A051924 -id:A051924 - cp's OEIS frontend

A329533 First differences of A051924, or second differences of Central binomial coefficients A000984.

3, 10, 36, 132, 490, 1836, 6930, 26312, 100386, 384540, 1478048, 5697720, 22019556, 85284920, 330961950, 1286562960, 5009003250, 19528599420, 76231136520, 297910080600, 1165429743660, 4563490674600, 17884841191620, 70148829799152, 275344923755700, 1081512966189656, 4250730282412320

Offset: 0

M. F. Hasler, Nov 15 2019

nonn

Cf. A000984, A051924, A163771, A094796.

Differences[#, 2] &@ Array[Binomial[2 #, #] &, 29, 0] (* Michael De Vlieger, Nov 15 2019 *)

C=vector(30,n,binomial(2*n--,n));C=C[^1]-C[^-1];C=C[^1]-C[^-1]

a(n) = A051924(n) - A051924(n-1) = A000984(n+2) - 2*A000984(n+1) + A000984(n).

a(n) = 3*(3*n+2)*(n+1)*binomial(2*n+4,n+2)/(4*(2*n+1)*(2*n+3)). - Alois P. Heinz, Sep 13 2024

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

Original entry on oeis.org

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.
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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

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1

Offset: 0

N. J. A. Sloane

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
  rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
  For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

T. D. Noe, Rows n=0..50 of triangle, flattened
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].
Quang T. Bach and Jeffrey B. Remmel, Generating functions for descents over permutations which avoid sets of consecutive patterns, arXiv:1510.04319 [math.CO], 2015 (see p.25).
M. Barnabei, F. Bonetti and M. Silimbani, Two permutation classes enumerated by the central binomial coefficients, arXiv preprint arXiv:1301.1790 [math.CO], 2013 and J. Int. Seq. 16 (2013) #13.3.8
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry and A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010), Article 10.8.2, example 15.
Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
Paul Barry, Comparing two matrices of generalized moments defined by continued fraction expansions, arXiv preprint arXiv:1311.7161 [math.CO], 2013 and J. Int. Seq. 17 (2014) # 14.5.1.
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 15, 29.
Paul Barry, d-orthogonal polynomials, Fuss-Catalan matrices and lattice paths, arXiv:2505.16718 [math.CO], 2025. See p. 12.
Jonathan E. Beagley and Paul Drube, Combinatorics of Tableau Inversions, Electron. J. Combin., 22 (2015), #P2.44.
S. Chakravarty and Y. Kodama, A generating function for the N-soliton solutions of the Kadomtsev-Petviashvili II equation, arXiv preprint arXiv:0802.0524v2 [nlin.SI], 2008.
Wun-Seng Chou, Tian-Xiao He, and Peter J.-S. Shiue, On the Primality of the Generalized Fuss-Catalan Numbers, J. Int. Seqs., Vol. 21 (2018), #18.2.1.
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 11.
Paul Drube, Generating Functions for Inverted Semistandard Young Tableaux and Generalized Ballot Numbers, arXiv:1606.04869 [math.CO], 2016.
Paul Drube, Generalized Path Pairs and Fuss-Catalan Triangles, arXiv:2007.01892 [math.CO], 2020. See Figure 4 p. 8.
T.-X. He and L. W. Shapiro, Fuss-Catalan matrices, their weighted sums, and stabilizer subgroups of the Riordan group, Lin. Alg. Applic. 532 (2017) 25-41, example p 32.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Thomas Koshy, Lobb's generalization of Catalan's parenthesization problem, The College Mathematics Journal 40 (2), March 2009, 99-107, DOI:10.1080/07468342.2009.11922344.
Huyile Liang, Jeffrey Remmel, and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 11.
Andrew Lobb, Deriving the n-th Catalan number, Mathematical Gazette, Vol. 83, No. 496 (March 1999), 109-110.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174. See page 161.
Pedro J. Miana, Hideyuki Ohtsuka, and Natalia Romero, Sums of powers of Catalan triangle numbers, arXiv:1602.04347 [math.NT], 2016 (see 2.8).
A. Papoulis, A new method of inversion of the Laplace transform, Quart. Appl. Math 14 (1957), 405-414. [Annotated scan of selected pages]
Athanasios Papoulis, A new method of inversion of the Laplace transform, Quart. Appl. Math., Vol. 14, No. 4 (1957), 405-414: 124. [Note: there is a typo]
J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp. 29 (129) (1975) 215-222
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Luping Ma, Minors of a class of Riordan arrays related to weighted partial Motzkin paths. Eur. J. Comb. 39, 157-169 (2014), Table 2.2.
Wikipedia, Lobb number
W.-J. Woan, L. Shapiro and D. G. Rogers, The Catalan numbers, the Lebesgue integral and 4^{n-2}, Amer. Math. Monthly, 104 (1997), 926-931.
Sheng-Liang Yang, Yan-Ni Dong, and Tian-Xiao He, Some matrix identities on colored Motzkin paths, Discrete Mathematics 340.12 (2017), 3081-3091.

Columns give: A000108, A000245, A000344, A000588, A001392, A000589, A000590, A000012, A005408, A014107(n>1).
Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).
Cf. A008313, A039598, A084938, A000007, A067311, A321620.

/* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015

T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023

Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)

a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
trianglerows(10) \\ Felix Fröhlich, Jun 24 2016

# Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle
def A039599_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True ; h = 1
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        if b : print([D[z] for z in (1..h-1)])
        b = not b
A039599_triangle(10)  # Peter Luschny, May 01 2012

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A030662 Number of combinations of n things from 1 to n at a time, with repeats allowed.

Original entry on oeis.org

1, 5, 19, 69, 251, 923, 3431, 12869, 48619, 184755, 705431, 2704155, 10400599, 40116599, 155117519, 601080389, 2333606219, 9075135299, 35345263799, 137846528819, 538257874439, 2104098963719, 8233430727599, 32247603683099, 126410606437751, 495918532948103

Offset: 1

Donald Mintz (djmintz(AT)home.com)

Add terms of an increasingly bigger diamond-shaped part of Pascal's triangle:
.......................... 1
............ 1 .......... 1 1
.. 1 ...... 1 1 ........ 1 2 1
. 1 1 =5 . 1 2 1 =19 .. 1 3 3 1 =69
.. 2 ...... 3 3 ........ 4 6 4
............ 6 ......... 10 10
.......................... 20
-  Ralf Stephan, May 17 2004
The prime p divides a((p-1)/2) for p in A002144 (Pythagorean primes). - Alexander Adamchuk, Jul 04 2006
Also, number of square submatrices of a square matrix. - Jono Henshaw (jjono(AT)hotmail.com), Apr 22 2008
Partial sums of A051924. - J. M. Bergot, Jun 22 2013
Number of partitions with Ferrers diagrams that fit in an n X n box (excluding the empty partition of 0). - Michael Somos, Jun 02 2014
Also number of non-descending sequences with length and last number are less or equal to n, and also the number of integer partitions (of any positive integer) with length and largest part are less or equal to n. - Zlatko Damijanic, Dec 06 2024

			G.f. = x + 5*x^2 + 19*x^3 + 69*x^4 + 251*x^5 + 923*x^6 + 3431*x^7 + ...

T. D. Noe, Table of n, a(n) for n = 1..500
Narcisse G. Bell Bogmis, Guy R. Biyogmam, Hesam Safa, and Calvin Tcheka, Upper bounds on the dimension of the Schur Lie-multiplier of Lie-nilpotent Leibniz n-algebras, arXiv:2403.14884 [math.RA], 2024. See p. 7.
Joseph D. Horton and Andrew Kurn, Counting sequences with complete increasing subsequences, Congress Numerantium, 33 (1981), 75-80. MR 681905
Milan Janjic and Boris Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
Milan Janjic and Boris Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
Raimundas Vidunas, Counting derangements and Nash equilibria, Ann. Comb. 21, No. 1, 131-152 (2017).
Jianqiang Zhao, Uniform Approach to Double Shuffle and Duality Relations of Various q-Analogs of Multiple Zeta Values via Rota-Baxter Algebras, arXiv preprint arXiv:1412.8044 [math.NT], 2014.

Cf. A000984, A001700, A002144, A051924, A091908, A144660, A322938.
Column k=2 of A047909.
Central column of triangle A014473.
Right-hand column 2 of triangle A102541.

[(n+1)*Catalan(n)-1: n in [1..40]]; // G. C. Greubel, Apr 07 2024

seq(sum((binomial(n,m))^2,m=1..n),n=1..23); # Zerinvary Lajos, Jun 19 2008
f:=n->add( add( binomial(i+j,i), i=0..n),j=0..n); [seq(f(n),n=0..12)]; # N. J. A. Sloane, Jan 31 2009

Table[Sum[Sum[(2n-i-j)!/(n-i)!/(n-j)!,{i,1,n}],{j,1,n}],{n,1,20}] (* Alexander Adamchuk, Jul 04 2006 *)
a[n_] := 2*(2*n-1)!/(n*(n-1)!^2)-1; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Oct 11 2012, from first formula *)

a(n)=binomial(2*n,n)-1 \\ Charles R Greathouse IV, Jun 26 2013

from math import comb
def a(n): return comb(2*n, n) - 1
print([a(n) for n in range(1, 27)]) # Michael S. Branicky, Jul 11 2023

def a(n) : return binomial(2*n,n) - 1
[a(n) for n in (1..26)] # Peter Luschny, Apr 21 2012

a(n) = A000984(n) - 1.
a(n) = 2*A001700(n-1) - 1.
a(n) = 2*(2*n-1)!/(n!*(n-1)!)-1.
a(n) = Sum_{k=1..n} binomial(n, k)^2. - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{j=0..n} Sum_{i=j..n+j} binomial(i, j). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jul 23 2003
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-1} binomial(i+j, i). - N. J. A. Sloane, Jan 31 2009
Also for n>1: a(n)=(2*n)!/(n!)^2-1. - Hugo Pfoertner, Feb 10 2004
a(n) = Sum_{j=1..n} Sum_{i=1..n} (2n-i-j)!/((n-i)!*(n-j)!). - Alexander Adamchuk, Jul 04 2006
a(n) = A115112(n) + 1. - Jono Henshaw (jjono(AT)hotmail.com), Apr 22 2008
G.f.: Q(0)*(1-4*x)/x - 1/x/(1-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
D-finite with recurrence: n*a(n) +2*(-3*n+2)*a(n-1) +(9*n-14)*a(n-2) +2*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Jun 25 2013
0 = a(n)*(+16*a(n+1) - 70*a(n+2) + 68*a(n+3) - 14*a(n+4)) + a(n+1)*(-2*a(n+1) + 61*a(n+2) - 96*a(n+3) + 23*a(n+4)) + a(n+2)*(-6*a(n+2) + 31*a(n+3)  - 10*a(n+4)) + a(n+3)*(-2*a(n+3) + a(n+4)) for all n in Z. - Michael Somos, Jun 02 2014
From Ilya Gutkovskiy, Jan 25 2017: (Start)
O.g.f.: (1 - x - sqrt(1 - 4*x))/((1 - x)*sqrt(1 - 4*x)).
E.g.f.: exp(x)*(exp(x)*BesselI(0,2*x) - 1). (End)
a(n) = 3*n*Sum_{k=1..n} (-1)^(k+1)/(2*n+k)*binomial(2*n+k,n-k). - Vladimir Kruchinin, Jul 29 2025
a(n) = n * binomial(2*n, n) * Sum_{k = 1..n} 1/(k*binomial(n+k, k)). - Peter Bala, Aug 05 2025

A263297 The greater of bigomega(n) and maximal prime index in the prime factorization of n.

Original entry on oeis.org

0, 1, 2, 2, 3, 2, 4, 3, 2, 3, 5, 3, 6, 4, 3, 4, 7, 3, 8, 3, 4, 5, 9, 4, 3, 6, 3, 4, 10, 3, 11, 5, 5, 7, 4, 4, 12, 8, 6, 4, 13, 4, 14, 5, 3, 9, 15, 5, 4, 3, 7, 6, 16, 4, 5, 4, 8, 10, 17, 4, 18, 11, 4, 6, 6, 5, 19, 7, 9, 4, 20, 5, 21, 12, 3, 8, 5, 6, 22, 5

Offset: 1

Alexei Kourbatov, Oct 13 2015

nonn

Also: minimal m such that n is the product of at most m primes not exceeding prime(m). (Here the primes do not need to be distinct; cf. A263323.)
By convention, a(1)=0, as 1 is the empty product.
Those n with a(n) <= k fill a k-simplex whose 1-sides span from 0 to k.
For a similar construction with distinct primes (hypercube), see A263323.
Each nonnegative integer occurs finitely often; in particular:
- Terms a(n) <= k occur A000984(k) = (2*k)!/(k!)^2 times.
- The term a(n) = 0 occurs exactly once.
- The term a(n) = k > 0 occurs exactly A051924(k) = (3*k-2)*C(k-1) times, where C(k)=A000108(k) are Catalan numbers.

			a(6)=2 because 6 is the product of 2 primes (2*3), each not exceeding prime(2)=3.
a(8)=3 because 8 is the product of 3 primes (2*2*2), each not exceeding prime(3)=5.
a(11)=5 because 11 is prime(5).

Antti Karttunen, Table of n, a(n) for n = 1..20000
Index entries for sequences computed from indices in prime factorization

Cf. A000108, A000984, A001222, A051924, A061395, A263323, A325225, A331296 (ordinal transform), A331297.

seq(`if`(n=1,0,max(pi(max(factorset(n))),bigomega(n))),n=1..80); # Peter Luschny, Oct 15 2015

f[n_] := Max[ PrimePi[ Max @@ First /@ FactorInteger@n], Plus @@ Last /@ FactorInteger@n]; Array[f, 80]

a(n)=if(n<2, return(0)); my(f=factor(n)); max(vecsum(f[,2]), primepi(f[#f~,1])) \\ Charles R Greathouse IV, Oct 13 2015

a(n) = max(A001222(n), A061395(n)).

a(n) <= pi(n), with equality when n is 1 or prime.

A097613 a(n) = binomial(2n-3,n-1) + binomial(2n-2,n-2).

Original entry on oeis.org

1, 2, 7, 25, 91, 336, 1254, 4719, 17875, 68068, 260338, 999362, 3848222, 14858000, 57500460, 222981435, 866262915, 3370764540, 13135064250, 51250632510, 200205672810, 782920544640, 3064665881940, 12007086477750, 47081501377326, 184753963255176, 725510446350004

Offset: 1

David Callan, Sep 20 2004

nonn

a(n) is the number of Dyck (2n-1)-paths with maximum pyramid size = n. A pyramid in a Dyck path is a maximal subpath of the form k upsteps immediately followed by k downsteps and its size is k.
a(n) is the total number of runs of peaks in all Dyck (n+1)-paths. A run of peaks is a maximal subpath of the form (UD)^k with k>=1. For example, a(2)=7 because the 5 Dyck 3-paths contain a total of 7 runs of peaks (in uppercase type): uuUDdd, uUDUDd, uUDdUD, UDuUDd, UDUDUD. - David Callan, Jun 07 2006
Binomial transform of A113682. - Paul Barry, Aug 21 2007
If Y is a fixed 2-subset of a (2n+1)-set X then a(n+1) is the number of n-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
Equals the Catalan sequence, A000108, convolved with A051924 prefaced with a 1: (1, 1, 4, 14, 50, ...). - Gary W. Adamson, May 15 2009
Central terms of triangle A209561. - Reinhard Zumkeller, Dec 26 2012
Also the number of compositions of 2*(n-1) in which the odd parts appear as many times in odd as in even positions. - Alois P. Heinz, May 26 2018

			a(2) = 2 because UUDDUD and UDUUDD each have maximum pyramid size = 2.

Michael De Vlieger, Table of n, a(n) for n = 1..1664
Paul Barry, Extensions of Riordan Arrays and Their Applications, Mathematics (2025) Vol. 13, No. 2, 242. See p. 16.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014.
Milan Janjic, Two Enumerative Functions
Toufik Mansour and I. L. Ramirez, Enumerations of polyominoes determined by Fuss-Catalan words, Australas. J. Combin. 81 (3) (2021) 447-457, table 2.
Lin Yang and Shengliang Yang, Protected Branches in Ordered Trees, J. Math. Study (2023) Vol. 56, No. 1, 1-17.

Same as A024482 except for first term.

Cf. A000108, A000712, A026010, A051924, A113682, A209561.

Flat(List([1..30], n->Binomial(2*n-3, n-1)+Binomial(2*n-2, n-2))); # Stefano Spezia, Oct 27 2018

a097613 n = a209561 (2 * n - 1) n  -- Reinhard Zumkeller, Dec 26 2012

[((3*n-2)*Catalan(n-1)+0^(n-1))/2: n in [1..40]]; // G. C. Greubel, Apr 04 2024

Z:=(1-z-sqrt(1-4*z))/sqrt(1-4*z)/2: Zser:=series(Z, z=0, 32): seq (ceil(coeff(Zser, z, n)), n=1..22); # Zerinvary Lajos, Jan 16 2007
a := n -> `if`(n=1, 1, (2-3*n)/(4-8*n)*binomial(2*n, n)):
seq(a(n), n=1..27); # Peter Luschny, Sep 06 2014

a[1]=1; a[n_] := (3n-2)(2n-3)!/(n!(n-2)!); Array[a, 27] (* Jean-François Alcover, Oct 27 2018 *)

a(n)=binomial(2*n-3,n-1)+binomial(2*n-2,n-2) \\ Charles R Greathouse IV, Aug 05 2013

@CachedFunction
def A097613(n):
    if n < 3: return n
    return (6*n-4)*(2*n-3)*A097613(n-1)/(n*(3*n-5))
[A097613(n) for n in (1..27)] # Peter Luschny, Sep 06 2014

G.f.: (x-1)*(1 - 1/sqrt(1-4*x))/2.
a(n) = ceiling(A051924(n)/2). - Zerinvary Lajos, Jan 16 2007
Integral representation as n-th moment of a signed weight function W(x) = W_a(x) + W_c(x), where W_a(x) = Dirac(x)/2 is the discrete (atomic) part, and  W_c(x) = (1/(2*Pi))*((x-1))*sqrt(1/(x*(4-x))) is the continuous part of W(x) defined on (0,4): a(n) = Integral_{x=-eps..eps} x^n*W_a(x) + Integral_{x=0..4} x^n*W_c(x) for any eps > 0, n >= 0. W_c(0) = -infinity, W_c(1) = 0 and W_c(4) = infinity. For 0 < x < 1, W_c(x) < 0, and for 1 < x < 4, W_c(x) > 0. - Karol A. Penson, Aug 05 2013
From Peter Luschny, Sep 06 2014: (Start)
a(n) = ((2-3*n)/(4-8*n))*binomial(2*n,n) for n >= 2.
D-finite with recurrence: a(n) = (6*n-4)*(2*n-3)*a(n-1)/(n*(3*n-5)) for n >= 3. (End)
a(n) ~ 3*2^(2*n-3)/sqrt(n*Pi). - Stefano Spezia, May 09 2023
From G. C. Greubel, Apr 04 2024: (Start)
a(n) = (1/2)*( (3*n-2)*A000108(n-1) + [n=1]).
E.g.f.: (1/2)*(-1+x + exp(2*x)*((1-x)*BesselI(0,2*x) + x*BesselI(1,2*x) )). (End)

A071724 a(n) = 3*binomial(2n, n-1)/(n+2), n > 0, with a(0)=1.

Original entry on oeis.org

1, 1, 3, 9, 28, 90, 297, 1001, 3432, 11934, 41990, 149226, 534888, 1931540, 7020405, 25662825, 94287120, 347993910, 1289624490, 4796857230, 17902146600, 67016296620, 251577050010, 946844533674, 3572042254128, 13505406670700

Offset: 0

N. J. A. Sloane, Jun 06 2002

Number of standard tableaux of shape (n+1,n-1) (n>=1). - Emeric Deutsch, May 30 2004
From Gus Wiseman, Apr 12 2019: (Start)
Also the number of integer partitions (of any positive integer) such that n is the maximum number of unit steps East or South in the Young diagram starting from the upper-left square and ending in a boundary square in the lower-right quadrant. Also the number of integer partitions fitting in a triangular partition of length n but not of length n - 1. For example, the a(0) = 1 through a(4) = 9 partitions are:
  ()  (1)  (2)   (3)
           (11)  (22)
           (21)  (31)
                 (32)
                 (111)
                 (211)
                 (221)
                 (311)
                 (321)
(End)
The sequence (-1)^(n+1)*a(n), for n >= 1 and +1 for n = 0, is the so-called Z-sequence of the Riordan triangle A158909. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Oct 22 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Joerg Arndt, The a(4)=28 Young tableaux of shape [5, 3].
FindStat, St000384: The maximal part of the shifted composition of an integer partition.
Spencer J. Franks, Pamela E. Harris, Kimberly Harry, Jan Kretschmann, and Megan Vance, Counting Parking Sequences and Parking Assortments Through Permutations, arXiv:2301.10830 [math.CO], 2023.
Ricardo Gómez Aíza, Trees with flowers: A catalog of integer partition and integer composition trees with their asymptotic analysis, arXiv:2402.16111 [math.CO], 2024. See p. 21.
Zhicong Lin, David G.L. Wang, and Tongyuan Zhao, A decomposition of ballot permutations, pattern avoidance and Gessel walks, arXiv:2103.04599 [math.CO], 2021.
Murray Tannock, Equivalence classes of mesh patterns with a dominating pattern, MSc Thesis, Reykjavik Univ., May 2016.
Gus Wiseman, Young diagrams of all integer partitions fitting in a triangular partition of length n but not of length n - 1, n = 1...4.

Cf. A000245, A001622, A002421, A030237, A051924, A096771, A115720, A158909.
Cf. A325169, A325188, A325193, A325195, A325200.
Number of times n appears in A065770.
Column sums of A325189.
Row sums of A030237.

[1] cat [3*Binomial(2*n,n-1)/(n+2): n in [1..29]]; // Vincenzo Librandi, Jul 12 2017

A071724:= n-> 3*binomial(2*n, n-1)/(n+2); 1,seq(A071724(n), n=1..30); # G. C. Greubel, Mar 17 2021

Join[{1}, Table[3Binomial[2n, n-1]/(n+2), {n,1,30}]] (* Vincenzo Librandi, Jul 12 2017 *)
nn=7;
otbmax[ptn_]:=Max@@MapIndexed[#1+#2[[1]]-1&,Append[ptn,0]];
allip=Join@@Table[IntegerPartitions[n],{n,0,nn*(nn+1)/2}];
Table[Length[Select[allip,otbmax[#]==n&]],{n,0,nn}] (* Gus Wiseman, Apr 12 2019 *)

a(n)=if(n<1,n==0,3*(2*n)!/(n+2)!/(n-1)!)

[1]+[3*n*catalan_number(n)/(n+2) for n in (1..30)] # G. C. Greubel, Mar 17 2021

a(n) = A000245(n), n>0.
G.f.: (C(x)-1)*(1-x)/x = (1 + x^2 * C(x)^3)*C(x), where C(x) is g.f. for Catalan numbers, A000108.
G.f.: ((1-sqrt(1-4*x))/(2*x)-1)*(1-x)/x = A(x) satisfies x^2*A(x)^2 + (x-1)*(2*x-1)*A(x) + (x-1)^2 = 0.
G.f.: 1 + x*C(x)^3, where C(x) is g.f. for the Catalan numbers (A000108). Sequence without the first term is the 3-fold convolution of the Catalan sequence. - Emeric Deutsch, May 30 2004
a(n) is the n-th moment of the function defined on the segment (0, 4) of x axis: a(n) = Integral_{x=0..4} x^n*(-x^(1/2)*cos(3*arcsin((1/2)*x^(1/2)))/Pi) dx, n=0, 1... . - Karol A. Penson, Sep 29 2004
D-finite with recurrence -(n+2)*(n-1)*a(n) + 2*n*(2*n-1)*a(n-1) = 0. - R. J. Mathar, Jul 10 2017
a(n) ~ c*2^(2*n)*n^(-3/2), where c = 3/sqrt(Pi). - Stefano Spezia, Sep 23 2022
From Amiram Eldar, Sep 29 2022: (Start)
Sum_{n>=0} 1/a(n) = 14*(Pi/(3*sqrt(3)) + 1)/9.
Sum_{n>=0} (-1)^n/a(n) = 18/25 - 164*log(phi)/(75*sqrt(5)), where phi is the golden ratio (A001622). (End)

A237981 Array: row n gives the NW partitions of n; see Comments.

Original entry on oeis.org

1, 2, 3, 4, 3, 1, 5, 4, 1, 6, 5, 1, 4, 2, 7, 6, 1, 5, 2, 8, 7, 1, 6, 2, 5, 3, 9, 8, 1, 7, 2, 6, 3, 5, 3, 1, 10, 9, 1, 8, 2, 7, 3, 6, 4, 6, 3, 1, 11, 10, 1, 9, 2, 8, 3, 7, 4, 7, 3, 1, 6, 4, 1, 12, 11, 1, 10, 2, 9, 3, 8, 4, 8, 3, 1, 7, 5, 7, 4, 1, 6, 4, 2, 13

Offset: 1

Clark Kimberling and Peter J. C. Moses, Feb 23 2014

Suppose that p is a partition of n, and let m = max{greatest part of p, number of parts of p}.  Write the Ferrers graph of p with 1's as nodes, and pad the graph with 0's to form an m X m square matrix, which is introduced here as the Ferrers matrix of p, denoted by f(p).  Four kinds of partitions are defined from f(p); they will be described by referring to the example of a 3 X 3 matrix, as follows:
...
a .. b .. c
d .. e .. f
g .. h .. i
...
Writing summands in clockwise order, the four directional partitions of p are by
NW(p) = [g + d + a + b + c, h + e + f, i]
NE(p) = [a + b + c + f + i, d + e + h, g]
SE(p) = [c + f + i + h + g, b + e + d, a]
SW(p) = [i + h + g + d + a, f + e + b, c].
The order in which the parts appear does not change the partition, but it is common to list them in nondecreasing order, as in Example 1.
...
Note that "Ferrers matrix" can be defined without reference to Ferrers graphs, as follows: an m X m matrix (x(i,j)) of 0's and 1's satisfying three properties: (1) x(1,m) = 1 or x(m,1) = 1; (2) x(i,j+1) >= x(i,j) for j=1..m-1 and i = 1..m; and (3) x(i+1,j) >= x(i,j) for i=1..m-1 and j=1..m.  The number of Ferrers matrices of order m is given by A051924.
The number of NW partitions of n is A003114(n) for n >=1. - Clark Kimberling, Mar 20 2014

			Example 1.  Let p = {6,3,3,3,1), a partition of 16.  Then NW(p) = [10, 4, 2], NE(p) = [6,3,3,3,1], SE(p) = [5, 4, 3, 2, 1, 1], SW(p) = [5,4,4,1,1,1].
...
Example 2.
The first 9 rows of the array of NW partitions:
1
2
3
4 .. 3 .. 1
5 .. 4 .. 1
6 .. 5 .. 1 .. 4 .. 2
7 .. 6 .. 1 .. 5 .. 2
8 .. 7 .. 1 .. 6 .. 2 .. 5 .. 3
9 .. 8 .. 1 .. 7 .. 2 .. 6 .. 3 .. 5 .. 3 .. 1
Row 9, for example, represents the 5 NW partitions of 9 as follows:  [9], [8,1], [7,2], [6,3], [5,3,1], listed in "Mathematica order".

Clark Kimberling, Table of n, a(n) for n = 1..1000
Tewodros Amdeberhan, George E. Andrews, and Cristina Ballantine, Hook length and symplectic content in partitions, arXiv:2205.07322 [math.CO], 2022.
Clark Kimberling and Peter J. C. Moses, Ferrers Matrices and Related Partitions of Integers

Cf. A237982, A237983, A237985, A238325, A238326.

z = 10; ferrersMatrix[list_] := PadRight[Map[Table[1, {#}] &, #], {#, #} &[Max[#, Length[#]]]] &[list]; cornerPart[list_] := Module[{f = ferrersMatrix[list], u, l, ur, lr, nw, ne, se, sw}, {u, l} = {UpperTriangularize[#, 1], LowerTriangularize[#]} &[f]; {ur, lr} = {UpperTriangularize[#, 1], LowerTriangularize[#]} &[Reverse[f]]; {nw, ne, se, sw} =
{Total[Transpose[u]] + Total[l], Total[ur] + Total[Transpose[lr]], Total[u] + Total[Transpose[l]], Total[Transpose[ur]] + Total[lr]};    Map[DeleteCases[Reverse[Sort[#]], 0] &, {nw, ne, se, sw}]]; cornerParts[n_] :=
Map[#[[Reverse[Ordering[PadRight[#]]]]] &, Map[DeleteDuplicates[#] &,    Transpose[Map[cornerPart, IntegerPartitions[n]]]]]; cP = Map[cornerParts, Range[z]];
Flatten[Map[cP[[#, 1]] &, Range[Length[cP]]]](*NW corner: A237981*)
Flatten[Map[cP[[#, 2]] &, Range[Length[cP]]]](*NE corner: A237982*)
Flatten[Map[cP[[#, 3]] &, Range[Length[cP]]]](*SE corner: A237983*)
Flatten[Map[cP[[#, 4]] &, Range[Length[cP]]]](*SW corner: A237982*)
(* Peter J. C. Moses, Feb 25 2014 *)

A096771 Triangle read by rows: T(n,m) is the number of partitions of n that (just) fit inside an m X m box, but not in an (m-1) X (m-1) box. Partitions of n with Max(max part, length) = m.

Original entry on oeis.org

1, 0, 2, 0, 1, 2, 0, 1, 2, 2, 0, 0, 3, 2, 2, 0, 0, 3, 4, 2, 2, 0, 0, 2, 5, 4, 2, 2, 0, 0, 1, 7, 6, 4, 2, 2, 0, 0, 1, 6, 9, 6, 4, 2, 2, 0, 0, 0, 7, 11, 10, 6, 4, 2, 2, 0, 0, 0, 5, 14, 13, 10, 6, 4, 2, 2, 0, 0, 0, 5, 15, 19, 14, 10, 6, 4, 2, 2, 0, 0, 0, 3, 17, 22, 21, 14, 10, 6, 4, 2, 2, 0, 0, 0, 2, 17, 29

Offset: 1

Wouter Meeussen, Aug 21 2004

Row sums are A000041. Columns are finite and sum to A051924. The final floor(n/2) terms of each row are the reverse of the initial terms of 2*A000041.

			T(5,3)=3, counting 32, 311 and 221.
From _Gus Wiseman_, Apr 12 2019: (Start)
Triangle begins:
  1
  0  2
  0  1  2
  0  1  2  2
  0  0  3  2  2
  0  0  3  4  2  2
  0  0  2  5  4  2  2
  0  0  1  7  6  4  2  2
  0  0  1  6  9  6  4  2  2
  0  0  0  7 11 10  6  4  2  2
  0  0  0  5 14 13 10  6  4  2  2
  0  0  0  5 15 19 14 10  6  4  2  2
  0  0  0  3 17 22 21 14 10  6  4  2  2
  0  0  0  2 17 29 27 22 14 10  6  4  2  2
  0  0  0  1 17 33 36 29 22 14 10  6  4  2  2
  0  0  0  1 15 39 45 41 30 22 14 10  6  4  2  2
  0  0  0  0 14 41 57 52 43 30 22 14 10  6  4  2  2
  0  0  0  0 11 47 67 69 57 44 30 22 14 10  6  4  2  2
  0  0  0  0  9 46 81 85 76 59 44 30 22 14 10  6  4  2  2
(End)

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)
Findstat, FindStat - St000784: The maximum of the length and the largest part of the integer partition

A version with reflected rows is A338621.
Cf. A051924, A096272, A096597, A115994, A252464, A263297, A325193, A368985.
Related triangles are A115720, A325188, A325189, A325192, A325200, with Heinz-encoded versions A257990, A325169, A065770, A325178, A325195.

Table[Count[Partitions[n], q_List /; Max[Length[q], Max[q]]===k], {n, 16}, {k, n}]

row(n)={my(r=vector(n)); forpart(p=n, r[max(#p,p[#p])]++); r} \\ Andrew Howroyd, Jan 12 2024

Sum_{k>=1} k*T(n,k) = A368985(n). - Andrew Howroyd, Jan 12 2024

A026010 a(n) = number of (s(0), s(1), ..., s(n)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| = 1 for i = 1,2,...,n and s(0) = 2. Also a(n) = sum of numbers in row n+1 of array T defined in A026009.

Original entry on oeis.org

1, 2, 4, 7, 14, 25, 50, 91, 182, 336, 672, 1254, 2508, 4719, 9438, 17875, 35750, 68068, 136136, 260338, 520676, 999362, 1998724, 3848222, 7696444, 14858000, 29716000, 57500460, 115000920, 222981435, 445962870, 866262915, 1732525830, 3370764540

Offset: 0

Clark Kimberling

nonn

Conjecture: a(n) is the number of integer compositions of n + 2 in which the even parts appear as often at even positions as at odd positions (confirmed up to n = 19). - Gus Wiseman, Mar 17 2018

			The a(3) = 7 compositions of 5 in which the even parts appear as often at even positions as at odd positions are (5), (311), (131), (113), (221), (122), (11111). Missing are (41), (14), (32), (23), (212), (2111), (1211), (1121), (1112). - _Gus Wiseman_, Mar 17 2018

Michael De Vlieger, Table of n, a(n) for n = 0..3326
Christian Krattenthaler, Daniel Yaqubi, Some determinants of path generating functions, II, arXiv:1802.05990 [math.CO], 2018; Adv. Appl. Math. 101 (2018), 232-265.

First differences of A050168. Pairwise sums of A037952.

Cf. A000712, A001405, A005774, A045931, A063886, A097613, A130780, A171966, A239241, A299926, A300061, A300787, A300788, A300789.

[(&+[Binomial(Floor((n+k)/2), Floor(k/2)): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Nov 08 2018

Array[Sum[Binomial[Floor[(# + k)/2], Floor[k/2]], {k, 0, #}] &, 34, 0] (* Michael De Vlieger, May 16 2018 *)
Table[2^(-1 + n)*(((2 + 3*#)*Gamma[(1 + #)/2])/(Sqrt[Pi]*Gamma[2 + #/2]) &[n + Mod[n, 2]]), {n,0,40}] (* Peter Pein, Nov 08 2018 *)
Table[(1/2)^((5 - (-1)^n)/2)*(6*n + 7 - 3*(-1)^n)*CatalanNumber[(2*n + 1 - (-1)^n)/4], {n, 0, 40}] (* G. C. Greubel, Nov 08 2018 *)

vector(40, n, n--; sum(k=0,n, binomial(floor((n+k)/2), floor(k/2)))) \\ G. C. Greubel, Nov 08 2018

a(2*n) = ((3*n + 1)/(2*n + 1))*C(2*n + 1, n)= A051924(1+n), n>=0, a(2*n-1) = a(2*n)/2 = A097613(1+n), n >= 1. - Herbert Kociemba, May 08 2004
a(n) = Sum_{k=0..n} binomial(floor((n+k)/2), floor(k/2)). - Paul Barry, Jul 15 2004
Inverse binomial transform of A005774: (1, 3, 9, 26, 75, 216, ...). - Gary W. Adamson, Oct 22 2007
Conjecture: (n+3)*a(n) - 2*a(n-1) + (-5*n-3)*a(n-2) + 2*a(n-3) + 4*(n-3)*a(n-4) = 0. - R. J. Mathar, Jun 20 2013
a(n) = (1/2)^((5 - (-1)^n)/2)*(6*n + 7 - 3*(-1)^n)*Catalan((2*n + 1 - (-1)^n)/4), where Catalan is the Catalan number = A000108. - G. C. Greubel, Nov 08 2018

cp's OEIS Frontend

A329533 First differences of A051924, or second differences of Central binomial coefficients A000984.

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

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A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

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A030662 Number of combinations of n things from 1 to n at a time, with repeats allowed.

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A263297 The greater of bigomega(n) and maximal prime index in the prime factorization of n.

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A097613 a(n) = binomial(2n-3,n-1) + binomial(2n-2,n-2).

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