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

A000110 Bell or exponential numbers: number of ways to partition a set of n labeled elements.

Original entry on oeis.org

1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 1382958545, 10480142147, 82864869804, 682076806159, 5832742205057, 51724158235372, 474869816156751, 4506715738447323, 44152005855084346, 445958869294805289, 4638590332229999353, 49631246523618756274
Offset: 0

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Comments

The leading diagonal of its difference table is the sequence shifted, see Bernstein and Sloane (1995). - N. J. A. Sloane, Jul 04 2015
Also the number of equivalence relations that can be defined on a set of n elements. - Federico Arboleda (federico.arboleda(AT)gmail.com), Mar 09 2005
a(n) = number of nonisomorphic colorings of a map consisting of a row of n+1 adjacent regions. Adjacent regions cannot have the same color. - David W. Wilson, Feb 22 2005
If an integer is squarefree and has n distinct prime factors then a(n) is the number of ways of writing it as a product of its divisors. - Amarnath Murthy, Apr 23 2001
Consider rooted trees of height at most 2. Letting each tree 'grow' into the next generation of n means we produce a new tree for every node which is either the root or at height 1, which gives the Bell numbers. - Jon Perry, Jul 23 2003
Begin with [1,1] and follow the rule that [1,k] -> [1,k+1] and [1,k] k times, e.g., [1,3] is transformed to [1,4], [1,3], [1,3], [1,3]. Then a(n) is the sum of all components: [1,1] = 2; [1,2], [1,1] = 5; [1,3], [1,2], [1,2], [1,2], [1,1] = 15; etc. - Jon Perry, Mar 05 2004
Number of distinct rhyme schemes for a poem of n lines: a rhyme scheme is a string of letters (e.g., 'abba') such that the leftmost letter is always 'a' and no letter may be greater than one more than the greatest letter to its left. Thus 'aac' is not valid since 'c' is more than one greater than 'a'. For example, a(3)=5 because there are 5 rhyme schemes: aaa, aab, aba, abb, abc; also see example by Neven Juric. - Bill Blewett, Mar 23 2004
In other words, number of length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(0)=0 and s(k) <= 1 + max(prefix) for k >= 1, see example (cf. A080337 and A189845). - Joerg Arndt, Apr 30 2011
Number of partitions of {1, ..., n+1} into subsets of nonconsecutive integers, including the partition 1|2|...|n+1. E.g., a(3)=5: there are 5 partitions of {1,2,3,4} into subsets of nonconsecutive integers, namely, 13|24, 13|2|4, 14|2|3, 1|24|3, 1|2|3|4. - Augustine O. Munagi, Mar 20 2005
Triangle (addition) scheme to produce terms, derived from the recurrence, from Oscar Arevalo (loarevalo(AT)sbcglobal.net), May 11 2005:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
... [This is Aitken's array A011971]
With P(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j,i) = the j-th part of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, one has: a(n) = Sum_{i=1..P(n)} (n!/(Product_{j=1..p(i)} p(i,j)!)) * (1/(Product_{j=1..d(i)} m(i,j)!)). - Thomas Wieder, May 18 2005
a(n+1) is the number of binary relations on an n-element set that are both symmetric and transitive. - Justin Witt (justinmwitt(AT)gmail.com), Jul 12 2005
If the rule from Jon Perry, Mar 05 2004, is used, then a(n-1) = [number of components used to form a(n)] / 2. - Daniel Kuan (dkcm(AT)yahoo.com), Feb 19 2006
a(n) is the number of functions f from {1,...,n} to {1,...,n,n+1} that satisfy the following two conditions for all x in the domain: (1) f(x) > x; (2) f(x)=n+1 or f(f(x))=n+1. E.g., a(3)=5 because there are exactly five functions that satisfy the two conditions: f1={(1,4),(2,4),(3,4)}, f2={(1,4),(2,3),(3,4)}, f3={(1,3),(2,4),(3,4)}, f4={(1,2),(2,4),(3,4)} and f5={(1,3),(2,3),(3,4)}. - Dennis P. Walsh, Feb 20 2006
Number of asynchronic siteswap patterns of length n which have no zero-throws (i.e., contain no 0's) and whose number of orbits (in the sense given by Allen Knutson) is equal to the number of balls. E.g., for n=4, the condition is satisfied by the following 15 siteswaps: 4444, 4413, 4242, 4134, 4112, 3441, 2424, 1344, 2411, 1313, 1241, 2222, 3131, 1124, 1111. Also number of ways to choose n permutations from identity and cyclic permutations (1 2), (1 2 3), ..., (1 2 3 ... n) so that their composition is identity. For n=3 we get the following five: id o id o id, id o (1 2) o (1 2), (1 2) o id o (1 2), (1 2) o (1 2) o id, (1 2 3) o (1 2 3) o (1 2 3). (To see the bijection, look at Ehrenborg and Readdy paper.) - Antti Karttunen, May 01 2006
a(n) is the number of permutations on [n] in which a 3-2-1 (scattered) pattern occurs only as part of a 3-2-4-1 pattern. Example: a(3) = 5 counts all permutations on [3] except 321. See "Eigensequence for Composition" reference a(n) = number of permutation tableaux of size n (A000142) whose first row contains no 0's. Example: a(3)=5 counts {{}, {}, {}}, {{1}, {}}, {{1}, {0}}, {{1}, {1}}, {{1, 1}}. - David Callan, Oct 07 2006
From Gottfried Helms, Mar 30 2007: (Start)
This sequence is also the first column in the matrix-exponential of the (lower triangular) Pascal-matrix, scaled by exp(-1): PE = exp(P) / exp(1) =
1
1 1
2 2 1
5 6 3 1
15 20 12 4 1
52 75 50 20 5 1
203 312 225 100 30 6 1
877 1421 1092 525 175 42 7 1
First 4 columns are A000110, A033306, A105479, A105480. The general case is mentioned in the two latter entries. PE is also the Hadamard-product Toeplitz(A000110) (X) P:
1
1 1
2 1 1
5 2 1 1
15 5 2 1 1 (X) P
52 15 5 2 1 1
203 52 15 5 2 1 1
877 203 52 15 5 2 1 1
(End)
The terms can also be computed with finite steps and precise integer arithmetic. Instead of exp(P)/exp(1) one can compute A = exp(P - I) where I is the identity-matrix of appropriate dimension since (P-I) is nilpotent to the order of its dimension. Then a(n)=A[n,1] where n is the row-index starting at 1. - Gottfried Helms, Apr 10 2007
When n is prime, a(n) == 2 (mod n), but the converse is not always true. Define a Bell pseudoprime to be a composite number n such that a(n) == 2 (mod n). W. F. Lunnon recently found the Bell pseudoprimes 21361 = 41*521 and C46 = 3*23*16218646893090134590535390526854205539989357 and conjectured that Bell pseudoprimes are extremely scarce. So the second Bell pseudoprime is unlikely to be known with certainty in the near future. I confirmed that 21361 is the first. - David W. Wilson, Aug 04 2007 and Sep 24 2007
This sequence and A000587 form a reciprocal pair under the list partition transform described in A133314. - Tom Copeland, Oct 21 2007
Starting (1, 2, 5, 15, 52, ...), equals row sums and right border of triangle A136789. Also row sums of triangle A136790. - Gary W. Adamson, Jan 21 2008
This is the exponential transform of A000012. - Thomas Wieder, Sep 09 2008
From Abdullahi Umar, Oct 12 2008: (Start)
a(n) is also the number of idempotent order-decreasing full transformations (of an n-chain).
a(n) is also the number of nilpotent partial one-one order-decreasing transformations (of an n-chain).
a(n+1) is also the number of partial one-one order-decreasing transformations (of an n-chain). (End)
From Peter Bala, Oct 19 2008: (Start)
Bell(n) is the number of n-pattern sequences [Cooper & Kennedy]. An n-pattern sequence is a sequence of integers (a_1,...,a_n) such that a_i = i or a_i = a_j for some j < i. For example, Bell(3) = 5 since the 3-pattern sequences are (1,1,1), (1,1,3), (1,2,1), (1,2,2) and (1,2,3).
Bell(n) is the number of sequences of positive integers (N_1,...,N_n) of length n such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (see the comment by B. Blewett above). It is interesting to note that if we strengthen the latter condition to N_(i+1) <= 1 + N_i we get the Catalan numbers A000108 instead of the Bell numbers.
(End)
Equals the eigensequence of Pascal's triangle, A007318; and starting with offset 1, = row sums of triangles A074664 and A152431. - Gary W. Adamson, Dec 04 2008
The entries f(i, j) in the exponential of the infinite lower-triangular matrix of binomial coefficients b(i, j) are f(i, j) = b(i, j) e a(i - j). - David Pasino, Dec 04 2008
Equals lim_{k->oo} A071919^k. - Gary W. Adamson, Jan 02 2009
Equals A154107 convolved with A014182, where A014182 = expansion of exp(1-x-exp(-x)), the eigensequence of A007318^(-1). Starting with offset 1 = A154108 convolved with (1,2,3,...) = row sums of triangle A154109. - Gary W. Adamson, Jan 04 2009
Repeated iterates of (binomial transform of [1,0,0,0,...]) will converge upon (1, 2, 5, 15, 52, ...) when each result is prefaced with a "1"; such that the final result is the fixed limit: (binomial transform of [1,1,2,5,15,...]) = (1,2,5,15,52,...). - Gary W. Adamson, Jan 14 2009
From Karol A. Penson, May 03 2009: (Start)
Relation between the Bell numbers B(n) and the n-th derivative of 1/Gamma(1+x) evaluated at x=1:
a) produce a number of such derivatives through seq(subs(x=1, simplify((d^n/dx^n)GAMMA(1+x)^(-1))), n=1..5);
b) leave them expressed in terms of digamma (Psi(k)) and polygamma (Psi(k,n)) functions and unevaluated;
Examples of such expressions, for n=1..5, are:
n=1: -Psi(1),
n=2: -(-Psi(1)^2 + Psi(1,1)),
n=3: -Psi(1)^3 + 3*Psi(1)*Psi(1,1) - Psi(2,1),
n=4: -(-Psi(1)^4 + 6*Psi(1)^2*Psi(1,1) - 3*Psi(1,1)^2 - 4*Psi(1)*Psi(2,1) + Psi(3, 1)),
n=5: -Psi(1)^5 + 10*Psi(1)^3*Psi(1,1) - 15*Psi(1)*Psi(1,1)^2 - 10*Psi(1)^2*Psi(2,1) + 10*Psi(1,1)*Psi(2,1) + 5*Psi(1)*Psi(3,1) - Psi(4,1);
c) for a given n, read off the sum of absolute values of coefficients of every term involving digamma or polygamma functions.
This sum is equal to B(n). Examples: B(1)=1, B(2)=1+1=2, B(3)=1+3+1=5, B(4)=1+6+3+4+1=15, B(5)=1+10+15+10+10+5+1=52;
d) Observe that this decomposition of the Bell number B(n) apparently does not involve the Stirling numbers of the second kind explicitly.
(End)
The numbers given above by Penson lead to the multinomial coefficients A036040. - Johannes W. Meijer, Aug 14 2009
Column 1 of A162663. - Franklin T. Adams-Watters, Jul 09 2009
Asymptotic expansions (0!+1!+2!+...+(n-1)!)/(n-1)! = a(0) + a(1)/n + a(2)/n^2 + ... and (0!+1!+2!+...+n!)/n! = 1 + a(0)/n + a(1)/n^2 + a(2)/n^3 + .... - Michael Somos, Jun 28 2009
Starting with offset 1 = row sums of triangle A165194. - Gary W. Adamson, Sep 06 2009
a(n+1) = A165196(2^n); where A165196 begins: (1, 2, 4, 5, 7, 12, 14, 15, ...). such that A165196(2^3) = 15 = A000110(4). - Gary W. Adamson, Sep 06 2009
The divergent series g(x=1,m) = 1^m*1! - 2^m*2! + 3^m*3! - 4^m*4! + ..., m >= -1, which for m=-1 dates back to Euler, is related to the Bell numbers. We discovered that g(x=1,m) = (-1)^m * (A040027(m) - A000110(m+1) * A073003). We observe that A073003 is Gompertz's constant and that A040027 was published by Gould, see for more information A163940. - Johannes W. Meijer, Oct 16 2009
a(n) = E(X^n), i.e., the n-th moment about the origin of a random variable X that has a Poisson distribution with (rate) parameter, lambda = 1. - Geoffrey Critzer, Nov 30 2009
Let A000110 = S(x), then S(x) = A(x)/A(x^2) when A(x) = A173110; or (1, 1, 2, 5, 15, 52, ...) = (1, 1, 3, 6, 20, 60, ...) / (1, 0, 1, 0, 3, 0, 6, 0, 20, ...). - Gary W. Adamson, Feb 09 2010
The Bell numbers serve as the upper limit for the number of distinct homomorphic images from any given finite universal algebra. Every algebra homomorphism is determined by its kernel, which must be a congruence relation. As the number of possible congruence relations with respect to a finite universal algebra must be a subset of its possible equivalence classes (given by the Bell numbers), it follows naturally. - Max Sills, Jun 01 2010
For a proof of the o.g.f. given in the R. Stephan comment see, e.g., the W. Lang link under A071919. - Wolfdieter Lang, Jun 23 2010
Let B(x) = (1 + x + 2x^2 + 5x^3 + ...). Then B(x) is satisfied by A(x)/A(x^2) where A(x) = polcoeff A173110: (1 + x + 3x^2 + 6x^3 + 20x^4 + 60x^5 + ...) = B(x) * B(x^2) * B(x^4) * B(x^8) * .... - Gary W. Adamson, Jul 08 2010
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 without choosing any two colors the same positive number of times. (See related comments for A000108, A008277, A016098.) - Matthew Vandermast, Nov 22 2010
A binary counter with faulty bits starts at value 0 and attempts to increment by 1 at each step. Each bit that should toggle may or may not do so. a(n) is the number of ways that the counter can have the value 0 after n steps. E.g., for n=3, the 5 trajectories are 0,0,0,0; 0,1,0,0; 0,1,1,0; 0,0,1,0; 0,1,3,0. - David Scambler, Jan 24 2011
No Bell number is divisible by 8, and no Bell number is congruent to 6 modulo 8; see Theorem 6.4 and Table 1.7 in Lunnon, Pleasants and Stephens. - Jon Perry, Feb 07 2011, clarified by Eric Rowland, Mar 26 2014
a(n+1) is the number of (symmetric) positive semidefinite n X n 0-1 matrices. These correspond to equivalence relations on {1,...,n+1}, where matrix element M[i,j] = 1 if and only if i and j are equivalent to each other but not to n+1. - Robert Israel, Mar 16 2011
a(n) is the number of monotonic-labeled forests on n vertices with rooted trees of height less than 2. We note that a labeled rooted tree is monotonic-labeled if the label of any parent vertex is greater than the label of any offspring vertex. See link "Counting forests with Stirling and Bell numbers". - Dennis P. Walsh, Nov 11 2011
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A000772 and A094198. - Peter Bala, Nov 25 2011
B(n) counts the length n+1 rhyme schemes without repetitions. E.g., for n=2 there are 5 rhyme schemes of length 3 (aaa, aab, aba, abb, abc), and the 2 without repetitions are aba, abc. This is basically O. Munagi's result that the Bell numbers count partitions into subsets of nonconsecutive integers (see comment above dated Mar 20 2005). - Eric Bach, Jan 13 2012
Right and left borders and row sums of A212431 = A000110 or a shifted variant. - Gary W. Adamson, Jun 21 2012
Number of maps f: [n] -> [n] where f(x) <= x and f(f(x)) = f(x) (projections). - Joerg Arndt, Jan 04 2013
Permutations of [n] avoiding any given one of the 8 dashed patterns in the equivalence classes (i) 1-23, 3-21, 12-3, 32-1, and (ii) 1-32, 3-12, 21-3, 23-1. (See Claesson 2001 reference.) - David Callan, Oct 03 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
Sum_{n>=0} a(n)/n! = e^(e-1) = 5.57494152476..., see A234473. - Richard R. Forberg, Dec 26 2013 (This is the e.g.f. for x=1. - Wolfdieter Lang, Feb 02 2015)
Sum_{j=0..n} binomial(n,j)*a(j) = (1/e)*Sum_{k>=0} (k+1)^n/k! = (1/e) Sum_{k=1..oo} k^(n+1)/k! = a(n+1), n >= 0, using the Dobinski formula. See the comment by Gary W. Adamson, Dec 04 2008 on the Pascal eigensequence. - Wolfdieter Lang, Feb 02 2015
In fact it is not really an eigensequence of the Pascal matrix; rather the Pascal matrix acts on the sequence as a shift. It is an eigensequence (the unique eigensequence with eigenvalue 1) of the matrix derived from the Pascal matrix by adding at the top the row [1, 0, 0, 0 ...]. The binomial sum formula may be derived from the definition in terms of partitions: label any element X of a set S of N elements, and let X(k) be the number of subsets of S containing X with k elements. Since each subset has a unique coset, the number of partitions p(N) of S is given by p(N) = Sum_{k=1..N} (X(k) p(N-k)); trivially X(k) = N-1 choose k-1. - Mason Bogue, Mar 20 2015
a(n) is the number of ways to nest n matryoshkas (Russian nesting dolls): we may identify {1, 2, ..., n} with dolls of ascending sizes and the sets of a set partition with stacks of dolls. - Carlo Sanna, Oct 17 2015
Number of permutations of [n] where the initial elements of consecutive runs of increasing elements are in decreasing order. a(4) = 15: `1234, `2`134, `23`14, `234`1, `24`13, `3`124, `3`2`14, `3`24`1, `34`12, `34`2`1, `4`123, `4`2`13, `4`23`1, `4`3`12, `4`3`2`1. - Alois P. Heinz, Apr 27 2016
Taking with alternating signs, the Bell numbers are the coefficients in the asymptotic expansion (Ramanujan): (-1)^n*(A000166(n) - n!/exp(1)) ~ 1/n - 2/n^2 + 5/n^3 - 15/n^4 + 52/n^5 - 203/n^6 + O(1/n^7). - Vladimir Reshetnikov, Nov 10 2016
Number of treeshelves avoiding pattern T231. See A278677 for definitions and examples. - Sergey Kirgizov, Dec 24 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. - N. J. A. Sloane, Feb 09 2017
a(n) = Sum(# of standard immaculate tableaux of shape m, m is a composition of n), where this sum is over all integer compositions m of n > 0. This formula is easily seen to hold by identifying standard immaculate tableaux of size n with set partitions of { 1, 2, ..., n }. For example, if we sum over integer compositions of 4 lexicographically, we see that 1+1+2+1+3+3+3+1 = 15 = A000110(4). - John M. Campbell, Jul 17 2017
a(n) is also the number of independent vertex sets (and vertex covers) in the (n-1)-triangular honeycomb bishop graph. - Eric W. Weisstein, Aug 10 2017
Even-numbered entries represent the numbers of configurations of identity and non-identity for alleles of a gene in n diploid individuals with distinguishable maternal and paternal alleles. - Noah A Rosenberg, Jan 28 2019
Number of partial equivalence relations (PERs) on a set with n elements (offset=1), i.e., number of symmetric, transitive (not necessarily reflexive) relations. The idea is to add a dummy element D to the set, and then take equivalence relations on the result; anything equivalent to D is then removed for the partial equivalence relation. - David Spivak, Feb 06 2019
Number of words of length n+1 with no repeated letters, when letters are unlabeled. - Thomas Anton, Mar 14 2019
Named by Becker and Riordan (1948) after the Scottish-American mathematician and writer Eric Temple Bell (1883 - 1960). - Amiram Eldar, Dec 04 2020
Also the number of partitions of {1,2,...,n+1} with at most one n+1 singleton. E.g., a(3)=5: {13|24, 12|34, 123|4, 14|23, 1234}. - Yuchun Ji, Dec 21 2020
a(n) is the number of sigma algebras on the set of n elements. Note that each sigma algebra is generated by a partition of the set. For example, the sigma algebra generated by the partition {{1}, {2}, {3,4}} is {{}, {1}, {2}, {1,2}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. - Jianing Song, Apr 01 2021
a(n) is the number of P_3-free graphs on n labeled nodes. - M. Eren Kesim, Jun 04 2021
a(n) is the number of functions X:([n] choose 2) -> {+,-} such that for any ordered 3-tuple abc we have X(ab)X(ac)X(bc) not in {+-+,++-,-++}. - Robert Lauff, Dec 09 2022
From Manfred Boergens, Mar 11 2025: (Start)
The partitions in the definition can be described as disjoint covers of the set. "Covers" in general give rise to the following amendments:
For disjoint covers which may include one empty set see A186021.
For arbitrary (including non-disjoint) covers see A003465.
For arbitrary (including non-disjoint) covers which may include one empty set see A000371. (End)

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 203*x^6 + 877*x^7 + 4140*x^8 + ...
From Neven Juric, Oct 19 2009: (Start)
The a(4)=15 rhyme schemes for n=4 are
  aaaa, aaab, aaba, aabb, aabc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, abcd
The a(5)=52 rhyme schemes for n=5 are
  aaaaa, aaaab, aaaba, aaabb, aaabc, aabaa, aabab, aabac, aabba, aabbb, aabbc, aabca, aabcb, aabcc, aabcd, abaaa, abaab, abaac, ababa, ababb, ababc, abaca, abacb, abacc, abacd, abbaa, abbab, abbac, abbba, abbbb, abbbc, abbca, abbcb, abbcc, abbcd, abcaa, abcab, abcac, abcad, abcba, abcbb, abcbc, abcbd, abcca, abccb, abccc, abccd, abcda, abcdb, abcdc, abcdd, abcde
(End)
From _Joerg Arndt_, Apr 30 2011: (Start)
Restricted growth strings (RGS):
For n=0 there is one empty string;
for n=1 there is one string [0];
for n=2 there are 2 strings [00], [01];
for n=3 there are a(3)=5 strings [000], [001], [010], [011], and [012];
for n=4 there are a(4)=15 strings
1: [0000], 2: [0001], 3: [0010], 4: [0011], 5: [0012], 6: [0100], 7: [0101], 8: [0102], 9: [0110], 10: [0111], 11: [0112], 12: [0120], 13: [0121], 14: [0122], 15: [0123].
These are one-to-one with the rhyme schemes (identify a=0, b=1, c=2, etc.).
(End)
Consider the set S = {1, 2, 3, 4}. The a(4) = 1 + 3 + 6 + 4 + 1 = 15 partitions are: P1 = {{1}, {2}, {3}, {4}}; P21 .. P23 = {{a,4}, S\{a,4}} with a = 1, 2, 3; P24 .. P29 = {{a}, {b}, S\{a,b}} with 1 <= a < b <= 4;  P31 .. P34 = {S\{a}, {a}} with a = 1 .. 4; P4 = {S}. See the Bottomley link for a graphical illustration. - _M. F. Hasler_, Oct 26 2017
		

References

  • Stefano Aguzzoli, Brunella Gerla and Corrado Manara, Poset Representation for Goedel and Nilpotent Minimum Logics, in Symbolic and Quantitative Approaches to Reasoning with Uncertainty, Lecture Notes in Computer Science, Volume 3571/2005, Springer-Verlag. [Added by N. J. A. Sloane, Jul 08 2009]
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  • J. Balogh, B. Bollobas and D. Weinreich, A jump to the Bell numbers for hereditary graph properties, J. Combin. Theory Ser. B 95 (2005), no. 1, 29-48.
  • R. E. Beard, On the coefficients in the expansion of exp(exp(t)) and exp(-exp(t)), J. Institute Actuaries, 76 (1951), 152-163.
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  • E. T. Bell, The iterated exponential numbers, Ann. Math., 39 (1938), 539-557.
  • C. M. Bender, D. C. Brody and B. K. Meister, Quantum Field Theory of Partitions, J. Math. Phys., 40,7 (1999), 3239-45.
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  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 92-93.
  • John H. Conway et al., The Symmetries of Things, Peters, 2008, p. 207.
  • Colin Defant, Highly sorted permutations and Bell numbers, ECA 1:1 (2021) Article S2R6.
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  • H. W. Gould, Research bibliography of two special number sequences, Mathematica Monongaliae, Vol. 12, 1971.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, 2nd ed., p. 493.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 26.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.5 (p. 418).
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  • J. Levine and R. E. Dalton, Minimum periods, modulo p, of first-order Bell exponential integers, Math. Comp., 16 (1962), 416-423.
  • Levinson, H.; Silverman, R. Topologies on finite sets. II. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 699--712, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561090 (81c:54006)
  • S. Linusson, The number of M-sequences and f-vectors, Combinatorica, 19 (1999), 255-266.
  • L. Lovasz, Combinatorial Problems and Exercises, North-Holland, 1993, pp. 14-15.
  • M. Meier, On the number of partitions of a given set, Amer. Math. Monthly, 114 (2007), p. 450.
  • Merris, Russell, and Stephen Pierce. "The Bell numbers and r-fold transitivity." Journal of Combinatorial Theory, Series A 12.1 (1972): 155-157.
  • Moser, Leo, and Max Wyman. An asymptotic formula for the Bell numbers. Trans. Royal Soc. Canada, 49 (1955), 49-53.
  • A. Murthy, Generalization of partition function, introducing Smarandache factor partition, Smarandache Notions Journal, Vol. 11, No. 1-2-3, Spring 2000.
  • Amarnath Murthy and Charles Ashbacher, Generalized Partitions and Some New Ideas on Number Theory and Smarandache Sequences, Hexis, Phoenix; USA 2005. See Section 1.4,1.8.
  • P. Peart, Hankel determinants via Stieltjes matrices. Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000). Congr. Numer. 144 (2000), 153-159.
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  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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  • R. P. Stanley, Enumerative Combinatorics, Cambridge; see Section 1.4 and Example 5.2.4.
  • Abdullahi Umar, On the semigroups of order-decreasing finite full transformations, Proc. Roy. Soc. Edinburgh 120A (1992), 129-142.
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Crossrefs

Equals row sums of triangle A008277 (Stirling subset numbers).
Partial sums give A005001. a(n) = A123158(n, 0).
See A061462 for powers of 2 dividing a(n).
Rightmost diagonal of triangle A121207. A144293 gives largest prime factor.
Equals row sums of triangle A152432.
Row sums, right and left borders of A212431.
A diagonal of A011971. - N. J. A. Sloane, Jul 31 2012
Diagonal of A102661. - Manfred Boergens, Mar 11 2025
Cf. A054767 (period of this sequence mod n).
Row sums are A048993. - Wolfdieter Lang, Oct 16 2014
Sequences in the Erné (1974) paper: A000110, A000798, A001035, A001927, A001929, A006056, A006057, A006058, A006059.
Bell polynomials B(n,x): A001861 (x=2), A027710 (x=3), A078944 (x=4), A144180 (x=5), A144223 (x=6), A144263 (x=7), A221159 (x=8).
Cf. A243991 (sum of reciprocals), A085686 (inv. Euler Transf.).

Programs

  • Haskell
    type N = Integer
    n_partitioned_k :: N -> N -> N
    1 `n_partitioned_k` 1 = 1
    1 `n_partitioned_k` _ = 0
    n `n_partitioned_k` k = k * (pred n `n_partitioned_k` k) + (pred n `n_partitioned_k` pred k)
    n_partitioned :: N -> N
    n_partitioned 0 = 1
    n_partitioned n = sum $ map (\k -> n `n_partitioned_k` k) $ [1 .. n]
    -- Felix Denis, Oct 16 2012
    
  • Haskell
    a000110 = sum . a048993_row -- Reinhard Zumkeller, Jun 30 2013
    
  • Julia
    function a(n)
        t = [zeros(BigInt, n+1) for _ in 1:n+1]
        t[1][1] = 1
        for i in 2:n+1
            t[i][1] = t[i-1][i-1]
            for j in 2:i
                t[i][j] = t[i-1][j-1] + t[i][j-1]
            end
        end
        return [t[i][1] for i in 1:n+1]
    end
    print(a(28)) # Paul Muljadi, May 07 2024
    
  • Magma
    [Bell(n): n in [0..40]]; // Vincenzo Librandi, Feb 07 2011
    
  • Maple
    A000110 := proc(n) option remember; if n <= 1 then 1 else add( binomial(n-1,i)*A000110(n-1-i),i=0..n-1); fi; end: # version 1
    A := series(exp(exp(x)-1),x,60): A000110 := n->n!*coeff(A,x,n): # version 2
    A000110:= n-> add(Stirling2(n, k), k=0..n): seq(A000110(n), n=0..22); # version 3, from Zerinvary Lajos, Jun 28 2007
    A000110 := n -> combinat[bell](n): # version 4, from Peter Luschny, Mar 30 2011
    spec:= [S, {S=Set(U, card >= 1), U=Set(Z, card >= 1)}, labeled]: G:={P=Set(Set(Atom, card>0))}: combstruct[gfsolve](G, unlabeled, x): seq(combstruct[count]([P, G, labeled], size=i), i=0..22);  # version 5, Zerinvary Lajos, Dec 16 2007
    BellList := proc(m) local A, P, n; A := [1, 1]; P := [1]; for n from 1 to m - 2 do
    P := ListTools:-PartialSums([A[-1], op(P)]); A := [op(A), P[-1]] od; A end: BellList(29); # Peter Luschny, Mar 24 2022
  • Mathematica
    f[n_] := Sum[ StirlingS2[n, k], {k, 0, n}]; Table[ f[n], {n, 0, 40}] (* Robert G. Wilson v *)
    Table[BellB[n], {n, 0, 40}] (* Harvey P. Dale, Mar 01 2011 *)
    B[0] = 1; B[n_] := 1/E Sum[k^(n - 1)/(k-1)!, {k, 1, Infinity}] (* Dimitri Papadopoulos, Mar 10 2015, edited by M. F. Hasler, Nov 30 2018 *)
    BellB[Range[0,40]] (* Eric W. Weisstein, Aug 10 2017 *)
    b[1] = 1; k = 1; Flatten[{1, Table[Do[j = k; k += b[m]; b[m] = j;, {m, 1, n-1}]; b[n] = k, {n, 1, 40}]}] (* Vaclav Kotesovec, Sep 07 2019 *)
    Table[j! Coefficient[Series[Exp[Exp[x] - 1], {x, 0, 20}], x, j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 01 2023 *)
    Table[(D[Exp[Exp[x]], {x, n}] /. x -> 0)/E, {n, 0, 20}] (* Joan Ludevid, Nov 05 2024 *)
  • Maxima
    makelist(belln(n),n,0,40); /* Emanuele Munarini, Jul 04 2011 */
    
  • PARI
    {a(n) = my(m); if( n<0, 0, m = contfracpnqn( matrix(2, n\2, i, k, if( i==1, -k*x^2, 1 - (k+1)*x))); polcoeff(1 / (1 - x + m[2,1] / m[1,1]) + x * O(x^n), n))}; /* Michael Somos */
    
  • PARI
    {a(n) = polcoeff( sum( k=0, n, prod( i=1, k, x / (1 - i*x)), x^n * O(x)), n)}; /* Michael Somos, Aug 22 2004 */
    
  • PARI
    a(n)=round(exp(-1)*suminf(k=0,1.0*k^n/k!)) \\ Gottfried Helms, Mar 30 2007 - WARNING! For illustration only: Gives silently a wrong result for n = 42 and an error for n > 42, with standard precision of 38 digits. - M. F. Hasler, Nov 30 2018
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( exp( x + x * O(x^n)) - 1), n))}; /* Michael Somos, Jun 28 2009 */
    
  • PARI
    Vec(serlaplace(exp(exp('x+O('x^66))-1))) \\ Joerg Arndt, May 26 2012
    
  • PARI
    A000110(n)=sum(k=0,n,stirling(n,k,2)) \\ M. F. Hasler, Nov 30 2018
    
  • Perl
    use bignum;sub a{my($n)=@;my@t=map{[(0)x($n+1)]}0..$n;$t[0][0]=1;for my$i(1..$n){$t[$i][0]=$t[$i-1][$i-1];for my$j(1..$i){$t[$i][$j]=$t[$i-1][$j-1]+$t[$i][$j-1]}}return map{$t[$][0]}0..$n-1}print join(", ",a(28)),"\n" # Paul Muljadi, May 08 2024
  • Python
    # The objective of this implementation is efficiency.
    # m -> [a(0), a(1), ..., a(m)] for m > 0.
    def A000110_list(m):
        A = [0 for i in range(m)]
        A[0] = 1
        R = [1, 1]
        for n in range(1, m):
            A[n] = A[0]
            for k in range(n, 0, -1):
                A[k-1] += A[k]
            R.append(A[0])
        return R
    A000110_list(40) # Peter Luschny, Jan 18 2011
    
  • Python
    # requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
    from itertools import accumulate
    A000110, blist, b = [1,1], [1], 1
    for _ in range(20):
        blist = list(accumulate([b]+blist))
        b = blist[-1]
        A000110.append(b) # Chai Wah Wu, Sep 02 2014, updated Chai Wah Wu, Sep 19 2014
    
  • Python
    from sympy import bell
    print([bell(n) for n in range(27)]) # Michael S. Branicky, Dec 15 2021
    
  • Python
    from functools import cache
    @cache
    def a(n, k=0): return int(n < 1) or k*a(n-1, k) + a(n-1, k+1)
    print([a(n) for n in range(27)])  # Peter Luschny, Jun 14 2022
    
  • Sage
    from sage.combinat.expnums import expnums2; expnums2(30, 1) # Zerinvary Lajos, Jun 26 2008
    
  • Sage
    [bell_number(n) for n in (0..40)] # G. C. Greubel, Jun 13 2019
    

Formula

E.g.f.: exp(exp(x) - 1).
Recurrence: a(n+1) = Sum_{k=0..n} a(k)*binomial(n, k).
a(n) = Sum_{k=0..n} Stirling2(n, k).
a(n) = Sum_{j=0..n-1} (1/(n-1)!)*A000166(j)*binomial(n-1, j)*(n-j)^(n-1). - André F. Labossière, Dec 01 2004
G.f.: (Sum_{k>=0} 1/((1-k*x)*k!))/exp(1) = hypergeom([-1/x], [(x-1)/x], 1)/exp(1) = ((1-2*x)+LaguerreL(1/x, (x-1)/x, 1)+x*LaguerreL(1/x, (2*x-1)/x, 1))*Pi/(x^2*sin(Pi*(2*x-1)/x)), where LaguerreL(mu, nu, z) = (gamma(mu+nu+1)/(gamma(mu+1)*gamma(nu+1)))* hypergeom([-mu], [nu+1], z) is the Laguerre function, the analytic extension of the Laguerre polynomials, for mu not equal to a nonnegative integer. This generating function has an infinite number of poles accumulating in the neighborhood of x=0. - Karol A. Penson, Mar 25 2002
a(n) = exp(-1)*Sum_{k >= 0} k^n/k! [Dobinski]. - Benoit Cloitre, May 19 2002
a(n) is asymptotic to n!*(2 Pi r^2 exp(r))^(-1/2) exp(exp(r)-1) / r^n, where r is the positive root of r exp(r) = n. See, e.g., the Odlyzko reference.
a(n) is asymptotic to b^n*exp(b-n-1/2)*sqrt(b/(b+n)) where b satisfies b*log(b) = n - 1/2 (see Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed., p. 493). - Benoit Cloitre, Oct 23 2002, corrected by Vaclav Kotesovec, Jan 06 2013
Lovasz (Combinatorial Problems and Exercises, North-Holland, 1993, Section 1.14, Problem 9) gives another asymptotic formula, quoted by Mezo and Baricz. - N. J. A. Sloane, Mar 26 2015
G.f.: Sum_{k>=0} x^k/(Product_{j=1..k} (1-j*x)) (see Klazar for a proof). - Ralf Stephan, Apr 18 2004
a(n+1) = exp(-1)*Sum_{k>=0} (k+1)^(n)/k!. - Gerald McGarvey, Jun 03 2004
For n>0, a(n) = Aitken(n-1, n-1) [i.e., a(n-1, n-1) of Aitken's array (A011971)]. - Gerald McGarvey, Jun 26 2004
a(n) = Sum_{k=1..n} (1/k!)*(Sum_{i=1..k} (-1)^(k-i)*binomial(k, i)*i^n + 0^n). - Paul Barry, Apr 18 2005
a(n) = A032347(n) + A040027(n+1). - Jon Perry, Apr 26 2005
a(n) = (2*n!/(Pi*e))*Im( Integral_{x=0..Pi} e^(e^(e^(ix))) sin(nx) dx ) where Im denotes imaginary part [Cesaro]. - David Callan, Sep 03 2005
O.g.f.: 1/(1-x-x^2/(1-2*x-2*x^2/(1-3*x-3*x^2/(.../(1-n*x-n*x^2/(...)))))) (continued fraction due to Ph. Flajolet). - Paul D. Hanna, Jan 17 2006
From Karol A. Penson, Jan 14 2007: (Start)
Representation of Bell numbers B(n), n=1,2,..., as special values of hypergeometric function of type (n-1)F(n-1), in Maple notation: B(n)=exp(-1)*hypergeom([2,2,...,2],[1,1,...,1],1), n=1,2,..., i.e., having n-1 parameters all equal to 2 in the numerator, having n-1 parameters all equal to 1 in the denominator and the value of the argument equal to 1.
Examples:
B(1)=exp(-1)*hypergeom([],[],1)=1
B(2)=exp(-1)*hypergeom([2],[1],1)=2
B(3)=exp(-1)*hypergeom([2,2],[1,1],1)=5
B(4)=exp(-1)*hypergeom([2,2,2],[1,1,1],1)=15
B(5)=exp(-1)*hypergeom([2,2,2,2],[1,1,1,1],1)=52
(Warning: this formula is correct but its application by a computer may not yield exact results, especially with a large number of parameters.)
(End)
a(n+1) = 1 + Sum_{k=0..n-1} Sum_{i=0..k} binomial(k,i)*(2^(k-i))*a(i). - Yalcin Aktar, Feb 27 2007
a(n) = [1,0,0,...,0] T^(n-1) [1,1,1,...,1]', where T is the n X n matrix with main diagonal {1,2,3,...,n}, 1's on the diagonal immediately above and 0's elsewhere. [Meier]
a(n) = ((2*n!)/(Pi * e)) * ImaginaryPart(Integral[from 0 to Pi](e^e^e^(i*theta))*sin(n*theta) dtheta). - Jonathan Vos Post, Aug 27 2007
From Tom Copeland, Oct 10 2007: (Start)
a(n) = T(n,1) = Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * Lag(n,-1,j-n) = Sum_{j=0..n} [ E(n,j)/n! ] * [ n!*Lag(n,-1, j-n) ] where T(n,x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m. Note that E(n,j)/n! = E(n,j) / (Sum_{k=0..n} E(n,k)).
The Eulerian numbers count the permutation ascents and the expression [n!*Lag(n,-1, j-n)] is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to n!*a(n) = Sum_{j=0..n} E(n,j) * [n!*Lag(n,-1, j-n)].
(End)
Define f_1(x), f_2(x), ... such that f_1(x)=e^x and for n=2,3,... f_{n+1}(x) = (d/dx)(x*f_n(x)). Then for Bell numbers B_n we have B_n=1/e*f_n(1). - Milan Janjic, May 30 2008
a(n) = (n-1)! Sum_{k=1..n} a(n-k)/((n-k)! (k-1)!) where a(0)=1. - Thomas Wieder, Sep 09 2008
a(n+k) = Sum_{m=0..n} Stirling2(n,m) Sum_{r=0..k} binomial(k,r) m^r a(k-r). - David Pasino (davepasino(AT)yahoo.com), Jan 25 2009. (Umbrally, this may be written as a(n+k) = Sum_{m=0..n} Stirling2(n,m) (a+m)^k. - N. J. A. Sloane, Feb 07 2009)
Sum_{k=1..n-1} a(n)*binomial(n,k) = Sum_{j=1..n}(j+1)*Stirling2(n,j+1). - [Zhao] - R. J. Mathar, Jun 24 2024
From Thomas Wieder, Feb 25 2009: (Start)
a(n) = Sum_{k_1=0..n+1} Sum_{k_2=0..n}...Sum_{k_i=0..n-i}...Sum_{k_n=0..1}
delta(k_1,k_2,...,k_i,...,k_n)
where delta(k_1,k_2,...,k_i,...,k_n) = 0 if any k_i > k_(i+1) and k_(i+1) <> 0
and delta(k_1,k_2,...,k_i,...,k_n) = 1 otherwise.
(End)
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>=1, a(n)=det(A). - Milan Janjic, Jul 08 2010
G.f. satisfies A(x) = (x/(1-x))*A(x/(1-x)) + 1. - Vladimir Kruchinin, Nov 28 2011
G.f.: 1 / (1 - x / (1 - 1*x / (1 - x / (1 - 2*x / (1 - x / (1 - 3*x / ... )))))). - Michael Somos, May 12 2012
a(n+1) = Sum_{m=0..n} Stirling2(n, m)*(m+1), n >= 0. Compare with the third formula for a(n) above. Here Stirling2 = A048993. - Wolfdieter Lang, Feb 03 2015
G.f.: (-1)^(1/x)*((-1/x)!/e + (!(-1-1/x))/x) where z! and !z are factorial and subfactorial generalized to complex arguments. - Vladimir Reshetnikov, Apr 24 2013
The following formulas were proposed during the period Dec 2011 - Oct 2013 by Sergei N. Gladkovskii: (Start)
E.g.f.: exp(exp(x)-1) = 1 + x/(G(0)-x); G(k) = (k+1)*Bell(k) + x*Bell(k+1) - x*(k+1)*Bell(k)*Bell(k+2)/G(k+1) (continued fraction).
G.f.: W(x) = (1-1/(G(0)+1))/exp(1); G(k) = x*k^2 + (3*x-1)*k - 2 + x - (k+1)*(x*k+x-1)^2/G(k+1); (continued fraction Euler's kind, 1-step).
G.f.: W(x) = (1 + G(0)/(x^2-3*x+2))/exp(1); G(k) = 1 - (x*k+x-1)/( ((k+1)!) - (((k+1)!)^2)*(1-x-k*x+(k+1)!)/( ((k+1)!)*(1-x-k*x+(k+1)!) - (x*k+2*x-1)*(1-2*x-k*x+(k+2)!)/G(k+1))); (continued fraction).
G.f.: A(x) = 1/(1 - x/(1-x/(1 + x/G(0)))); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1); (continued fraction, 1-step).
G.f.: -1/U(0) where U(k) = x*k - 1 + x - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + x/U(0) where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + 1/(U(0) - x) where U(k) = 1 + x - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1 + x/(U(0)-x) where U(k) = 1 - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1/G(0) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/G(k+1) ))); (continued fraction).
G.f.: G(0)/(1+x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-1) - x*(2*k+1)*(2*k+3)*(2*x*k-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+x-1)/G(k+1) )); (continued fraction).
G.f.: -(1+2*x) * Sum_{k >= 0} x^(2*k)*(4*x*k^2-2*k-2*x-1) / ((2*k+1) * (2*x*k-1)) * A(k) / B(k) where A(k) = Product_{p=0..k} (2*p+1), B(k) = Product_{p=0..k} (2*p-1) * (2*x*p-x-1) * (2*x*p-2*x-1).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: 1 + x*(S-1) where S = Sum_{k>=0} ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k/Product_{i=0..k-1} (1-x-x*i)/(1-x).
G.f.: (G(0) - 2)/(2*x-1) where G(k) = 2 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: -G(0) where G(k) = 1 - (x*k - 2)/(x*k - 1 - x*(x*k - 1)/(x + (x*k - 2)/G(k+1) )); (continued fraction).
G.f.: G(0) where G(k) = 2 - (2*x*k - 1)/(x*k - 1 - x*(x*k - 1)/(x + (2*x*k - 1)/G(k+1) )); (continued fraction).
G.f.: (G(0) - 1)/(1+x) where G(k) = 1 + 1/(1-k*x)/(1-x/(x+1/G(k+1) )); (continued fraction).
G.f.: 1/(x*(1-x)*G(0)) - 1/x where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )); (continued fraction).
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 + x/( x*k - 1 )/Q(k+1); (continued fraction).
G.f.: 1+x/Q(0), where Q(k) = 1 - x - x/(1 - x*(k+1)/Q(k+1)); (continued fraction).
G.f.: 1/(1-x*Q(0)), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))); (continued fraction).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x*(k+1))*(1-x*(k+2))/Q(k+1) ); (continued fraction).
(End)
a(n) ~ exp(exp(W(n))-n-1)*n^n/W(n)^(n+1/2), where W(x) is the Lambert W-function. - Vladimir Reshetnikov, Nov 01 2015
a(n) ~ n^n * exp(n/W(n)-1-n) / (sqrt(1+W(n)) * W(n)^n). - Vaclav Kotesovec, Nov 13 2015
From Natalia L. Skirrow, Apr 13 2025: (Start)
By taking logarithmic derivatives of the equivalent to Kotesovec's asymptotic for Bell polynomials at x=1, we obtain properties of the nth row of A008277 as a statistical distribution (where W=W(n),X=W(n)+1)
a(n+1)/a(n) ~ n/W + W/(2*(W+1)^2) is 1 more than the expectation.
(2*a(n+1)+a(n+2))/a(n) - (a(n+1)/a(n))^2 - a(n+2)/a(n+1) ~ n/(W*X)+1/(2*X^2)-3/(2*X^3)+1/X^4 is 1 more than the variance.
(This is a complete asymptotic characterisation, since they converge to normal distributions; see Harper, 1967)
(End)
a(n) are the coefficients in the asymptotic expansion of -exp(-1)*(-1)^x*x*Gamma(-x,0,-1), where Gamma(a,z0,z1) is the generalized incomplete Gamma function. - Vladimir Reshetnikov, Nov 12 2015
a(n) = 1 + floor(exp(-1) * Sum_{k=1..2*n} k^n/k!). - Vladimir Reshetnikov, Nov 13 2015
a(p^m) == m+1 (mod p) when p is prime and m >= 1 (see Lemma 3.1 in the Hurst/Schultz reference). - Seiichi Manyama, Jun 01 2016
a(n) = Sum_{k=0..n} hypergeom([1, -k], [], 1)*Stirling2(n+1, k+1) = Sum_{k=0..n} A182386(k)*Stirling2(n+1, k+1). - Mélika Tebni, Jul 02 2022
For n >= 1, a(n) = Sum_{i=0..n-1} a(i)*A074664(n-i). - Davide Rotondo, Apr 21 2024
a(n) is the n-th derivative of e^e^x divided by e at point x=0. - Joan Ludevid, Nov 05 2024

Extensions

Edited by M. F. Hasler, Nov 30 2018

A002293 Number of dissections of a polygon: binomial(4*n, n)/(3*n + 1).

Original entry on oeis.org

1, 1, 4, 22, 140, 969, 7084, 53820, 420732, 3362260, 27343888, 225568798, 1882933364, 15875338990, 134993766600, 1156393243320, 9969937491420, 86445222719724, 753310723010608, 6594154339031800, 57956002331347120, 511238042454541545
Offset: 0

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Comments

The number of rooted loopless n-edge maps in the plane (planar with a distinguished outside face). - Valery A. Liskovets, Mar 17 2005
Number of lattice paths from (1,0) to (3*n+1,n) which, starting from (1,0), only utilize the steps +(1,0) and +(0,1) and additionally, the paths lie completely below the line y = (1/3)*x (i.e., if (a,b) is in the path, then b < a/3). - Joseph Cooper (jecooper(AT)mit.edu), Feb 07 2006
Number of length-n restricted growth strings (RGS) [s(0), s(1), ..., s(n-1)] where s(0) = 0 and s(k) <= s(k-1) + 3, see fxtbook link below. - Joerg Arndt, Apr 08 2011
From Wolfdieter Lang, Sep 14 2007: (Start)
a(n), n >= 1, enumerates quartic trees (rooted, ordered, incomplete) with n vertices (including the root).
Pfaff-Fuss-Catalan sequence C^{m}_n for m = 4. See the Graham et al. reference, p. 347. eq. 7.66. (Second edition, p. 361, eq. 7.67.) See also the Pólya-Szegő reference.
Also 4-Raney sequence. See the Graham et al. reference, pp. 346-347.
(End)
Bacher: "We describe the statistics of checkerboard triangulations obtained by coloring black every other triangle in triangulations of convex polygons." The current sequence (A002293) occurs on p. 12 as one of two "extremal sequences" of an array of coefficients of polynomials, whose generating functions are given in terms of hypergeometric functions. - Jonathan Vos Post, Oct 05 2007
A generating function in terms of a (labyrinthine) solution to a depressed quartic equation is given in the Copeland link for signed A005810. With D(z,t) that g.f., a g.f. for signed A002293 is {[-1+1/D(z,t)]/(4t)}^(1/3). - Tom Copeland, Oct 10 2012
For a relation to the inviscid Burgers's equation, see A001764. - Tom Copeland, Feb 15 2014
For relations to compositional inversion, the Legendre transform, and convex geometry, see the Copeland, the Schuetz and Whieldon, and the Gross (p. 58) links. - Tom Copeland, Feb 21 2017 (See also Gross et al. in A062994. - Tom Copeland, Dec 24 2019)
This is the number of A'Campo bicolored forests of degree n and co-dimension 0. This can be shown using generating functions or a combinatorial approach. See Combe and Jugé link below. - Noemie Combe, Feb 28 2017
Conjecturally, a(n) is the number of 3-uniform words over the alphabet [n] that avoid the patterns 231 and 221 (see the Defant and Kravitz link). - Colin Defant, Sep 26 2018
The compositional inverse o.g.f. pair in Copeland's comment above are related to a pair of quantum fields in Balduf's thesis by Theorem 4.2 on p. 92. Cf. A001764. - Tom Copeland, Dec 13 2019
a(n) is the total number of down steps before the first up step in all 3_1-Dyck paths of length 4*n. A 3_1-Dyck path is a lattice path with steps (1, 3), (1, -1) that starts and ends at y = 0 and stays above the line y = -1. - Sarah Selkirk, May 10 2020
a(n) is the number of pairs (A<=B) of noncrossing partitions of [2n] such that every block of A has exactly two elements. In fact, it is proved that a(n) is the number of planar tied arc diagrams with n arcs (see Aicardi link below). A planar diagram with n arcs represents a noncrossing partition A of [2n] with n blocks, each block containing the endpoints of one arc; each tie connects two arcs, so that the ties define a partition B >= A: the endpoints of two arcs connected by a tie belong to the same block of B. Ties do not cross arcs nor other ties iff B has a planar diagram, i.e., B is a noncrossing partition. - Francesca Aicardi, Nov 07 2022
Dropping the initial 1 (starting 1, 4, 22 with offset 1) yields the REVERT transformation 1, -4 ,10, -20, 35.. essentially A000292 without leading 0. - R. J. Mathar, Aug 17 2023
Number of rooted polyominoes composed of n pentagonal cells of the hyperbolic regular tiling with Schläfli symbol {5,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {5,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
This is instance k = 4 of the generalized Catalan family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment of A130564. - _Wolfdieter Lang, Feb 05 2024
a(n) is the cardinality of the planar ramified Jones monoid PR(J_n). - Diego Arcis, Nov 21 2024

Examples

			There are a(2) = 4 quartic trees (vertex degree <= 4 and 4 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these four trees yields 4*4 + 6 = 22 = a(3) such trees.
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 23.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347.
  • Peter Hilton and Jean Pedersen, Catalan numbers, their generalization, and their uses, Math. Intelligencer 13 (1991), no. 2, 64-75.
  • V. A. Liskovets and T. R. Walsh, Enumeration of unrooted maps on the plane, Rapport technique, UQAM, No. 2005-01, Montreal, Canada, 2005.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, Heidelberg, New York, 2 vols., 1972, Vol. 1, problem 211, p. 146 with solution on p. 348.
  • 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

Column k=3 of triangle A062993 and A070914.
Cf. A000260, A002295, A002296, A027836, A062994, A346646 (binomial transform), A346664 (inverse binomial transform).
Polyominoes: A005038 (oriented), A005040 (unoriented), A369471 (chiral), A369472 (achiral), A001764 {4,oo}, A002294 {6,oo}.
Cf. A130564 (for generalized Catalan C(k, n), for = 4).

Programs

  • GAP
    List([0..22],n->Binomial(4*n,n)/(3*n+1)); # Muniru A Asiru, Nov 01 2018
  • Magma
    [ Binomial(4*n,n)/(3*n+1): n in [0..50]]; // Vincenzo Librandi, Apr 19 2011
    
  • Maple
    series(RootOf(g = 1+x*g^4, g),x=0,20); # Mark van Hoeij, Nov 10 2011
    seq(binomial(4*n, n)/(3*n+1),n=0..20); # Robert FERREOL, Apr 02 2015
    # Using the integral representation above:
    Digits:=6;
    R:=proc(x)((I + sqrt(3))*(4*sqrt(256 - 27*x) - 12*I*sqrt(3)*sqrt(x))^(1/3))/16 - ((I - sqrt(3))*(4*sqrt(256 - 27*x) + 12*I*sqrt(3)*sqrt(x))^(1/3))/16;end;
    W:=proc(x) x^(-3/4) * sqrt(4*R(x) - 3^(3/4)*x^(1/4)/sqrt(R(x)))/(2*3^(1/4)*Pi);end;
    # Attention: W(x) is singular at x = 0. Integration is done from  a very small positive x to x = 256/27.
    # For a(8):  ... gives 420732
    evalf(int(x^8*W(x),x=0.000001..256/27));
    # Karol A. Penson, Jul 05 2024
  • Mathematica
    CoefficientList[InverseSeries[ Series[ y - y^4, {y, 0, 60}], x], x][[Range[2, 60, 3]]]
    Table[Binomial[4n,n]/(3n+1),{n,0,25}] (* Harvey P. Dale, Apr 18 2011 *)
    CoefficientList[1 + InverseSeries[Series[x/(1 + x)^4, {x, 0, 60}]], x] (* Gheorghe Coserea, Aug 12 2015 *)
    terms = 22; A[] = 0; Do[A[x] = 1 + x*A[x]^4 + O[x]^terms, terms];
    CoefficientList[A[x], x] (* Jean-François Alcover, Jan 13 2018 *)
  • PARI
    a(n)=binomial(4*n,n)/(3*n+1) /* Charles R Greathouse IV, Jun 16 2011 */
    
  • PARI
    my(x='x+O('x^33)); Vec(1 + serreverse(x/(1+x)^4)) \\ Gheorghe Coserea, Aug 12 2015
    
  • Python
    A002293_list, x = [1], 1
    for n in range(100):
        x = x*4*(4*n+3)*(4*n+2)*(4*n+1)//((3*n+2)*(3*n+3)*(3*n+4))
        A002293_list.append(x) # Chai Wah Wu, Feb 19 2016
    

Formula

O.g.f. satisfies: A(x) = 1 + x*A(x)^4 = 1/(1 - x*A(x)^3).
a(n) = binomial(4*n,n-1)/n, n >= 1, a(0) = 1. From the Lagrange series of the o.g.f. A(x) with its above given implicit equation.
From Karol A. Penson, Apr 02 2010: (Start)
Integral representation as n-th Hausdorff power moment of a positive function on the interval [0, 256/27]:
a(n) = Integral_{x=0..256/27}(x^n((3/256) * sqrt(2) * sqrt(3) * ((2/27) * 3^(3/4) * 27^(1/4) * 256^(/4) * hypergeom([-1/12, 1/4, 7/12], [1/2, 3/4], (27/256)*x)/(sqrt(Pi) * x^(3/4)) - (2/27) * sqrt(2) * sqrt(27) * sqrt(256) * hypergeom([1/6, 1/2, 5/6], [3/4, 5/4], (27/256)*x)/ (sqrt(Pi) * sqrt(x)) - (1/81) * 3^(1/4) * 27^(3/4) * 256^(1/4) * hypergeom([5/12, 3/4, 13/12], [5/4, 3/2], (27/256)*x/(sqrt(Pi)*x^(1/4)))/sqrt(Pi))).
This representation is unique as it represents the solution of the Hausdorff moment problem.
O.g.f.: hypergeom([1/4, 1/2, 3/4], [2/3, 4/3], (256/27)*x);
E.g.f.: hypergeom([1/4, 1/2, 3/4], [2/3, 1, 4/3], (256/27)*x). (End)
a(n) = upper left term in M^n, M = the production matrix:
1, 1
3, 3, 1
6, 6, 3, 1
...
(where 1, 3, 6, 10, ...) is the triangular series. - Gary W. Adamson, Jul 08 2011
O.g.f. satisfies g = 1+x*g^4. If h is the series reversion of x*g, so h(x*g)=x, then (x-h(x))/x^2 is the o.g.f. of A006013. - Mark van Hoeij, Nov 10 2011
a(n) = binomial(4*n+1, n)/(4*n+1) = A062993(n+2,2). - Robert FERREOL, Apr 02 2015
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-1-i} Sum_{k=0..n-1-i-j} a(i)*a(j)*a(k)*a(n-1-i-j-k) for n>=1; and a(0) = 1. - Robert FERREOL, Apr 02 2015
a(n) ~ 2^(8*n+1/2) / (sqrt(Pi) * n^(3/2) * 3^(3*n+3/2)). - Vaclav Kotesovec, Jun 03 2015
From Peter Bala, Oct 16 2015: (Start)
A(x)^2 is o.g.f. for A069271; A(x)^3 is o.g.f. for A006632;
A(x)^5 is o.g.f. for A196678; A(x)^6 is o.g.f. for A006633;
A(x)^7 is o.g.f. for A233658; A(x)^8 is o.g.f. for A233666;
A(x)^9 is o.g.f. for A006634; A(x)^10 is o.g.f. for A233667. (End)
D-finite with recurrence: a(n+1) = a(n)*4*(4*n + 3)*(4*n + 2)*(4*n + 1)/((3*n + 2)*(3*n + 3)*(3*n + 4)). - Chai Wah Wu, Feb 19 2016
E.g.f.: F([1/4, 1/2, 3/4], [2/3, 1, 4/3], 256*x/27), where F is the generalized hypergeometric function. - Stefano Spezia, Dec 27 2019
x*A'(x)/A(x) = (A(x) - 1)/(- 3*A(x) + 4) = x + 7*x^2 + 55*x^3 + 455*x^4 + ... is the o.g.f. of A224274. Cf. A001764 and A002294 - A002296. - Peter Bala, Feb 04 2022
a(n) = hypergeom([1 - n, -3*n], [2], 1). Row sums of A173020. - Peter Bala, Aug 31 2023
G.f.: t*exp(4*t*hypergeom([1, 1, 5/4, 3/2, 7/4], [4/3, 5/3, 2, 2], (256*t)/27))+1. - Karol A. Penson, Dec 20 2023
From Karol A. Penson, Jul 03 2024: (Start)
a(n) = Integral_{x=0..256/27} x^(n)*W(x)dx, n>=0, where W(x) = x^(-3/4) * sqrt(4*R(x) - 3^(3/4)*x^(1/4)/sqrt(R(x)))/(2*3^(1/4)*Pi), with R(x) = ((i + sqrt(3))*(4*sqrt(256 - 27*x) -12*i*sqrt(3*x))^(1/3))/16 - ((i - sqrt(3))*(4*sqrt(256 - 27*x) + 12*i* sqrt(3*x))^(1/3))/16, where i is the imaginary unit.
The elementary function W(x) is positive on the interval x = (0, 256/27) and is equal to the combination of hypergeometric functions in my formula from 2010; see above.
(Pi*W(x))^6 satisfies an algebraic equation of order 6, with integer polynomials as coefficients. (End)
G.f.: (Sum_{n >= 0} binomial(4*n+1, n)*x^n) / (Sum_{n >= 0} binomial(4*n, n)*x^n). - Peter Bala, Dec 14 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^7). - Seiichi Manyama, Jun 16 2025

A007556 Number of 8-ary trees with n vertices.

Original entry on oeis.org

1, 1, 8, 92, 1240, 18278, 285384, 4638348, 77652024, 1329890705, 23190029720, 410333440536, 7349042994488, 132969010888280, 2426870706415800, 44627576949364104, 826044435409399800, 15378186970730687400, 287756293703544823872, 5409093674555090316300
Offset: 0

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Comments

Shifts left when convolved three times.
From Wolfdieter Lang, Sep 14 2007: (Start)
a(n), n >= 1, enumerates octic (8-ary) trees (rooted, ordered, incomplete) with n vertices (including the root).
Pfaff-Fuss-Catalan sequence C^{m}_n for m = 8. See the Graham et al. reference, p. 347. eq. 7.66. See also the Pólya-Szegő reference.
Also 8-Raney sequence. See the Graham et al. reference, p. 346-7.
(End)
This is instance k = 8 of the generalized Catalan family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment of A130564. - _Wolfdieter Lang, Feb 05 2024

Examples

			There are a(2) = 8 octic trees (vertex degree less than or equal to 8 and 8 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these 8 trees yields 8*8 + binomial(8, 2) = 92 = a(3) such trees.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, Heidelberg, New York, 2 vols., 1972, Vol. 1, problem 211, p. 146 with solution on p. 348.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Seventh column of triangle A062993.
Cf. A130564.

Programs

  • Haskell
    a007556 0 = 1
    a007556 n = a007318' (8 * n) (n - 1) `div` n
    -- Reinhard Zumkeller, Jul 30 2013
    
  • Magma
    [Binomial(8*n, n)/(7*n+1): n in [0..20]]; // Vincenzo Librandi, Apr 02 2015
    
  • Maple
    seq(binomial(8*n+1,n)/(8*n+1),n=0..30); # Robert FERREOL, Apr 01 2015
    n:=30: G:=series(RootOf(g = 1+x*g^8, g),x=0,n+1): seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 01 2015
  • Mathematica
    Table[Binomial[8n, n]/(7n + 1), {n, 0, 20}] (* Harvey P. Dale, Dec 24 2012 *)
  • PARI
    vector(100, n, n--; binomial(8*n, n)/(7*n+1)) \\ Altug Alkan, Oct 14 2015

Formula

a(n) = binomial(8*n, n)/(7*n+1) = binomial(8*n+1, n)/(8*n+1) = A062993(n+6,6).
O.g.f.: A(x) = 1 + x*A(x)^8 = 1/(1-x*A(x)^7).
a(0) = 1; a(n) = Sum_{i1 + i2 + .. i8 = n - 1} a(i1)*a(i2)*...*a(i8) for n >= 1. - Robert FERREOL, Apr 01 2015
a(n) = binomial(8*n, n - 1)/n for n >= 1, a(0) = 1 (from the Lagrange series of the o.g.f. A(x) with its above given implicit equation).
From Karol A. Penson, Mar 26 2015: (Start)
In Maple notation,
e.g.f.: hypergeom([1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8], [2/7, 3/7, 4/7, 5/7, 6/7, 1, 8/7],(2^24/7^7)*z);
o.g.f.: hypergeom([1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8], [2/7, 3/7, 4/7, 5/7, 6/7, 8/7],(2^24/7^7)*z);
a(n) are special values of Jacobi polynomials, in Maple notation:
a(n) = JacobiP(n - 1, 7*n + 1, -n, 1)/n, n = 1, 2, ...
(End)
From Peter Bala, Oct 14 2015: (Start)
A(x)^2 is o.g.f. for A234461; A(x)^3 is o.g.f. for A234462;
A(x)^4 is o.g.f. for A234463; A(x)^5 is o.g.f. for A234464;
A(x)^6 is o.g.f. for A234465; A(x)^7 is o.g.f. for A234466;
A(x)^9 is o.g.f. for A234467. (End)
a(n) ~ 2^(24*n + 1)/(sqrt(Pi)*7^(7*n + 3/2)*n^(3/2)). - Ilya Gutkovskiy, Feb 07 2017
D-finite with recurrence: 7*n*(7*n-3)*(7*n+1)*(7*n-2)*(7*n-5)*(7*n-1)*(7*n-4)*a(n) -128*(8*n-5)*(4*n-1)*(8*n-7)*(2*n-1)*(8*n-1)*(4*n-3)*(8*n-3)*a(n-1)=0. - R. J. Mathar, Feb 20 2020
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^15). - Seiichi Manyama, Jun 16 2025

A086971 Number of semiprime divisors of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 2, 0, 1, 1, 1, 0, 2, 0, 2, 1, 1, 0, 2, 1, 1, 1, 2, 0, 3, 0, 1, 1, 1, 1, 3, 0, 1, 1, 2, 0, 3, 0, 2, 2, 1, 0, 2, 1, 2, 1, 2, 0, 2, 1, 2, 1, 1, 0, 4, 0, 1, 2, 1, 1, 3, 0, 2, 1, 3, 0, 3, 0, 1, 2, 2, 1, 3, 0, 2, 1, 1, 0, 4, 1, 1, 1, 2, 0, 4, 1, 2, 1, 1, 1, 2, 0, 2, 2, 3, 0, 3
Offset: 1

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Author

Reinhard Zumkeller, Sep 22 2003

Keywords

Comments

Inverse Moebius transform of A064911. - Jonathan Vos Post, Dec 08 2004

References

  • G. H. Hardy and E. M. Wright, Section 17.10 in An Introduction to the Theory of Numbers, 5th ed., Oxford, England: Clarendon Press, 1979.

Crossrefs

Programs

  • Haskell
    a086971 = sum . map a064911 . a027750_row
    -- Reinhard Zumkeller, Dec 14 2012
  • Maple
    a:= proc(n) local l, m; l:=ifactors(n)[2]; m:=nops(l);
           m*(m-1)/2 +add(`if`(i[2]>1, 1, 0), i=l)
        end:
    seq(a(n), n=1..120);  # Alois P. Heinz, Jul 18 2013
  • Mathematica
    semiPrimeQ[n_] := PrimeOmega@ n == 2; f[n_] := Length@ Select[Divisors@ n, semiPrimeQ@# &]; Array[f, 105] (* Zak Seidov, Mar 31 2011 and modified by Robert G. Wilson v, Dec 08 2012 *)
    a[n_] := Count[e = FactorInteger[n][[;; , 2]], ?(# > 1 &)] + (o = Length[e])*(o - 1)/2; Array[a, 100] (* _Amiram Eldar, Jun 30 2022 *)
  • PARI
    /* The following definitions of a(n) are equivalent. */
    a(n) = sumdiv(n,d,bigomega(d)==2)
    a(n) = f=factor(n); j=matsize(f)[1]; sum(m=1,j,f[m,2]>=2) + binomial(j,2)
    a(n) = f=factor(n); j=omega(n); sum(m=1,j,f[m,2]>=2) + binomial(j,2)
    a(n) = omega(n/core(n)) + binomial(omega(n),2)
    /* Rick L. Shepherd, Mar 06 2006 */
    

Formula

a(n) = A106404(n) + A106405(n). - Reinhard Zumkeller, May 02 2005
a(n) = omega(n/core(n)) + binomial(omega(n),2) = A001221(n/A007913(n)) + binomial(A001221(n),2) = A056170(n) + A079275(n). - Rick L. Shepherd, Mar 06 2006
From Reinhard Zumkeller, Dec 14 2012: (Start)
a(n) = Sum_{k=1..A000005(n)} A064911(A027750(n,k)).
a(A220264(n)) = n and a(m) <> n for m < A220264(n); a(A008578(n)) = 0; a(A002808(n)) > 0; for n > 1: a(A102466(n)) <= 1 and a(A102467(n)) > 1; A066247(n) = A057427(a(n)). (End)
G.f.: Sum_{k = p*q, p prime, q prime} x^k/(1 - x^k). - Ilya Gutkovskiy, Jan 25 2017

Extensions

Entry revised by N. J. A. Sloane, Mar 28 2006

A007437 Inverse Moebius transform of triangular numbers.

Original entry on oeis.org

1, 4, 7, 14, 16, 31, 29, 50, 52, 74, 67, 119, 92, 137, 142, 186, 154, 247, 191, 294, 266, 323, 277, 455, 341, 446, 430, 553, 436, 686, 497, 714, 634, 752, 674, 1001, 704, 935, 878, 1150, 862, 1298, 947, 1323, 1222, 1361, 1129, 1767, 1254, 1674, 1486, 1834
Offset: 1

Views

Author

Keywords

References

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

Crossrefs

Programs

  • Maple
    with(numtheory):
    a:= proc(n) option remember;
          add(d*(d+1)/2, d=divisors(n))
        end:
    seq(a(n), n=1..60);  # Alois P. Heinz, Feb 09 2011
  • Mathematica
    a[n_] := (DivisorSigma[1, n] + DivisorSigma[2, n])/2; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Mar 10 2014, after Benoit Cloitre *)
  • PARI
    a(n)=if(n<1,1,sumdiv(n,d,(d^2+d))/2); /* Joerg Arndt, Aug 14 2012 */
    
  • PARI
    my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, binomial(k+1, 2)*x^k/(1-x^k))) \\ Seiichi Manyama, Apr 19 2021

Formula

Coefficients in expansion of Sum_{n >= 1} x^n/(1-x^n)^3.
G.f.: Sum_{n>=1} (n*(n+1)/2) * x^n/(1-x^n). - Joerg Arndt, Jan 30 2011
a(n) = Sum_{d|n} d*(d+1)/2 = (1/2)*(sigma(n) + sigma_2(n)) = (1/2)*(A000203(n) + A001157(n)). - Benoit Cloitre, Apr 08 2002
Row sums of triangles A134544 and A134545. - Gary W. Adamson, Oct 31 2007
Row sums of triangle A134839 - Gary W. Adamson, Nov 12 2007
Dirichlet g.f. zeta(s)*(zeta(s-1) + zeta(s-2))/2. - Franklin T. Adams-Watters, Nov 05 2009
L.g.f.: -log(Product_{k>=1} (1 - x^k)^((k+1)/2)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, Mar 12 2018
Sum_{k=1..n} a(k) ~ Zeta(3) * n^3 / 6. - Vaclav Kotesovec, Nov 06 2018
a(n) = Sum_{i=1..n} i*A135539(n,i). - Ridouane Oudra, Jul 22 2022

Extensions

More terms from Christian G. Bower.

A007562 Number of planted trees where non-root, non-leaf nodes an even distance from root are of degree 2.

Original entry on oeis.org

1, 1, 1, 2, 3, 6, 10, 20, 36, 72, 137, 275, 541, 1098, 2208, 4521, 9240, 19084, 39451, 82113, 171240, 358794, 753460, 1587740, 3353192, 7100909, 15067924, 32044456, 68272854, 145730675, 311575140, 667221030, 1430892924, 3072925944, 6607832422, 14226665499
Offset: 1

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Author

Keywords

Comments

There is no planted tree on one node by definition.
Column k=2 of A144018. - Alois P. Heinz, Oct 17 2012
It appears that a(n) is also the number of locally non-intersecting unlabeled rooted trees with n nodes, where a tree is locally non-intersecting if the branches directly under of any non-leaf node have empty intersection. - Gus Wiseman, Aug 22 2018

Examples

			G.f. = x + x^2 + x^3 + 2*x^4 + 3*x^5 + 6*x^6 + 10*x^7 + 20*x^8 + 36*x^9 + ...
From _Joerg Arndt_, Jun 23 2014: (Start)
The a(8) = 20 such trees have the following level sequences:
01:  [ 0 1 2 3 4 3 2 1 ]
02:  [ 0 1 2 3 3 3 2 1 ]
03:  [ 0 1 2 3 3 2 2 1 ]
04:  [ 0 1 2 3 3 2 1 1 ]
05:  [ 0 1 2 3 2 3 2 1 ]
06:  [ 0 1 2 3 2 2 2 1 ]
07:  [ 0 1 2 3 2 2 1 1 ]
08:  [ 0 1 2 3 2 1 2 1 ]
09:  [ 0 1 2 3 2 1 1 1 ]
10:  [ 0 1 2 2 2 2 2 1 ]
11:  [ 0 1 2 2 2 2 1 1 ]
12:  [ 0 1 2 2 2 1 2 1 ]
13:  [ 0 1 2 2 2 1 1 1 ]
14:  [ 0 1 2 2 1 2 2 1 ]
15:  [ 0 1 2 2 1 2 1 1 ]
16:  [ 0 1 2 2 1 1 1 1 ]
17:  [ 0 1 2 1 2 1 2 1 ]
18:  [ 0 1 2 1 2 1 1 1 ]
19:  [ 0 1 2 1 1 1 1 1 ]
20:  [ 0 1 1 1 1 1 1 1 ]
Successive levels change by at most 1 and the last level is 1, compare to the example in A000081.
(End)
From _Gus Wiseman_, Aug 22 2018: (Start)
The a(7) = 10 locally non-intersecting trees:
  (o(o(oo)))
  (o(oo(o)))
  (o(oooo))
  (oo(o(o)))
  (oo(ooo))
  (o(o)(oo))
  (ooo(oo))
  (oo(o)(o))
  (oooo(o))
  (oooooo)
(End)
		

References

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

Crossrefs

Programs

  • Maple
    with(numtheory): etr:= proc(p) local b; b:= proc(n) option remember; if n=0 then 1 else (add(d*p(d), d=divisors(n)) +add(add(d*p(d), d= divisors(j)) *b(n-j), j=1..n-1))/n fi end end: b:= etr(a): a:= n-> `if`(n<=1, n, b(n-2)): seq(a(n), n=1..40);  # Alois P. Heinz, Sep 06 2008
  • Mathematica
    etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, (Sum[ Sum[ d*p[d], {d, Divisors[j]}]*b[n-j], {j, 1, n-1}] + Sum[ d*p[d], {d, Divisors[n]}])/n]; b]; b = etr[a]; a[n_] := If[n <= 1, n, b[n-2]]; Table[a[n], {n, 1, 36}] (* Jean-François Alcover, Aug 01 2013, after Alois P. Heinz *)
    purt[n_]:=If[n==1,{{}},Join@@Table[Select[Union[Sort/@Tuples[purt/@ptn]],Intersection@@#=={}&],{ptn,IntegerPartitions[n-1]}]];
    Table[Length[purt[n]],{n,10}] (* Gus Wiseman, Aug 22 2018 *)
  • PARI
    {a(n) = local(A); if( n<2, n>0, A = x / (1 - x) + O(x^n); for(k=2, n-2, A /= (1 - x^k + O(x^n))^polcoeff(A, k-1)); polcoeff(A, n-1))}; /* Michael Somos, Oct 06 2003 */

Formula

Shifts left 2 places under Euler transform.
G.f.: x + x^2 / (Product_{k>0} (1 - x^k)^a(k)). - Michael Somos, Oct 06 2003
a(n) ~ c * d^n / n^(3/2), where d = 2.246066877341161662499621547921... and c = 0.68490297576105466417608032... . - Vaclav Kotesovec, Jun 23 2014
G.f. A(x) satisfies: A(x) = x + x^2 * exp(A(x) + A(x^2)/2 + A(x^3)/3 + A(x^4)/4 + ...). - Ilya Gutkovskiy, Jun 11 2021

Extensions

Better description from Christian G. Bower, May 15 1998

A007476 Shifts 2 places left under binomial transform.

Original entry on oeis.org

1, 1, 1, 2, 4, 9, 23, 65, 199, 654, 2296, 8569, 33825, 140581, 612933, 2795182, 13298464, 65852873, 338694479, 1805812309, 9963840219, 56807228074, 334192384460, 2026044619017, 12642938684817, 81118550133657, 534598577947465, 3615474317688778, 25070063421597484
Offset: 0

Views

Author

Keywords

Comments

Starting (1, 2, 4, 9, 23, ...) = row sums of triangle A153859. - Gary W. Adamson, Jan 02 2009
Binomial transform of the sequence starting (1, 1, 2, 4, 9, ...) = first differences of (1, 2, 4, 9, 23, ...); that is, (1, 2, 5, 14, 42, 134, 455, 1642, ...). - Gary W. Adamson, May 20 2013
Row sums of triangle A256161. - Margaret A. Readdy, Mar 16 2015
RG-words corresponding to set partitions of {1, ..., n} with every even entry appearing exactly once. - Margaret A. Readdy, Mar 16 2015
a(n) is the number of partitions of [n] whose blocks can be written such that the smallest elements form an increasing sequence and the largest elements form a decreasing sequence. a(5) = 9: 12345, 1235|4, 1245|3, 125|34, 1345|2, 135|24, 145|23, 15|234, 15|24|3. - Alois P. Heinz, Apr 24 2023

References

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

Crossrefs

Row sums of A246118.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, 1,
          add(a(j)*binomial(n-2, j), j=0..n-2))
        end:
    seq(a(n), n=0..31);  # Alois P. Heinz, Jul 29 2019
  • Mathematica
    a[0] = a[1] = 1; a[n_] := a[n] = Sum[Binomial[n-2, k] a[k], {k, 0, n-2}]; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Aug 08 2012, after Ralf Stephan *)
  • PARI
    a(n)=if(n<2, 1, sum(k=0, n-2, binomial(n-2, k)*a(k))) /* Ralf Stephan; corrected by Manuel Blum, May 22 2010 */

Formula

G.f.: Sum_{k>=0} x^(2k)/(Product_{m=0..k-1} (1-mx) * Product_{m=0..k+1} (1-mx)).
G.f. A(x) satisfies A(x) = 1 + x + (x^2/(1-x))*A(x/(1-x)). - Vladimir Kruchinin, Nov 28 2011
a(n) = A000994(n) + A000995(n). - Peter Bala, Jan 27 2015

Extensions

Spelling correction by Jason G. Wurtzel, Aug 22 2010

A007564 Shifts left when INVERT transform applied thrice.

Original entry on oeis.org

1, 1, 4, 19, 100, 562, 3304, 20071, 124996, 793774, 5120632, 33463102, 221060008, 1473830308, 9904186192, 67015401391, 456192667396, 3122028222934, 21467769499864, 148246598341018, 1027656663676600, 7148588698592956, 49884553176689584
Offset: 0

Views

Author

Keywords

Comments

More generally, coefficients of (1+m*x-sqrt(m^2*x^2-(2*m+2)*x+1))/(2*m*x) are given by a(n) = Sum_{k=0..n} (m+1)^k*N(n,k) where N(n,k) = (1/n)*binomial(n,k)*binomial(n,k+1) are the Narayana numbers (A001263). - Benoit Cloitre, May 24 2003
If y = x*A(x) then 3*y^2 - (1+2*x)*y + x = 0 and x = y*(1-3*y)/(1-2*y). - Michael Somos, Sep 28 2003
The sequence 0,1,4,19,... with g.f. (1-4*x-sqrt(1-8*x+4*x^2))/(6*x) and has a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1,2k)*C(k)*4^(n-1-2*k)*3^k. a(n+1) = Sum_{k=0..floor(n/2)} binomial(n,2*k)*C(k)*4^(n-2*k)*3^k counts Motzkin paths of length n in which the level steps have 4 colors and the up steps have 3. It is the binomial transform of A107264 and corresponds to the series reversion of x/(1+4*x+3*x^2). - Paul Barry, May 18 2005
The Hankel transform of this sequence is 3^binomial(n+1,2). - Philippe Deléham, Oct 29 2007
a(n) is the number of Schroder paths of semilength n in which there are no (2,0)-steps at level 0 and at a higher level they come in 2 colors. Example: a(2)=4 because we have UDUD, UUDD, UBD, and URD, where U=(1,1), D=(1,-1), while B (R) is a blue (red) (2,0)-step. - Emeric Deutsch, May 02 2011
a(n) is the number of Schroder paths of semilength n-1 in which the (2,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=19 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 3^2 = 9 paths of shape HH, 3 paths of shape HUD, 3 paths of shape UDH, 2 paths of shape UHD, and 1 path of each of the shapes UDUD and UUDD. - Emeric Deutsch, May 02 2011
From David Callan, Jun 21 2013: (Start)
a(n) = number of (left) planted binary trees with n edges in which each vertex has a designated favorite neighbor. Planted binary trees are counted by the Catalan numbers A000108.
Example: for n=2, there are 2 planted binary trees: edges LL and LR from the root (L=left, R=right). Each has just one vertex with 2 neighbors, and so a(2)=4.
Proof outline: each vertex has 1,2 or 3 neighbors. Let X (resp. Y) denote the number of vertices with 2 (resp. 3) neighbors. Then X + 2Y = n - 1 (split the non-root edges into pairs with a common parent vertex and singletons). Thus the number of choices for designating favorite neighbors is 2^X * 3^Y = 2^(n-1)(3/4)^Y. The distribution for Y is known because, under the rotation correspondence, a.k.a. the deBruijn-Morselt bijection, vertices with 2 children in an n-edge planted binary tree correspond to DDUs in a Dyck path, and DDUs have the Touchard distribution (A091894) with gf F(x,y) = (1-2x+2xy - sqrt(1-4x+4x^2-4x^2 y))/(2xy). The desired g.f., Sum_{n>=1} a(n)*x^n, is therefore 1/2*(F(2x,3/4)-1). (End)

Examples

			G.f. = 1 + x + 4*x^2 + 19*x^3 + 100*x^4 + 562*x^5 + 3304*x^6 + 20071*x^7 + 124996*x^8 + ...
		

References

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

Crossrefs

Programs

  • Maple
    A007564_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
    for w from 1 to n do a[w] := a[w-1]+3*add(a[j]*a[w-j-1],j=1..w-1) od;
    convert(a, list) end: A007564_list(21); # Peter Luschny, May 19 2011
  • Mathematica
    a[0]=1; a[1]=1; a[n_]/;n>=2 := a[n] = a[n-1] + 3 Sum[a[k-1]a[n-k],{k,n-1}] ; Table[a[n],{n,0,10}] (* David Callan, Aug 25 2009 *)
    Table[Hypergeometric2F1[-n, 1 - n, 2, 3], {n, 0, 22}] (* Arkadiusz Wesolowski, Aug 13 2012 *)
    Table[(2^n (LegendreP[n+1, 2] - LegendreP[n-1, 2]) + 2 KroneckerDelta[n])/(6n+3), {n, 0, 20}] (* Vladimir Reshetnikov, Nov 01 2015 *)
    CoefficientList[Series[(1+2x-Sqrt[1-8x+4x^2])/(6x),{x,0,30}],x] (* Harvey P. Dale, Feb 07 2016 *)
  • PARI
    {a(n) = if( n<1, n==0, sum( k=0, n, 3^k * binomial( n, k) * binomial( n, k+1)) / n)} /* Michael Somos, Sep 28 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( x * (1 - 3*x) / (1 - 2*x) + x * O(x^n)), n))} /* Michael Somos, Sep 28 2003 */
    
  • PARI
    a(n) = (2^n*(pollegendre(n+1,2)-pollegendre(n-1,2)) + 2*(n==0))/(6*n+3); \\ Michel Marcus, Nov 02 2015
    
  • PARI
    x='x+O('x^100); Vec((1+2*x-sqrt(1-8*x+4*x^2))/(6*x)) \\ Altug Alkan, Nov 02 2015

Formula

G.f.: (1+2*x-sqrt(1-8*x+4*x^2))/(6*x). - Emeric Deutsch, Nov 03 2001
a(0)=1; for n>=1, a(n) = Sum_{k=0..n} 3^k*N(n,k) where N(n,k) = (1/n)*binomial(n, k)*binomial(n, k+1) are the Narayana numbers (A001263). - Benoit Cloitre, May 24 2003
a(n) = Sum_{k=0..n} A088617(n, k)*3^k*(-2)^(n-k). - Philippe Deléham, Jan 21 2004
With offset 1: a(1) = 1, a(n) = -2*a(n-1) + 3*Sum_{i=1..n-1} a(i)*a(n-i). - Benoit Cloitre, Mar 16 2004
D-finite with recurrence a(n) = (4*(2n-1)*a(n-1) - 4*(n-2)*a(n-2)) / (n+1) for n>=2, a(0) = a(1) = 1. - Philippe Deléham, Aug 19 2005
From Paul Barry, Dec 15 2008: (Start)
G.f.: 1/(1-x/(1-3x/(1-x/(1-3x/(1-x/(1-3x/(1-x/(1-3x........ (continued fraction).
The g.f. of a(n+1) is 1/(1-4x-3x^2/(1-4x-3x^2/(1-4x-3x^2/(1-4x-3x^2.... (continued fraction). (End)
a(0) = 1, for n>=1, 3a(n) = A047891(n). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
a(n) = upper left term in M^n, M = the production matrix:
1, 1
3, 3, 3
1, 1, 1, 1
3, 3, 3, 3, 3
1, 1, 1, 1, 1, 1
...
- Gary W. Adamson, Jul 08 2011
G.f.: A(x)= (1+2*x-sqrt(1-8*x+4*x^2))/(6*x)= 1/G(0); G(k)= 1 + 2*x - 3*x/G(k+1); (continued fraction, 1-step ). - Sergei N. Gladkovskii, Jan 05 2012
a(n) ~ sqrt(6+4*sqrt(3))*(4+2*sqrt(3))^n/(6*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 07 2012
a(n) = 2^n/sqrt(3)*LegendreP(n,-1,2) for n >= 1, where LegendreP is the associated Legendre function of the first kind, in Maple's notation. - Robert Israel, Mar 24 2015

A007477 Shifts 2 places left when convolved with itself.

Original entry on oeis.org

1, 1, 1, 2, 3, 6, 11, 22, 44, 90, 187, 392, 832, 1778, 3831, 8304, 18104, 39666, 87296, 192896, 427778, 951808, 2124135, 4753476, 10664458, 23981698, 54045448, 122041844, 276101386, 625725936, 1420386363, 3229171828, 7351869690, 16760603722, 38258956928, 87437436916, 200057233386, 458223768512, 1050614664580
Offset: 0

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Keywords

Comments

Words of length n in language defined by L = 1 + a + (L)L: L(0)=1, L(1)=a, L(2)=(), L(3)=(a)+()a, L(4)=(())+(a)a+()(), ...
Series reversion of x*A(x) is x*A082582(-x). - Michael Somos, Jul 22 2003
a(n) = number of Motzkin n-paths (A001006) in which no flatstep (F) is immediately followed by either an upstep (U) or a flatstep, in other words, each flatstep is either followed by a downstep (D) or ends the path. For example, a(4)=3 counts UDUD, UFDF, UUDD. - David Callan, Jun 07 2006
a(n) = number of Dyck (n+1)-paths (A000108) containing no UDUs and no subpaths of the form UUPDD where P is a nonempty Dyck path. For example, a(4)=3 counts UUUDDUUDDD, UUDDUUDDUD, UUUDDUDDUD. - David Callan, Sep 25 2006
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_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-1}*a_0 for n >= t. For example phi([1]) is the Catalan numbers A000108. The present sequence is (essentially) phi([0,1,1]). - Gary W. Adamson and R. J. Mathar, Oct 27 2008
The Kn21(n) triangle sums of A175136 lead to A007477(n+1), while the Kn22(n) = A007477(n+3)-1, Kn23(n) = A007477(n+5)-(4+n) and Kn3(n) = A007477(2*n+1) triangle sums of A175136 are related to the sequence given above. For the definition of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011
For n>=2, a(n) gives number of possible, ways to parse an English sentence consisting of just n+1 copies of word "buffalo", with one particular "plausible" grammar. See the Wikipedia page and my Python source at OEIS Wiki. Whether these are really intelligible sentences is of course debatable. See A213705 for a more plausible example in the Finnish language. - Antti Karttunen, Sep 14 2012

References

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

Crossrefs

Programs

  • Haskell
    a007477 n = a007477_list !! n
    a007477_list = 1 : 1 : f [1,1] where
       f xs = y : f (y:xs) where y = sum $ zipWith (*) (tail xs) (reverse xs)
    -- Reinhard Zumkeller, Apr 09 2012
    
  • Maple
    A007477 := proc(n) option remember; local k; if n <= 1 then 1 else add(A007477(k)*A007477(n-k-2),k=0..n-2); fi; end;
    unprotect(phi);
    phi:=proc(t,u,M) local i,a;
    a:=Array(0..M); for i from 0 to t-1 do a[i]:=u[i+1]; od:
    for i from t to M do a[i]:=add(a[j]*a[i-1-j],j=0..i-1); od:
    [seq(a[i],i=0..M)]; end;
    phi(3,[0,1,1],30);
    # N. J. A. Sloane, Nov 02 2008
  • Mathematica
    f[x_] := (1 - Sqrt[1 - 4x^2 - 4x^3])/2; Drop[ CoefficientList[ Series[f[x], {x, 0, 32}], x], 2] (* Jean-François Alcover, Nov 22 2011, after Pari *)
    a[n_] := Sum[Binomial[2*k+2, n-k-2]*Binomial[n-k-2, k]/(k+1), {k, 0, n-2}]; a[0] = a[1] = 1; Array[a, 40, 0] (* Jean-François Alcover, Mar 04 2016, after Vladimir Kruchinin *)
  • Maxima
    a(n):=if n<2 then 1 else sum((binomial(2*k+2,n-k-2)*binomial(n-k-2,k))/(k+1),k,0,n-2); /* Vladimir Kruchinin, Nov 22 2014 */
  • PARI
    a(n)=polcoeff((1-sqrt(1-4*x^2-4*x^3+x^3*O(x^n)))/2,n+2)
    

Formula

a(n) = sum( a(k) * a(n-2-k) ), n>1.
G.f. A(x) satisfies the equation 0 = 1 + x - A(x) + (x*A(x))^2.
The g.f. satisfies A(x)-x^2*A(x)^2 = 1+x. - Ralf Stephan, Jun 30 2003
G.f.: (1-sqrt(1-4x^2-4x^3))/(2x^2).
G.f.: (1+x)c(x^2(1+x)) where c(x) is g.f. of A000108. - Paul Barry, May 31 2006
G.f.: 1/(1-x/(1-x^2/(1-x^2/(1-x/(1-x^2/(1-x^2/(1-x/(1-x^2/(1-x^2/(1-... (continued fraction). - Paul Barry, Jul 30 2010
D-finite with recurrence: (n+2)*a(n) +(n+1)*a(n-1) +4*(-n+1)*a(n-2) +2*(-4*n+9)*a(n-3) +2*(-2*n+7)*a(n-4)=0. - R. J. Mathar, Dec 02 2012
a(n) = Sum_{k=0..n-2} binomial(2*k+2,n-k-2)*binomial(n-k-2,k)/(k+1), n>1, a(0)=1, a(1)=1. - Vladimir Kruchinin, Nov 22 2014
a(n) = Sum_{k=0..n-1} (-1)^(n-1-k)*binomial(n-1,k)*A082582(k+2), for n>0. - Thomas Baruchel, Jan 22 2015
a(n) ~ sqrt(3 - 4*r^2) * (4*r)^n * (1+r)^(n+1) / (sqrt(Pi)*n^(3/2)), where r = 0.41964337760708056627592628232664330021208937304879612338939... is the root of the equation 4*r^2*(1+r) = 1. - Vaclav Kotesovec, Jul 03 2021

Extensions

Additional comments from Michael Somos, Aug 03 2000
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