A059481 -id:A059481 - cp's OEIS frontend

A171379 Triangle, read by rows, T(n, k) = A059481(n,k)*(A059481(n,k) - 1)/2.

0, 1, 3, 3, 15, 45, 6, 45, 190, 595, 10, 105, 595, 2415, 7875, 15, 210, 1540, 7875, 31626, 106491, 21, 378, 3486, 21945, 106491, 426426, 1471470, 28, 630, 7140, 54285, 313236, 1471470, 5887596, 20701395, 36, 990, 13530, 122265, 827541, 4507503, 20701395, 82812015, 295475895

Offset: 1

Roger L. Bagula, Dec 07 2009

Row sums are: {0, 4, 63, 836, 11000, 147757, 2030217, 28435780, 404461170, 5824442504, ...}.

The sequence is the number of connections between figurate numbers A059481 as points page 25 Riordan.

			Triangle begins as:
   0;
   1,   3;
   3,  15,   45;
   6,  45,  190,   595;
  10, 105,  595,  2415,   7875;
  15, 210, 1540,  7875,  31626,  106491;
  21, 378, 3486, 21945, 106491,  426426, 1471470;
  28, 630, 7140, 54285, 313236, 1471470, 5887596, 20701395;

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 25.

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A059481, A143418.

Flat(List([1..10], n-> List([1..n], k-> Binomial(Binomial(n+k-1, k), 2) ))); # G. C. Greubel, Nov 28 2019

[Binomial(Binomial(n+k-1, k), 2): k in [1..n], n in [1..10]]; // G. C. Greubel, Nov 28 2019

seq(seq( binomial(binomial(n+k-1, k), 2), k=1..n), n=1..10); # G. C. Greubel, Nov 28 2019

Table[Binomial[Binomial[n+k-1, k], 2], {n,10}, {k,n}]//Flatten (* modified by G. C. Greubel, Nov 28 2019 *)

T(n,k) = binomial(binomial(n+k-1, k), 2); \\ G. C. Greubel, Nov 28 2019

[[binomial(binomial(n+k-1, k), 2) for k in (1..n)] for n in (1..10)] # G. C. Greubel, Nov 28 2019

T(n,k) = binomial(n+k-1, k)*(binomial(n+k-1, k) - 1)/2.

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.
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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

A009766 Catalan's triangle T(n,k) (read by rows): each term is the sum of the entries above and to the left, i.e., T(n,k) = Sum_{j=0..k} T(n-1,j).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 5, 5, 1, 4, 9, 14, 14, 1, 5, 14, 28, 42, 42, 1, 6, 20, 48, 90, 132, 132, 1, 7, 27, 75, 165, 297, 429, 429, 1, 8, 35, 110, 275, 572, 1001, 1430, 1430, 1, 9, 44, 154, 429, 1001, 2002, 3432, 4862, 4862, 1, 10, 54, 208, 637, 1638, 3640, 7072, 11934

Offset: 0

Wouter Meeussen

The entries in this triangle (in its many forms) are often called ballot numbers.
T(n,k) = number of standard tableaux of shape (n,k) (n > 0, 0 <= k <= n). Example: T(3,1) = 3 because we have 134/2, 124/3 and 123/4. - Emeric Deutsch, May 18 2004
T(n,k) is the number of full binary trees with n+1 internal nodes and jump-length k. In the preorder traversal of a full binary tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump; the positive difference of the levels is called the jump distance; the sum of the jump distances in a given ordered tree is called the jump-length. - Emeric Deutsch, Jan 18 2007
The k-th diagonal from the right (k=1, 2, ...) gives the sequence obtained by asking in how many ways can we toss a fair coin until we first get k more heads than tails. The k-th diagonal has formula k(2m+k-1)!/(m!(m+k)!) and g.f. (C(x))^k where C(x) is the generating function for the Catalan numbers, (1-sqrt(1-4*x))/(2*x). - Anthony C Robin, Jul 12 2007
T(n,k) is also the number of order-decreasing and order-preserving full transformations (of an n-element chain) of waist k (waist (alpha) = max(Im(alpha))). - Abdullahi Umar, Aug 25 2008
Formatted as an upper right triangle (see tables) a(c,r) is the number of different triangulated planar polygons with c+2 vertices, with triangle degree c-r+1 for the same vertex X (c=column number, r=row number, with c >= r >= 1). - Patrick Labarque, Jul 28 2010
The triangle sums, see A180662 for their definitions, link Catalan's triangle, its mirror is A033184, with several sequences, see crossrefs. - Johannes W. Meijer, Sep 22 2010
The m-th row of Catalan's triangle consists of the unique nonnegative differences of the form binomial(m+k,m)-binomial(m+k,m+1) with 0 <= k <= m (See Links). - R. J. Cano, Jul 22 2014
T(n,k) is also the number of nondecreasing parking functions of length n+1 whose maximum element is k+1. For example T(3,2) = 5 because we have (1,1,1,3), (1,1,2,3), (1,2,2,3), (1,1,3,3), (1,2,3,3). - Ran Pan, Nov 16 2015
T(n,k) is the number of Dyck paths from (0,0) to (n+2,n+2) which start with n-k+2 east steps and touch the diagonal y=x only on the last north step. - Felipe Rueda, Sep 18 2019
T(n-1,k) for k < n is number of well-formed strings of n parenthesis pairs with prefix of exactly n-k opening parenthesis; T(n,n) = T(n,n-1). - Hermann Stamm-Wilbrandt, May 02 2021

			Triangle begins in row n=0 with 0 <= k <= n:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  5,   5;
  1, 4,  9,  14,  14;
  1, 5, 14,  28,  42,   42;
  1, 6, 20,  48,  90,  132,  132;
  1, 7, 27,  75, 165,  297,  429,  429;
  1, 8, 35, 110, 275,  572, 1001, 1430, 1430;
  1, 9, 44, 154, 429, 1001, 2002, 3432, 4862, 4862;

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Paweł J. Szabłowski, Yet another way of calculating moments of the Kesten's distribution and its consequences, arXiv:2106.10461 [math.CO], 2021.
Benjamin Testart, Completing the enumeration of inversion sequences avoiding one or two patterns of length 3, arXiv:2407.07701 [math.CO], 2024. See pp. 9, 29.
Rafael Vazquez and M. Krstic, Boundary control of a singular reaction-diffusion equation on a disk, arXiv:1601.02010 [math.OC], 2016.
Luis Verde-Star, A Matrix Approach to Generalized Delannoy and Schröder Arrays, J. Int. Seq., Vol. 24 (2021), Article 21.4.1.
Eric Weisstein's World of Mathematics, Catalan's Triangle
Eric Weisstein's World of Mathematics, Nonnegative Partial Sum
Tad White, Quota Trees, arXiv:2401.01462 [math.CO], 2024. See p. 17.
Martha Yip, Rook Placements and Jordan Forms of Upper-Triangular Nilpotent Matrices, arXiv:1703.00057 [math.CO], 2017; The Electronic Journal of Combinatorics 25(1) (2018), #P1.68.

The following are all versions of (essentially) the same Catalan triangle: A009766, A008315, A028364, A030237, A047072, A053121, A059365, A062103, A099039, A106566, A130020, A140344.
Cf. A062745, A214292.
Sums of the shallow diagonals give A001405, central binomial coefficients: 1=1, 1=1, 1+1=2, 1+2=3, 1+3+2=6, 1+4+5=10, 1+5+9+5=20, ...
Row sums as well as the sums of the squares of the shallow diagonals give Catalan numbers (A000108).
Reflected version of A033184.
Diagonals give A000108, A000108, A000245, A002057, A000344, A003517, A000588, A003518, A003519, A001392, ...
Triangle sums (see the comments): A000108 (Row1), A000957 (Row2), A001405 (Kn11), A014495 (Kn12), A194124 (Kn13), A030238 (Kn21), A000984 (Kn4), A000958 (Fi2), A165407 (Ca1), A026726 (Ca4), A081696 (Ze2).

Flat(List([0..10],n->List([0..n],m->Binomial(n+m,n)*(n-m+1)/(n+1)))); # Muniru A Asiru, Feb 18 2018

a009766 n k = a009766_tabl !! n !! k
a009766_row n = a009766_tabl !! n
a009766_tabl = iterate (\row -> scanl1 (+) (row ++ [0])) [1]
-- Reinhard Zumkeller, Jul 12 2012

[[Binomial(n+k,n)*(n-k+1)/(n+1): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 07 2019

A009766 := proc(n,k) binomial(n+k,n)*(n-k+1)/(n+1); end proc:
seq(seq(A009766(n,k), k=0..n), n=0..10); # R. J. Mathar, Dec 03 2010

Flatten[NestList[Append[Accumulate[#], Last[Accumulate[#]]] &, {1}, 9]] (* Birkas Gyorgy, May 19 2012 *)
T[n_, k_] := T[n, k] = Which[k == 0, 1, k>n, 0, True, T[n-1, k] + T[n, k-1] ]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 07 2016 *)

{T(n, k) = if( k<0 || k>n, 0, binomial(n+1+k, k) * (n+1-k) / (n+1+k) )}; /* Michael Somos, Oct 17 2006 */

b009766=(n1=0,n2=100)->{my(q=if(!n1,0,binomial(n1+1,2)));for(w=n1,n2,for(k=0,w,write("b009766.txt",1*q" "1*(binomial(w+k,w)-binomial(w+k,w+1)));q++))} \\ R. J. Cano, Jul 22 2014

from math import comb, isqrt
def A009766(n): return comb((a:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)))+(b:=n-comb(a+1,2)),b)*(a-b+1)//(a+1) # Chai Wah Wu, Nov 12 2024

@CachedFunction
def ballot(p,q):
    if p == 0 and q == 0: return 1
    if p < 0 or p > q: return 0
    S = ballot(p-2, q) + ballot(p, q-2)
    if q % 2 == 1: S += ballot(p-1, q-1)
    return S
A009766 = lambda n, k: ballot(2*k, 2*n)
for n in (0..7): [A009766(n, k) for k in (0..n)]  # Peter Luschny, Mar 05 2014

[[binomial(n+k,n)*(n-k+1)/(n+1) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 07 2019

T(n, m) = binomial(n+m, n)*(n-m+1)/(n+1), 0 <= m <= n.
G.f. for column m: (x^m)*N(2; m-1, x)/(1-x)^(m+1), m >= 0, with the row polynomials from triangle A062991 and N(2; -1, x) := 1.
G.f.: C(t*x)/(1-x*C(t*x)) = 1 + (1+t)*x + (1+2*t+2*t^2)*x^2 + ..., where C(x) = (1-sqrt(1-4*x))/(2*x) is the Catalan function. - Emeric Deutsch, May 18 2004
Another version of triangle T(n, k) given by [1, 0, 0, 0, 0, 0, ...] DELTA [0, 1, 1, 1, 1, 1, 1, ...] = 1; 1, 0; 1, 1, 0; 1, 2, 2, 0; 1, 3, 5, 5, 0; 1, 4, 9, 14, 14, 0; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 16 2005
O.g.f. (with offset 1) is the series reversion of x*(1+x*(1-t))/(1+x)^2 = x - x^2*(1+t) + x^3*(1+2*t) - x^4*(1+3*t) + ... . - Peter Bala, Jul 15 2012
Sum_{k=0..floor(n/2)} T(n-k+p-1, k+p-1) = A001405(n+2*p-2) - C(n+2*p-2, p-2), p >= 1. - Johannes W. Meijer, Oct 03 2013
Let A(x,t) denote the o.g.f. Then 1 + x*d/dx(A(x,t))/A(x,t) = 1 + (1 + t)*x + (1 + 2*t + 3*t^2)*x^2 + (1 + 3*t + 6*t^2 + 10*t^3)*x^3 + ... is the o.g.f. for A059481. - Peter Bala, Jul 21 2015
The n-th row polynomial equals the n-th degree Taylor polynomial of the function (1 - 2*x)/(1 - x)^(n+2) about 0. For example, for n = 4, (1 - 2*x)/(1 - x)^6 = 1 + 4*x + 9*x^2 + 14*x^3 + 14*x^4 + O(x^5). - Peter Bala, Feb 18 2018
T(n,k) = binomial(n + k + 1, k) - 2*binomial(n + k, k - 1), for 0 <= k <= n. - David Callan, Jun 15 2022

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

Original entry on oeis.org

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

Offset: 0

Thomas Wieder, Sep 29 2009

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

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

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

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

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

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

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

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

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

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

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

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

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

A014300 Number of nodes of odd outdegree in all ordered rooted (planar) trees with n edges.

Original entry on oeis.org

1, 2, 7, 24, 86, 314, 1163, 4352, 16414, 62292, 237590, 909960, 3497248, 13480826, 52097267, 201780224, 783051638, 3044061116, 11851853042, 46208337584, 180383564228, 704961896036, 2757926215742, 10799653176704, 42326626862636, 166021623024584, 651683311373788

Offset: 1

Emeric Deutsch

Also total number of blocks of odd size in all Catalan(n) possible noncrossing partitions of [n].
Convolution of the sequence of central binomial coefficients 1,2,6,20,70,... (A000984) and of the sequence of Fine numbers 1,0,1,2,6,18,... (A000957).
Row sums of A119307. - Paul Barry, May 13 2006
Hankel transform is A079935. - Paul Barry, Jul 17 2009
Also for n>=1 the number of unimodal functions f:[n]->[n] with f(i)<>f(i+1). a(3) = 7: [1,2,1], [1,2,3], [1,3,1], [1,3,2], [2,3,1], [2,3,2], [3,2,1]. - Alois P. Heinz, May 23 2013
Also, number of sets of n rational numbers on [0,1) such that if x belongs to the set, the fractional part of 2x also belongs to it. - Jianing Song and Andrew Howroyd, May 18 2018
Let A(i, j) denote the infinite array such that the i-th row of this array is the sequence obtained by applying the partial sum operator i times to the function ((-1)^(n + 1) + 1)/2 for n > 0. Then A(n, n) equals a(n) for all n > 0. - John M. Campbell, Jan 20 2019
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p >= 3 and positive integers n and k. - Peter Bala, Jan 07 2022

Alois P. Heinz, Table of n, a(n) for n = 1..500
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Hacène Belbachir and Abdelghani Mehdaoui, Diagonal sums in Pascal pyramid (1, 2, r), Les Annales RECITS (2019) Vol. 6, 45-52.
N. Dershowitz and S. Zaks, Ordered trees and non-crossing partitions, Discrete Math., 62 (1986), 215-218.
Emeric Deutsch and L. Shapiro, A survey of the Fine numbers, Discrete Math., 241 (2001), 241-265.
Index entries for sequences related to rooted trees

Cf. A059481, A000957, A000984, A072547, A119259, A119307, A079935, A333564.

[(&+[(-1)^(n-k)*Binomial(n+k-1, k-1): k in [0..n]]): n in [1..30]]; // G. C. Greubel, Feb 19 2019

a:= proc(n) a(n):= `if`(n<3, n, ((12-40*n+21*n^2) *a(n-1)+
       2*(3*n-1)*(2*n-3) *a(n-2))/ (2*(3*n-4)*n))
    end:
seq(a(n), n=1..30);  # Alois P. Heinz, Oct 30 2012

Rest[CoefficientList[Series[2x/(1-4x+(1+2x)Sqrt[1-4x]),{x,0,40}],x]]  (* Harvey P. Dale, Apr 25 2011 *)
a[n_] := Sum[Binomial[2k, n-1], {k, 0, n-1}]; Array[a, 30] (* Jean-François Alcover, Dec 25 2015, after Paul Barry *)

a(n) = n--; sum(k=0, n, binomial(2*k,n)); \\ Michel Marcus, May 18 2018

from itertools import count, islice
def A014300_gen(): # generator of terms
    yield from (1,2)
    a, c = 1, 1
    for n in count(1):
        yield (a:=(3*n+5)*(c:=c*((n<<2)+2)//(n+2))-a>>1)
A014300_list = list(islice(A014300_gen(),20)) # Chai Wah Wu, Apr 26 2023

[sum((-1)^(n-k)*binomial(n+k-1, k-1) for k in (0..n)) for n in (1..30)] # G. C. Greubel, Feb 19 2019

a(n) = (binomial(2*n, n) + A000957(n))/3; [simplified by Alexander Burstein, Nov 24 2023]
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k-1, k-1). - Vladeta Jovovic, Aug 28 2002
G.f.: 2*z/(1-4*z+(1+2*z)*sqrt(1-4*z)).
a(n) = Sum_{j=0..floor((n-1)/2)} binomial(2*n-2*j-2, n-1).
2*a(n) + a(n-1) = (3*n-1)*Catalan(n-1). - Vladeta Jovovic, Dec 03 2004
a(n) = (-1)^n*Sum_{i=0..n} Sum_{j=n..2*n} (-1)^(i+j)*binomial(j, i). - Benoit Cloitre, Jun 18 2005
a(n) = Sum_{k=0..n} C(2*k,n) [offset 0]. - Paul Barry, May 13 2006
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(n+k-1,k-1). - Paul Barry, Jul 18 2006
From Paul Barry, Jul 17 2009: (Start)
a(n) = Sum_{k=0..n} C(2*n-k,n-k)*(1+(-1)^k)/2.
a(n) = Sum_{k=0..n} C(n+k,k)*(1+(-1)^(n-k))/2. (End)
a(n) is the coefficient of x^(n+1)*y^(n+1) in 1/(1- x^2*y/((1-2*x)*(1-y))). - Ira M. Gessel, Oct 30 2012
a(n) = -binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2], 2). - Peter Luschny, Jul 25 2014
a(n) = [x^n] x/((1 - x^2)*(1 - x)^n). - Ilya Gutkovskiy, Oct 25 2017
a(n) ~ 4^n / (3*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 25 2017
D-finite with recurrence: 2*n*a(n) +(-3*n-4)*a(n-1) +2*(-9*n+19)*a(n-2) +4*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Feb 20 2020
a(n) = A333564(n)/2^n. - Peter Bala, Apr 09 2020
a(n) = (1/2)*(binomial(2*n,n) - A072547(n)). - Peter Bala, Mar 28 2023

A062319 Number of divisors of n^n, or of A000312(n).

Original entry on oeis.org

1, 1, 3, 4, 9, 6, 49, 8, 25, 19, 121, 12, 325, 14, 225, 256, 65, 18, 703, 20, 861, 484, 529, 24, 1825, 51, 729, 82, 1653, 30, 29791, 32, 161, 1156, 1225, 1296, 5329, 38, 1521, 1600, 4961, 42, 79507, 44, 4005, 4186, 2209, 48, 9457, 99, 5151, 2704, 5565, 54

Offset: 0

Jason Earls, Jul 05 2001

From Gus Wiseman, May 02 2021: (Start)
Conjecture: The number of divisors of n^n equals the number of pairwise coprime ordered n-tuples of divisors of n. Confirmed up to n = 30. For example, the a(1) = 1 through a(5) = 6 tuples are:
  (1)  (1,1)  (1,1,1)  (1,1,1,1)  (1,1,1,1,1)
       (1,2)  (1,1,3)  (1,1,1,2)  (1,1,1,1,5)
       (2,1)  (1,3,1)  (1,1,1,4)  (1,1,1,5,1)
              (3,1,1)  (1,1,2,1)  (1,1,5,1,1)
                       (1,1,4,1)  (1,5,1,1,1)
                       (1,2,1,1)  (5,1,1,1,1)
                       (1,4,1,1)
                       (2,1,1,1)
                       (4,1,1,1)
The unordered case (pairwise coprime n-multisets of divisors of n) is counted by A343654.
(End)

			From _Gus Wiseman_, May 02 2021: (Start)
The a(1) = 1 through a(5) = 6 divisors:
  1  1  1   1    1
     2  3   2    5
     4  9   4    25
        27  8    125
            16   625
            32   3125
            64
            128
            256
(End)

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Harry J. Smith)

Cf. A046798, A077592, A035116. - Enrique Pérez Herrero, Nov 09 2010
Number of divisors of A000312(n).
Taking Omega instead of sigma gives A066959.
Positions of squares are A173339.
Diagonal n = k of the array A343656.
A000005 counts divisors.
A059481 counts k-multisets of elements of {1..n}.
A334997 counts length-k strict chains of divisors of n.
A343658 counts k-multisets of divisors.
Pairwise coprimality:
- A018892 counts coprime pairs of divisors.
- A084422 counts pairwise coprime subsets of {1..n}.
- A100565 counts pairwise coprime triples of divisors.
- A225520 counts pairwise coprime sets of divisors.
- A343652 counts maximal pairwise coprime sets of divisors.
- A343653 counts pairwise coprime non-singleton sets of divisors > 1.
- A343654 counts pairwise coprime sets of divisors > 1.
Cf. A000169, A000272, A002064, A002109, A009998, A048691, A143773, A146291, A176029, A327527, A343657.

[NumberOfDivisors(n^n): n in  [0..60]]; // Vincenzo Librandi, Nov 09 2014

A062319[n_IntegerQ]:=DivisorSigma[0,n^n]; (* Enrique Pérez Herrero, Nov 09 2010 *)
Join[{1},DivisorSigma[0,#^#]&/@Range[60]] (* Harvey P. Dale, Jun 06 2024 *)

je=[]; for(n=0,200,je=concat(je,numdiv(n^n))); je

{ for (n=0, 1000, write("b062319.txt", n, " ", numdiv(n^n)); ) } \\ Harry J. Smith, Aug 04 2009

a(n)=local(fm);fm=factor(n);prod(k=1,matsize(fm)[1],fm[k,2]*n+1) \\ Franklin T. Adams-Watters, May 03 2011

a(n) = if(n==0, 1, sumdiv(n, d, n^omega(d))); \\ Seiichi Manyama, May 12 2021

from math import prod
from sympy import factorint
def A062319(n): return prod(n*d+1 for d in factorint(n).values()) # Chai Wah Wu, Jun 03 2021

a(n) = A000005(A000312(n)). - Enrique Pérez Herrero, Nov 09 2010
a(2^n) = A002064(n). - Gus Wiseman, May 02 2021
a(prime(n)) = prime(n) + 1. - Gus Wiseman, May 02 2021
a(n) = Product_{i=1..s} (1 + n * m_i) where (m_1,...,m_s) is the sequence of prime multiplicities (prime signature) of n. - Gus Wiseman, May 02 2021
a(n) = Sum_{d|n} n^omega(d) for n > 0. - Seiichi Manyama May 12 2021

A009998 Triangle in which j-th entry in i-th row is (j+1)^(i-j).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 9, 4, 1, 1, 16, 27, 16, 5, 1, 1, 32, 81, 64, 25, 6, 1, 1, 64, 243, 256, 125, 36, 7, 1, 1, 128, 729, 1024, 625, 216, 49, 8, 1, 1, 256, 2187, 4096, 3125, 1296, 343, 64, 9, 1, 1, 512, 6561, 16384, 15625, 7776, 2401, 512, 81, 10, 1

Offset: 0

N. J. A. Sloane

Read as a square array this is the Hilbert transform of triangle A123125 (see A145905 for the definition of this term). For example, the fourth row of A123125 is (0,1,4,1) and the expansion (x + 4*x^2 + x^3)/(1-x)^4 = x + 8*x^2 + 27*x^3 + 64*x^4 + ... generates the entries in the fourth row of this array read as a square. - Peter Bala, Oct 28 2008

			Triangle begins:
  1;
  1,  1;
  1,  2,  1;
  1,  4,  3,  1;
  1,  8,  9,  4,  1;
  1, 16, 27, 16,  5,  1;
  1, 32, 81, 64, 25,  6,  1;
  ...
From _Gus Wiseman_, May 01 2021: (Start)
The rows of the triangle are obtained by reading antidiagonals upward in the following table of A(k,n) = n^k, with offset k = 0, n = 1:
         n=1:     n=2:     n=3:     n=4:     n=5:     n=6:
   k=0:   1        1        1        1        1        1
   k=1:   1        2        3        4        5        6
   k=2:   1        4        9       16       25       36
   k=3:   1        8       27       64      125      216
   k=4:   1       16       81      256      625     1296
   k=5:   1       32      243     1024     3125     7776
   k=6:   1       64      729     4096    15625    46656
   k=7:   1      128     2187    16384    78125   279936
   k=8:   1      256     6561    65536   390625  1679616
   k=9:   1      512    19683   262144  1953125 10077696
  k=10:   1     1024    59049  1048576  9765625 60466176
(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. 24.

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].
Peter Luschny, Figurate number - a very short introduction

Row sums give A026898.
Cf. A088956, A123125, A179927.
Column n = 2 of the array is A000079.
Column n = 3 of the array is A000244.
Row k = 2 of the array is A000290.
Row k = 3 of the array is A000578.
Diagonal n = k of the array is A000312.
Diagonal n = k + 1 of the array is A000169.
Diagonal n = k + 2 of the array is A000272.
The transpose of the array is A009999.
The numbers of divisors of the entries are A343656 (row sums: A343657).
A007318 counts k-sets of elements of {1..n}.
A059481 counts k-multisets of elements of {1..n}.
Cf. A002109, A062319, A066959, A143773, A176029, A204688, A326358, A327527.

a009998 n k = (k + 1) ^ (n - k)
a009998_row n = a009998_tabl !! n
a009998_tabl = map reverse a009999_tabl
-- Reinhard Zumkeller, Feb 02 2014

E := (n,x) -> `if`(n=0,1,x*(1-x)*diff(E(n-1,x),x)+E(n-1,x)*(1+(n-1)*x));
G := (n,x) -> E(n,x)/(1-x)^(n+1);
A009998 := (n,k) -> coeff(series(G(n-k,x),x,18),x,k);
seq(print(seq(A009998(n,k),k=0..n)),n=0..6);
# Peter Luschny, Aug 02 2010

Flatten[Table[(j+1)^(i-j),{i,0,20},{j,0,i}]] (* Harvey P. Dale, Dec 25 2012 *)

T(i,j)=(j+1)^(i-j) \\ Charles R Greathouse IV, Feb 06 2017

T(n,n) = 1; T(n,k) = (k+1)*T(n-1,k) for k=0..n-1. - Reinhard Zumkeller, Feb 02 2014

T(n,m) = (m+1)*Sum_{k=0..n-m}((n+1)^(k-1)*(n-m)^(n-m-k)*(-1)^(n-m-k)*binomial(n-m-1,k-1)). - Vladimir Kruchinin, Sep 12 2015

a(62) corrected to 512 by T. D. Noe, Dec 20 2007

A014301 Number of internal nodes of even outdegree in all ordered rooted trees with n edges.

Original entry on oeis.org

0, 1, 3, 11, 40, 148, 553, 2083, 7896, 30086, 115126, 442118, 1703052, 6577474, 25461493, 98759971, 383751472, 1493506534, 5820778858, 22714926826, 88745372992, 347087585824, 1358789148058, 5324148664846, 20878676356240, 81937643449468, 321786401450268

Offset: 1

Emeric Deutsch

nonn

Number of protected vertices in all ordered rooted trees with n edges. A protected vertex in an ordered tree is a vertex at least 2 edges away from its leaf descendants. - Emeric Deutsch, Aug 20 2008
1,3,11,... gives the diagonal sums of A111418. Hankel transform of a(n) is A128834. Hankel transform of a(n+1) is A187340. - Paul Barry, Mar 08 2011
a(n) = A035317(2*n-1,n-1) for n > 0. - Reinhard Zumkeller, Jul 19 2012
Apparently the number of peaks in all Dyck paths of semilength n+1 that are the same height as the preceding peak. - David Scambler, Apr 22 2013
Define an infinite triangle by T(n,0)=A001045(n) (the first column) and T(r,c) = Sum_{k=c-1..r} T(k,c-1) (the sum of all the terms in the preceding column down to row r). Then T(n,n)=a(n+1). The triangle is 0; 1,1; 1,2,3; 3,5,8,11; 5,10,18,29,40; 11,21,39,68,108,148;... Example: T(5,2)=39=the sum of the terms in column 1 from T(1,1) to T(5,1), namely, 1+2+5+10+21. - J. M. Bergot, May 17 2013
Also for n>=1 the number of unimodal functions f:[n]->[n] with f(1)<>1 and f(i)<>f(i+1). a(4) = 11: [2,3,2,1], [2,3,4,1], [2,3,4,2], [2,3,4,3], [2,4,2,1], [2,4,3,1], [2,4,3,2], [3,4,2,1], [3,4,3,1], [3,4,3,2], [4,3,2,1]. - Alois P. Heinz, May 23 2013

G. C. Greubel, Table of n, a(n) for n = 1..1000
Gi-Sang Cheon and Louis W. Shapiro, Protected points in ordered trees, Appl. Math. Letters, 21, 2008, 516-520.
Sergi Elizalde, Symmetric peaks and symmetric valleys in Dyck paths, arXiv:2008.05669 [math.CO], 2020, Corollary 3.4.
Torleiv Kløve, Spheres of Permutations under the Infinity Norm - Permutations with limited displacement, Reports in Informatics, Department of Informatics, University of Bergen, Norway, no. 376, November 2008.
Index entries for sequences related to rooted trees

Cf. A059481, A026641, A143362, A143363.

[(1/2)*(&+[(-1)^(n-k)*Binomial(n+k-1,k): k in [0..n]]): n in [1..30]]; // G. C. Greubel, Jan 15 2018

Rest[CoefficientList[Series[(1-2*x-Sqrt[1-4*x])/(3*Sqrt[1-4*x]-1+4*x), {x, 0, 50}], x]] (* G. C. Greubel, Jan 15 2018 *)

x='x+O('x^30); Vec((1-2*x-sqrt(1-4*x))/(3*sqrt(1-4*x)-1+4*x)) \\ G. C. Greubel, Jan 15 2018

from itertools import count, islice
def A014301_gen(): # generator of terms
    yield from (0,1)
    a, b, c = 0, 3, 1
    for n in count(1):
        yield ((b:=b*((n<<1)+3<<1)//(n+2))-(a:=(c:=c*((n<<2)+2)//(n+2))-a>>1))//3
A014301_list = list(islice(A014301_gen(),20)) # Chai Wah Wu, Apr 27 2023

a(n) = binomial(2*n-1, n)/3 - A000957(n)/3;
a(n) = (1/2)*Sum_{k=0..n} (-1)^(n-k)*binomial(n+k-1, k). - Vladeta Jovovic, Aug 28 2002
From Emeric Deutsch, Jan 26 2004: (Start)
G.f.: (1-2*z-sqrt(1-4*z))/(3*sqrt(1-4*z)-1+4*z).
a(n) = [A026641(n) - A026641(n-1)]/3 for n>1. (End)
a(n) = (1/2)*Sum_{j=0..floor(n/2)} binomial(2n-2j-2, n-2).
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(n+k,k-1). - Paul Barry, Jul 18 2006
D-finite with recurrence: 2*n*a(n) +(-9*n+8)*a(n-1) +(3*n-16)*a(n-2) +2*(2*n-5)*a(n-3)=0. - R. J. Mathar, Dec 03 2012

A100100 Triangle T(n,k) = binomial(2*n-k-1, n-k) read by rows.

Original entry on oeis.org

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

Offset: 0

Paul Barry, Nov 08 2004

Sometimes called a Catalan triangle, although there are many other triangles that carry that name - see A009766, A008315, A028364, A033184, A053121, A059365, A062103.
Number of nodes of outdegree k in all ordered trees with n edges. Equivalently, number of ascents of length k in all Dyck paths of semilength n. Example: T(3,2) = 3 because the Dyck paths of semilength 3 are UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and UUUDDD, where U = (1,1), D = (1,-1), the ascents of length 2 being shown between parentheses. - Emeric Deutsch, Nov 19 2006
Riordan array (f(x), x*g(x)) where f(x) is the g.f. of A088218 and g(x) is the g.f. of A000108. - Philippe Deléham, Jan 23 2010
T(n,k) is the number of nonnegative paths of upsteps U = (1,1) and downsteps D = (1,-1) of length 2*n with k returns to  ground level, the horizontal line through the initial vertex. Example: T(2,1) = 2 counts UDUU, UUDD. Also, T(n,k) = number of these paths whose last descent has length k, that is, k downsteps follow the last upstep. Example: T(2,1) = 2 counts UUUD, UDUD. - David Callan, Nov 21 2011
Belongs to the hitting-time subgroup of the Riordan group. Multiplying this triangle by the square Pascal matrix gives A092392 read as a square array. See the example below. - Peter Bala, Nov 03 2015

			From _Paul Barry_, Mar 15 2010: (Start)
Triangle begins in row n=0 with columns 0<=k<=n as:
    1;
    1,   1;
    3,   2,   1;
   10,   6,   3,  1;
   35,  20,  10,  4,  1;
  126,  70,  35, 15,  5, 1;
  462, 252, 126, 56, 21, 6, 1;
Production matrix begins
  1, 1;
  2, 1, 1;
  3, 1, 1, 1;
  4, 1, 1, 1, 1;
  5, 1, 1, 1, 1, 1;
  6, 1, 1, 1, 1, 1, 1;
  7, 1, 1, 1, 1, 1, 1, 1;
(End)
A092392 as a square array = A100100 * square Pascal matrix:
/1   1  1  1 ...\   / 1          \/1 1  1  1 ...\
|2   3  4  5 ...|   | 1 1        ||1 2  3  4 ...|
|6  10 15 21 ...| = | 3 2 1      ||1 3  6 10 ...|
|20 35 56 84 ...|   |10 6 3 1    ||1 4 10 20 ...|
|70 ...         |   |35 ...      ||1 ...        |
- _Peter Bala_, Nov 03 2015

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, An algebraic structure for Faber polynomials, Lin. Alg. Applic. 433 (2010) 1170-1179.
Tian-Xiao He and Renzo Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. 309 (2009), no. 12, 3962-3974.

Row sums are A000984. Equivalent to A092392, to which A088218 has been added as a first column. Columns include A088218, A000984, A001700, A001791, A002054, A002694. Diagonal sums are A100217. Matrix inverse is A100218.

Cf. A059481 (mirrored). Cf. A033184, A094527, A113955.

a100100 n k = a100100_tabl !! n !! n
a100100_row n = a100100_tabl !! n
a100100_tabl = [1] : f a092392_tabl where
   f (us : wss'@(vs : wss)) = (vs !! 1 : us) : f wss'
-- Reinhard Zumkeller, Jan 15 2014

/* As triangle */ [[Binomial(2*n - k - 1, n - k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 21 2018

A100100 := proc(n,k)
    binomial(2*n-k-1,n-1) ;
end proc:
seq(seq(A100100(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Flatten[Table[Binomial[2 n - k - 1, n - k], {n, 0, 11}, {k, 0, n}]] (* Vincenzo Librandi, Nov 21 2018 *)

T(n,k)=binomial(2*n-k-1,n-k) \\ Charles R Greathouse IV, Jan 16 2012

From Peter Bala, Sep 06 2015: (Start)
Matrix product A094527 * P^(-1) = A113955 * P^(-2), where P denotes Pascal's triangle A007318.
Essentially, the logarithmic derivative of A033184. (End)
Let column(k) = [T(n, k), n >= k], then the generating function for column(k) is (2/(sqrt(1-4*x)+1))^(k-1)/sqrt(1-4*x). - Peter Luschny, Mar 19 2021
O.g.f. row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, i.e. o.g.f. of the triangle, G(z,x) = 1/((2 - c(z))*(1 - x*z*c(z))), with c the o.g.f. of A000108 (Catalan). See the Riordan coomment by Philippe Deléham above. - Wolfdieter Lang, Apr 06 2021

A343656 Array read by antidiagonals where A(n,k) is the number of divisors of n^k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 3, 3, 1, 1, 5, 4, 5, 2, 1, 1, 6, 5, 7, 3, 4, 1, 1, 7, 6, 9, 4, 9, 2, 1, 1, 8, 7, 11, 5, 16, 3, 4, 1, 1, 9, 8, 13, 6, 25, 4, 7, 3, 1, 1, 10, 9, 15, 7, 36, 5, 10, 5, 4, 1, 1, 11, 10, 17, 8, 49, 6, 13, 7, 9, 2, 1, 1, 12, 11, 19, 9, 64, 7, 16, 9, 16, 3, 6, 1

Offset: 1

Gus Wiseman, Apr 28 2021

First differs from A343658 at A(4,2) = 5, A343658(4,2) = 6.

As a triangle, T(n,k) = number of divisors of k^(n-k).

			Array begins:
       k=0 k=1 k=2 k=3 k=4 k=5 k=6 k=7
  n=1:  1   1   1   1   1   1   1   1
  n=2:  1   2   3   4   5   6   7   8
  n=3:  1   2   3   4   5   6   7   8
  n=4:  1   3   5   7   9  11  13  15
  n=5:  1   2   3   4   5   6   7   8
  n=6:  1   4   9  16  25  36  49  64
  n=7:  1   2   3   4   5   6   7   8
  n=8:  1   4   7  10  13  16  19  22
  n=9:  1   3   5   7   9  11  13  15
Triangle begins:
  1
  1  1
  1  2  1
  1  3  2  1
  1  4  3  3  1
  1  5  4  5  2  1
  1  6  5  7  3  4  1
  1  7  6  9  4  9  2  1
  1  8  7 11  5 16  3  4  1
  1  9  8 13  6 25  4  7  3  1
  1 10  9 15  7 36  5 10  5  4  1
  1 11 10 17  8 49  6 13  7  9  2  1
  1 12 11 19  9 64  7 16  9 16  3  6  1
  1 13 12 21 10 81  8 19 11 25  4 15  2  1
For example, row n = 8 counts the following divisors:
  1  64  243  256  125  36  7  1
     32  81   128  25   18  1
     16  27   64   5    12
     8   9    32   1    9
     4   3    16        6
     2   1    8         4
     1        4         3
              2         2
              1         1

Seiichi Manyama, Antidiagonals n = 1..140, flattened

Columns k=1..9 of the array give A000005, A048691, A048785, A344327, A344328, A344329, A343526, A344335, A344336.
Row n = 6 of the array is A000290.
Diagonal n = k of the array is A062319.
Array antidiagonal sums (row sums of the triangle) are A343657.
Dominated by A343658.
A000312 = n^n.
A007318 counts k-sets of elements of {1..n}.
A009998(n,k) = n^k (as an array, offset 1).
A059481 counts k-multisets of elements of {1..n}.
Cf. A000169, A000272, A002064, A002109, A066959, A143773, A146291, A176029, A251683, A282935, A326358, A327527, A334996, A343652.

Table[DivisorSigma[0,k^(n-k)],{n,10},{k,n}]

A(n, k) = numdiv(n^k); \\ Seiichi Manyama, May 15 2021

A(n,k) = A000005(A009998(n,k)), where A009998(n,k) = n^k is the interpretation as an array.

A(n,k) = Sum_{d|n} k^omega(d). - Seiichi Manyama, May 15 2021

cp's OEIS Frontend

A171379 Triangle, read by rows, T(n, k) = A059481(n,k)*(A059481(n,k) - 1)/2.

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

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A009766 Catalan's triangle T(n,k) (read by rows): each term is the sum of the entries above and to the left, i.e., T(n,k) = Sum_{j=0..k} T(n-1,j).

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

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A014300 Number of nodes of odd outdegree in all ordered rooted (planar) trees with n edges.

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