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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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.
  • D. Foata and D. Zeilberger, A classic proof of a recurrence for a very classical sequence, J. Comb Thy A 80 380-384 1997.
  • H. G. Forder, Some problems in combinatorics, Math. Gazette, vol. 45, 1961, 199-201.
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Crossrefs

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. 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

A005043 Riordan numbers: a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1).

Original entry on oeis.org

1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, 4213, 11298, 30537, 83097, 227475, 625992, 1730787, 4805595, 13393689, 37458330, 105089229, 295673994, 834086421, 2358641376, 6684761125, 18985057351, 54022715451, 154000562758, 439742222071, 1257643249140
Offset: 0

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Comments

Also called Motzkin summands or ring numbers.
The old name was "Motzkin sums", used in certain publications. The sequence has the property that Motzkin(n) = A001006(n) = a(n) + a(n+1), e.g., A001006(4) = 9 = 3 + 6 = a(4) + a(5).
Number of 'Catalan partitions', that is partitions of a set 1,2,3,...,n into parts that are not singletons and whose convex hulls are disjoint when the points are arranged on a circle (so when the parts are all pairs we get Catalan numbers). - Aart Blokhuis (aartb(AT)win.tue.nl), Jul 04 2000
Number of ordered trees with n edges and no vertices of outdegree 1. For n > 1, number of dissections of a convex polygon by nonintersecting diagonals with a total number of n+1 edges. - Emeric Deutsch, Mar 06 2002
Number of Motzkin paths of length n with no horizontal steps at level 0. - Emeric Deutsch, Nov 09 2003
Number of Dyck paths of semilength n with no peaks at odd level. Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUDUDD, where U=(1,1), D=(1,-1). Number of Dyck paths of semilength n with no ascents of length 1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUUDDD. - Emeric Deutsch, Dec 05 2003
Arises in Schubert calculus as follows. Let P = complex projective space of dimension n+1. Take n projective subspaces of codimension 3 in P in general position. Then a(n) is the number of lines of P intersecting all these subspaces. - F. Hirzebruch, Feb 09 2004
Difference between central trinomial coefficient and its predecessor. Example: a(6) = 15 = 141 - 126 and (1 + x + x^2)^6 = ... + 126*x^5 + 141*x^6 + ... (Catalan number A000108(n) is the difference between central binomial coefficient and its predecessor.) - David Callan, Feb 07 2004
a(n) = number of 321-avoiding permutations on [n] in which each left-to-right maximum is a descent (i.e., is followed by a smaller number). For example, a(4) counts 4123, 3142, 2143. - David Callan, Jul 20 2005
The Hankel transform of this sequence give A000012 = [1, 1, 1, 1, 1, 1, 1, ...]; example: Det([1, 0, 1, 1; 0, 1, 1, 3; 1, 1, 3, 6; 1, 3, 6, 15]) = 1. - Philippe Deléham, May 28 2005
The number of projective invariants of degree 2 for n labeled points on the projective line. - Benjamin J. Howard (bhoward(AT)ima.umn.edu), Nov 24 2006
Define a random variable X=trA^2, where A is a 2 X 2 unitary symplectic matrix chosen from USp(2) with Haar measure. The n-th central moment of X is E[(X+1)^n] = a(n). - Andrew V. Sutherland, Dec 02 2007
Let V be the adjoint representation of the complex Lie algebra sl(2). The dimension of the invariant subspace of the n-th tensor power of V is a(n). - Samson Black (sblack1(AT)uoregon.edu), Aug 27 2008
Starting with offset 3 = iterates of M * [1,1,1,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 08 2009
a(n) has the following standard-Young-tableaux (SYT) interpretation: binomial(n+1,k)*binomial(n-k-1,k-1)/(n+1)=f^(k,k,1^{n-2k}) where f^lambda equals the number of SYT of shape lambda. - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 02 2010
a(n) is also the sum of the numbers of standard Young tableaux of shapes (k,k,1^{n-2k}) for all 1 <= k <= floor(n/2). - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 10 2010
a(n) is the number of derangements of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(3)=1 because p=231=(123) is the only derangement of {1,2,3} with genus 0. Indeed, cp'=231*312=123=(1)(2)(3) and so g(p) = (1/2)(3+1-1-3)=0. - Emeric Deutsch, May 29 2010
Apparently: Number of Dyck 2n-paths with all ascents length 2 and no descent length 2. - David Scambler, Apr 17 2012
This is true. Proof: The mapping "insert a peak (UD) after each upstep (U)" is a bijection from all Dyck n-paths to those Dyck (2n)-paths in which each ascent is of length 2. It sends descents of length 1 in the n-path to descents of length 2 in the (2n)-path. But Dyck n-paths with no descents of length 1 are equinumerous with Riordan n-paths (Motzkin n-paths with no flatsteps at ground level) as follows. Given a Dyck n-path with no descents of length 1, split it into consecutive step pairs, then replace UU with U, DD with D, UD with a blue flatstep (F), DU with a red flatstep, and concatenate the new steps to get a colored Motzkin path. Each red F will be (immediately) preceded by a blue F or a D. In the latter case, transfer the red F so that it precedes the matching U of the D. Finally, erase colors to get the required Riordan path. For example, with lowercase f denoting a red flatstep, U^5 D^2 U D^4 U^4 D^3 U D^2 -> (U^2, U^2, UD, DU, D^2, D^2, U^2, U^2 D^2, DU, D^2) -> UUFfDDUUDfD -> UUFFDDUFUDD. - David Callan, Apr 25 2012
From Nolan Wallach, Aug 20 2014: (Start)
Let ch[part1, part2] be the value of the character of the symmetric group on n letters corresponding to the partition part1 of n on the conjucgacy class given by part2. Let A[n] be the set of (n+1) partitions of 2n with parts 1 or 2. Then deleting the first term of the sequence one has a(n) = Sum_{k=1..n+1} binomial(n,k-1)*ch[[n,n], A[n][[k]]])/2^n. This via the Frobenius Character Formula can be interpreted as the dimension of the SL(n,C) invariants in tensor^n (wedge^2 C^n).
Explanation: Let p_j denote sum (x_i)^j the sum in k variables. Then the Frobenius formula says then (p_1)^j_1 (p_2)^j_2 ... (p_r)^j_r is equal to sum(lambda, ch[lambda, 1^j_12^j_2 ... r^j_r] S_lambda) with S_lambda the Schur function corresponding to lambda. This formula implies that the coefficient of S([n,n]) in (((p_1)^1+p_2)/2)^n in its expansion in terms of Schur functions is the right hand side of our formula. If we specialize the number of variables to 2 then S[n,n](x,y)=(xy)^n. Which when restricted to y=x^(-1) is 1. That is it is 1 on SL(2).
On the other hand ((p_1)^2+p_2)/2 is the complete homogeneous symmetric function of degree 2 that is tr(S^2(X)). Thus our formula for a(n) is the same as that of Samson Black above since his V is the same as S^2(C^2) as a representation of SL(2). On the other hand, if we multiply ch(lambda) by sgn you get ch(Transpose(lambda)). So ch([n,n]) becomes ch([2,...,2]) (here there are n 2's). The formula for a(n) is now (1/2^n)*Sum_{j=0..n} ch([2,..,2], 1^(2n-2j) 2^j])*(-1)^j)*binomial(n,j), which calculates the coefficient of S_(2,...,2) in (((p_1)^2-p_2)/2)^n. But ((p_1)^2-p_2)/2 in n variables is the second elementary symmetric function which is the character of wedge^2 C^n and S_(2,...,2) is 1 on SL(n).
(End)
a(n) = number of noncrossing partitions (A000108) of [n] that contain no singletons, also number of nonnesting partitions (A000108) of [n] that contain no singletons. - David Callan, Aug 27 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Various relations among the o.g.f.s may be easily constructed, such as Fib[-Mot(-x)] = -P[P(-x)] = x/(1-2*x) a generating fct for 2^n.
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)
Consistent with David Callan's comment above, A249548, provides a refinement of the Motzkin sums into the individual numbers for the non-crossing partitions he describes. - Tom Copeland, Nov 09 2014
The number of lattice paths from (0,0) to (n,0) that do not cross below the x-axis and use up-step=(1,1) and down-steps=(1,-k) where k is a positive integer. For example, a(4) = 3: [(1,1)(1,1)(1,-1)(1,-1)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,1)(1,-3)]. - Nicholas Ham, Aug 19 2015
A series created using 2*(a(n) + a(n+1)) + (a(n+1) + a(n+2)) has Hankel transform of F(2n), offset 3, F being a Fibonacci number, A001906 (Empirical observation). - Tony Foster III, Jul 30 2016
The series a(n) + A001006(n) has Hankel transform F(2n+1), offset n=1, F being the Fibonacci bisection A001519 (empirical observation). - Tony Foster III, Sep 05 2016
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n. Moreover, the number of such algebras having the double centraliser property with respect to a minimal faithful projective-injective module equals the number of Riordan paths, that is, Motzkin paths without level-steps at height zero, of length n." - Eric M. Schmidt, Dec 16 2017
A connection to the Thue-Morse sequence: (-1)^a(n) = (-1)^A010060(n) * (-1)^A010060(n+1) = A106400(n) * A106400(n+1). - Vladimir Reshetnikov, Jul 21 2019
Named by Bernhart (1999) after the American mathematician John Riordan (1903-1988). - Amiram Eldar, Apr 15 2021

Examples

			a(5)=6 because the only dissections of a polygon with a total number of 6 edges are: five pentagons with one of the five diagonals and the hexagon with no diagonals.
G.f. = 1 + x^2 + x^3 + 3*x^4 + 6*x^5 + 15*x^6 + 36*x^7 + 91*x^8 + 232*x^9 + ...
From _Gus Wiseman_, Nov 15 2022: (Start)
The a(0) = 1 through a(6) = 15 lone-child-avoiding (no vertices of outdegree 1) ordered rooted trees with n + 1 vertices (ranked by A358376):
  o  .  (oo)  (ooo)  (oooo)   (ooooo)   (oooooo)
                     ((oo)o)  ((oo)oo)  ((oo)ooo)
                     (o(oo))  ((ooo)o)  ((ooo)oo)
                              (o(oo)o)  ((oooo)o)
                              (o(ooo))  (o(oo)oo)
                              (oo(oo))  (o(ooo)o)
                                        (o(oooo))
                                        (oo(oo)o)
                                        (oo(ooo))
                                        (ooo(oo))
                                        (((oo)o)o)
                                        ((o(oo))o)
                                        ((oo)(oo))
                                        (o((oo)o))
                                        (o(o(oo)))
(End)
		

References

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

Crossrefs

Row sums of triangle A020474, first differences of A082395.
First diagonal of triangular array in A059346.
Binomial transform of A126930. - Philippe Deléham, Nov 26 2009
The Hankel transform of a(n+1) is A128834. The Hankel transform of a(n+2) is floor((2*n+4)/3) = A004523(n+2). - Paul Barry, Mar 08 2011
The Kn11 triangle sums of triangle A175136 lead to A005043(n+2), while the Kn12(n) = A005043(n+4)-2^(n+1), Kn13(n) = A005043(n+6)-(n^2+9*n+56)*2^(n-2) and the Kn4(n) = A005043(2*n+2) = A099251(n+1) triangle sums are related to the sequence given above. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Cf. A187306 (self-convolution), A348210 (column 1).
Bisections: A099251, A099252.

Programs

  • Haskell
    a005043 n = a005043_list !! n
    a005043_list = 1 : 0 : zipWith div
       (zipWith (*) [1..] (zipWith (+)
           (map (* 2) $ tail a005043_list) (map (* 3) a005043_list))) [3..]
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    A005043 := proc(n) option remember; if n <= 1 then 1-n else (n-1)*(2*A005043(n-1)+3*A005043(n-2))/(n+1); fi; end;
    Order := 20: solve(series((x-x^2)/(1-x+x^2),x)=y,x); # outputs g.f.
  • Mathematica
    a[0]=1; a[1]=0; a[n_]:= a[n] = (n-1)*(2*a[n-1] + 3*a[n-2])/(n+1); Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jun 14 2005 *)
    Table[(-3)^(1/2)/6 * (-1)^n*(3*Hypergeometric2F1[1/2,n+1,1,4/3]+ Hypergeometric2F1[1/2,n+2,1,4/3]), {n,0,32}] (* cf. Mark van Hoeij in A001006 *) (* Wouter Meeussen, Jan 23 2010 *)
    RecurrenceTable[{a[0]==1,a[1]==0,a[n]==(n-1) (2a[n-1]+3a[n-2])/(n+1)},a,{n,30}] (* Harvey P. Dale, Sep 27 2013 *)
    a[ n_]:= SeriesCoefficient[2/(1+x +Sqrt[1-2x-3x^2]), {x, 0, n}]; (* Michael Somos, Aug 21 2014 *)
    a[ n_]:= If[n<0, 0, 3^(n+3/2) Hypergeometric2F1[3/2, n+2, 2, 4]/I]; (* Michael Somos, Aug 21 2014 *)
    Table[3^(n+3/2) CatalanNumber[n] (4(5+2n)Hypergeometric2F1[3/2, 3/2, 1/2-n, 1/4] -9 Hypergeometric2F1[3/2, 5/2, 1/2 -n, 1/4])/(4^(n+3) (n+1)), {n, 0, 31}] (* Vladimir Reshetnikov, Jul 21 2019 *)
    Table[Sqrt[27]/8 (3/4)^n CatalanNumber[n] Hypergeometric2F1[1/2, 3/2, 1/2 - n, 1/4], {n, 0, 31}] (* Jan Mangaldan, Sep 12 2021 *)
  • Maxima
    a[0]:1$
    a[1]:0$
    a[n]:=(n-1)*(2*a[n-1]+3*a[n-2])/(n+1)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( (x - x^3) / (1 + x^3) + x * O(x^n)), n))}; /* Michael Somos, May 31 2005 */
    
  • PARI
    my(N=66); Vec(serreverse(x/(1+x*sum(k=1,N,x^k))+O(x^N))) \\ Joerg Arndt, Aug 19 2012
    
  • Python
    from functools import cache
    @cache
    def A005043(n: int) -> int:
        if n <= 1: return 1 - n
        return (n - 1) * (2 * A005043(n - 1) + 3 * A005043(n - 2)) // (n + 1)
    print([A005043(n) for n in range(32)]) # Peter Luschny, Nov 20 2022
  • Sage
    A005043 = lambda n: (-1)^n*jacobi_P(n,1,-n-3/2,-7)/(n+1)
    [simplify(A005043(n)) for n in (0..29)]
    # Peter Luschny, Sep 23 2014
    
  • Sage
    def ms():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        while True:
            yield -b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)/((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)/n
    A005043 = ms()
    print([next(A005043) for  in range(32)]) # _Peter Luschny, May 16 2016
    

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k). a(n) = (1/(n+1)) * Sum_{k=0..ceiling(n/2)} binomial(n+1, k)*binomial(n-k-1, k-1), for n > 1. - Len Smiley. [Comment from Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 02 2010: the latter sum should be over the range k=1..floor(n/2).]
G.f.: (1 + x - sqrt(1-2*x-3*x^2))/(2*x*(1+x)).
G.f.: 2/(1+x+sqrt(1-2*x-3*x^2)). - Paul Peart (ppeart(AT)fac.howard.edu), May 27 2000
a(n+1) + (-1)^n = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). - Bernhart
a(n) = (1/(n+1)) * Sum_{i} (-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i). - Bernhart
G.f. A(x) satisfies A = 1/(1+x) + x*A^2.
E.g.f.: exp(x)*(BesselI(0, 2*x) - BesselI(1, 2*x)). - Vladeta Jovovic, Apr 28 2003
a(n) = A001006(n-1) - a(n-1).
a(n+1) = Sum_{k=0..n} (-1)^k*A026300(n, k), where A026300 is the Motzkin triangle.
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
a(n) = Sum_{k>=0} A086810(n-k, k). - Philippe Deléham, May 30 2005
a(n+2) = Sum_{k>=0} A064189(n-k, k). - Philippe Deléham, May 31 2005
Moment representation: a(n) = (1/(2*Pi))*Int(x^n*sqrt((1+x)(3-x))/(1+x),x,-1,3). - Paul Barry, Jul 09 2006
Inverse binomial transform of A000108 (Catalan numbers). - Philippe Deléham, Oct 20 2006
a(n) = (2/Pi)* Integral_{x=0..Pi} (4*cos(x)^2-1)^n*sin(x)^2 dx. - Andrew V. Sutherland, Dec 02 2007
G.f.: 1/(1-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 22 2009
G.f.: 1/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: 1/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-... (continued fraction). - Paul Barry, Mar 02 2010
a(n) = -(-1)^n * hypergeom([1/2, n+2],[2],4/3) / sqrt(-3). - Mark van Hoeij, Jul 02 2010
a(n) = (-1)^n*hypergeometric([-n,1/2],[2],4). - Peter Luschny, Aug 15 2012
Let A(x) be the g.f., then x*A(x) is the reversion of x/(1 + x^2*Sum_{k>=0} x^k); see A215340 for the correspondence to Dyck paths without length-1 ascents. - Joerg Arndt, Aug 19 2012 and Apr 16 2013
a(n) ~ 3^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 02 2012
G.f.: 2/(1+x+1/G(0)), where G(k) = 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 05 2013
D-finite (an alternative): (n+1)*a(n) = 3*(n-2)*a(n-3) + (5*n-7)*a(n-2) + (n-2)*a(n-1), n >= 3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) = (3^(n+2)/(sqrt(3*n*Pi)*(8*n)))*(1-21/(16*n) + O(1/n^2)) (with contribution by Vaclav Kotesovec). - Fung Lam, Mar 22 2014
a(n) = T(2*n-1,n)/n, where T(n,k) = triangle of A180177. - Vladimir Kruchinin, Sep 23 2014
a(n) = (-1)^n*JacobiP(n,1,-n-3/2,-7)/(n+1). - Peter Luschny, Sep 23 2014
a(n) = Sum_{k=0..n} C(n,k)*(C(k,n-k)-C(k,n-k-1)). - Peter Luschny, Oct 01 2014
Conjecture: a(n) = A002426(n) - A005717(n), n > 0. - Mikhail Kurkov, Feb 24 2019 [The conjecture is true. - Amiram Eldar, May 17 2024]
a(n) = A309303(n) + A309303(n+1). - Vladimir Reshetnikov, Jul 22 2019
From Peter Bala, Feb 11 2022: (Start)
a(n) = A005773(n+1) - 2*A005717(n) for n >= 1.
Conjectures: for n >= 1, n divides a(2*n+1) and 2*n-1 divides a(2*n). (End)

Extensions

Thanks to Laura L. M. Yang (yanglm(AT)hotmail.com) for a correction, Aug 29 2004
Name changed to Riordan numbers following a suggestion from Ira M. Gessel. - N. J. A. Sloane, Jul 24 2020

A000296 Set partitions without singletons: number of partitions of an n-set into blocks of size > 1. Also number of cyclically spaced (or feasible) partitions.

Original entry on oeis.org

1, 0, 1, 1, 4, 11, 41, 162, 715, 3425, 17722, 98253, 580317, 3633280, 24011157, 166888165, 1216070380, 9264071767, 73600798037, 608476008122, 5224266196935, 46499892038437, 428369924118314, 4078345814329009, 40073660040755337, 405885209254049952, 4232705122975949401
Offset: 0

Views

Author

Keywords

Comments

a(n+2) = p(n+1) where p(x) is the unique degree-n polynomial such that p(k) = A000110(k) for k = 0, 1, ..., n. - Michael Somos, Oct 07 2003
Number of complete rhyming schemes.
Whereas the Bell number B(n) (A000110(n)) is the number of terms in the polynomial that expresses the n-th moment of a probability distribution as a function of the first n cumulants, these numbers give the number of terms in the corresponding expansion of the central moment as a function of the first n cumulants. - Michael Hardy (hardy(AT)math.umn.edu), Jan 26 2005
a(n) is the number of permutations on [n] for which the left-to-right maxima coincide with the descents (entries followed by a smaller number). For example, a(4) counts 2143, 3142, 3241, 4123. - David Callan, Jul 20 2005
From Gus Wiseman, Feb 10 2019: (Start)
Also the number of stable partitions of an n-cycle, where a stable partition of a graph is a set partition of the vertex set such that no edge has both ends in the same block. A bijective proof is given in David Callan's article. For example, the a(5) = 11 stable partitions are:
{{1},{2},{3},{4},{5}}
{{1},{2},{3,5},{4}}
{{1},{2,4},{3},{5}}
{{1},{2,5},{3},{4}}
{{1,3},{2},{4},{5}}
{{1,4},{2},{3},{5}}
{{1},{2,4},{3,5}}
{{1,3},{2,4},{5}}
{{1,3},{2,5},{4}}
{{1,4},{2},{3,5}}
{{1,4},{2,5},{3}}
(End)
Also number of partitions of {1, 2, ..., n-1} with singletons. E.g., a(4) = 4: {1|2|3, 12|3, 13|2, 1|23}. Also number of cyclical adjacencies partitions of {1, 2, ..., n-1}. E.g., a(4) = 4: {12|3, 13|2, 1|23, 123}. The two partitions can be mapped by a Kreweras bijection. - Yuchun Ji, Feb 22 2021
Also the k-th central moment of a Poisson random variable with mean 1. a(n) = E[(X-1)^n, X~Poisson(1)]. - Thomas Dybdahl Ahle, Dec 14 2022

Examples

			a(4) = card({{{1, 2}, {3, 4}}, {{1, 4}, {2, 3}}, {{1, 3}, {2, 4}}, {{1, 2, 3, 4}}}) = 4.
		

References

  • Martin Gardner in Sci. Amer. May 1977.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.5 (p. 436).
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, NY, 2 vols., 1972, Vol. 1, p. 228.
  • J. Riordan, A budget of rhyme scheme counts, pp. 455-465 of Second International Conference on Combinatorial Mathematics, New York, April 4-7, 1978. Edited by Allan Gewirtz and Louis V. Quintas. Annals New York Academy of Sciences, 319, 1979.
  • J. Shallit, A Triangle for the Bell numbers, in V. E. Hoggatt, Jr. and M. Bicknell-Johnson, A Collection of Manuscripts Related to the Fibonacci Sequence, 1980, pp. 69-71.
  • 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).

Crossrefs

A diagonal of triangle in A106436.
Row sums of the triangle of associated Stirling numbers of second kind A008299. - Philippe Deléham, Feb 10 2005
Row sums of the triangle of basic multinomial coefficients A178866. - Johannes W. Meijer, Jun 21 2010
Row sums of A105794. - Peter Bala, Jan 14 2015
Row sums of A261139, main diagonal of A261137.
Column k=0 of A216963.
Column k=0 of A124323.

Programs

  • Magma
    [1,0] cat [ n le 1 select 1 else Bell(n)-Self(n-1) : n in [1..40]]; // Vincenzo Librandi, Jun 22 2015
    
  • Maple
    spec := [ B, {B=Set(Set(Z,card>1))}, labeled ]; [seq(combstruct[count](spec, size=n), n=0..30)];
    with(combinat): A000296 :=n->(-1)^n + add((-1)^(j-1)*bell(n-j),j=1..n): seq(A000295(n),n=0..30); # Emeric Deutsch, Oct 29 2006
    f:=exp(exp(x)-1-x): fser:=series(f, x=0, 31): 1, seq(n!*coeff(fser, x^n), n=1..23); # Zerinvary Lajos, Nov 22 2006
    G:={P=Set(Set(Atom,card>=2))}: combstruct[gfsolve](G,unlabeled,x): seq(combstruct[count]([P,G,labeled], size=i), i=0..23); # Zerinvary Lajos, Dec 16 2007
    # [a(0),a(1),..,a(n)]
    A000296_list := proc(n)
    local A, R, i, k;
    if n = 0 then return 1 fi;
    A := array(0..n-1);
    A[0] := 1; R := 1;
    for i from 0 to n-2 do
       A[i+1] := A[0] - A[i];
       A[i] := A[0];
       for k from i by -1 to 1 do
          A[k-1] := A[k-1] + A[k] od;
       R := R,A[i+1];
    od;
    R,A[0]-A[i] end:
    A000296_list(100);  # Peter Luschny, Apr 09 2011
  • Mathematica
    nn = 25; Range[0, nn]! CoefficientList[Series[Exp[Exp[x] - 1 - x], {x, 0, nn}], x]
    (* Second program: *)
    a[n_] := a[n] = If[n==0, 1, Sum[Binomial[n-1, i]*a[n-i-1], {i, 1, n-1}]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Feb 06 2016, after Vladimir Kruchinin *)
    spsu[,{}]:={{}};spsu[foo,set:{i_,_}]:= Join@@Function[s,Prepend[#,s]&/@spsu[ Select[foo,Complement[#, Complement[set,s]]=={}&], Complement[set,s]]]/@Cases[foo,{i,_}];
    Table[Length[spsu[Select[Subsets[Range[n]],Select[Partition[Range[n],2,1,1], Function[ed,Complement[ed,#]=={}]]=={}&],Range[n]]],{n,8}] (* Gus Wiseman, Feb 10 2019 *)
    s = 1; Join[{1}, Table[s = BellB[n] - s, {n, 0, 25}]] (* Vaclav Kotesovec, Jun 20 2022 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(binomial(n-1,i)*a(n-i-1),i,1,n-1); /* Vladimir Kruchinin, Feb 22 2015 */
    
  • PARI
    a(n) = if(n<2, n==0, subst( polinterpolate( Vec( serlaplace( exp( exp( x+O(x^n)/x )-1 ) ) ) ), x, n) )
    
  • Python
    from itertools import accumulate, islice
    def A000296_gen():
        yield from (1,0)
        blist, a, b = (1,), 0, 1
        while True:
            blist = list(accumulate(blist, initial = (b:=blist[-1])))
            yield (a := b-a)
    A000296_list = list(islice(A000296_gen(),20)) # Chai Wah Wu, Jun 22 2022

Formula

E.g.f.: exp(exp(x) - 1 - x).
B(n) = a(n) + a(n+1), where B = A000110 = Bell numbers [Becker].
Inverse binomial transform of Bell numbers (A000110).
a(n)= Sum_{k>=-1} (k^n/(k+1)!)/exp(1). - Vladeta Jovovic and Karol A. Penson, Feb 02 2003
a(n) = Sum_{k=0..n} ((-1)^(n-k))*binomial(n, k)*Bell(k) = (-1)^n + Bell(n) - A087650(n), with Bell(n) = A000110(n). - Wolfdieter Lang, Dec 01 2003
O.g.f.: A(x) = 1/(1-0*x-1*x^2/(1-1*x-2*x^2/(1-2*x-3*x^2/(1-... -(n-1)*x-n*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
a(n) = Sum_{k = 0..n} {(-1)^(n-k) * Sum_{j = 0..k}((-1)^j * binomial(k,j) * (1-j)^n)/ k!} = sum over row n of A105794. - Tom Copeland, Jun 05 2006
a(n) = (-1)^n + Sum_{j=1..n} (-1)^(j-1)*B(n-j), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Oct 29 2006
Let A be the upper Hessenberg matrix of order n defined by: A[i, i-1] = -1, A[i,j] = binomial(j-1, i-1), (i <= j), and A[i, j] = 0 otherwise. Then, for n >= 2, a(n) = (-1)^(n)charpoly(A,1). - Milan Janjic, Jul 08 2010
From Sergei N. Gladkovskii, Sep 20 2012, Oct 11 2012, Dec 19 2012, Jan 15 2013, May 13 2013, Jul 20 2013, Oct 19 2013, Jan 25 2014: (Start)
Continued fractions:
G.f.: (2/E(0) - 1)/x where E(k) = 1 + 1/(1 + 2*x/(1 - 2*(k+1)*x/E(k+1))).
G.f.: 1/U(0) where U(k) = 1 - x*k - x^2*(k+1)/U(k+1).
G.f.: G(0)/(1+2*x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-x-1) - x*(2*k+1)*(2*k+3)*(2*x*k-x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k-1)/G(k+1))).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1+x-k*x)/(1-x/(x-1/G(k+1))).
G.f.: 1 + x^2/(1+x)/Q(0) where Q(k) = 1-x-x/(1-x*(2*k+1)/(1-x-x/(1-x*(2*k+2)/Q(k+1)))).
G.f.: 1/(x*Q(0)) where Q(k) = 1 + 1/(x + x^2/(1 - x - (k+1)/Q(k+1))).
G.f.: -(1+(2*x+1)/G(0))/x where G(k) = x*k - x - 1 - (k+1)*x^2/G(k+1).
G.f.: T(0) where T(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x*k)*(1-x-x*k)/T(k+1)).
G.f.: (1 + x * Sum_{k>=0} (x^k / Product_{p=0..k}(1 - p*x))) / (1 + x). (End)
a(n) = Sum_{i=1..n-1} binomial(n-1,i)*a(n-i-1), a(0)=1. - Vladimir Kruchinin, Feb 22 2015
G.f. A(x) satisfies: A(x) = (1/(1 + x)) * (1 + x * A(x/(1 - x)) / (1 - x)). - Ilya Gutkovskiy, May 21 2021
a(n) ~ exp(n/LambertW(n) - n - 1) * n^(n-1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n-1)). - Vaclav Kotesovec, Jun 28 2022

Extensions

More terms, new description from Christian G. Bower, Nov 15 1999
a(23) corrected by Sean A. Irvine, Jun 22 2015

A074679 Signature permutation of a Catalan automorphism: Rotate binary tree left if possible, otherwise swap its sides.

Original entry on oeis.org

0, 1, 3, 2, 6, 7, 8, 4, 5, 14, 15, 16, 17, 18, 19, 20, 21, 9, 10, 22, 11, 12, 13, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 23, 24, 59, 25, 26, 27, 60, 61, 62, 28, 29, 63, 30, 31, 32, 64, 33, 34, 35, 36, 107, 108, 109, 110, 111
Offset: 0

Views

Author

Antti Karttunen, Sep 11 2002

Keywords

Comments

This automorphism effects the following transformation on the unlabeled rooted plane binary trees (letters A, B, C refer to arbitrary subtrees located on those nodes and () stands for an implied terminal node.)
...B...C.......A...B
....\./.........\./
.A...x....-->....x...C.................A..().........()..A..
..\./.............\./...................\./....-->....\./...
...x...............x.....................x.............x....
(a . (b . c)) -> ((a . b) . c) ____ (a . ()) --> (() . a)
That is, we rotate the binary tree left, in case it is possible and otherwise (if the right hand side of a tree is a terminal node) swap the left and right subtree (so that the terminal node ends to the left hand side), i.e., apply the automorphism *A069770. Look at the example in A069770 to see how this will produce the given sequence of integers.
This is the first multiclause nonrecursive automorphism in table A089840 and the first one whose order is not finite, i.e., the maximum size of cycles in this permutation is not bounded (see A089842). The cycle counts in range [A014137(n-1)..A014138(n)] of this permutation is given by A001683(n+1), which is otherwise the same sequence as for Catalan automorphisms *A057161/*A057162, but shifted once right. For an explanation, please see the notes in OEIS Wiki.

Crossrefs

This automorphism has several variants, where the first clause is same (rotate binary tree to the left, if possible), but something else is done (than just swapping sides), in case the right hand side is empty: A082335, A082349, A123499, A123695. The following automorphisms can be derived recursively from this one: A057502, A074681, A074683, A074685, A074687, A074690, A089865, A120706, A122321, A122332. See also somewhat similar ones: A069773, A071660, A071656, A071658, A072091, A072095, A072093.
Inverse: A074680.
Row 12 of A089840.
Occurs also in A073200 as row 557243 because a(n) = A073283(A073280(A072796(n))). a(n) = A083927(A123498(A057123(n))).
Number of cycles: LEFT(A001683). Number of fixed points: LEFT(A019590). Max. cycle size & LCM of all cycle sizes: A089410 (in range [A014137(n-1)..A014138(n)] of this permutation).

Extensions

Description clarified Oct 10 2006

A074680 Signature permutation of the seventeenth nonrecursive Catalan automorphism in table A089840. (Rotate binary tree right if possible, otherwise swap its sides.)

Original entry on oeis.org

0, 1, 3, 2, 7, 8, 4, 5, 6, 17, 18, 20, 21, 22, 9, 10, 11, 12, 13, 14, 15, 16, 19, 45, 46, 48, 49, 50, 54, 55, 57, 58, 59, 61, 62, 63, 64, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 47, 51, 52, 53, 56, 60, 129, 130, 132, 133, 134
Offset: 0

Views

Author

Antti Karttunen, Sep 11 2002

Keywords

Comments

This automorphism effects the following transformation on the unlabeled rooted plane binary trees (letters A, B, C refer to arbitrary subtrees located on those nodes and () stands for an implied terminal node.)
A...B..............B...C
.\./................\./
..x...C..-->.....A...x................()..B.......B..()
...\./............\./..................\./...-->...\./.
....x..............x....................x...........x..
((a . b) . c) -> (a . (b . c)) __ (() . b) --> (b . ())
That is, we rotate the binary tree right, in case it is possible and otherwise (if the left hand side of a tree is a terminal node) swap the right and left subtree (so that the terminal node ends to the right hand side), i.e. apply the automorphism *A069770. Look at the example in A069770 to see how this will produce the given sequence of integers.
See also the comments at A074679.

References

  • A. Karttunen, paper in preparation, draft available by e-mail.

Crossrefs

This automorphism has several variants, where the first clause is same (rotate binary tree to the right, if possible), but something else is done (than just swapping sides), in case the left hand side is empty: A082336, A082350, A123500, A123696. The following automorphisms can be derived recursively from this one: A057501, A074682, A074684, A074686, A074688, A074689, A089866, A120705, A122322, A122331. See also somewhat similar ones: A069774, A071659, A071655, A071657, A072090, A072094, A072092.
Inverse: A074679. Row 17 of A089840. Occurs also in A073200 as row 2156396687 as a(n) = A072796(A073280(A073282(n))). a(n) = A083927(A123497(A057123(n))).
Number of cycles: LEFT(A001683). Number of fixed points: LEFT(A019590). Max. cycle size & LCM of all cycle sizes: A089410 (in range [A014137(n-1)..A014138(n-1)] of this permutation).

Extensions

Description clarified Oct 10 2006

A000207 Number of inequivalent ways of dissecting a regular (n+2)-gon into n triangles by n-1 non-intersecting diagonals under rotations and reflections; also the number of (unlabeled) maximal outerplanar graphs on n+2 vertices.

Original entry on oeis.org

1, 1, 1, 3, 4, 12, 27, 82, 228, 733, 2282, 7528, 24834, 83898, 285357, 983244, 3412420, 11944614, 42080170, 149197152, 531883768, 1905930975, 6861221666, 24806004996, 90036148954, 327989004892, 1198854697588, 4395801203290, 16165198379984, 59609171366326, 220373278174641
Offset: 1

Views

Author

Keywords

Comments

Also a(n) is the number of hexaflexagons of order n+2. - Mike Godfrey (m.godfrey(AT)umist.ac.uk), Feb 25 2002 (see the Kosters paper).
Number of normally non-isomorphic realizations of the associahedron of type II with dimension n in Ceballos et al. - Tom Copeland, Oct 19 2011
Number of polyforms with n cells in the hyperbolic tiling with Schläfli symbol {3,oo}, not distinguishing enantiomorphs. - Thomas Anton, Jan 16 2019
A stereographic projection of the {3,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 20 2024
A maximal outerplanar graph (MOP) has a plane embedding with all vertices on the exterior region and interior regions triangles. - Allan Bickle, Feb 25 2024

Examples

			E.g., a square (4-gon, n=2) could have either diagonal drawn, C(3)=2, but with essentially only one result. A pentagon (5-gon, n=3) gives C(4)=5, but they each have 2 diags emanating from 1 of the 5 vertices and are essentially the same. A hexagon can have a nuclear disarmament sign (6 ways), an N (3 ways and 3 reflections) or a triangle (2 ways) of diagonals, 6 + 6 + 2 = 14 = C(5), but only 3 essentially different. - _R. K. Guy_, Mar 06 2004
G.f. = x + x^2 + x^3 + 3*x^4 + 4*x^5 + 12*x^6 + 27*x^7 + 82*x^8 + ...
		

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 in Math. Annalen, 191 (1971), 87-98.
  • Cameron, Peter J. Some treelike objects. Quart. J. Math. Oxford Ser. (2) 38 (1987), no. 150, 155--183. MR0891613 (89a:05009). See pp. 155, 163, but note that the formulas on p. 163, lines 5 and 6, contain typos. See the correct formulas given here. - N. J. A. Sloane, Apr 18 2014
  • B. N. Cyvin, E. Brendsdal, J. Brunvoll and S. J. Cyvin, Isomers of polyenes attached to benzene, Croatica Chemica Acta, 68 (1995), 63-73.
  • S. J. Cyvin, J. Brunvoll, E. Brendsdal, B. N. Cyvin and E. K. Lloyd, Enumeration of polyene hydrocarbons: a complete mathematical solution, J. Chem. Inf. Comput. Sci., 35 (1995) 743-751.
  • C. F. Earl and L. J. March, Architectural applications of graph theory, pp. 327-355 of R. J. Wilson and L. W. Beineke, editors, Applications of Graph Theory. Academic Press, NY, 1979.
  • R. K. Guy, "Dissecting a polygon into triangles," Bull. Malayan Math. Soc., Vol. 5, pp. 57-60, 1958.
  • R. K. Guy, Dissecting a polygon into triangles, Research Paper #9, Math. Dept., Univ. Calgary, 1967.
  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 79, Table 3.5.1 (the entries for n=16 and n=21 appear to be incorrect).
  • M. Kosters, A theory of hexaflexagons, Nieuw Archief Wisk., 17 (1999), 349-362.
  • 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).
  • P. K. Stockmeyer, The charm bracelet problem and its applications, pp. 339-349 of Graphs and Combinatorics (Washington, Jun 1973), Ed. by R. A. Bari and F. Harary. Lect. Notes Math., Vol. 406. Springer-Verlag, 1974.

Crossrefs

Column k=3 of A295260.
A row or column of the array in A169808.
Polyominoes: A001683(n+2) (oriented), A369314 (chiral), A208355(n-1) (achiral), A005036 {4,oo}, A007173 {3,3,oo}.
Cf. A097998, A097999, A098000 (labeled outerplanar graphs).
Cf. A111563, A111564, A111758, A111759, A111757 (unlabeled outerplanar graphs).

Programs

  • Maple
    A000108 := proc(n) if n >= 0 then binomial(2*n,n)/(n+1) ; else 0; fi; end:
    A000207 := proc(n) option remember: local k, it1, it2;
    if n mod 2 = 0 then k := n/2+2 else k := (n+3)/2 fi:
    if n mod 2 <> 0 then it1 := 0 else it1 := 1 fi:
    if (n+2) mod 3 <> 0 then it2 := 0 else it2 := 1 fi:
    RETURN(A000108(n)/(2*n+4) + it1*A000108(n/2)/4 + A000108(k-2)/2 + it2*A000108((n-1)/3)/3)
    end:
    seq(A000207(n),n=1..30) ; # (Revised Maple program from R. J. Mathar, Apr 19 2009)
    A000207 := proc(n) option remember: local k,it1,it2; if n mod 2 = 0 then k := n/2+1 else k := (n+1)/2 fi: if n mod 2 <> 0 then it1 := 0 else it1 := 1 fi: if n mod 3 <> 0 then it2 := 0 else it2 := 1 fi: RETURN(A000108(n-2)/(2*n) + it1*A000108(n/2+1-2)/4 + A000108(k-2)/2 + it2*A000108(n/3+1-2)/3) end:
    A000207 := n->(A000108(n)/(n+2)+A000108(floor(n/2))*((1+(n+1 mod 2) /2)))/2+`if`(n mod 3=1,A000108(floor((n-1)/3))/3,0); # Peter Luschny, Apr 19 2009 and M. F. Hasler, Apr 19 2009
    G:=(12*(1+x-2*x^2)+(1-4*x)^(3/2)-3*(3+2*x)*(1-4*x^2)^(1/2)-4*(1-4*x^3)^(1/2))/24/x^2: Gser:=series(G,x=0,35): seq(coeff(Gser,x^n),n=1..31); # Emeric Deutsch, Dec 19 2004
  • Mathematica
    p=3; Table[(Binomial[(p-1)n, n]/(((p-2)n+1)((p-2)n+2)) + If[OddQ[n], If[OddQ[p], Binomial[(p-1)n/2, (n-1)/2]/n, (p+1)Binomial[((p-1)n-1)/2, (n-1)/2]/((p-2)n+2)], 3Binomial[(p-1)n/2, n/2]/((p-2)n+2)]+Plus @@ Map[EulerPhi[ # ]Binomial[((p-1)n+1)/#, (n-1)/# ]/((p-1)n+1)&, Complement[Divisors[GCD[p, n-1]], {1, 2}]])/2, {n, 1, 20}] (* Robert A. Russell, Dec 11 2004 *)
    a[n_] := (CatalanNumber[n]/(n+2) + CatalanNumber[ Quotient[n, 2]] *((1 + Mod[n-1, 2]/2)))/2 + If[Mod[n, 3] == 1, CatalanNumber[ Quotient[n-1, 3]]/3, 0] ; Table[a[n], {n, 1, 28}] (* Jean-François Alcover, Sep 08 2011, after PARI *)
  • PARI
    A000207(n)=(A000108(n)/(n+2)+A000108(n\2)*if(n%2,1,3/2))/2+if(n%3==1,A000108(n\3)/3) \\ M. F. Hasler, Apr 19 2009

Formula

a(n) = C(n)/(2*n) + C(n/2+1)/4 + C(k)/2 + C(n/3+1)/3 where C(n) = A000108(n-2) if n is an integer, 0 otherwise and k = (n+1)/2 if n is odd, k = n/2+1 if n is even. Thus C(2), C(3), C(4), C(5), ... are 1, 1, 2, 5, ...
G.f.: (12*(1+x-2*x^2) + (1-4*x)^(3/2) - 3*(3+2*x)*(1-4*x^2)^(1/2) - 4*(1-4*x^3)^(1/2))/(24*x^2). - Emeric Deutsch, Dec 19 2004, from the S. J. Cyvin et al. reference.
a(n) ~ A000108(n)/(2*n+4) ~ 4^n / (2 sqrt(n Pi)*(n + 1)*(n + 2)). - M. F. Hasler, Apr 19 2009
a(n) = A001683(n+2) - A369314(n) = (A001683(n+2) + A208355(n-1)) / 2 = A369314(n) + A208355(n-1). - Robert A. Russell, Jan 19 2024
Beineke and Pippert have an explicit formula with six cases (based on the value of n mod 6). - Allan Bickle, Feb 25 2024

Extensions

More terms from James Sellers, Jul 10 2000

A007173 Number of simplicial 3-clusters with n cells.

Original entry on oeis.org

1, 1, 1, 4, 10, 40, 171, 831, 4147, 21822, 117062, 642600, 3582322, 20256885, 115888201, 669911568, 3907720521, 22979343010, 136107859377, 811430160282, 4866004426320, 29337068299728, 177738920836446, 1081668278379000, 6609923004626478, 40546403939165805
Offset: 1

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Author

Keywords

Comments

Also arises in enumeration of stereoisomers of alkane systems.
"A simplicial d-cluster may be informally described as being constructed by gluing regular d-simplexes together facet-by-facet, at each stage gluing a new simplex to exactly one facet of a cluster already constructed. The equivalence classes of such clusters under rigid motions are in one-to-one correspondence with the combinatorial types of stack polytopes." [Hering et al., 1982] - Jonathan Vos Post, Apr 22 2011
The Hering article has an error in the 14th term. - Robert A. Russell, Apr 11 2012
Also same as A027610 with mirror-image not treated as equivalence. - Brendan McKay, Mar 08 2014
Number of oriented polyominoes composed of n tetrahedral cells of the hyperbolic regular tiling with Schläfli symbol {3,3,oo}. For oriented polyominoes, chiral pairs are counted as two. - Robert A. Russell, Mar 20 2024

References

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

Crossrefs

Sum of achiral symmetry types (A047775, A047773, A047760, A047754, A047753, A047751, A047771, A047766 [type N], A047765, A047764) plus twice sum of chiral symmetry types (A047776, A047774, A047762, A047758, A047752, A047769, A047766 [type O]) in Beineke article.
Cf. A027610 (unoriented), A371350 (chiral), A371351 (achiral), A001764 (rooted), A001683(n+2) {3,oo}, A007175 {3,3,3,oo}.

Programs

  • Mathematica
    Table[Binomial[3 n, n]/(3 (2 n + 1) (2 n + 2)) + If[OddQ[n], Binomial[3 (n - 1)/2 + 1, n]/(n + 1), Binomial[3 n/2, n]/(n + 1)]/2 + 2 Switch[Mod[n, 3], 0, 0, 1, Binomial[n, (n - 1)/3]/n, 2, Binomial[n, (n - 2)/3]/n]/3, {n, 1, 30}] (* Robert A. Russell, Apr 11 2012 *)

Formula

From Robert A. Russell, Mar 20 2024: (Start)
a(n) = C(3n,n)/(3*(2n+1)*(2n+2)) + ([0==n mod 2]*C(3n/2,n) + [1==n mod 2]*C((3n-1)/2,(n-1)/2)) / (2n+2) + 2*([1==n mod 3]*C(n,(n-1)/3) + [2==n mod 3]*C(n,(n-2)/3)) / (3n).
a(n) = A027610(n) + A371350(n) = 2*A027610(n) - A371351(n) = 2*A371350(n) + A371351(n).
a(n) = H(3,n) in Table 8 of Hering link.
G.f.: (-8 + 4*G(z) - 2*G(z)^2 + z*G(z)^4 + 6*G(z^2) + 3z*G(z^2)^2 + 8z*G(z^3) + 4z^2*G(z^3)^2)/12, where G(z) = 1 + z*G(z)^3 is the g.f. for A001764. (End)

Extensions

a(14) corrected and additional terms from Robert A. Russell, Apr 11 2012

A057161 Signature-permutation of a Catalan Automorphism: rotate one step counterclockwise the triangulations of polygons encoded by A014486.

Original entry on oeis.org

0, 1, 3, 2, 7, 8, 5, 6, 4, 17, 18, 20, 21, 22, 12, 13, 15, 16, 19, 10, 11, 14, 9, 45, 46, 48, 49, 50, 54, 55, 57, 58, 59, 61, 62, 63, 64, 31, 32, 34, 35, 36, 40, 41, 43, 44, 47, 52, 53, 56, 60, 26, 27, 29, 30, 33, 38, 39, 42, 51, 24, 25, 28, 37, 23, 129, 130, 132, 133, 134
Offset: 0

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Author

Antti Karttunen, Aug 18 2000; entry revised Jun 06 2014

Keywords

Comments

This is a permutation of natural numbers induced when Euler's triangulation of convex polygons, encoded by the sequence A014486 in a straightforward way (via binary trees, cf. the illustration of the rotation of a triangulated pentagon, given in the Links section) are rotated counterclockwise.
The number of cycles in range [A014137(n-1)..A014138(n)] of this permutation is given by A001683(n+2), otherwise the same sequence as for Catalan bijections *A074679/*A074680, but shifted once left (for an explanation, see the related notes in OEIS Wiki).
E.g., in range [A014137(0)..A014138(1)] = [1,1] there is one cycle (as a(1)=1), in range [A014137(1)..A014138(2)] = [2,3] there is one cycle (as a(2)=3 and a(3)=2), in range [A014137(2)..A014138(3)] = [4,8] there is also one cycle (as a(4) = 7, a(7) = 6, a(6) = 5, a(5) = 8 and a(8) = 4), and in range [A014137(3)..A014138(4)] = [9,22] there are A001683(4+2) = 4 cycles.
From the recursive forms of A057161 and A057503 it is seen that both can be viewed as a convergent limits of a process where either the left or right side argument of A085201 in formula for A057501 is "iteratively recursivized", and on the other hand, both of these can then in turn be made to converge towards A057505 by the same method, when the other side of the formula is also "recursivized".

Crossrefs

Inverse: A057162.
Also, a "SPINE"-transform of A069774, and thus occurs as row 12 of A130403.
Other related permutations: A057163, A057164, A057501, A057504, A057505.
Cf. A001683 (cycle counts), A057544 (max cycle lengths).

Programs

  • Maple
    a(n) = CatalanRankGlobal(RotateTriangularization(A014486[n]))
    CatalanRankGlobal given in A057117 and the other Maple procedures in A038776.
    NextSubBinTree := proc(nn) local n,z,c; n := nn; c := 0; z := 0; while(c < 1) do z := 2*z + (n mod 2); c := c + (-1)^n; n := floor(n/2); od; RETURN(z); end;
    BinTreeLeftBranch := n -> NextSubBinTree(floor(n/2));
    BinTreeRightBranch := n -> NextSubBinTree(floor(n/(2^(1+binwidth(BinTreeLeftBranch(n))))));
    RotateTriangularization := proc(nn) local n,s,z,w; n := binrev(nn); z := 0; w := 0; while(1 = (n mod 2)) do s := BinTreeRightBranch(n); z := z + (2^w)*s; w := w + binwidth(s); z := z + (2^w); w := w + 1; n := floor(n/2); od; RETURN(z); end;

Formula

a(0) = 0, and for n>=1, a(n) = A085201(a(A072771(n)), A057548(A072772(n))). [This formula reflects the S-expression implementation given first in the Program section: A085201 is a 2-ary function corresponding to 'append', A072771 and A072772 correspond to 'car' and 'cdr' (known also as first/rest or head/tail in some languages), and A057548 corresponds to unary form of function 'list'.]
As a composition of related permutations:
a(n) = A069767(A069769(n)).
a(n) = A057163(A057162(A057163(n))).
a(n) = A057164(A057504(A057164(n))). [For a proof, see pp. 53-54 in the "Introductory survey ..." draft]

A262586 Square array T(n,m) (n>=0, m>=0) read by antidiagonals downwards giving number of rooted triangulations of type [n,m] up to orientation-preserving isomorphisms.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 4, 5, 6, 5, 6, 16, 21, 26, 24, 19, 48, 88, 119, 147, 133, 49, 164, 330, 538, 735, 892, 846, 150, 559, 1302, 2310, 3568, 4830, 5876, 5661, 442, 1952, 5005, 9882, 16500, 24596, 33253, 40490, 39556, 1424, 6872, 19504, 41715, 75387, 120582, 176354, 237336, 290020, 286000, 4522
Offset: 0

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Author

N. J. A. Sloane, Oct 20 2015

Keywords

Examples

			Array begins:
 ==============================================================
 n\k |    0     1      2       3       4        5         6 ...
 ----+---------------------------------------------------------
   0 |    1     1      1       4       6       19        49 ...
   1 |    1     2      5      16      48      164       559 ...
   2 |    1     6     21      88     330     1302      5005 ...
   3 |    5    26    119     538    2310     9882     41715 ...
   4 |   24   147    735    3568   16500    75387    338685 ...
   5 |  133   892   4830   24596  120582   578622   2730728 ...
   6 |  846  5876  33253  176354  900240  4493168  22037055 ...
   7 | 5661 40490 237336 1298732 6849810 35286534 178606610 ...
   ...
The first few antidiagonals are:
  1,
  1,1,
  1,2,1,
  4,5,6,5,
  6,16,21,26,24,
  19,48,88,119,147,133,
  49,164,330,538,735,892,846,
  ...
		

Crossrefs

Columns 0..2 are A002709, A002710, A002711.
Rows 0..3 are A001683, A210696, A005498, A005499.
Antidiagonal sums are A341855.
Cf. A169808 (unoriented), A169809 (achiral).

Programs

  • Maple
    A262586 := proc(n,m)
        BrownG(n,m) ; # procedure in A210696
    end proc:
    for d from 0 to 12 do
        for n from 0 to d do
            printf("%d,",A262586(n,d-n)) ;
        end do:
    end do: # R. J. Mathar, Oct 21 2015
  • Mathematica
    (* See LINKS section. *)
  • PARI
    \\ See Links in A169808 for PARI program file.
    { for(n=0, 7, for(k=0, 7, print1(OrientedTriangs(n,k), ", ")); print) } \\ Andrew Howroyd, Nov 23 2024

Formula

Brown (Eq. 6.3) gives a formula.

A005034 Number of nonequivalent dissections of a polygon into n quadrilaterals by nonintersecting diagonals up to rotation.

Original entry on oeis.org

1, 1, 1, 2, 7, 25, 108, 492, 2431, 12371, 65169, 350792, 1926372, 10744924, 60762760, 347653944, 2009690895, 11723100775, 68937782355, 408323229930, 2434289046255, 14598011263089, 88011196469040, 533216750567280, 3245004785069892, 19829768942544276, 121639211516546668
Offset: 0

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Author

Keywords

Comments

Also, with a different offset, number of colored quivers in the 2-mutation class of a quiver of Dynkin type A_n. - N. J. A. Sloane, Jan 22 2013
Closed formula is given in my paper linked below. - Nikos Apostolakis, Aug 01 2018
Number of oriented polyominoes composed of n square cells of the hyperbolic regular tiling with Schläfli symbol {4,oo}. A stereographic projection of this tiling on the Poincaré disk can be obtained via the Christensson link. For oriented polyominoes, chiral pairs are counted as two. - Robert A. Russell, Jan 20 2024

References

  • F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Camb. 1998, p. 290.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=4 of A295224.
Polyominoes: A005036 (unoriented), A369315 (chiral), A047749 (achiral), A385149 (asymmetric), A001764 (rooted), A001683(n+2) {3,oo}, A005038 {5,oo}.

Programs

  • Mathematica
    p=4; Table[Binomial[(p-1)n, n]/(((p-2)n+1)((p-2)n+2)) +If[OddQ[n], 0, Binomial[(p-1)n/2, n/2]/((p-2)n+2)]+Plus @@ Map[EulerPhi[ # ]Binomial[((p-1)n+1)/#, (n-1)/# ]/((p-1)n+1)&, Complement[Divisors[GCD[p, n-1]], {1}]], {n, 0, 20}] (* Robert A. Russell, Dec 11 2004 *)
    Table[(3Binomial[3n,n]/(2n+1)-Binomial[3n+1,n]/(n+1)-If[OddQ[n],-2Binomial[(3n-1)/2,(n-1)/2]-If[1==Mod[n,4],4Binomial[(3n-3)/4,(n-1)/4],0],-2Binomial[3n/2,n/2]]/(n+1))/4,{n,0,30}] (* Robert A. Russell, Jun 19 2025 *)

Formula

a(n) ~ 3^(3*n + 1/2) / (sqrt(Pi) * n^(5/2) * 2^(2*n + 3)). - Vaclav Kotesovec, Mar 13 2016
a(n) = A005036(n) + A369315(n) = 2*A005036(n) - A047749(n) = 2*A369315(n) + A047749(n). - Robert A. Russell, Jan 19 2024
G.f.: (3*G(z) - G(z)^2 + 2*G(z^2) + z*G(z^2)^2 + 2z*G(z^4)) / 4, where G(z)=1+z*G(z)^3 is the g.f. for A001764. - Robert A. Russell, Jun 19 2025

Extensions

Name clarified by Andrew Howroyd, Nov 20 2017
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