,Q are arbitrary Dyck paths), because the involution is fixed-point-free on Dyck (2n)-paths and contains one path of the type being counted in each orbit.
a(n) = Sum_{k=0..n-1} C(2n-1-2k)*C(2k). This identity has the following combinatorial interpretation:
a(n) is the number of odd-GL-marked Dyck (2n-1)-paths. An odd-GL vertex is a vertex at location (2i,0) for some odd i >= 1 (path starts at origin). An odd-GL-marked Dyck path is a Dyck path with one of its odd-GL vertices marked. For example, a(2)=7 counts UUUDDD*, UUDUDD*, UD*UUDD, UDUUDD*, UD*UDUD, UDUDUD*, UUDDUD* (the * denotes the marked odd-GL vertex). (End)
a(n+1) = Sum_{k=0..n} C(k)*C(2*n+1-k), n >= 0, with C(n) = A000108(n), also gives the odd part of the bisection of the half-convolution of the Catalan sequence A000108 with itself. For the definition of the half-convolution of a sequence with itself see a comment on A201204. There one also finds the rule for the o.g.f. given below in the formula section. The even part of this bisection is found under A201205. - Wolfdieter Lang, Jan 05 2012
From Peter Bala, Dec 01 2015: (Start)
Let x = p/q be a positive rational in reduced form with p,q > 0. Define Cat(x) = (1/(2*p + q))*binomial(2*p + q, p). Then Cat(n) = Catalan(n). This sequence is Cat(n + 1/2) = (1/(4*n + 4))*binomial(4*n + 4, 2*n + 1). Cf. A265101 (Cat(n + 1/3)), A265102 (Cat(n + 1/4)) and A265103 (Cat(n + 1/5)).
Number of maximal faces of the rational associahedron Ass(2*n + 1, 2*n + 3). Number of lattice paths from (0, 0) to (2*n + 3, 2*n + 1) using steps of the form (1, 0) and (0, 1) and staying above the line y = (2*n + 1)/(2*n + 3)*x. See Armstrong et al. (End)
Also the number of ordered rooted trees with 2n nodes, most of which are leaves, i.e., the odd bisection of A358585. This follows from Callan's formula below. - Gus Wiseman, Nov 27 2022

			G.f.: 1 + x + 7*x^2 + 66*x^3 + 715*x^4 + 8398*x^5 + 104006*x^6 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..834 (terms n = 1..100 from Harry J. Smith)
D. Armstrong, B. Rhoades, and N. Williams, Rational associahedra and noncrossing partitions arxiv:1305.7286 [math.CO], 2008.
Emeric Deutsch, An involution on Dyck paths and its consequences, Discrete Math., 204 (1999), no. 1-3, 163-166.

Cf. A003150 (for analog with consecutive Fibonacci numbers).
Equals (1/2) A048990. Cf. A000108, A001795, A201204, A201205, A265101, A265102, A265103.
Cf. A000891, A001263, A001622, A090181, A185650, A358585.

[Binomial(4*n-1, 2*n-1)/(2*n+1): n in [1..20]]; // Vincenzo Librandi, Dec 09 2015

seq(binomial(4*n-1,2*n-1)/(2*n+1), n=0..30); # Robert Israel, Dec 08 2015

a[ n_] := If[ n < 1, 0, Binomial[ 4 n - 1, 2 n - 1] / (2 n + 1)]; (* Michael Somos, Oct 25 2014 *)

combinat::dyckWords::count(2*n)/2 $ n = 1..26 // Zerinvary Lajos, Apr 25 2007

a(n) = { if(n==0, 1, (4*n - 1)!/((2*n + 1)!*(2*n - 1)!)) } \\ Harry J. Smith, Oct 07 2009

vector(20, n, binomial(4*n-1, 2*n-1)/(2*n+1)) \\ Altug Alkan, Dec 08 2015

A065097 = lambda n: hypergeometric([1-2*n,-2*n],[2],1)/2
[Integer(A065097(n).n(500)) for n in (1..20)] # Peter Luschny, Sep 22 2014

a(n) = binomial(4*n-1, 2*n-1)/(2*n+1).
a(n) = C(2n)/2 where C(n) is the Catalan number A000108. - David Callan, Jun 01 2006
G.f.: 1/2 + (sqrt(2)/2)/sqrt(1+sqrt(1-16*x)). - Vladeta Jovovic, Sep 26 2003
G.f.: 1 + 3F2([1, 5/4, 7/4], [2, 5/2], 16*x). - Olivier Gérard, Feb 16 2011
O.g.f.: (1 + (cata(sqrt(x)) + cata(-sqrt(x)))/2)/2, with the o.g.f. cata(x) of the Catalan numbers. See the W. Lang comment above. - Wolfdieter Lang, Jan 05 2012
a(n) = hypergeometric([1-2*n,-2*n],[2],1)/2. - Peter Luschny, Sep 22 2014
a(n) = A001448(n) / (4*n + 2) if n>0. - Michael Somos, Oct 25 2014
n*(2*n+1)*a(n) - 2*(4*n-1)*(4*n-3)*a(n-1) = 0. - R. J. Mathar, Oct 31 2015
O.g.f. is 1 + Revert( x*(1 + x)/(1 + 2*x)^4 ). - Peter Bala, Dec 01 2015
Sum_{n>=0} 1/a(n) = 39/25 + 4*Pi/(9*sqrt(3)) - 24*log(phi)/(25*sqrt(5)), where phi is the golden ratio (A001622). - Amiram Eldar, Mar 02 2023
From Peter Bala, Apr 29 2024: (Start)
For n >= 1, a(n) = (1/8)*Sum_{k = 0..2*n-1} (-1)^k * 4^(2*n-k)*binomial(2*n-1, k)*Catalan(k+1).
For n >= 1, a(n) = (1/8)*(16^n)*hypergeom([1 - 2*n, 3/2], [3], 1). (End)

a(0)=1 prepended by Alois P. Heinz, Nov 28 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

A002458 a(n) = binomial(4n+1, 2n).

Original entry on oeis.org

1, 10, 126, 1716, 24310, 352716, 5200300, 77558760, 1166803110, 17672631900, 269128937220, 4116715363800, 63205303218876, 973469712824056, 15033633249770520, 232714176627630544, 3609714217008132870, 56093138908331422716, 873065282167813104916

Offset: 0

N. J. A. Sloane

			1 + 10*x + 126*x^2 + 1716*x^3 + 24310*x^4 + 352716*x^5 + 5200300*x^6 + ...

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1982, (3.109), page 35.

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

Cf. A000984, A001448, A001700, A024492, A100033, A187364, A187365.

Row sums of A067001.

A002458:=n->binomial(4*n+1,2*n): seq(A002458(n), n=0..30); # Wesley Ivan Hurt, Jan 17 2017

Table[Binomial[4n+1,2n],{n,0,30}] (* Harvey P. Dale, Apr 04 2011 *)
4^Range[0, 22] Simplify[ CoefficientList[ Series[ Sqrt[2]/(((Sqrt[1 - 4 x] + 1)^(1/2))*Sqrt[1 - 4 x]), {x, 0, 22}], x]] (* Robert G. Wilson v, Aug 08 2011 *)

PARI
```
a(n) = binomial( 4*n + 1, 2*n)
```

a(n) = Sum_{k=0..n} 4^k * binomial( n + k, n) * binomial( 2*n - 2*k, n - k). - Michael Somos, Feb 25 2012
a(n) = A001700(2*n) = (n+1)*A000108(2*n+1).
G.f.: (4 - (1+4*y)*c(y) - (1-4*y)*c(-y))/(2*(1 - (4*y)^2)) with y^2 = x, c(y) = g.f. for A000108 (Catalan). - Wolfdieter Lang, Dec 13 2001
a(n) ~ 2^(1/2)*Pi^(-1/2)*n^(-1/2)*2^(4*n)*{1 - 5/16*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
a(n) = A024492(n)*(n+1). - R. J. Mathar, Aug 10 2015
G.f.: 2F1(3/4,5/4; 3/2; 16*x). - R. J. Mathar, Aug 10 2015
D-finite with recurrence n*(2*n + 1)*a(n) - 2*(4*n - 1)*(4*n + 1)*a(n-1) = 0. - R. J. Mathar, Aug 10 2015
From Peter Bala, Nov 04 2015: (Start)
a(n) = 4^n*binomial(2*n + 1/2, n).
O.g.f.: sqrt(c(4*x)/(1 - 16*x)) = sqrt(2/(1 - 16*x)/(1 + sqrt(1 - 16*x))), where
c(y) = g.f. for A000108 (Catalan). In general, c(x)^k/sqrt(1 - 4*x) is the o.g.f. for the sequence binomial(2*n + k, n). (End) [Edited by Petros Hadjicostas, May 25 2020]
From Ilya Gutkovskiy, Jan 17 2017: (Start)
E.g.f.: 2F2(3/4,5/4; 1,3/2; 16*x).
Sum_{n>=0} 1/a(n) = 3F2(1,1,3/2; 3/4,5/4; 1/16) = 1.108563435104316693... (End)
From Peter Bala, Mar 16 2018: (Start)
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} 4^(n-k) * C(2*n+1, 2*k) * C(2*k, k) = a(n).
a(n) = 4^n*hypergeom([-n, -n-1/2], [1], 1). (End)
From Peter Bala, Mar 20 2023: (Start)
a(n) = Sum_{k = 0..n} binomial(2*n+1,k)^2.
a(n) = (1/2)*hypergeom([-1 - 2*n, -1 - 2*n], [1], 1). (End)

A091527 a(n) = ((3n)!/n!^2)(Gamma(1+n/2)/Gamma(1+3n/2)).

Original entry on oeis.org

1, 4, 30, 256, 2310, 21504, 204204, 1966080, 19122246, 187432960, 1848483780, 18320719872, 182327718300, 1820797698048, 18236779032600, 183120225632256, 1842826521244230, 18581317012684800, 187679234340049620, 1898554215471513600, 19232182592635611060

Offset: 0

Michael Somos, Jan 18 2004

Sequence terms are given by [x^n] ( (1 + x)^(k+2)/(1 - x)^k )^n for k = 1. See the crossreferences for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

Let a > b be nonnegative integers. Then the ratio of factorials ((2*a + 1)*n)!*((b + 1/2)*n)!/(((a + 1/2)*n)!*((2*b + 1)*n)!*((a - b)*n)!) is an integer for n >= 0. This is the case a = 1, b = 0. - Peter Bala, Aug 28 2016

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Michael De Vlieger, Table of n, a(n) for n = 0..985
Peter Bala, Some integer ratios of factorials
Paul Barry, On the Central Coefficients of Bell Matrices, J. Int. Seq. 14 (2011) # 11.4.3.
Karl Dilcher, Armin Straub, and Christophe Vignat, Identities for Bernoulli polynomials related to multiple Tornheim zeta functions, arXiv:1903.11759 [math.NT], 2019. See p. 11.
I. M. Gessel, A short proof of the Deutsch-Sagan congruence for connected non crossing graphs, arXiv preprint arXiv:1403.7656 [math.CO], 2014.
Bernhard Heim and Markus Neuhauser, Asymptotic Distribution of the Zeros of recursively defined Non-Orthogonal Polynomials, arXiv:2107.05013 [math.CA], 2021.
W. Mlotkowski and K. A. Penson, Probability distributions with binomial moments, arXiv preprint arXiv:1309.0595 [math.PR], 2013.

Cf. A061162(n) = a(2n), A007297, A000984 (k = 0), A001448 (k = 2), A262732 (k = 3), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A276098, A276099.

a := n -> 4^n * `if`(n<2, 1, (2*(n+1)*binomial((3*n-1)/2, n + 1))/(n-1)):
seq(a(n), n=0..18); # Peter Luschny, Feb 03 2020

Table[((3 n)!/n!^2) Gamma[1 + n/2]/Gamma[1 + 3 n/2], {n, 0, 18}] (* Michael De Vlieger, Oct 02 2015 *)
Table[4^n Sum[Binomial[k - 1 + (n - 1)/2, k], {k, 0, n}], {n, 0, 18}] (* Michael De Vlieger, Aug 28 2016 *)

B(x):=(-1/3+(2/3)*sqrt(1+9*x)*sin((1/3)*asin((2+27*x+54*x^2)/2/(1+9*x)^(3/2))))/x-1;
taylor(x*diff(B(x),x)/B(x),x,0,10); /* Vladimir Kruchinin, Oct 02 2015 */

a(n)=4^n*sum(i=0,n,binomial(i-1+(n-1)/2,i))

vector(30, n, sum(k=0, n, binomial(3*n-3, k)*binomial(2*n-k-3, n-k-1))) \\ Altug Alkan, Oct 04 2015

from math import factorial
from sympy import factorial2
def A091527(n): return int((factorial(3*n)*factorial2(n)<Chai Wah Wu, Aug 10 2023

D-finite with recurrence n*(n - 1)*a(n) = 12*(3*n - 1)*(3*n - 5)*a(n-2).
From Peter Bala, Sep 29 2015: (Start)
a(n) = Sum_{i = 0..n} binomial(3*n,i) * binomial(2*n-i-1,n-i).
a(n) = [x^n] ( (1 + x)^3/(1 - x) )^n.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 4*x + 23*x^2 + 156*x^3 + 1162*x^4 + 9192*x^5 + ... is the o.g.f. for A007297 (but with an offset of 0). (End)
a(n) = (n+1)*A078531(n). [Barry, JIS (2011)]
G.f.: x*B'(x)/B(x), where x*B(x)+1 is g.f. of A007297. - Vladimir Kruchinin, Oct 02 2015
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..floor(n/2)} binomial(4*n,n-2*k)*binomial(n+k-1,k).
O.g.f.: A(x) = Hypergeom([5/6, 1/6], [1/2], 108*x^2) + 4*x*Hypergeom([4/3, 2/3], [3/2], 108*x^2).
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^3/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
a(n) ~ 2^n*3^(3*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Aug 22 2016
a(n) = 4^n*2*(n+1)*binomial((3*n-1)/2, n+1)/(n-1) for n >= 2. - Peter Luschny, Feb 03 2020
From Peter Bala, Mar 04 2022: (Start)
The o.g.f. A(x) satisfies the algebraic equation (1 - 108*x^2)*A(x)^3 - A(x) = 8*x. Cf. A244039.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for primes p >= 5 and positive integers n and k. (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^n).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((3*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((n+1)/2).
a(n) = [x^n] (1+4*x)^((3*n-1)/2). (End)

A211419 a(n) = (6n)!(2n)!/((4n)!(3n)!*n!).

Original entry on oeis.org

1, 10, 198, 4420, 104006, 2521260, 62300700, 1560167752, 39457579590, 1005490725148, 25776935824948, 664048851069240, 17175945353271068, 445775181599116600, 11602978540817349240, 302767701121286251920, 7917664916276259668550, 207452338901630123085180

Offset: 0

Peter Bala, Apr 10 2012

This sequence is the particular case a = 3, b = 2 of the following result (see Bober, Theorem 1.2): Let a, b be nonnegative integers with a > b and gcd(a,b) = 1. Then (2*a*n)!*(b*n)!/((a*n)!*(2*b*n)!*((a-b)*n)!) is an integer for all integer n >= 0. Other cases include A061162 (a = 3, b = 1), A211420 (a = 4, b = 1), A211421 (a = 4, b = 3) and A061163 (a = 5, b = 1).
This is the case m = 3n in Catalan's formula (2m)!*(2n)!/(m!*(m+n)!*n!) - see Umberto Scarpis in References. - Bruno Berselli, Apr 27 2012
Sequence terms are given by the coefficient of x^n in the expansion of ((1 + x)^(k+2)/(1 - x)^k)^n when k = 4. See the cross references for related sequences obtained from other values of k. - Peter Bala, Sep 29 2015

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.

Seiichi Manyama, Table of n, a(n) for n = 0..500
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977v1 [math.NT], 2007; J. London Math. Soc., Vol. 79, Issue 2 (2009), 422-444.
F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Cf. A061162, A061163, A182400, A211420, A211421.

Cf. A000984 (k=0), A091527 (k=1), A001448 (k=2), A262732 (k=3), A262733 (k=5), A211421 (k=6), A262738.

[Factorial(6*n) * Factorial(2*n) / (Factorial(4*n) * Factorial(3*n) * Factorial(n)): n in [0..20]]; // Vincenzo Librandi, May 03 2018

A211419 := n-> (6*n)!*(2*n)!/((4*n)!*(3*n)!*n!):
seq(A211419(n), n=0..20);
# Using the o.g.f. from Karol A. Penson and Jean-Marie Maillard:
u := 27*x-1: c := (u^3*((3*x*u)^(1/2)*(12+81*x)-u^2+216*x-7))^(1/3):
gf := ((c^2-2*c*u+27*u*(7-81*x)*x-4*u)/(6*c*u^2))^(1/2):
ser := series(gf, x, 8); # Peter Luschny, May 03 2018
ogf := hypergeom([1/6, 1/2, 5/6], [1/4, 3/4], 27*z): ser := series(ogf, z, 20):
seq(coeff(ser, z, n), n = 0..17);  # Peter Luschny, Feb 22 2024

Table[(6 n)!*(2 n)!/((4 n)!*(3 n)!*n!), {n, 0, 16}] (* Michael De Vlieger, Oct 04 2015 *)
CoefficientList[Series[Sqrt[(4 + 7290 x^2 - 59049 x^3 + 2 (8 + 3 Sqrt[3] Sqrt[x (-1 + 27 x)^7 (4 + 27 x)^2] - 27 x (-2 + 27 x) (-17 + 27 x (19 + 27 x (-11 + 27 x))))^(1/3) + (8 + 3 Sqrt[3] Sqrt[x (-1 + 27 x)^7 (4 + 27 x)^2] - 27 x (-2 + 27 x) (-17 + 27 x (19 + 27 x (-11 + 27 x))))^(2/3) - 27 x (11 + 2 (8 + 3 Sqrt[3] Sqrt[x (-1 + 27 x)^7 (4 + 27 x)^2] - 27 x (-2 + 27 x) (-17 + 27 x (19 + 27 x (-11 + 27 x))))^(1/3)))/(6 (1 - 27 x)^2 (8 + 3 Sqrt[3] Sqrt[x (-1 + 27 x)^7 (4 + 27 x)^2] - 27 x (-2 + 27 x) (-17 + 27 x (19 + 27 x (-11 + 27 x))))^(1/3))],{x,0,16}],x] (* Karol A. Penson and Jean-Marie Maillard, May 02 2018 *)

a(n) = (6*n)!*(2*n)!/((4*n)!*(3*n)!*n!);
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 02 2015

The o.g.f. Sum_{n >= 1} a(n)*z^n is algebraic over the field of rational functions Q(z) (see Rodriguez-Villegas).
From Peter Bala, Sep 29 2015: (Start)
a(n) = Sum_{i = 0..n} binomial(6*n, i) * binomial(5*n-i-1, n-i).
a(n) = [x^n] ( (1 + x)^6/(1 - x)^4 )^n.
O.g.f. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 10*x + 149*x^2 + 2630*x^3 + 51002*x^4 + ... has integer coefficients and equals 1/x * series reversion of x*(1 - x)^4/ (1 + x)^6. See A262738. (End)
a(n) ~ 27^n/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
O.g.f.: sqrt((4 + 7290*x^2 - 59049*x^3 + 2*(8 + 3*sqrt(3)*sqrt(x*(-1 + 27*x)^7*(4 + 27*x)^2) - 27*x*(-2 + 27*x)*(-17 + 27*x*(19 + 27*x*(-11 + 27*x))))^(1/3) + (8 + 3*sqrt(3)*sqrt(x*(-1 + 27*x)^7*(4 + 27*x)^2) - 27*x*(-2 + 27*x)*(-17 + 27*x*(19 + 27*x*(-11 + 27*x))))^(2/3) - 27*x*(11 + 2*(8 + 3*sqrt(3)*sqrt(x*(-1 + 27*x)^7*(4 + 27*x)^2) - 27*x*(-2 + 27*x)*(-17 + 27*x*(19 + 27*x*(-11 + 27*x))))^(1/3)))/(6*(1 - 27*x)^2*(8 + 3*sqrt(3)*sqrt(x*(-1 + 27*x)^7*(4 + 27*x)^2) - 27*x*(-2 + 27*x)*(-17 + 27*x*(19 + 27*x*(-11 + 27*x))))^(1/3))). - Karol A. Penson and Jean-Marie Maillard, May 02 2018
Right-hand side of the binomial sum identity: Sum_{k = 0..2*n} (-1)^(n+k) * binomial(6*n, 2*n+k) * binomial(2*n, k) = (6*n)!*(2*n)!/((4*n)!*(3*n)!*n!). - Peter Bala, Jan 19 2020
a(n) = 6*(6*n - 1)*(2*n - 1)*(6*n - 5)*a(n-1)/(n*(4*n - 1)*(4*n - 3)). - Neven Sajko, Jul 19 2023
From Peter Luschny, Feb 22 2024: (Start)
a(n) = 4^n*(Gamma(3*n + 1/2)/Gamma(2*n + 1/2))/Gamma(n + 1).
O.g.f.: hypergeom([1/6, 1/2, 5/6], [1/4, 3/4], 27*z). (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(4*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(6*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(4*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = 4^n * binomial((6*n-1)/2,n).
a(n) = [x^n] 1/(1-4*x)^((4*n+1)/2).
a(n) = [x^n] (1+4*x)^((6*n-1)/2). (End)

A066357 Number of ordered (i.e., planar) trees on 2n edges with every subtree at the root having an even number of edges.

Original entry on oeis.org

1, 1, 6, 53, 554, 6362, 77580, 986253, 12927170, 173452334, 2370742868, 32892031042, 462030186916, 6557906929108, 93909078262808, 1355087936016957, 19684187540818866, 287612514032460070, 4224238030616082948, 62329883931236020470, 923519220367120779820

Offset: 0

Louis Shapiro, Feb 01 2002

Row sums of A078990. First column of A079513.
a(n) is the number of walks from (0,0) to (2n,2n) using steps (0,1) and (1,0) which never stray below the line y=x and which avoid the points (m,m) m odd. - Paul Boddington, Mar 14 2003
Series reversion of Sum_{n>0} -a(n)(-x)^n is g.f. of A005900.
a(n) is the number of linear extensions of the one-level grid poset G[(0^n), (1^(n-1)), (1^(n-1))]. The definition of a one-level grid poset can be found in the Pan links. - Ran Pan, Jul 05 2016
These numbers have the same parity as the Catalan numbers C(n), that is, a(n) is even except when n has the form 2^m - 1. This follows immediately from the formula a(n) = C(2*n+1) + 2*C(2*n) - 2^(2*n + 1)*C(n) given below by Callan. We conjecture that a(n) and C(n) have the same 2-adic valuation (checked up to n = 100). - Peter Bala, Aug 02 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..200
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Nantel Bergeron, Cesar Ceballos, Vincent Pilaud, Hopf dreams, arXiv:1807.03044 [math.CO], 2018. See p. 17.
A. de Mier and M. Noy, A solution to the tennis ball problem, arXiv:math/0311242 [math.CO], 2003.
J.-G. Luque and J.-Y. Thibon, Noncommutative Symmetric Functions Associated with a Code, Lazard Elimination and Witt Vectors, arXiv:math/0607254 [math.CO], 2006; Discrete Math. Theor. Comput. Sci. 9 (2007), no. 2, 59-72.
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (p. 333).
Ran Pan, Problem 1, Project P.
Ran Pan, Algorithmic Solution to Problem 1 (and linear extensions of general one-level grid-like posets), Project P.
A. Regev, Enumerating triangulations by parallel diagonals, arXiv:1208.3915 [math.CO], 2012, J. Int. Seq. 15 (2012) #12.8.5

Cf. A078990, A079513, A001448, A005900, A079489, A274644, A274763.

[1] cat [(&+[Binomial(4*n,k)*Binomial(3*n-k-2,n-k-1)/n: k in [0..n]]): n in [1..30]]; // G. C. Greubel, Jan 15 2019

gf := (1-sqrt(1-4*z)-sqrt(1+4*z)+sqrt(1-16*z^2))/(z*(sqrt(1-4*z)-sqrt(1+4*z))):s := series(gf, z, 80): for i from 0 to 50 by 2 do printf(`%d,`,coeff(s,z,i)) od: # James Sellers, Feb 11 2002
a := n -> `if`(n=0,1,binomial(3*n-2,n-1)*hypergeom([1-n,-4*n],[2-3*n], -1)/n): seq(simplify(a(n)),n=0..20); # Peter Luschny, Oct 15 2015

CoefficientList[Series[2/(1 + 4 Sqrt[x]/(Sqrt[1 + 4 Sqrt[x]] - Sqrt[1 - 4 Sqrt[x]])), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 21 2014 *)

a(n)=local(A); if(n<1,n==0,A=sqrt(1+4*x+O(x^(2*n+2))); A-=subst(A,x,-x); polcoeff(((2*A-8*x)/A^2)^2,2*n))

vector (100, n, n--; if(n<1, 1, sum(k=0, n, binomial(4*n,k)*binomial(3*n-k-2,n-k-1)/n))) \\ Altug Alkan, Oct 07 2015

[1] + [sum(binomial(4*n,k)*binomial(3*n-k-2,n-k-1)/n for k in (0..n)) for n in (1..30)] # G. C. Greubel, Jan 15 2019

For n>0, a(n) = Sum_{r=1..n} C(2*r-1)*a(n-r). Here C(2*r-1) is a Catalan number (A000108). - Paul Boddington, Mar 14 2003
G.f.: 2/(1+4*sqrt(x)/(sqrt(1+4*sqrt(x))-sqrt(1-4*sqrt(x)))).
D-finite with recurrence a(n)*(2*n-1)*(n+1)n = a(n-1)*(32*n^2 - 64*n + 39)*2*n - a(n-2)*(2*n-3)*(4*n-5)*(4*n-7)*16, n>1.
a(0) = 1,a(n) = (1/n)*Sum_{k=0..n} C(4*n,k)*C(3*n-k-2,n-k-1), n>1. - Paul Barry, Apr 09 2007
a(n) = ((2^(4*n))/Gamma(1/2)) * ((6*(2*n+1)*Gamma(2*n+1/2)/Gamma(2*n+3))-2*Gamma(n+1/2)/Gamma(n+2)). - David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009
Convolution of A079489 with itself: (1, 6, 53, 554, ...) = (1, 3, 22, 211, ...)*(1, 3, 22, 211, ...).
Proof. Working with Dyck paths, we must show that Dyck paths of size (semilength) 2n, all of whose components (constituent primitive Dyck paths) have even size, are equinumerous with ordered pairs of nonempty Dyck paths of total size 2n in each of which the first component is of odd size and all other components (if any) are of even size. Given a Dyck path P of the former class, use the first return decomposition to write P (uniquely) as the concatenation of U A_1 A_2 ... A_j O E D Q where U denotes upstep, D denotes downstep, A_1,...,A_j are all primitive Dyck paths of even size with j>=0, O is a primitive Dyck path of odd size, E is a Dyck path of even size, and Q is a Dyck path in which all components are of even size. Then P -> (O A_1 A_2 ... A_j, U E D Q) is the desired bijection. QED - David Callan, Apr 11 2012
a(n) = C(2*n+1) + 2*C(2*n) - 2^(2*n+1)*C(n), where C(n) is the Catalan number A000108. This formula can be obtained by manipulating generating functions. The equivalence of this formula and the Barry (Apr 09 2007) sum can be established by the WZ method with a second-order operator. A combinatorial interpretation of the Barry sum would be nice. - David Callan, Apr 10 2012
a(n) ~ (3-2*sqrt(2)) * 2^(4*n) / (n^(3/2) * sqrt(2*Pi)). - Vaclav Kotesovec, Mar 21 2014
exp( Sum_{n >= 1} binomial(4*n,2*n)*x^n/n ) = 1 + 6*x + 53*x^2 + 554*x^3 + ... is an o.g.f. for this sequence omitting the initial term. See A001448. - Peter Bala, Oct 02 2015
a(n) = binomial(3*n-2,n-1)*hypergeom([1-n,-4*n],[2-3*n],-1)/n for n>=1. - Peter Luschny, Oct 15 2015
a(n) = 3*(2*n+1) /(2*n+2) /(4*n+1) *binomial(4*n+2,2*n+1) -4^n /(2*n+1) *binomial(2*n+2,n+1) [Merlini et al F_n formula] - R. J. Mathar, Oct 01 2021

A262732 a(n) = (1/n!) * (5n)!/(5n/2)! * (3n/2)!/(3n)!.

Original entry on oeis.org

1, 8, 126, 2240, 41990, 811008, 15967980, 318636032, 6421422150, 130395668480, 2663825039876, 54684895150080, 1127155102890908, 23311847679590400, 483537022180231320, 10054732930602762240, 209536624110664757830, 4375058594685417160704, 91505601042318156186900

Offset: 0

Peter Bala, Sep 29 2015

Sequence terms are given by the coefficient of x^n in the expansion of ( (1 + x)^(k+2)/(1 - x)^k )^n when k = 3. See the cross references for related sequences obtained from other values of k.

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Michael De Vlieger, Table of n, a(n) for n = 0..751
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Peter Bala, Some integer ratios of factorials
Peter Bala, A supercongruence for A262732

Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A262737, A276098, A276099.

Cf. A115293.

a := n -> 1/n! * (5*n)!/GAMMA(1 + 5*n/2) * GAMMA(1 + 3*n/2)/(3*n)!:
seq(a(n), n = 0..18);

Table[1/n!*(5 n)!/(5 n/2)!*(3 n/2)!/(3 n)!, {n, 0, 18}] (* or *)
Table[Sum[Binomial[8 n, n - 2 k] Binomial[3 n + k - 1, k], {k, 0, Floor[n/2]}], {n, 0, 18}] (* Michael De Vlieger, Aug 28 2016 *)

a(n) = sum(k=0, n, binomial(5*n,k)*binomial(4*n-k-1,n-k));
vector(30, n, a(n-1)) \\ Altug Alkan, Oct 03 2015

from math import factorial
from sympy import factorial2
def A262732(n): return int((factorial(5*n)*factorial2(3*n)<Chai Wah Wu, Aug 10 2023

a(n) = Sum_{i = 0..n} binomial(5*n,i) * binomial(4*n-i-1,n-i).
a(n) = [x^n] ( (1 + x)^5/(1 - x)^3 )^n.
D-finite with recurrence a(n) = 20*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)/( n*(3*n - 1)*(3*n - 3)*(3*n - 5) ) * a(n-2).
The o.g.f. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 8*x + 95*x^2 + 1336*x^4 + ... has integer coefficients and equals (1/x) * (series reversion of x*(1 - x)^3/(1 + x)^5). See A262737.
a(n) ~ 2^n*3^(-3*n/2)*5^(5*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016
From Peter Bala, Aug 22 2016: (Start)
a(n) = Sum_{k = 0..floor(n/2)} binomial(8*n,n - 2*k) * binomial(3*n + k - 1,k).
O.g.f.: A(x) = Hypergeom([9/10, 7/10, 3/10, 1/10], [5/6, 1/2, 1/6], (12500/27)*x^2) + 8*x*Hypergeom([7/5, 6/5, 4/5, 3/5], [4/3, 3/2, 2/3], (12500/27)*x^2).
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^5/(1 - x)^3) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
From Karol A. Penson, Apr 26 2018: (Start)
Integral representation of a(n) as the n-th moment of a positive function w(x) on the support (0, sqrt(12500/27)):
a(n) = Integral_{x=0..sqrt(12500/27)} x^n*w(x) dx,
where w(x) = sqrt(5)*2^(3/5)*csc((1/5)*Pi)*sin((1/10)*Pi)*hypergeom([1/10, 4/15, 3/5, 14/15], [1/5, 2/5, 4/5], 27*x^2*(1/12500))/(10*Pi*x^(4/5)) + sqrt(5)*2^(4/5)*csc(2*Pi*(1/5))*sin(3*Pi*(1/10))*hypergeom([3/10, 7/15, 4/5, 17/15], [2/5, 3/5, 6/5], 27*x^2*(1/12500))/(50*Pi*x^(2/5)) + sqrt(5)*2^(1/5)*csc(2*Pi*(1/5))*sin(3*Pi*(1/10))*x^(2/5)*hypergeom([7/10, 13/15, 6/5, 23/15], [4/5, 7/5, 8/5], 27*x^2*(1/12500))/(625*Pi) + 11*sqrt(5)*2^(2/5)*csc((1/5)*Pi)*sin((1/10)*Pi)*x^(4/5)*hypergeom([9/10, 16/15, 7/5, 26/15], [6/5, 8/5, 9/5], 27*x^2*(1/12500))/(50000*Pi). The function w(x) involves four different hypergeometric functions of type 4F3. The function w(x) is singular at both ends of the support. It is the solution of the Hausdorff moment problem and as such it is unique. (End)
From Peter Bala, Sep 15 2021: (Start)
a(n) = [x^n] (1 + 4*x)^((5*n-1)/2) = 4^n*binomial((5*n-1)/2,n).
a(p) == a(1) (mod p^3) for prime p >= 5.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) for prime p >= 5 and positive integers n and k. (End)
From Seiichi Manyama, Aug 09 2025: (Start)
a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(3*n)).
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(5*n,k) * binomial(2*n-k,n-k).
a(n) = Sum_{k=0..n} 2^k * binomial(3*n+k-1,k) * binomial(2*n-k,n-k).
a(n) = [x^n] 1/(1-4*x)^((3*n+1)/2). (End)

cp's OEIS Frontend

A000984 Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.

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A001700 a(n) = binomial(2*n+1, n+1): number of ways to put n+1 indistinguishable balls into n+1 distinguishable boxes = number of (n+1)-st degree monomials in n+1 variables = number of monotone maps from 1..n+1 to 1..n+1.

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A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

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A065097 a(n) = ((2n+1) + (2n-1) - 1)!/((2n+1)!*(2n-1)!).

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

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A000984 Central binomial coefficients: binomial(2n,n) = (2n)!/(n!)^2.

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

A002458 a(n) = binomial(4n+1, 2n).

A091527 a(n) = ((3n)!/n!^2)(Gamma(1+n/2)/Gamma(1+3n/2)).

A211419 a(n) = (6n)!(2n)!/((4n)!(3n)!*n!).