cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000108 Catalan numbers: C(n) = binomial(2n,n)/(n+1) = (2n)!/(n!(n+1)!).

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152, 69533550916004, 263747951750360, 1002242216651368, 3814986502092304
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

These were formerly sometimes called Segner numbers.
A very large number of combinatorial interpretations are known - see references, esp. R. P. Stanley, "Catalan Numbers", Cambridge University Press, 2015. This is probably the longest entry in the OEIS, and rightly so.
The solution to Schröder's first problem: number of ways to insert n pairs of parentheses in a word of n+1 letters. E.g., for n=2 there are 2 ways: ((ab)c) or (a(bc)); for n=3 there are 5 ways: ((ab)(cd)), (((ab)c)d), ((a(bc))d), (a((bc)d)), (a(b(cd))).
Consider all the binomial(2n,n) paths on squared paper that (i) start at (0, 0), (ii) end at (2n, 0) and (iii) at each step, either make a (+1,+1) step or a (+1,-1) step. Then the number of such paths that never go below the x-axis (Dyck paths) is C(n). [Chung-Feller]
Number of noncrossing partitions of the n-set. For example, of the 15 set partitions of the 4-set, only [{13},{24}] is crossing, so there are a(4)=14 noncrossing partitions of 4 elements. - Joerg Arndt, Jul 11 2011
Noncrossing partitions are partitions of genus 0. - Robert Coquereaux, Feb 13 2024
a(n-1) is the number of ways of expressing an n-cycle (123...n) in the symmetric group S_n as a product of n-1 transpositions (u_1,v_1)*(u_2,v_2)*...*(u_{n-1},v_{n-1}) where u_iA000272. - Joerg Arndt and Greg Stevenson, Jul 11 2011
a(n) is the number of ordered rooted trees with n nodes, not including the root. See the Conway-Guy reference where these rooted ordered trees are called plane bushes. See also the Bergeron et al. reference, Example 4, p. 167. - Wolfdieter Lang, Aug 07 2007
As shown in the paper from Beineke and Pippert (1971), a(n-2)=D(n) is the number of labeled dissections of a disk, related to the number R(n)=A001761(n-2) of labeled planar 2-trees having n vertices and rooted at a given exterior edge, by the formula D(n)=R(n)/(n-2)!. - M. F. Hasler, Feb 22 2012
Shifts one place left when convolved with itself.
For n >= 1, a(n) is also the number of rooted bicolored unicellular maps of genus 0 on n edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 15 2001
Number of ways of joining 2n points on a circle to form n nonintersecting chords. (If no such restriction imposed, then the number of ways of forming n chords is given by (2n-1)!! = (2n)!/(n!*2^n) = A001147(n).)
Arises in Schubert calculus - see Sottile reference.
Inverse Euler transform of sequence is A022553.
With interpolated zeros, the inverse binomial transform of the Motzkin numbers A001006. - Paul Barry, Jul 18 2003
The Hankel transforms of this sequence or of this sequence with the first term omitted give A000012 = 1, 1, 1, 1, 1, 1, ...; example: Det([1, 1, 2, 5; 1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132]) = 1 and Det([1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132; 14, 42, 132, 429]) = 1. - Philippe Deléham, Mar 04 2004
a(n) equals the sum of squares of terms in row n of triangle A053121, which is formed from successive self-convolutions of the Catalan sequence. - Paul D. Hanna, Apr 23 2005
Also coefficients of the Mandelbrot polynomial M iterated an infinite number of times. Examples: M(0) = 0 = 0*c^0 = [0], M(1) = c = c^1 + 0*c^0 = [1 0], M(2) = c^2 + c = c^2 + c^1 + 0*c^0 = [1 1 0], M(3) = (c^2 + c)^2 + c = [0 1 1 2 1], ... ... M(5) = [0 1 1 2 5 14 26 44 69 94 114 116 94 60 28 8 1], ... - Donald D. Cross (cosinekitty(AT)hotmail.com), Feb 04 2005
The multiplicity with which a prime p divides C_n can be determined by first expressing n+1 in base p. For p=2, the multiplicity is the number of 1 digits minus 1. For p an odd prime, count all digits greater than (p+1)/2; also count digits equal to (p+1)/2 unless final; and count digits equal to (p-1)/2 if not final and the next digit is counted. For example, n=62, n+1 = 223_5, so C_62 is not divisible by 5. n=63, n+1 = 224_5, so 5^3 | C_63. - Franklin T. Adams-Watters, Feb 08 2006
Koshy and Salmassi give an elementary proof that the only prime Catalan numbers are a(2) = 2 and a(3) = 5. Is the only semiprime Catalan number a(4) = 14? - Jonathan Vos Post, Mar 06 2006
The answer is yes. Using the formula C_n = binomial(2n,n)/(n+1), it is immediately clear that C_n can have no prime factor greater than 2n. For n >= 7, C_n > (2n)^2, so it cannot be a semiprime. Given that the Catalan numbers grow exponentially, the above consideration implies that the number of prime divisors of C_n, counted with multiplicity, must grow without limit. The number of distinct prime divisors must also grow without limit, but this is more difficult. Any prime between n+1 and 2n (exclusive) must divide C_n. That the number of such primes grows without limit follows from the prime number theorem. - Franklin T. Adams-Watters, Apr 14 2006
The number of ways to place n indistinguishable balls in n numbered boxes B1,...,Bn such that at most a total of k balls are placed in boxes B1,...,Bk for k=1,...,n. For example, a(3)=5 since there are 5 ways to distribute 3 balls among 3 boxes such that (i) box 1 gets at most 1 ball and (ii) box 1 and box 2 together get at most 2 balls:(O)(O)(O), (O)()(OO), ()(OO)(O), ()(O)(OO), ()()(OOO). - Dennis P. Walsh, Dec 04 2006
a(n) is also the order of the semigroup of order-decreasing and order-preserving full transformations (of an n-element chain) - now known as the Catalan monoid. - Abdullahi Umar, Aug 25 2008
a(n) is the number of trivial representations in the direct product of 2n spinor (the smallest) representations of the group SU(2) (A(1)). - Rutger Boels (boels(AT)nbi.dk), Aug 26 2008
The invert transform appears to converge to the Catalan numbers when applied infinitely many times to any starting sequence. - Mats Granvik, Gary W. Adamson and Roger L. Bagula, Sep 09 2008, Sep 12 2008
Limit_{n->oo} a(n)/a(n-1) = 4. - Francesco Antoni (francesco_antoni(AT)yahoo.com), Nov 24 2008
Starting with offset 1 = row sums of triangle A154559. - Gary W. Adamson, Jan 11 2009
C(n) is the degree of the Grassmannian G(1,n+1): the set of lines in (n+1)-dimensional projective space, or the set of planes through the origin in (n+2)-dimensional affine space. The Grassmannian is considered a subset of N-dimensional projective space, N = binomial(n+2,2) - 1. If we choose 2n general (n-1)-planes in projective (n+1)-space, then there are C(n) lines that meet all of them. - Benji Fisher (benji(AT)FisherFam.org), Mar 05 2009
Starting with offset 1 = A068875: (1, 2, 4, 10, 18, 84, ...) convolved with Fine numbers, A000957: (1, 0, 1, 2, 6, 18, ...). a(6) = 132 = (1, 2, 4, 10, 28, 84) dot (18, 6, 2, 1, 0, 1) = (18 + 12 + 8 + 10 + 0 + 84) = 132. - Gary W. Adamson, May 01 2009
Convolved with A032443: (1, 3, 11, 42, 163, ...) = powers of 4, A000302: (1, 4, 16, ...). - Gary W. Adamson, May 15 2009
Sum_{k>=1} C(k-1)/2^(2k-1) = 1. The k-th term in the summation is the probability that a random walk on the integers (beginning at the origin) will arrive at positive one (for the first time) in exactly (2k-1) steps. - Geoffrey Critzer, Sep 12 2009
C(p+q)-C(p)*C(q) = Sum_{i=0..p-1, j=0..q-1} C(i)*C(j)*C(p+q-i-j-1). - Groux Roland, Nov 13 2009
Leonhard Euler used the formula C(n) = Product_{i=3..n} (4*i-10)/(i-1) in his 'Betrachtungen, auf wie vielerley Arten ein gegebenes polygonum durch Diagonallinien in triangula zerschnitten werden könne' and computes by recursion C(n+2) for n = 1..8. (Berlin, 4th September 1751, in a letter to Goldbach.) - Peter Luschny, Mar 13 2010
Let A179277 = A(x). Then C(x) is satisfied by A(x)/A(x^2). - Gary W. Adamson, Jul 07 2010
a(n) is also the number of quivers in the mutation class of type B_n or of type C_n. - Christian Stump, Nov 02 2010
From Matthew Vandermast, Nov 22 2010: (Start)
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1) equals the number of ways to choose 0 or more balls of each color while satisfying the following conditions: 1. No two colors are chosen the same positive number of times. 2. For any two colors (c, d) that are chosen at least once, color c is chosen more times than color d iff color c appears more times in the original set than color d.
If the second requirement is lifted, the number of acceptable ways equals A000110(n+1). See related comments for A016098, A085082. (End)
Deutsch and Sagan prove the Catalan number C_n is odd if and only if n = 2^a - 1 for some nonnegative integer a. Lin proves for every odd Catalan number C_n, we have C_n == 1 (mod 4). - Jonathan Vos Post, Dec 09 2010
a(n) is the number of functions f:{1,2,...,n}->{1,2,...,n} such that f(1)=1 and for all n >= 1 f(n+1) <= f(n)+1. For a nice bijection between this set of functions and the set of length 2n Dyck words, see page 333 of the Fxtbook (see link below). - Geoffrey Critzer, Dec 16 2010
Postnikov (2005) defines "generalized Catalan numbers" associated with buildings (e.g., Catalan numbers of Type B, see A000984). - N. J. A. Sloane, Dec 10 2011
Number of permutations in S(n) for which length equals depth. - Bridget Tenner, Feb 22 2012
a(n) is also the number of standard Young tableau of shape (n,n). - Thotsaporn Thanatipanonda, Feb 25 2012
a(n) is the number of binary sequences of length 2n+1 in which the number of ones first exceed the number of zeros at entry 2n+1. See the example below in the example section. - Dennis P. Walsh, Apr 11 2012
Number of binary necklaces of length 2*n+1 containing n 1's (or, by symmetry, 0's). All these are Lyndon words and their representatives (as cyclic maxima) are the binary Dyck words. - Joerg Arndt, Nov 12 2012
Number of sequences consisting of n 'x' letters and n 'y' letters such that (counting from the left) the 'x' count >= 'y' count. For example, for n=3 we have xxxyyy, xxyxyy, xxyyxy, xyxxyy and xyxyxy. - Jon Perry, Nov 16 2012
a(n) is the number of Motzkin paths of length n-1 in which the (1,0)-steps come in 2 colors. Example: a(4)=14 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 8 paths of shape HHH, 2 paths of shape UHD, 2 paths of shape UDH, and 2 paths of shape HUD. - José Luis Ramírez Ramírez, Jan 16 2013
If p is an odd prime, then (-1)^((p-1)/2)*a((p-1)/2) mod p = 2. - Gary Detlefs, Feb 20 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. - Zhi-Wei Sun, Mar 23 2013
a(n) is the size of the Jones monoid on 2n points (cf. A225798). - James Mitchell, Jul 28 2013
For 0 < p < 1, define f(p) = Sum_{n>=0} a(n)*(p*(1-p))^n, then f(p) = min{1/p, 1/(1-p)}, so f(p) reaches its maximum value 2 at p = 0.5, and p*f(p) is constant 1 for 0.5 <= p < 1. - Bob Selcoe, Nov 16 2013 [Corrected by Jianing Song, May 21 2021]
No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
From Alexander Adamchuk, Dec 27 2013: (Start)
Prime p divides a((p+1)/2) for p > 3. See A120303(n) = Largest prime factor of Catalan number.
Reciprocal Catalan Constant C = 1 + 4*sqrt(3)*Pi/27 = 1.80613.. = A121839.
Log(Phi) = (125*C - 55) / (24*sqrt(5)), where C = Sum_{k>=1} (-1)^(k+1)*1/a(k). See A002390 = Decimal expansion of natural logarithm of golden ratio.
3-d analog of the Catalan numbers: (3n)!/(n!(n+1)!(n+2)!) = A161581(n) = A006480(n) / ((n+1)^2*(n+2)), where A006480(n) = (3n)!/(n!)^3 De Bruijn's S(3,n). (End)
For a relation to the inviscid Burgers's, or Hopf, equation, see A001764. - Tom Copeland, Feb 15 2014
From Fung Lam, May 01 2014: (Start)
One class of generalized Catalan numbers can be defined by g.f. A(x) = (1-sqrt(1-q*4*x*(1-(q-1)*x)))/(2*q*x) with nonzero parameter q. Recurrence: (n+3)*a(n+2) -2*q*(2*n+3)*a(n+1) +4*q*(q-1)*n*a(n) = 0 with a(0)=1, a(1)=1.
Asymptotic approximation for q >= 1: a(n) ~ (2*q+2*sqrt(q))^n*sqrt(2*q*(1+sqrt(q))) /sqrt(4*q^2*Pi*n^3).
For q <= -1, the g.f. defines signed sequences with asymptotic approximation: a(n) ~ Re(sqrt(2*q*(1+sqrt(q)))*(2*q+2*sqrt(q))^n) / sqrt(q^2*Pi*n^3), where Re denotes the real part. Due to Stokes' phenomena, accuracy of the asymptotic approximation deteriorates at/near certain values of n.
Special cases are A000108 (q=1), A068764 to A068772 (q=2 to 10), A240880 (q=-3).
(End)
Number of sequences [s(0), s(1), ..., s(n)] with s(n)=0, Sum_{j=0..n} s(j) = n, and Sum_{j=0..k} s(j)-1 >= 0 for k < n-1 (and necessarily Sum_{j=0..n-1} s(j)-1 = 0). These are the branching sequences of the (ordered) trees with n non-root nodes, see example. - Joerg Arndt, Jun 30 2014
Number of stack-sortable permutations of [n], these are the 231-avoiding permutations; see the Bousquet-Mélou reference. - Joerg Arndt, Jul 01 2014
a(n) is the number of increasing strict binary trees with 2n-1 nodes that avoid 132. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
In a one-dimensional medium with elastic scattering (zig-zag walk), first recurrence after 2n+1 scattering events has the probability C(n)/2^(2n+1). - Joachim Wuttke, Sep 11 2014
The o.g.f. C(x) = (1 - sqrt(1-4x))/2, for the Catalan numbers, with comp. inverse Cinv(x) = x*(1-x) and the functions P(x) = x / (1 + t*x) and its inverse Pinv(x,t) = -P(-x,t) = x / (1 - t*x) form a group under composition that generates or interpolates among many classic arrays, such as the Motzkin (Riordan, A005043), Fibonacci (A000045), and Fine (A000957) numbers and polynomials (A030528), and enumerating arrays for Motzkin, Dyck, and Łukasiewicz lattice paths and different types of trees and non-crossing partitions (A091867, connected to sums of the refined Narayana numbers A134264). - Tom Copeland, Nov 04 2014
Conjecture: All the rational numbers Sum_{i=j..k} 1/a(i) with 0 < min{2,k} <= j <= k have pairwise distinct fractional parts. - Zhi-Wei Sun, Sep 24 2015
The Catalan number series A000108(n+3), offset n=0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset n=0 (empirical observation). - Tony Foster III, Sep 05 2016
Hankel transforms of the Catalan numbers with the first 2, 4, and 5 terms omitted give A001477, A006858, and A091962, respectively, without the first 2 terms in all cases. More generally, the Hankel transform of the Catalan numbers with the first k terms omitted is H_k(n) = Product_{j=1..k-1} Product_{i=1..j} (2*n+j+i)/(j+i) [see Cigler (2011), Eq. (1.14) and references therein]; together they form the array A078920/A123352/A368025. - Andrey Zabolotskiy, Oct 13 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. See S. J. Miller, ed., 2015, p. 5. - N. J. A. Sloane, Feb 09 2017
Coefficients of the generating series associated to the Magmatic and Dendriform operadic algebras. Cf. p. 422 and 435 of the Loday et al. paper. - Tom Copeland, Jul 08 2018
Let M_n be the n X n matrix with M_n(i,j) = binomial(i+j-1,2j-2); then det(M_n) = a(n). - Tony Foster III, Aug 30 2018
Also the number of Catalan trees, or planted plane trees (Bona, 2015, p. 299, Theorem 4.6.3). - N. J. A. Sloane, Dec 25 2018
Number of coalescent histories for a caterpillar species tree and a matching caterpillar gene tree with n+1 leaves (Rosenberg 2007, Corollary 3.5). - Noah A Rosenberg, Jan 28 2019
Finding solutions of eps*x^2+x-1 = 0 for eps small, that is, writing x = Sum_{n>=0} x_{n}*eps^n and expanding, one finds x = 1 - eps + 2*eps^2 - 5*eps^3 + 14*eps^3 - 42*eps^4 + ... with x_{n} = (-1)^n*C(n). Further, letting x = 1/y and expanding y about 0 to find large roots, that is, y = Sum_{n>=1} y_{n}*eps^n, one finds y = 0 - eps + eps^2 - 2*eps^3 + 5*eps^3 - ... with y_{n} = (-1)^n*C(n-1). - Derek Orr, Mar 15 2019
Permutations of length n that produce a bipartite permutation graph of order n [see Knuth (1973), Busch (2006), Golumbic and Trenk (2004)]. - Elise Anderson, R. M. Argus, Caitlin Owens, Tessa Stevens, Jun 27 2019
For n > 0, a random selection of n + 1 objects (the minimum number ensuring one pair by the pigeonhole principle) from n distinct pairs of indistinguishable objects contains only one pair with probability 2^(n-1)/a(n) = b(n-1)/A098597(n), where b is the 0-offset sequence with the terms of A120777 repeated (1,1,4,4,8,8,64,64,128,128,...). E.g., randomly selecting 6 socks from 5 pairs that are black, blue, brown, green, and white, results in only one pair of the same color with probability 2^(5-1)/a(5) = 16/42 = 8/21 = b(4)/A098597(5). - Rick L. Shepherd, Sep 02 2019
See Haran & Tabachnikov link for a video discussing Conway-Coxeter friezes. The Conway-Coxeter friezes with n nontrivial rows are generated by the counts of triangles at each vertex in the triangulations of regular n-gons, of which there are a(n). - Charles R Greathouse IV, Sep 28 2019
For connections to knot theory and scattering amplitudes from Feynman diagrams, see Broadhurst and Kreimer, and Todorov. Eqn. 6.12 on p. 130 of Bessis et al. becomes, after scaling, -12g * r_0(-y/(12g)) = (1-sqrt(1-4y))/2, the o.g.f. (expressed as a Taylor series in Eqn. 7.22 in 12gx) given for the Catalan numbers in Copeland's (Sep 30 2011) formula below. (See also Mizera p. 34, Balduf pp. 79-80, Keitel and Bartosch.) - Tom Copeland, Nov 17 2019
Number of permutations in S_n whose principal order ideals in the weak order are modular lattices. - Bridget Tenner, Jan 16 2020
Number of permutations in S_n whose principal order ideals in the weak order are distributive lattices. - Bridget Tenner, Jan 16 2020
Legendre gives the following formula for computing the square root modulo 2^m:
sqrt(1 + 8*a) mod 2^m = (1 + 4*a*Sum_{i=0..m-4} C(i)*(-2*a)^i) mod 2^m
as cited by L. D. Dickson, History of the Theory of Numbers, Vol. 1, 207-208. - Peter Schorn, Feb 11 2020
a(n) is the number of length n permutations sorted to the identity by a consecutive-132-avoiding stack followed by a classical-21-avoiding stack. - Kai Zheng, Aug 28 2020
Number of non-crossing partitions of a 2*n-set with n blocks of size 2. Also number of non-crossing partitions of a 2*n-set with n+1 blocks of size at most 3, and without cyclical adjacencies. The two partitions can be mapped by rotated Kreweras bijection. - Yuchun Ji, Jan 18 2021
Named by Riordan (1968, and earlier in Mathematical Reviews, 1948 and 1964) after the French and Belgian mathematician Eugène Charles Catalan (1814-1894) (see Pak, 2014). - Amiram Eldar, Apr 15 2021
For n >= 1, a(n-1) is the number of interpretations of x^n is an algebra where power-associativity is not assumed. For example, for n = 4 there are a(3) = 5 interpretations: x(x(xx)), x((xx)x), (xx)(xx), (x(xx))x, ((xx)x)x. See the link "Non-associate powers and a functional equation" from I. M. H. Etherington and the page "Nonassociative Product" from Eric Weisstein's World of Mathematics for detailed information. See also A001190 for the case where multiplication is commutative. - Jianing Song, Apr 29 2022
Number of states in the transition diagram associated with the Laplacian system over the complete graph K_N, corresponding to ordered initial conditions x_1 < x_2 < ... < x_N. - Andrea Arlette España, Nov 06 2022
a(n) is the number of 132-avoiding stabilized-interval-free permutations of size n+1. - Juan B. Gil, Jun 22 2023
Number of rooted polyominoes composed of n triangular cells of the hyperbolic regular tiling with Schläfli symbol {3,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {3,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
a(n) is the number of extremely lucky Stirling permutations of order n; i.e., the number of Stirling permutations of order n that have exactly n lucky cars. (see Colmenarejo et al. reference) - Bridget Tenner, Apr 16 2024

Examples

			From _Joerg Arndt_ and Greg Stevenson, Jul 11 2011: (Start)
The following products of 3 transpositions lead to a 4-cycle in S_4:
(1,2)*(1,3)*(1,4);
(1,2)*(1,4)*(3,4);
(1,3)*(1,4)*(2,3);
(1,4)*(2,3)*(2,4);
(1,4)*(2,4)*(3,4). (End)
G.f. = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + ...
For n=3, a(3)=5 since there are exactly 5 binary sequences of length 7 in which the number of ones first exceed the number of zeros at entry 7, namely, 0001111, 0010111, 0011011, 0100111, and 0101011. - _Dennis P. Walsh_, Apr 11 2012
From _Joerg Arndt_, Jun 30 2014: (Start)
The a(4) = 14 branching sequences of the (ordered) trees with 4 non-root nodes are (dots denote zeros):
01:  [ 1 1 1 1 . ]
02:  [ 1 1 2 . . ]
03:  [ 1 2 . 1 . ]
04:  [ 1 2 1 . . ]
05:  [ 1 3 . . . ]
06:  [ 2 . 1 1 . ]
07:  [ 2 . 2 . . ]
08:  [ 2 1 . 1 . ]
09:  [ 2 1 1 . . ]
10:  [ 2 2 . . . ]
11:  [ 3 . . 1 . ]
12:  [ 3 . 1 . . ]
13:  [ 3 1 . . . ]
14:  [ 4 . . . . ]
(End)

References

  • The large number of references and links demonstrates the ubiquity of the Catalan numbers.
  • R. Alter, Some remarks and results on Catalan numbers, pp. 109-132 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 2, edited R. C. Mullin et al., 1971.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, many references.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 53.
  • J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, 1995, ch. 4, pp. 96-106.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 183, 196, etc.).
  • Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, Non-contiguous pattern avoidance in binary trees. Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227.
  • E. Deutsch, Dyck path enumeration, Discrete Math., 204, 167-202, 1999.
  • E. Deutsch and L. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31, 31-38, 2001.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, 207-208.
  • Tomislav Doslic and Darko Veljan, Logarithmic behavior of some combinatorial sequences. Discrete Math. 308 (2008), no. 11, 2182-2212. MR2404544 (2009j:05019)
  • S. Dulucq and J.-G. Penaud, Cordes, arbres et permutations. Discrete Math. 117 (1993), no. 1-3, 89-105.
  • A. Errera, Analysis situs - Un problème d'énumération, Mémoires Acad. Bruxelles, Classe des sciences, Série 2, Vol. XI, Fasc. 6, No. 1421 (1931), 26 pp.
  • Ehrenfeucht, Andrzej; Haemer, Jeffrey; Haussler, David. Quasimonotonic sequences: theory, algorithms and applications. SIAM J. Algebraic Discrete Methods 8 (1987), no. 3, 410-429. MR0897739 (88h:06026)
  • I. M. H. Etherington, Non-associate powers and a functional equation. The Mathematical Gazette, 21 (1937): 36-39; addendum 21 (1937), 153.
  • I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
  • I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
  • K. Fan, Structure of a Hecke algebra quotient, J. Amer. Math. Soc., 10 (1997), 139-167.
  • Susanna Fishel, Myrto Kallipoliti and Eleni Tzanaki, Facets of the Generalized Cluster Complex and Regions in the Extended Catalan Arrangement of Type A, The electronic Journal of Combinatorics 20(4) (2013), #P7.
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Links

Crossrefs

Cf. A000142, A000245, A000344, A000588, A000957, A000984, A001392, A001453, A001791, A002057, A002420, A003046, A003517, A003518, A003519, A006480, A008276, A008549, A014137, A014138, A014140, A022553 (inv. Eul. trans.), A024492, A032357, A032443, A039599, A048990, A059288, A068875, A069640, A086117, A088327 (Eul. trans.), A094216, A094638, A094639, A098597, A099731, A119822, A120304, A124926, A129763, A137697, A154559, A161581, A167892, A167893, A179277, A211611, A275431 (multisets).
A row of A060854.
See A001003, A001190, A001699, A000081 for other ways to count parentheses.
Enumerates objects encoded by A014486.
A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Cf. A051168 (diagonal of the square array described).
Cf. A033552, A176137 (partitions into Catalan numbers).
Cf. A000753, A000736 (Boustrophedon transforms).
Cf. A120303 (largest prime factor of Catalan number).
Cf. A121839 (reciprocal Catalan constant), A268813.
Cf. A038003, A119861, A119908, A120274, A120275 (odd Catalan number).
Cf. A002390 (decimal expansion of natural logarithm of golden ratio).
Coefficients of square root of the g.f. are A001795/A046161.
For a(n) mod 6 see A259667.
For a(n) in base 2 see A264663.
Hankel transforms with first terms omitted: A001477, A006858, A091962, A078920, A123352, A368025.
Cf. A001147, A163960.
Cf. A332602 (conjectured production matrix).
Polyominoes: A001683(n+2) (oriented), A000207 (unoriented), A369314 (chiral), A208355(n-1) (achiral), A001764 {4,oo}.

Programs

  • GAP
    A000108:=List([0..30],n->Binomial(2*n,n)/(n+1)); # Muniru A Asiru, Feb 17 2018
  • Haskell
    import Data.List (genericIndex)
a000108 n = genericIndex a000108_list n
a000108_list = 1 : catalan [1] where
   catalan cs = c : catalan (c:cs) where
      c = sum $ zipWith (*) cs $ reverse cs
-- Reinhard Zumkeller, Nov 12 2011
a000108 = map last $ iterate (scanl1 (+) . (++ [0])) [1]
-- David Spies, Aug 23 2015
  • Magma
    C:= func< n | Binomial(2*n,n)/(n+1) >; [ C(n) : n in [0..60]];
  • Magma
    [Catalan(n): n in [0..40]]; // Vincenzo Librandi, Apr 02 2011
  • Maple
    A000108 := n->binomial(2*n,n)/(n+1);
G000108 := (1 - sqrt(1 - 4*x)) / (2*x);
spec := [ A, {A=Prod(Z,Sequence(A))}, unlabeled ]: [ seq(combstruct[count](spec, size=n+1), n=0..42) ];
with(combstruct): bin := {B=Union(Z,Prod(B,B))}: seq(count([B,bin,unlabeled],size=n+1), n=0..25); # Zerinvary Lajos, Dec 05 2007
gser := series(G000108, x=0, 42): seq(coeff(gser, x, n), n=0..41); # Zerinvary Lajos, May 21 2008
seq((2*n)!*coeff(series(hypergeom([],[2],x^2),x,2*n+2),x,2*n),n=0..30); # Peter Luschny, Jan 31 2015
A000108List := proc(m) local A, P, n; A := [1, 1]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), A[-1]]);
A := [op(A), P[-1]] od; A end: A000108List(31); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[(2 n)!/n!/(n + 1)!, {n, 0, 20}]
Table[4^n Gamma[n + 1/2]/(Sqrt[Pi] Gamma[n + 2]), {n, 0, 20}] (* Eric W. Weisstein, Oct 31 2024 *)
Table[Hypergeometric2F1[1 - n, -n, 2, 1], {n, 0, 20}] (* Richard L. Ollerton, Sep 13 2006 *)
Table[CatalanNumber @ n, {n, 0, 20}] (* Robert G. Wilson v, Feb 15 2011 *)
CatalanNumber[Range[0, 20]] (* Eric W. Weisstein, Oct 31 2024 *)
CoefficientList[InverseSeries[Series[x/Sum[x^n, {n, 0, 31}], {x, 0, 31}]]/x, x] (* Mats Granvik, Nov 24 2013 *)
CoefficientList[Series[(1 - Sqrt[1 - 4 x])/(2 x), {x, 0, 20}], x] (* Stefano Spezia, Aug 31 2018 *)
  • Maxima
    A000108(n):=binomial(2*n,n)/(n+1)$ makelist(A000108(n),n,0,30); /* Martin Ettl, Oct 24 2012 */
  • MuPAD
    combinat::dyckWords::count(n) $ n = 0..38 // Zerinvary Lajos, Apr 14 2007
  • PARI
    a(n)=binomial(2*n,n)/(n+1) \\ M. F. Hasler, Aug 25 2012
  • PARI
    a(n) = (2*n)! / n! / (n+1)!
  • PARI
    a(n) = my(A, m); if( n<0, 0, m=1; A = 1 + x + O(x^2); while(m<=n, m*=2; A = sqrt(subst(A, x, 4*x^2)); A += (A - 1) / (2*x*A)); polcoeff(A, n));
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( serreverse( x / (1 + x)^2 + x * O(x^n)), n))}; /* Michael Somos */
  • PARI
    (recur(a,b)=if(b<=2,(a==2)+(a==b)+(a!=b)*(1+a/2), (1+a/b)*recur(a,b-1))); a(n)=recur(n,n); \\ R. J. Cano, Nov 22 2012
  • PARI
    x='x+O('x^40); Vec((1-sqrt(1-4*x))/(2*x)) \\ Altug Alkan, Oct 13 2015
  • Python
    from gmpy2 import divexact
A000108 = [1, 1]
for n in range(1, 10**3):
    A000108.append(divexact(A000108[-1]*(4*n+2),(n+2))) # Chai Wah Wu, Aug 31 2014
  • Python
    # Works in Sage also.
A000108 = [1]
for n in range(1000):
    A000108.append(A000108[-1]*(4*n+2)//(n+2)) # Günter Rote, Nov 08 2023
  • Sage
    [catalan_number(i) for i in range(27)] # Zerinvary Lajos, Jun 26 2008
  • Sage
    # Generalized algorithm of L. Seidel
def A000108_list(n) :
    D = [0]*(n+1); D[1] = 1
    b = True; h = 1; R = []
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1; R.append(D[1])
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
    return R
A000108_list(31) # Peter Luschny, Jun 02 2012

Formula

a(n) = binomial(2*n, n)/(n+1) = (2*n)!/(n!*(n+1)!) = A000984(n)/(n+1).
Recurrence: a(n) = 2*(2*n-1)*a(n-1)/(n+1) with a(0) = 1.
Recurrence: a(n) = Sum_{k=0..n-1} a(k)a(n-1-k).
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x), and satisfies A(x) = 1 + x*A(x)^2.
a(n) = Product_{k=2..n} (1 + n/k).
a(n+1) = Sum_{i} binomial(n, 2*i)*2^(n-2*i)*a(i). - Touchard
It is known that a(n) is odd if and only if n=2^k-1, k=0, 1, 2, 3, ... - Emeric Deutsch, Aug 04 2002, corrected by M. F. Hasler, Nov 08 2015
Using the Stirling approximation in A000142 we get the asymptotic expansion a(n) ~ 4^n / (sqrt(Pi * n) * (n + 1)). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 13 2001
Integral representation: a(n) = (1/(2*Pi))*Integral_{x=0..4} x^n*sqrt((4-x)/x). - Karol A. Penson, Apr 12 2001
E.g.f.: exp(2*x)*(I_0(2*x)-I_1(2*x)), where I_n is Bessel function. - Karol A. Penson, Oct 07 2001
a(n) = polygorial(n, 6)/polygorial(n, 3). - Daniel Dockery (peritus(AT)gmail.com), Jun 24 2003
G.f. A(x) satisfies ((A(x) + A(-x)) / 2)^2 = A(4*x^2). - Michael Somos, Jun 27 2003
G.f. A(x) satisfies Sum_{k>=1} k(A(x)-1)^k = Sum_{n>=1} 4^{n-1}*x^n. - Shapiro, Woan, Getu
a(n+m) = Sum_{k} A039599(n, k)*A039599(m, k). - Philippe Deléham, Dec 22 2003
a(n+1) = (1/(n+1))*Sum_{k=0..n} a(n-k)*binomial(2k+1, k+1). - Philippe Deléham, Jan 24 2004
a(n) = Sum_{k>=0} A008313(n, k)^2. - Philippe Deléham, Feb 14 2004
a(m+n+1) = Sum_{k>=0} A039598(m, k)*A039598(n, k). - Philippe Deléham, Feb 15 2004
a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
Sum_{n>=0} 1/a(n) = 2 + 4*Pi/3^(5/2) = F(1,2;1/2;1/4) = A268813 = 2.806133050770763... (see L'Univers de Pi link). - Gerald McGarvey and Benoit Cloitre, Feb 13 2005
a(n) = Sum_{k=0..floor(n/2)} ((n-2*k+1)*binomial(n, n-k)/(n-k+1))^2, which is equivalent to: a(n) = Sum_{k=0..n} A053121(n, k)^2, for n >= 0. - Paul D. Hanna, Apr 23 2005
a((m+n)/2) = Sum_{k>=0} A053121(m, k)*A053121(n, k) if m+n is even. - Philippe Deléham, May 26 2005
E.g.f. Sum_{n>=0} a(n) * x^(2*n) / (2*n)! = BesselI(1, 2*x) / x. - Michael Somos, Jun 22 2005
Given g.f. A(x), then B(x) = x * A(x^3) satisfies 0 = f(x, B(X)) where f(u, v) = u - v + (u*v)^2 or B(x) = x + (x * B(x))^2 which implies B(-B(x)) = -x and also (1 + B^3) / B^2 = (1 - x^3) / x^2. - Michael Somos, Jun 27 2005
a(n) = a(n-1)*(4-6/(n+1)). a(n) = 2a(n-1)*(8a(n-2)+a(n-1))/(10a(n-2)-a(n-1)). - Franklin T. Adams-Watters, Feb 08 2006
Sum_{k>=1} a(k)/4^k = 1. - Franklin T. Adams-Watters, Jun 28 2006
a(n) = A047996(2*n+1, n). - Philippe Deléham, Jul 25 2006
Binomial transform of A005043. - Philippe Deléham, Oct 20 2006
a(n) = Sum_{k=0..n} (-1)^k*A116395(n,k). - Philippe Deléham, Nov 07 2006
a(n) = (1/(s-n))*Sum_{k=0..n} (-1)^k (k+s-n)*binomial(s-n,k) * binomial(s+n-k,s) with s a nonnegative free integer [H. W. Gould].
a(k) = Sum_{i=1..k} |A008276(i,k)| * (k-1)^(k-i) / k!. - André F. Labossière, May 29 2007
a(n) = Sum_{k=0..n} A129818(n,k) * A007852(k+1). - Philippe Deléham, Jun 20 2007
a(n) = Sum_{k=0..n} A109466(n,k) * A127632(k). - Philippe Deléham, Jun 20 2007
Row sums of triangle A124926. - Gary W. Adamson, Oct 22 2007
Limit_{n->oo} (1 + Sum_{k=0..n} a(k)/A004171(k)) = 4/Pi. - Reinhard Zumkeller, Aug 26 2008
a(n) = Sum_{k=0..n} A120730(n,k)^2 and a(k+1) = Sum_{n>=k} A120730(n,k). - Philippe Deléham, Oct 18 2008
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, the present sequence is Phi([1]) (also Phi([1,1])). - Gary W. Adamson, Oct 27 2008
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}...Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n) where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i < l_(i+1) and l_(i+1) <> 0 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
a(n) = A000680(n)/A006472(n+1). - Mark Dols, Jul 14 2010; corrected by M. F. Hasler, Nov 08 2015
Let A(x) be the g.f., then B(x)=x*A(x) satisfies the differential equation B'(x)-2*B'(x)*B(x)-1=0. - Vladimir Kruchinin, Jan 18 2011
Complement of A092459; A010058(a(n)) = 1. - Reinhard Zumkeller, Mar 29 2011
G.f.: 1/(1-x/(1-x/(1-x/(...)))) (continued fraction). - Joerg Arndt, Mar 18 2011
With F(x) = (1-2*x-sqrt(1-4*x))/(2*x) an o.g.f. in x for the Catalan series, G(x) = x/(1+x)^2 is the compositional inverse of F (nulling the n=0 term). - Tom Copeland, Sep 04 2011
With H(x) = 1/(dG(x)/dx) = (1+x)^3 / (1-x), the n-th Catalan number is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)), and H(x) is the o.g.f. for A115291. - Tom Copeland, Sep 04 2011
From Tom Copeland, Sep 30 2011: (Start)
With F(x) = (1-sqrt(1-4*x))/2 an o.g.f. in x for the Catalan series, G(x)= x*(1-x) is the compositional inverse and this relates the Catalan numbers to the row sums of A125181.
With H(x) = 1/(dG(x)/dx) = 1/(1-2x), the n-th Catalan number (offset 1) is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)). (End)
G.f.: (1-sqrt(1-4*x))/(2*x) = G(0) where G(k) = 1 + (4*k+1)*x/(k+1-2*x*(k+1)*(4*k+3)/(2*x*(4*k+3)+(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: exp(2*x)*(BesselI(0,2*x) - BesselI(1,2*x)) = G(0) where G(k) = 1 + (4*k+1)*x/((k+1)*(2*k+1)-x*(k+1)*(2*k+1)*(4*k+3)/(x*(4*k+3)+(k+1)*(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: Hypergeometric([1/2],[2],4*x) which coincides with the e.g.f. given just above, and also by Karol A. Penson further above. - Wolfdieter Lang, Jan 13 2012
A076050(a(n)) = n + 1 for n > 0. - Reinhard Zumkeller, Feb 17 2012
a(n) = A208355(2*n-1) = A208355(2*n) for n > 0. - Reinhard Zumkeller, Mar 04 2012
a(n+1) = A214292(2*n+1,n) = A214292(2*n+2,n). - Reinhard Zumkeller, Jul 12 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = k*(4*x+1) + 2*x + 2 - x*(2*k+3)*(2*k+4)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 20 2012
G.f.: hypergeom([1/2,1],[2],4*x). - Joerg Arndt, Apr 06 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,1,-1/2-n,-1)/(n+1). - Karol A. Penson, Jul 28 2013
For n > 0: a(n) = sum of row n in triangle A001263. - Reinhard Zumkeller, Oct 10 2013
a(n) = binomial(2n,n-1)/n and a(n) mod n = binomial(2n,n) mod n = A059288(n). - Jonathan Sondow, Dec 14 2013
a(n-1) = Sum_{t1+2*t2+...+n*tn=n} (-1)^(1+t1+t2+...+tn)*multinomial(t1+t2 +...+tn,t1,t2,...,tn)*a(1)^t1*a(2)^t2*...*a(n)^tn. - Mircea Merca, Feb 27 2014
a(n) = Sum_{k=1..n} binomial(n+k-1,n)/n if n > 0. Alexander Adamchuk, Mar 25 2014
a(n) = -2^(2*n+1) * binomial(n-1/2, -3/2). - Peter Luschny, May 06 2014
a(n) = (4*A000984(n) - A000984(n+1))/2. - Stanislav Sykora, Aug 09 2014
a(n) = A246458(n) * A246466(n). - Tom Edgar, Sep 02 2014
a(n) = (2*n)!*[x^(2*n)]hypergeom([],[2],x^2). - Peter Luschny, Jan 31 2015
a(n) = 4^(n-1)*hypergeom([3/2, 1-n], [3], 1). - Peter Luschny, Feb 03 2015
a(2n) = 2*A000150(2n); a(2n+1) = 2*A000150(2n+1) + a(n). - John Bodeen, Jun 24 2015
a(n) = Sum_{t=1..n+1} n^(t-1)*abs(Stirling1(n+1, t)) / Sum_{t=1..n+1} abs(Stirling1(n+1, t)), for n > 0, see (10) in Cereceda link. - Michel Marcus, Oct 06 2015
a(n) ~ 4^(n-2)*(128 + 160/N^2 + 84/N^4 + 715/N^6 - 10180/N^8)/(N^(3/2)*Pi^(1/2)) where N = 4*n+3. - Peter Luschny, Oct 14 2015
a(n) = Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*binomial(n+1-k,k)*a(n-k) if n > 0; and a(0) = 1. - David Pasino, Jun 29 2016
Sum_{n>=0} (-1)^n/a(n) = 14/25 - 24*arccsch(2)/(25*sqrt(5)) = 14/25 - 24*A002390/(25*sqrt(5)) = 0.353403708337278061333... - Ilya Gutkovskiy, Jun 30 2016
C(n) = (1/n) * Sum_{i+j+k=n-1} C(i)*C(j)*C(k)*(k+1), n >= 1. - Yuchun Ji, Feb 21 2016
C(n) = 1 + Sum_{i+j+kYuchun Ji, Sep 01 2016
a(n) = A001700(n) - A162551(n) = binomial(2*n+1,n+1). - 2*binomial(2*n,n-1). - Taras Goy, Aug 09 2018
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x) = 2F1(1/2,1;2;4*x). G.f. A(x) satisfies A = 1 + x*A^2. - R. J. Mathar, Nov 17 2018
C(n) = 1 + Sum_{i=0..n-1} A000245(i). - Yuchun Ji, Jan 10 2019
From A.H.M. Smeets, Apr 11 2020: (Start)
(1+sqrt(1+4*x))/2 = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4; and sqrt(x+sqrt(x+sqrt(x+...))) = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4 and x <> 0. (End)
a(3n+1)*a(5n+4)*a(15n+10) = a(3n+2)*a(5n+2)*a(15n+11). The first case of Catalan product equation of a triple partition of 23n+15. - Yuchun Ji, Sep 27 2020
a(n) = 4^n * (-1)^(n+1) * 3F2[{n + 1,n + 1/2,n}, {3/2,1}, -1], n >= 1. - Sergii Voloshyn, Oct 22 2020
a(n) = 2^(1 + 2 n) * (-1)^(n)/(1 + n) * 3F2[{n, 1/2 + n, 1 + n}, {1/2, 1}, -1], n >= 1. - Sergii Voloshyn, Nov 08 2020
a(n) = (1/Pi)*4^(n+1)*Integral_{x=0..Pi/2} cos(x)^(2*n)*sin(x)^2 dx. - Greg Dresden, May 30 2021
From Peter Bala, Aug 17 2021: (Start)
G.f. A(x) satisfies A(x) = 1/sqrt(1 - 4*x) * A( -x/(1 - 4*x) ) and (A(x) + A(-x))/2 = 1/sqrt(1 - 4*x) * A( -2*x/(1 - 4*x) ); these are the cases k = 0 and k = -1 of the general formula 1/sqrt(1 - 4*x) * A( (k-1)*x/(1 - 4*x) ) = Sum_{n >= 0} ((k^(n+1) - 1)/(k - 1))*Catalan(n)*x^n.
2 - sqrt(1 - 4*x)/A( k*x/(1 - 4*x) ) = 1 + Sum_{n >= 1} (1 + (k + 1)^n) * Catalan(n-1)*x^n. (End)
Sum_{n>=0} a(n)*(-1/4)^n = 2*(sqrt(2)-1) (A163960). - Amiram Eldar, Mar 22 2022
0 = a(n)*(16*a(n+1) - 10*a(n+2)) + a(n+1)*(2*a(n+1) + a(n+2)) for all n>=0. - Michael Somos, Dec 12 2022
G.f.: (offset 1) 1/G(x), with G(x) = 1 - 2*x - x^2/G(x) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 01 2023
a(n) = K^(2n+1, n, 1) for all n >= 0, where K^(n, s, x) is the Krawtchouk polynomial defined to be Sum_{k=0..s} (-1)^k * binomial(n-x, s-k) * binomial(x, k). - Vladislav Shubin, Aug 17 2023
From Peter Bala, Feb 03 2024: (Start)
The g.f. A(x) satisfies the following functional equations:
A(x) = 1 + x/(1 - 4*x) * A(-x/(1 - 4*x))^2,
A(x^2) = 1/(1 - 2*x) * A(- x/(1 - 2*x))^2 and, for arbitrary k,
1/(1 - k*x) * A(x/(1 - k*x))^2 = 1/(1 - (k+4)*x) * A(-x/(1 - (k+4)*x))^2. (End)
a(n) = A363448(n) + A363449(n). - Julien Rouyer, Jun 28 2024

A001813 Quadruple factorial numbers: a(n) = (2n)!/n!.

Original entry on oeis.org

1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 670442572800, 28158588057600, 1295295050649600, 64764752532480000, 3497296636753920000, 202843204931727360000, 12576278705767096320000, 830034394580628357120000, 58102407620643984998400000
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Counts binary rooted trees (with out-degree <= 2), embedded in plane, with n labeled end nodes of degree 1. Unlabeled version gives Catalan numbers A000108.
Define a "downgrade" to be the permutation which places the items of a permutation in descending order. We are concerned with permutations that are identical to their downgrades. Only permutations of order 4n and 4n+1 can have this property; the number of permutations of length 4n having this property are equinumerous with those of length 4n+1. If a permutation p has this property then the reversal of this permutation also has it. a(n) = number of permutations of length 4n and 4n+1 that are identical to their downgrades. - Eugene McDonnell (eemcd(AT)mac.com), Oct 26 2003
Number of broadcast schemes in the complete graph on n+1 vertices, K_{n+1}. - Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Hankel transform is A137565. - Paul Barry, Nov 25 2009
The e.g.f. of 1/a(n) = n!/(2*n)! is (exp(sqrt(x)) + exp(-sqrt(x)) )/2. - Wolfdieter Lang, Jan 09 2012
From Tom Copeland, Nov 15 2014: (Start)
Aerated with intervening zeros (1,0,2,0,12,0,120,...) = a(n) (cf. A123023 and A001147), the e.g.f. is e^(t^2), so this is the base for the Appell sequence with e.g.f. e^(t^2) e^(x*t) = exp(P(.,x),t) (reverse A059344, cf. A099174, A066325 also). P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for e^(-t^2)e^(x*t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), e.g., (P(.,t))^n = P(n,t).
Equals A000407*2 with leading 1 added. (End)
a(n) is also the number of square roots of any permutation in S_{4*n} whose disjoint cycle decomposition consists of 2*n transpositions. - Luis Manuel Rivera Martínez, Mar 04 2015
Self-convolution gives A076729. - Vladimir Reshetnikov, Oct 11 2016
For n > 1, it follows from the formula dated Aug 07 2013 that a(n) is a Zumkeller number (A083207). - Ivan N. Ianakiev, Feb 28 2017
For n divisible by 4, a(n/4) is the number of ways to place n points on an n X n grid with pairwise distinct abscissae, pairwise distinct ordinates, and 90-degree rotational symmetry. For n == 1 (mod 4), the number of ways is a((n-1)/4) because the center point can be considered "fixed". For 180-degree rotational symmetry see A006882, for mirror symmetry see A000085, A135401, and A297708. - Manfred Scheucher, Dec 29 2017

Examples

			The following permutations of order 8 and their reversals have this property:
  1 7 3 5 2 4 0 6
  1 7 4 2 5 3 0 6
  2 3 7 6 1 0 4 5
  2 4 7 1 6 0 3 5
  3 2 6 7 0 1 5 4
  3 5 1 7 0 6 2 4

References

  • D. E. Knuth, The Art of Computer Programming, Vol. 4, Section 7.2.1.6, Eq. 32.
  • L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181.
  • Eugene McDonnell, "Magic Squares and Permutations" APL Quote-Quad 7.3 (Fall, 1976)
  • R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
  • 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).

Links

Crossrefs

Cf. A037224, A048854, A001147, A007696, A008545, A122670 (essentially the same sequence), A000165, A047055, A047657, A084947, A084948, A084949, A010050, A000142, A008275, A000108, A000984, A008276, A000680, A094216.
Catalan(n-1)*M^(n-1)*n! for M=1,2,3,4,5,6: A001813, A052714 (or A144828), A221954, A052734, A221953, A221955.
Cf. A123023, A059344, A099174, A066325, A001700, A000407, A006882.
Cf. A000085, A135401, A297708. - Manfred Scheucher, Jan 07 2018

Programs

  • GAP
    List([0..20],n->Factorial(2*n)/Factorial(n)); # Muniru A Asiru, Nov 01 2018
  • Magma
    [Factorial(2*n)/Factorial(n): n in [0..20]]; // Vincenzo Librandi, Oct 09 2018
  • Maple
    A001813 := n->(2*n)!/n!;
A001813 := n -> mul(k, k = select(k-> k mod 4 = 2,[$1 .. 4*n])):
seq(A001813(n), n=0..16);  # Peter Luschny, Jun 23 2011
  • Mathematica
    Table[(2n)!/n!, {n,0,20}] (* Harvey P. Dale, May 02 2011 *)
  • Maxima
    makelist(binomial(n+n, n)*n!,n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    a(n)=binomial(n+n,n)*n! \\ Charles R Greathouse IV, Jun 15 2011
  • PARI
    first(n) = x='x+O('x^n); Vec(serlaplace((1 - 4*x)^(-1/2))) \\ Iain Fox, Jan 01 2018 (corrected by Iain Fox, Jan 11 2018)
  • Python
    from math import factorial
def A001813(n): return factorial(n<<1)//factorial(n) # Chai Wah Wu, Feb 14 2023
  • Sage
    [binomial(2*n,n)*factorial(n) for n in range(0, 17)] # Zerinvary Lajos, Dec 03 2009

Formula

E.g.f.: (1-4*x)^(-1/2).
a(n) = (2*n)!/n! = Product_{k=0..n-1} (4*k + 2) = A081125(2*n).
Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/4)/(sqrt(x)*2*sqrt(Pi)) dx, n >= 0. This representation is unique. - Karol A. Penson, Sep 18 2001
Define a'(1)=1, a'(n) = Sum_{k=1..n-1} a'(n-k)*a'(k)*C(n, k); then a(n)=a'(n+1). - Benoit Cloitre, Apr 27 2003
With interpolated zeros (1, 0, 2, 0, 12, ...) this has e.g.f. exp(x^2). - Paul Barry, May 09 2003
a(n) = A000680(n)/A000142(n)*A000079(n) = Product_{i=0..n-1} (4*i + 2) = 4^n*Pochhammer(1/2, n) = 4^n*GAMMA(n+1/2)/sqrt(Pi). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
For asymptotics, see the Robinson paper.
a(k) = (2*k)!/k! = Sum_{i=1..k+1} |A008275(i,k+1)| * k^(i-1). - André F. Labossière, Jun 21 2007
a(n) = 12*A051618(a) n >= 2. - Zerinvary Lajos, Feb 15 2008
a(n) = A000984(n)*A000142(n). - Zerinvary Lajos, Mar 25 2008
a(n) = A016825(n-1)*a(n-1). - Roger L. Bagula, Sep 17 2008
a(n) = (-1)^n*A097388(n). - D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
From Paul Barry, Jan 15 2009: (Start)
G.f.: 1/(1-2x/(1-4x/(1-6x/(1-8x/(1-10x/(1-... (continued fraction);
a(n) = (n+1)!*A000108(n). (End)
a(n) = Sum_{k=0..n} A132393(n,k)*2^(2n-k). - Philippe Deléham, Feb 10 2009
G.f.: 1/(1-2x-8x^2/(1-10x-48x^2/(1-18x-120x^2/(1-26x-224x^2/(1-34x-360x^2/(1-42x-528x^2/(1-... (continued fraction). - Paul Barry, Nov 25 2009
a(n) = A173333(2*n,n) for n>0; cf. A006963, A001761. - Reinhard Zumkeller, Feb 19 2010
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n, M = an infinite square production matrix as follows:
2, 2, 0, 0, 0, 0, ...
4, 4, 4, 0, 0, 0, ...
6, 6, 6, 6, 0, 0, ...
8, 8, 8, 8, 8, 0, ...
...
(End)
a(n) = (-2)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0), where Q(k) = 1 + x*(4*k+2) - x*(4*k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 18 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(8*k+4)/(x*(8*k+4) - 1 + 8*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x/(2*x + 1/(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
D-finite with recurrence: a(n) = (4*n-6)*a(n-2) + (4*n-3)*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 07 2013
Sum_{n>=0} 1/a(n) = (exp(1/4)*sqrt(Pi)*erf(1/2) + 2)/2 = 1 + A214869, where erf(x) is the error function. - Ilya Gutkovskiy, Nov 10 2016
Sum_{n>=0} (-1)^n/a(n) = 1 - sqrt(Pi)*erfi(1/2)/(2*exp(1/4)), where erfi(x) is the imaginary error function. - Amiram Eldar, Feb 20 2021
a(n) = 1/([x^n] hypergeom([1], [1/2], x/4)). - Peter Luschny, Sep 13 2024
a(n) = 2^n*n!*JacobiP(n, -1/2, -n, 3). - Peter Luschny, Jan 22 2025
G.f.: 2F0(1,1/2;;4x). - R. J. Mathar, Jun 07 2025

Extensions

More terms from James Sellers, May 01 2000

A000407 a(n) = (2*n+1)! / n!.

Original entry on oeis.org

1, 6, 60, 840, 15120, 332640, 8648640, 259459200, 8821612800, 335221286400, 14079294028800, 647647525324800, 32382376266240000, 1748648318376960000, 101421602465863680000, 6288139352883548160000, 415017197290314178560000
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The e.g.f. of 1/a(n) = n!/(2*n+1)! is (exp(sqrt(x)) - exp(-sqrt(x)))/(2*sqrt(x)). - Wolfdieter Lang, Jan 09 2012
Product of the larger parts of the partitions of 2n+2 into exactly two parts. - Wesley Ivan Hurt, Jun 15 2013
For n > 0, a(n-1) = (2n-1)!/(n-1)!, the number of ways n people can line up in n labeled queues. The derivation is straightforward. Person 1 has (2n-1) choices - be first in line in one of the queues or get behind one of the other people. Person 2 has (2n-2) choices - choose one of the n queues or get behind one of the remaining n-2 people. Continuing in this fashion, we finally find that person n has to choose one of the n queues. - Dennis P. Walsh, Mar 24 2016
For n > 0, a(n-1) is the number of functions f:[n]->[2n] that are acyclic and injective. Note that f is acyclic if, for all x in [n], x is not a member of the set {f(x),f(f(x)), f(f(f(x))), ...}. - Dennis P. Walsh, Mar 25 2016
a(n) is the number of labeled maximal outerplanar graphs with n-3 vertices. - Allan Bickle, Feb 19 2024

Examples

			G.f. = 1 + 6*x + 60*x^2 + 840*x^3 + 15120*x^4 + 332640*x^5 + 8648640*x^6 + ...
For n=1 the a(1)=6 ways for 2 people to line up in 2 queues are as follows: Q1<P1,P2> Q2<>, Q1<P2,P1> Q2<>, Q1<P1> Q2<P2>, Q1<P2> Q2<P1>, Q1<> Q2<P1,P2>, Q1<> Q2<P2,P1>. - _Dennis P. Walsh_, Mar 24 2016
For the unique maximal outerplanar graph with 4 vertices, there are C(4,2)=6 ways to label the two degree 3 vertices, and the other two labels are forced.  Thus a(1) = 6.

References

  • L. W. Beineke and R. E. Pippert, Enumerating labeled k-dimensional trees and ball dissections, pp. 12-26 of Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and Its Applications, University of North Carolina, Chapel Hill, 1970. Reprinted with a slightly different title in Math. Annalen, 191 (1971), 87-98.
  • L. B. W. Jolley, Summation of Series, Dover, 1961.
  • Loren C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181.
  • 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).

Links

Crossrefs

Cf. A001761-A001763, A007696, A024199, A073010.
A100622 is the "Number of topologically distinct solutions to the clone ordering problem for n clones" without the restriction that they be in a single contig (see [Newberg] for definition of contig).
Column m=0 of A292219.

Programs

  • Magma
    [Factorial(2*n+1) / Factorial(n): n in [0..20]]; // Vincenzo Librandi, Jun 16 2015
  • Maple
    For Maple program see A000903.
a := n -> pochhammer(n+1,n+1); (for n>=0) # Peter Luschny, Feb 14 2009
  • Mathematica
    Table[(2n + 1)!/n!, {n, 0, 30}] (* Stefan Steinerberger, Apr 08 2006 *)
a[ n_] := If[ n < 0, 1/2, 1] Pochhammer[ n + 1, n + 1]; (* Michael Somos, Jan 03 2015 *)
a[ n_] := Which[ n < -1, -(-1)^n / (4 a[-n - 2]), n == -1, 1/2, True, (2 n + 1)! / n!]; (* Michael Somos, Jan 03 2015 *)
  • Maxima
    A000407(n):=(2*n+1)!/n!$
makelist(A000407(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    a(n)=(2*n+1)!/n! \\ Charles R Greathouse IV, Jan 12 2012
  • PARI
    {a(n) = if( n<-1, -(-1)^n / (4 * a(-n-2)), n==-1, 1/2, (2*n + 1)! / n!)}; /* Michael Somos, Jan 03 2015 */

Formula

E.g.f.: (1 - 4*x)^(-3/2). - Michael Somos, Jan 03 2015
E.g.f.: Sum_{k>=0} a(k+2) * x^k / k! = (1 - 2*x - sqrt(1 - 4*x)) / 4.
E.g.f. for a(n-1), n >= 0, with a(-1) := 0 is (-1+1/(1-4*x)^(1/2))/2. 2*a(n) = (4*n+2)(!^4) := Product_{j=0..n} (4*j + 2), (one half of 4-factorial numbers). - Wolfdieter Lang
a(n) = C(n+1)*(n+2)!/2 for all n>=0. - Paul Barry, Feb 16 2005
For n>1, a(n) = (1/2)*A001813(n+1). - Zerinvary Lajos, Jun 06 2007
For asymptotics see the Robinson paper.
Sum_{n >=0} n!/a(n) = 2*Pi/3^(3/2) = 1.2091995761... = A248897 [Jolley eq 261]
G.f.: 1 / (1 - 6*x / (1 - 4*x / (1 - 10*x / (1 - 8*x / (1 - 14*x / ... ))))). - Michael Somos, May 12 2012
G.f.: 1/Q(0), where Q(k) = 1 + 2*(2*k-1)*x - 4*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x/(2*x + 1/(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 02 2013
a(n) = -(-1)^n / (4 * a(-2-n)) = a(n-1) * (4*n+2) for all n in Z. - Michael Somos, Jan 03 2015
a(n) = A087299(2*n + 1). - Michael Somos, Jan 03 2015
From Peter Bala, Feb 16 2015: (Start)
Recurrence equation: a(n) = 4*a(n-1) + 4*(2*n - 1)^2*a(n-2) with a(0) = 1 and a(1) = 6.
The integer sequence b(n) := a(n)*Sum_{k = 0..n} (-1)^k/(2*k + 1), beginning [1, 4, 52, 608, 12624, ...], satisfies the same second-order recurrence equation. This leads to Brouncker's generalized continued fraction expansion Sum_{k >= 0} (-1)^k/(2*k + 1) = Pi/4 = 1/(1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... )))). Note b(n) = 2^n*A024199(n+1).
Recurrence equation: a(n) = (5*n + 2)*a(n-1) - 2*n*(2*n - 1)^2*a(n-2) with a(0) = 1 and a(1) = 6.
The integer sequence c(n) := a(n)*Sum_{k = 0..n} k!^2/(2*k + 1)!, beginning [1, 7, 72, 1014, 18276, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion Sum_{k >= 0} k!^2/(2*k + 1)! = 2*Pi/sqrt(27) = 2*A073010 = 1/(1 - 1/(7 - 12/(12 - 30/(17 - ... - 2*n*(2*n - 1)/((5*n + 2) - ... ))))). (End)
a(n) = Product_{k=n+1..(2*n+1)} k. - Carlos Eduardo Olivieri, Jun 03 2015
From Ilya Gutkovskiy, Jan 17 2017: (Start)
a(n) ~ 2^(2*n+3/2)*n^(n+1)/exp(n).
Sum_{n>=0} 1/a(n) = exp(1/4)*sqrt(Pi)*erf(1/2) = 1.184593072938653151..., where erf() is the error function. (End)
Sum_{n>=0} (-1)^n/a(n) = exp(-1/4)*sqrt(Pi)*erfi(1/2), where erfi() is the imaginary error function. - Amiram Eldar, Jan 18 2021
It follows from the comments above that we have a(n) = a(n-1)*(4*n+2), with a(1) = 6, a(0) = 1.
a(n) = A081125(2*n+1). - R. J. Mathar, Jun 07 2025

A006963 Number of planar embedded labeled trees with n nodes: (2*n-3)!/(n-1)! for n >= 2, a(1) = 1.

Original entry on oeis.org

1, 1, 3, 20, 210, 3024, 55440, 1235520, 32432400, 980179200, 33522128640, 1279935820800, 53970627110400, 2490952020480000, 124903451312640000, 6761440164390912000, 393008709555221760000, 24412776311194951680000, 1613955767240110694400000
Offset: 1

Views

Author

Simon Plouffe and N. J. A. Sloane

Keywords

Comments

For n>1: central terms of the triangle in A173333; cf. A001761, A001813. - Reinhard Zumkeller, Feb 19 2010
Can be obtained from the Vandermonde permanent of the first n positive integers; see A093883. - Clark Kimberling, Jan 02 2012
All trees can be embedded in the plane, but "planar embedded" means that orientation matters but rotation doesn't. For example, the n-star with n-1 edges has n! ways to label it, but rotation removes a factor of n-1. Another example, the n-path has n! ways to label it, but rotation removes a factor of 2. - Michael Somos, Aug 19 2014

Examples

			G.f. = x + x^2 + 3*x^3 + 20*x^4 + 210*x^5 + 3024*x^6 + 55440*x^7 + 1235520*x^8 + ...
a(5) = 210 = 30 + 60 + 120 where 30 is for the star, 60 for the path, and 120 for the tree with one trivalent vertex. - _Michael Somos_, Aug 19 2014

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000407, A001761, A001813, A173333.

Programs

  • Magma
    [1] cat [Factorial(2*n-3)/Factorial(n-1): n in [2..20]]; // Vincenzo Librandi, Nov 12 2011
  • Maple
    1, seq((2*n-3)!/(n-1)!,n=2..30); # Robert Israel, Aug 20 2014
  • Mathematica
    Join[{1},Table[(2n-3)!/(n-1)!,{n,2,20}]] (* Harvey P. Dale, Nov 03 2011 *)
a[ n_] := With[{m = n - 1}, If[m < 1, Boole[m == 0], m! SeriesCoefficient[ -Log[(1 + Sqrt[1 - 4 x]) / 2], {x, 0, m}]]] (* Michael Somos, Jul 01 2013 *)
a[ n_] := If[n < 2, Boole[n == 1], (2 n - 3)! / (n - 1)!]; (* Michael Somos, Aug 19 2014 *)
a[1] := 1; a[n_] := (-1)^(n - 1)*Sum[(-1)^k*Binomial[2*n - 3, n + k - 2]*StirlingS1[n + k - 1, k + 1], {k, 1, n - 1}]; Flatten[Table[a[n], {n, 1, 19}]] (* Detlef Meya, Jan 18 2024 *)
  • PARI
    {a(n) = n--; if( n<1, n==0, n! * polcoeff( -log( (1 + sqrt(1 - 4*x + x * O(x^n))) / 2), n))}; /* Michael Somos, Jul 01 2013 */
  • SageMath
    def A006963(n): return 1 if n==1 else factorial(2*n-3)/factorial(n-1)
[A006963(n) for n in range(1,31)] # G. C. Greubel, May 23 2023

Formula

E.g.f. for a(n+1), n >= 1, log(c(x)); c(x) = g.f. for Catalan numbers A000108. - Wolfdieter Lang
Integral representation as n-th moment of a positive function on a positive half-axis, in Maple notation: a(n) = int(x^n * erfc(sqrt(x)/2)/2, x=0..infinity), n=0, 1..., where erfc(x) is the complementary error function. - Karol A. Penson, Sep 27 2001
a(n) ~ 2^(-5/2)*n^-2*2^(2*n)*e^-n*n^n. - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
a(n+1) = (n+1)*(n+2)*...*(2n-1) for n>=2. - Jaroslav Krizek, Nov 09 2010
E.g.f. (A(x)-1) is reversion of exp(-x)-exp(-2*x). - Vladimir Kruchinin, Jan 30 2012
G.f.: 1 + x*G(0) where G(k) = 1 + x*(2*k+1)*(4*k+3)/(k + 1 - 4*x*(k+1)^2*(4*k+5)/(4*x*(k+1)*(4*k+5) + (2*k+3)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 02 2013
E.g.f.: 1 + x*E(0) where E(k) = 1 + x*(2*k+1)*(4*k+3)/(2*(k + 1)^2 - 8*x*(k+1)^3*(4*k+5)/(4*x*(k+1)*(4*k+5) + (2*k+3)^2/E(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 02 2013
E.g.f: sqrt(1-4*x)/4 - 1/4 + 3*x/2 - x*log((1+sqrt(1-4*x))/2). - Robert Israel, Aug 20 2014
D-finite with recurrence (-n+1)*a(n) +2*(2*n-3)*(n-2)*a(n-1)=0. - R. J. Mathar, Jan 03 2018
From Amiram Eldar, Apr 03 2022: (Start)
Sum_{n>=1} 1/a(n) = 3/2 + 3*exp(1/4)*sqrt(Pi)*erf(1/2)/4, where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/2 - sqrt(Pi)*erfi(1/2)/(4*exp(1/4)), where erfi is the imaginary error function. (End)
a(n) = A000407(n-2)/(n-1). - R. J. Mathar, Mar 30 2023
a(1) = 1; a(n) = (-1)^(n - 1)*Sum_{k=1..n - 1} (-1)^k*binomial(2*n - 3, n + k - 2)*Stirling1(n + k - 1, k + 1). - Detlef Meya, Jan 18 2024

A173333 Triangle read by rows: T(n, k) = n! / k!, 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 24, 12, 4, 1, 120, 60, 20, 5, 1, 720, 360, 120, 30, 6, 1, 5040, 2520, 840, 210, 42, 7, 1, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 19 2010

Keywords

Comments

From Wolfdieter Lang, Jun 27 2012: (Start)
T(n-1,k), k=1,...,n-1, gives the number of representative necklaces with n beads (C_N symmetry) of n+1-k distinct colors, say c[1],c[2],...,c[n-k+1], corresponding to the color signature determined by the partition k,1^(n-k) of n. The representative necklaces have k beads of color c[1]. E.g., n=4, k=2: partition 2,1,1, color signature (parts as exponents) c[1]c[1]c[2]c[3], 3=T(3,2) necklaces (write j for color c[j]): cyclic(1123), cyclic(1132) and cyclic(1213). See A212359 for the numbers for general partitions or color signatures. (End)

Examples

			Triangle starts:
n\k      1       2      3      4     5    6   7  8  9 10 ...
1        1
2        2       1
3        6       3      1
4       24      12      4      1
5      120      60     20      5     1
6      720     360    120     30     6    1
7     5040    2520    840    210    42    7   1
8    40320   20160   6720   1680   336   56   8  1
9   362880  181440  60480  15120  3024  504  72  9  1
10 3628800 1814400 604800 151200 30240 5040 720 90 10  1
... - _Wolfdieter Lang_, Jun 27 2012

Links

Crossrefs

Row sums give A002627.
Central terms give A006963:
T(2*n-1,n) = A006963(n+1).
T(2*n,n) = A001813(n).
T(2*n,n+1) = A001761(n).
1 < k <= n: T(n,k) = T(n,k-1) / k.
1 <= k <= n: T(n+1,k) = A119741(n,n-k+1).
1 <= k <= n: T(n+1,k+1) = A162995(n,k).
T(n,1) = A000142(n).
T(n,2) = A001710(n) for n>1.
T(n,3) = A001715(n) for n>2.
T(n,4) = A001720(n) for n>3.
T(n,5) = A001725(n) for n>4.
T(n,6) = A001730(n) for n>5.
T(n,7) = A049388(n-7) for n>6.
T(n,8) = A049389(n-8) for n>7.
T(n,9) = A049398(n-9) for n>8.
T(n,10) = A051431(n) for n>9.
T(n,n-7) = A159083(n+1) for n>7.
T(n,n-6) = A053625(n+1) for n>6.
T(n,n-5) = A052787(n) for n>5.
T(n,n-4) = A052762(n) for n>4.
T(n,n-3) = A007531(n) for n>3.
T(n,n-2) = A002378(n-1) for n>2.
T(n,n-1) = A000027(n) for n>1.
T(n,n) = A000012(n).
Cf. A138533, A002627.

Programs

  • Haskell
    a173333 n k = a173333_tabl !! (n-1) !! (k-1)
a173333_row n = a173333_tabl !! (n-1)
a173333_tabl = map fst $ iterate f ([1], 2)
   where f (row, i) = (map (* i) row ++ [1], i + 1)
-- Reinhard Zumkeller, Jul 04 2012
  • Mathematica
    Table[n!/k!, {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 01 2019 *)

Formula

E.g.f.: (exp(x*y) - 1)/(x*(1 - y)). - Olivier Gérard, Jul 07 2011
T(n,k) = A094587(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

A132382 Lower triangular array T(n,k) generator for group of arrays related to A001147 and A102625.

Original entry on oeis.org

1, -1, 1, -1, -2, 1, -3, -3, -3, 1, -15, -12, -6, -4, 1, -105, -75, -30, -10, -5, 1, -945, -630, -225, -60, -15, -6, 1, -10395, -6615, -2205, -525, -105, -21, -7, 1, -135135, -83160, -26460, -5880, -1050, -168, -28, -8, 1, -2027025, -1216215, -374220, -79380, -13230, -1890, -252, -36, -9, 1
Offset: 0

Views

Author

Tom Copeland, Nov 11 2007, Nov 12 2007, Nov 19 2007, Dec 04 2007, Dec 06 2007

Keywords

Comments

Let b(n) = LPT[ A001147 ] = -A001147(n-1) for n > 0 and 1 for n=0, where LPT represents the action of the list partition transform described in A133314.
Then T(n,k) = binomial(n,k) * b(n-k) .
Form the matrix of polynomials TB(n,k,t) = T(n,k) * t^(n-k) = binomial(n,k) * b(n-k) * t^(n-k) = binomial(n,k) * Pb(n-k,t),
beginning as
1;
-1, 1;
-1*t, -2, 1;
-3*t^2, -3*t, -3, 1;
-15*t^3, -12*t^2, -6*t, -4, 1;
-105*t^4, -75*t^3, -30*t^2, -10*t, -5, 1;
Let Pc(n,t) = LPT(Pb(.,t)).
Then [TB(t)]^(-1) = TC(t) = [ binomial(n,k) * Pc(n-k,t) ] = LPT(TB),
whose first column is
Pc(0,t) = 1
Pc(1,t) = 1
Pc(2,t) = 2 + t
Pc(3,t) = 6 + 6*t + 3*t^2
Pc(4,t) = 24 + 36*t + 30*t^2 + 15*t^3
Pc(5,t) = 120 + 240*t + 270*t^2 + 210*t^3 + 105*t^4.
The coefficients of these polynomials are given by the reverse of A102625 with the highest order coefficients given by A001147 with an additional leading 1.
Note this is not the complete matrix TC. The complete matrix is formed by multiplying along the diagonal of the lower triangular Pascal matrix by these polynomials, embedding trees of coefficients in the matrix.
exp[Pb(.,t)*x] = 1 + [(1-2t*x)^(1/2) - 1] / (t-0) = [1 + a finite diff. of [(1-2t*x)^(1/2)] with step t] = e.g.f. of the first column of TB.
exp[Pc(.,t)*x] = 1 / { 1 + [(1-2t*x)^(1/2) - 1] / t } = 1 / exp[Pb(.,t)*x) = e.g.f. of the first column of TC.
TB(t) and TC(t), being inverse to each other, are the generators of an Abelian group.
TB(0) and TC(0) are generators for a subgroup representing the iterated Laguerre operator described in A132013 and A132014.
Let sb(t,m) and sc(t,m) be the associated sequences under the LPT to TB(t)^m = B(t,m) and TC(t)^m = C(t,m).
Let Esb(t,m) and Esc(t,m) be e.g.f.'s for sb(t,m) and sc(t,m), rB(t,m) and rC(t,m) be the row sums of B(t,m) and C(t,m) and aB(t,m) and aC(t,m) be the alternating row sums.
Then B(t,m) is the inverse of C(t,m), Esb(t,m) is the reciprocal of Esc(t,m) and sb(t,m) and sc(t,m) form a reciprocal pair under the LPT. Similar relations hold among the row sums and the alternating sign row sums and associated quantities.
All the group members have the form B(t,m) * C(u,p) = TB(t)^m * TC(u)^p = [ binomial(n,k) * s(n-k) ]
with associated e.g.f. Es(x) = exp[m * Pb(.,t) * x] * exp[p * Pc(.,u) * x] for the first column of the matrix, with terms s(n), so group multiplication is isomorphic to matrix multiplication and to multiplication of the e.g.f.'s for the associated sequences (see examples).
These results can be extended to other groups of integer-valued arrays by replacing the 2 by any natural number in the expression for exp[Pb(.,t)*x].
More generally,
[ G.f. for M = Product_{i=0..j} B[s(i),m(i)] * C[t(i),n(i)] ]
= exp(u*x) * Product_{i=0..j} { exp[m(i) * Pb(.,s(i)) * x] * exp[n(i) * Pc(.,t(i)) * x] }
= exp(u*x) * Product_{i=0..j} { 1 + [ (1 - 2*s(i)*x)^(1/2) - 1 ] / s(i) }^m(i) / { 1 + [ (1 - 2*t(i)*x)^(1/2) - 1 ] / t(i) }^n(i)
= exp(u*x) * H(x)
[ E.g.f. for M ] = I_o[2*(u*x)^(1/2)] * H(x).
M is an integer-valued matrix for m(i) and n(i) positive integers and s(i) and t(i) integers. To invert M, change B to C in Product for M.
H(x) is the e.g.f. for the first column of M and diagonally multiplying the Pascal matrix by the terms of this column generates M. See examples.
The G.f. for M, i.e., the e.g.f. for the row polynomials of M, implies that the row polynomials form an Appell sequence (see Wikipedia and Mathworld). - Tom Copeland, Dec 03 2013

Examples

			Some group members and associated arrays are
(t,m) :: Array :: Asc. Matrix :: Asc. Sequence :: E.g.f. for sequence
..............................................................................
(0,1).::.B..::..A132013.::.(1,-1,0,0,0,0,...).....::.s(x).=.1-x
(0,1).::.C..::..A094587.::.(0!,1!,2!,3!,...)......::.1./.s(x)
(0,1).::.rB.::.~A055137.::.(1,0,-1,-2,-3,-4,...)..::.exp(x).*.s(x)
(0,1).::.rC.::....-.....::..A000522...............::.exp(x)./.s(x)
(0,1).::.aB.::....-.....::.(1,-2,3,-4,5,-6,...)...::.exp(-x).*.s(x)
(0,1).::.aC.::..A008290.::..A000166...............::.exp(-x)./.s(x)
..............................................................................
(0,2).::.B..::..A132014.::.(1,-2,2,0,0,0,0...)....::.s(x).=.(1-x)^2
(0,2).::.C..::..A132159.::.(1!,2!,3!,4!,...)......::..1./.s(x).
(0,2).::.rB.::...-......::.(1,-1,-1,1,5,11,19,29,)::.exp(x).*.s(x).
(0,2).::.rC.::...-......::..A001339...............::.exp(x)./.s(x).
(0,2).::.aB.::...-......::.(-1)^n.A002061(n+1)....::.exp(-x).*.s(x).
(0,2).::.aC.::...-......::..A000255...............::.exp(-x)./.s(x).
..............................................................................
(1,1).::.B..::..T.......::.(1,-A001147(n-1))......::.s(x).=.(1-2x)^(1/2)
(1,1).::.C..::.~A113278.::..A001147...............::.1./.s(x)...
(1,1).::.rB.::...-......::..A055142...............::.exp(x).*.s(x).
(1,1).::.rC.::...-......::..A084262...............::.exp(x)./.s(x).
(1,1).::.aB.::...-......::.(1,-2,2,-4,-4,-56,...).::.exp(-x).*.s(x).
(1,1).::.aC.::...-......::..A053871...............::.exp(-x)./.s(x).
..............................................................................
(2,1).::.B..::...-......::.(1,-A001813)...........::.s=[1+(1-4x)^(1/2)]/2....
(2,1).::.C..::...-......::..A001761...............::.1./.s(x)..
(2,1).::.rB.::...-......::.(1,0,-3,-20,-183,...)..::.exp(x).*.s(x)..
(2,1).::.rC.::...-......::.(1,2,7,46,485,...).....::.exp(x)./.s(x).
(2,1).::.aB.::...-......::.(1,-2,1,-10,-79,...)...::.exp(-x).*.s(x).
(2,1).::.aC.::...-......::.(1,0,3,20,237,...).....::.exp(-x)./.s(x)
..............................................................................
(1,2).::.B..::.~A134082.::.(1,-2,0,0,0,0,...).....::.s(x).=.1.-.2x
(1,2).::.C..::....-.....::..A000165...............::.1./.s(x)..
(1,2).::.rB.::....-.....::.(1,-1,-3,-5,-7,-9,...).::.exp(x).*.s(x).
(1,2).::.rC.::....-.....::..A010844...............::.exp(x)./.s(x)..
(1,2).::.aB.::....-.....::.(1,-3,5,-7,9,-11,...)..::.exp(-x).*.s(x).
(1,2).::.aC.::....-.....::..A000354...............::.exp(-x)./.s(x).
..............................................................................
(The tilde indicates the match is not exact--specifically, there are differences in signs from the true matrices.)
Note the row sums correspond to binomial transforms of s(x) and the alternating row sums, to inverse binomial transforms, or, finite differences.
Some additional examples:
C(1,2)*B(0,1) = B(1,-2)*C(0,-1) = [ binomial(n,k)*A002866(n-k) ] with asc. e.g.f. (1-x) / (1-2x).
B(1,2)*C(0,1) = C(1,-2)*B(0,-1) = 2I - A094587 with asc. e.g.f. (1-2x) / (1-x).

Formula

[G.f. for TB(n,k,t)] = GTB(u,x,t) = exp(u*x) * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pb(.,t))*x] where TB(n,k,t) = (D_x)^n (D_u)^k /k! GTB(u,x,t) eval. at u=x=0.
[G.f. for TC(n,k,t)] = GTC(u,x,t) = exp(u*x) / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pc(.,t))*x] where TC(n,k,t) = (D_x)^n (D_u)^k /k! GTC(u,x,t) eval. at u=x=0.
[E.g.f. for TB(n,k,t)] = I_o[2*(u*x)^(1/2)] * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } and
[E.g.f. for TC(n,k,t)] = I_o[2*(u*x)^(1/2)] / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t }
where I_o is the zeroth modified Bessel function of the first kind, i.e.,
I_o[2*(u*x)^(1/2)] = Sum_{j>=0} (u^j/j!) * (x^j/j!).
So [e.g.f. for TB(n,k)] = I_o[2*(u*x)^(1/2)] * (1 - 2x)^(1/2).

Extensions

More terms from Tom Copeland, Dec 05 2007

A255982 Number T(n,k) of partitions of the k-dimensional hypercube resulting from a sequence of n bisections, each of which splits any part perpendicular to any of the axes, such that each axis is used at least once; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 2, 4, 0, 5, 29, 30, 0, 14, 184, 486, 336, 0, 42, 1148, 5880, 9744, 5040, 0, 132, 7228, 64464, 192984, 230400, 95040, 0, 429, 46224, 679195, 3279060, 6792750, 6308280, 2162160, 0, 1430, 300476, 7043814, 51622600, 165293700, 259518600, 196756560, 57657600
Offset: 0

Views

Author

Alois P. Heinz, Mar 13 2015

Keywords

Examples

			A(3,1) = 5:
  [||-|---], [-|||---], [-|-|-|-], [---|||-], [---|-||].
.
A(2,2) = 4:
  ._______.  ._______.  ._______.  ._______.
  |   |   |  |   |   |  |   |   |  |       |
  |___|   |  |   |___|  |___|___|  |_______|
  |   |   |  |   |   |  |       |  |   |   |
  |___|___|  |___|___|  |_______|  |___|___|.
.
Triangle T(n,k) begins:
  1
  0,   1;
  0,   2,     4;
  0,   5,    29,     30;
  0,  14,   184,    486,     336;
  0,  42,  1148,   5880,    9744,    5040;
  0, 132,  7228,  64464,  192984,  230400,   95040;
  0, 429, 46224, 679195, 3279060, 6792750, 6308280, 2162160;
  ...

Links

Crossrefs

Columns k=0-10 give: A000007, A000108 (for n>0), A258416, A258417, A258418, A258419, A258420, A258421, A258422, A258423, A258424.
Main diagonal gives A001761.
Row sums give A258425.
T(2n,n) give A258426.
Cf. A237018, A256061, A258427.

Programs

  • Maple
    b:= proc(n, k, t) option remember; `if`(t=0, 1, `if`(t=1,
       A(n-1, k), add(A(j, k)*b(n-j-1, k, t-1), j=0..n-2)))
    end:
A:= proc(n, k) option remember; `if`(n=0, 1,
      -add(binomial(k, j)*(-1)^j*b(n+1, k, 2^j), j=1..k))
    end:
T:= (n, k)-> add(A(n, k-i)*(-1)^i*binomial(k, i), i=0..k):
seq(seq(T(n, k), k=0..n), n=0..10);
  • Mathematica
    b[n_, k_, t_] := b[n, k, t] = If[t == 0, 1, If[t == 1, A[n-1, k], Sum[ A[j, k]*b[n-j-1, k, t-1], {j, 0, n-2}]]];
A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[Binomial[k, j]*(-1)^j*b[n+1, k, 2^j], {j, 1, k}]];
T[n_, k_] := Sum[A[n, k-i]*(-1)^i*Binomial[k, i], {i, 0, k}]; Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Feb 20 2016, after Alois P. Heinz *)

Formula

T(n,k) = Sum_{i=0..k} (-1)^i * C(k,i) * A237018(n,k-i).

A065866 a(n) = n! * Catalan(n+1).

Original entry on oeis.org

1, 2, 10, 84, 1008, 15840, 308880, 7207200, 196035840, 6094932480, 213322636800, 8303173401600, 355850288640000, 16653793508352000, 845180020548864000, 46236318771202560000, 2712530701243883520000, 169890080762116915200000, 11314679378756986552320000
Offset: 0

Views

Author

Len Smiley, Dec 06 2001

Keywords

Comments

From Noam Zeilberger, Mar 19 2019: (Start)
a(n) is the number of flags in the associahedron of dimension n. For example, there are a(2) = 10 flags in the associahedron of dimension 2, a pentagon. (In this case a flag corresponds to a triple v:e:f of a mutually incident vertex v, edge e, and face f, with f necessarily the unique face of the pentagon.)
Equivalently, a(n) is the number of maximal sequences of consistent parenthesizations of a string of n + 2 letters, starting with n + 1 pairs of parentheses, then removing one pair, and so on, ending with the trivial (outermost) parenthesization. For example, (a(b(cd))):(ab(cd)):(abcd) and (a(b(cd))):(a(bcd)):(abcd) are two of the a(2) = 10 maximal sequences of consistent parenthesizations of the letters abcd. (End)

Examples

			G.f. = 1 + 2*x + 10*x^2 + 84*x^3 + 1008*x^4 + 15840*x^5 + 308880*x^6 + ...

References

  • R. L. Graham, D. E. Knuth, and O. Patashnik, "Concrete Mathematics", Addison-Wesley, 1994, pp. 200-204.

Links

Crossrefs

Cf. A000108, A001761.
Cf. A370055, A370058.
Equals 2 * A102693(n+1), n > 0.
Main diagonal of A256116.

Programs

  • GAP
    List([0..20], n-> 2*Factorial(2*n+1)/Factorial(n+2)); # G. C. Greubel, Mar 19 2019
  • Magma
    [Factorial(n)*Catalan(n+1): n in [0..20]]; // G. C. Greubel, Mar 19 2019
  • Maple
    with(combstruct): ZL:=[T, {T=Union(Z, Prod(Epsilon, Z, T), Prod(T, Z, Epsilon), Prod(T, T, Z))}, labeled]: seq(count(ZL, size=i+1)/(i+1), i=0..18); # Zerinvary Lajos, Dec 16 2007
a := n -> (2^(2*n+2)*GAMMA(n+3/2))/(sqrt(Pi)*(n+1)*(n+2)):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Mar 20 2019
  • Mathematica
    Table[2*(2n+1)!/(n+2)!, {n,0,20}] (* G. C. Greubel, Mar 19 2019 *)
Table[n! CatalanNumber[n+1],{n,0,20}] (* Harvey P. Dale, Feb 02 2023 *)
  • PARI
    { for (n = 0, 100, a = 2 * (2*n + 1)!/(n + 2)!; write("b065866.txt", n, " ", a) ) } \\ Harry J. Smith, Nov 02 2009
  • Sage
    [factorial(n)*catalan_number(n+1) for n in (0..20)] # G. C. Greubel, Mar 19 2019

Formula

a(n) = 2 * (2n+1)!/(n+2)!.
E.g.f.: (1-2*x-sqrt(1-4*x))/(2*x^2) = (O.g.f. for A000108)^2 = B_2(x)^2 (cf. GKP reference).
0 = a(n)*(-7200*a(n+2) + 2700*a(n+3) + 246*a(n+4) - 33*a(n+5)) + a(n+1)*(+36*a(n+2) + 372*a(n+3) + 36*a(n+4) - a(n+5)) + a(n+2)*(-18*a(n+2) + 9*a(n+3) + a(n+4)) for n >= 0. - Michael Somos, Apr 14 2015
The e.g.f. A(x) satisfies 0 = -2 + A(x) * (6*x - 2) + A'(x) * (4*x^2 - x). - Michael Somos, Apr 14 2015
(n+2)*a(n) - 2*n*(2*n+1)*a(n-1) = 0. - R. J. Mathar, Oct 31 2015
a(n) ~ 4^n*exp(-n)*n^(n - 2)*sqrt(2)*(24*n - 61)/6. - Peter Luschny, Mar 20 2019
Sum_{n>=0} 1/a(n) = (25*exp(1/4)*sqrt(Pi)*erf(1/2) + 22)/32, where erf is the error function. - Amiram Eldar, Dec 04 2022
a(n) = 2 * Sum_{k=0..n} (n+2)^(k-1) * |Stirling1(n,k)|. - Seiichi Manyama, Aug 31 2024
E.g.f.: (1/x) * Series_Reversion( x/(1 + x)^2 ). - Seiichi Manyama, Feb 06 2025

A124320 Triangle read by rows: T(n,k) = k!*binomial(n+k-1,k) (n >= 0, 0 <= k <= n), rising factorial power, Pochhammer symbol.

Original entry on oeis.org

1, 1, 1, 1, 2, 6, 1, 3, 12, 60, 1, 4, 20, 120, 840, 1, 5, 30, 210, 1680, 15120, 1, 6, 42, 336, 3024, 30240, 332640, 1, 7, 56, 504, 5040, 55440, 665280, 8648640, 1, 8, 72, 720, 7920, 95040, 1235520, 17297280, 259459200, 1, 9, 90, 990, 11880, 154440, 2162160, 32432400, 518918400, 8821612800
Offset: 0

Views

Author

Emeric Deutsch, Oct 26 2006

Keywords

Comments

This is the Pochhammer function which is defined P(x,n) = x*(x+1)*...*(x+n-1). By convention P(0,0) = 1. Also known as the rising factorial power. - Peter Luschny, Jan 09 2011

Examples

			Triangle starts:
[0]  1;
[1]  1, 1;
[2]  1, 2,   6;
[3]  1, 3,  12,  60;
[4]  1, 4,  20, 120,  840;
[5]  1, 5,  30, 210, 1680, 15120;
[6]  1, 6,  42, 336, 3024, 30240, 332640;
[7]  1, 7,  56, 504, 5040, 55440, 665280, 8648640;
Array starts:
[0] 1,  1,   6,   60,   840,   15120,   332640,   8648640, ... A000407
[1] 1,  2,  12,  120,  1680,   30240,   665280,  17297280, ... A001813
[2] 1,  3,  20,  210,  3024,   55440,  1235520,  32432400, ... A006963
[3] 1,  4,  30,  336,  5040,   95040,  2162160,  57657600, ... A001761
[4] 1,  5,  42,  504,  7920,  154440,  3603600,  98017920, ... A102693
[5] 1,  6,  56,  720, 11880,  240240,  5765760, 160392960, ... A093197
[6] 1,  7,  72,  990, 17160,  360360,  8910720, 253955520, ... A203473
[7] 1,  8,  90, 1320, 24024,  524160, 13366080, 390700800, ...
[8] 1,  9, 110, 1716, 32760,  742560, 19535040, 586051200, ...
[9] 1, 10, 132, 2184, 43680, 1028160, 27907200, 859541760, ...

References

  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1994.

Links

Crossrefs

Rows of array: A000407, A001813, A006963, A001761, A102693, A093197, A203473.
Cf. A123680 (row sums), A352601 (array main diagonal), A123680, A068424 (falling factorial power).

Programs

  • Maple
    T:=proc(n,k) if k<=n then binomial(n+k-1,k)*k! else 0 fi end: for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
A124320 := (n,k)-> `if`(n=0 and k=0,1,pochhammer(n,k)); seq(print(seq(A124320(n,k),k=0..n)),n=0..5); # Peter Luschny, Jan 09 2011
  • Mathematica
    Table[Pochhammer[n,k], {n,0,5},{k,0,n}]//Flatten (* Peter Luschny, Jan 09 2011 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(if(n==0 && k==0, 1, (n+k-1)!/(n-1)!), ", "))) \\ G. C. Greubel, Nov 19 2017
  • Sage
    for n in (0..5) : [rising_factorial(n, k) for k in (0..n)] # Peter Luschny, Jan 09 2011

Formula

T(n,k) = GAMMA(n+k)/GAMMA(n) for n>0. - Peter Luschny, Jan 09 2011

A268437 Triangle read by rows, T(n,k) = (-1)^k*(2*n)!*P[n,k](1/(n+1)) where P is the P-transform, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, 4, 6, 0, 30, 120, 90, 0, 336, 2800, 5040, 2520, 0, 5040, 80640, 264600, 302400, 113400, 0, 95040, 2827440, 15190560, 29937600, 24948000, 7484400, 0, 2162160, 118198080, 983782800, 2986663680, 4162158000, 2724321600, 681080400
Offset: 0

Views

Author

Peter Luschny, Mar 07 2016

Keywords

Comments

The P-transform is defined in the link. Compare also the Sage and Maple implementations below.

Examples

			[1],
[0, 1],
[0, 4, 6],
[0, 30, 120, 90],
[0, 336, 2800, 5040, 2520],
[0, 5040, 80640, 264600, 302400, 113400],
[0, 95040, 2827440, 15190560, 29937600, 24948000, 7484400].

Links

Crossrefs

Cf. A000680, A001761, A268438, A269939, A269941.

Programs

  • Maple
    A268437 := proc(n,k) local F,T;
  F := proc(n,k) option remember;
  `if`(n=0 and k=0, 1,`if`(n=k, (4*n-2)*F(n-1,k-1),
  F(n-1,k)*(n+k))) end;
  T := proc(n,k) option remember;
  `if`(k=0 and n=0, 1,`if`(k<=0 or k>n, 0,
  (4*n-2)*n*(k*T(n-1,k)+(n+k-1)*T(n-1,k-1)))) end;
T(n,k)/F(n,k) end:
for n from 0 to 6 do seq(A268437(n,k), k=0..n) od;
# Alternatively, with the function PTrans defined in A269941:
A268437_row := n -> PTrans(n, n->1/(n+1),(n,k)->(-1)^k*(2*n)!):
seq(print(A268437_row(n)),n=0..8);
  • Mathematica
    T[n_, k_] := (2n)!/FactorialPower[n+k, n] Sum[(-1)^(m+k) Binomial[n+k, n+m] StirlingS2[n+m, m], {m, 0, k}];
Table[T[n, k], {n, 0, 7}, {k, 0, n}] (* Jean-François Alcover, Jun 15 2019 *)
  • Sage
    A268437 = lambda n, k: (factorial(2*n)/falling_factorial(n+k, n))*sum((-1)^(m+k)* binomial(n+k, n+m)*stirling_number2(n+m, m) for m in (0..k))
for n in (0..7): print([A268437(n, m) for m in (0..n)])
  • Sage
    # uses[PtransMatrix from A269941]
PtransMatrix(8, lambda n: 1/(n+1), lambda n, k: (-1)^k* factorial(2*n))

Formula

T(n,k) = ((2*n)!/FF(n+k,n))*Sum_{m=0..k}(-1)^(m+k)*C(n+k,n+m)*Stirling2(n+m,m) where FF denotes the falling factorial function.
T(n,k) = ((2*n)!/FF(n+k,n))*A269939(n,k).
T(n,1) = (2*n)!/(n+1)! = A001761(n) for n>=1.
T(n,n) = (2*n)!/2^n = A000680(n) for n>=0.
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