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A000108 Catalan numbers: C(n) = binomial(2n,n)/(n+1) = (2n)!/(n!(n+1)!). Also called Segner numbers.
(Formerly M1459 N0577)
2434
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

The solution to Schroeder's first problem. A very large number of combinatorial interpretations are known - see references, esp. Stanley, Enumerative Combinatorics, Volume 2. This is probably the longest entry in the OEIS, and rightly so.

Number of ways to insert n pairs of parentheses in a word of n+1 letters. E.g., 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

a(n-1) is the number of ways of expressing an n-cycle 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 uk<=uj and vk<=vj for k<j; see example.  If the condition is dropped, one obtains A000272. - 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

Ways of joining 2n points on a circle to form n nonintersecting chords. (If no such restriction imposed, then 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

c(n) = C(2*n-2,n-1)/n = (1/n!) * [ n^(n-1) + { C(n-2,1) +C(n-2,2) }*n^(n-2) + { 2*C(n-3,1) +7*C(n-3,2) +8*C(n-3,3) +3*C(n-3,4) }*n^(n-3) + { 6*C(n-4,1) +38*C(n-4,2) +93*C(n-4,3) +111*C(n-4,4) +65*C(n-4,5) +15*C(n-4,6) }*n^(n-4) + ..... ]. - André F. Labossière, Nov 10 2004

Sum_{n=0..infinity} 1/a(n) = 2 + 4*Pi/3^(5/2) = F(1,2;1/2;1/4) = 2.806133050770763... (see L'Univers de Pi link). - Gerald McGarvey and Benoit Cloitre, Feb 13 2005

a(n) equals 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 = C(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

lim(a(n)/a(n-1): n->infinity) = 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 Grassmanian 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 Grassmanian 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...Infinity,c(k-1)/2^(2k-1)}=1. The k-th term in the summation is the probability that a random walk on the integers (begining 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(C(i)*C(j)*C(p+q-i-j-1), i=0..(p-1), j=0..(q-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) = A000680(n)/A006472(n). - Mark Dols (markdols99(AT)yahoo.com), Jul 14 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

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. - Matthew Vandermast, Nov 22 2010

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).

Complement of A092459; A010058(a(n)) = 1. - Reinhard Zumkeller, Mar 29 2011

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

A076050(a(n)) = n + 1 for n > 0. - Reinhard Zumkeller, Feb 17 2012

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

a(n+1) = A214292(2*n+1,n) = A214292(2*n+2,n). - Reinhard Zumkeller, Jul 12 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 1012

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

Given Probability (p): sum(n=0..inf) a(n)*(1-p)^n*p^(n+1) = sum(n=1..inf) p^n = p/(1-p). E.g., at p=0.4: 0.4 + 0.6*0.4^2 + 2*0.6^2*0.4^3 + 5*0.6^3*0.4^4 + 14*0.6^4*0.4^5 +... = 0.4 + 0.096 + 0.04608 + 0.027648 + 0.018579456... = 2/3. Since p/(1-p) is itself a probability, it therefore has a maximum value of 1 when p >= 0.5. - Bob Selcoe, Nov 16 2013

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 = 1.8061330507707... See A121839 = Decimal expansion of sum(k>=1, 1/a(k)).

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', 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 non-zero 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

REFERENCES

The large number of references and links demonstrates the ubiquity of the Catalan numbers.

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Index entries for "core" sequences

Index entries for sequences related to rooted trees

Index entries for sequences related to parenthesizing

Index entries for sequences related to necklaces

FORMULA

a(n)) = A000984(n)/(n+1) = binomial(2*n, n)/(n+1) = (2*n)!/(n!*(n+1)!).

a(n) = binomial(2*n, n)-binomial(2*n, n-1).

a(n) = Sum_{k=0..n-1} a(k)a(n-1-k).

a(n) = Prod_{k=2..n} (1 + n/k), if n>1.

G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x). G.f. A(x) satisfies A = 1 + x*A^2.

a(n+1) = Sum_{i} binomial(n, 2*i)*2^(n-2*i)*a(i). - Touchard

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

It is known that a(n) is odd if and only if n=2^k-1, k=1, 2, 3, ... - Emeric Deutsch, Aug 04 2002

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) = int(x^n*sqrt((4-x)/x), x=0..4)/(2*Pi). - 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

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

a(n) = Sum_{k=0..[n/2]} ((n-2*k+1)*C(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}^{infinity} 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<=k<=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<=k<=n} A129818(n,k) * A007852(k+1). - Philippe Deléham, Jun 20 2007

a(n) = Sum_{k, 0<=k<=n} A109466(n,k) * A127632(k). - Philippe Deléham, Jun 20 2007

Row sums of triangle A124926. - Gary W. Adamson, Oct 22 2007 lim(1+Sum(a(k)/A004171(k): 0<=k<=n): n->infinity) = 4/Pi. - Reinhard Zumkeller, Aug 26 2008

a(n) = Sum_{k, 0<=k<=n}A120730(n,k)^2 and a(k+1) = Sum_{n, 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

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

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

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. - Tom Copeland, Sep 30 2011

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)). - Tom Copeland, Sep 30 2011

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

a(n) = A208355(2*n-1) = A208355(2*n) for n > 0. - Reinhard Zumkeller, Mar 04 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

EXAMPLE

G.f. = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + ...

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)

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)

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), n=0..42) ];

with(combstruct):bin := {B=Union(Z, Prod(B, B))}: seq (count([B, bin, unlabeled], size=n), n=1..25); # Zerinvary Lajos, Dec 05 2007

Z[0]:=0: for k to 42 do Z[k]:=simplify(1/(1-z*Z[k-1])) od: g:=sum((Z[j]-Z[j-1]), j=1..42): gser:=series(g, z=0, 42): seq(coeff(gser, z, n), n=0..41); # Zerinvary Lajos, May 21 2008

MATHEMATICA

A000108[ n_ ] := (2 n)!/n!/(n+1)!

A000108[n_] := Hypergeometric2F1[1 - n, -n, 2, 1] (* Richard L. Ollerton, Sep 13 2006 *)

Table[ CatalanNumber@ n, {n, 0, 24}] (* Robert G. Wilson v, Feb 15 2011 *)

CoefficientList[InverseSeries[Series[x/Sum[x^n, {n, 0, 31}], {x, 0, 31}]]/x, x] (* Mats Granvik, Nov 24 2013 *)

PROG

(MAGMA) C:= func< n | Binomial(2*n, n)/(n+1) >; [ C(n) : n in [0..60]];

(PARI) a(n)=binomial(2*n, n)/(n+1) \\ M. F. Hasler, Aug 25 2012

(PARI) {a(n) = if( n<0, 0, (2*n)! / n! / (n+1)!)};

(PARI) {a(n) = local(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

(Mupad) combinat::dyckWords::count(n) $ n = 0..38 // Zerinvary Lajos, Apr 14 2007

(Sage) [catalan_number(i) for i in range(27)] # Zerinvary Lajos, Jun 26 2008

(Sage) [binomial(2*i, i)-binomial(2*i, i-1) for i in xrange(0, 25)] # Zerinvary Lajos, May 17 2009

(MAGMA) [Catalan(n): n in [0..40]]; // Vincenzo Librandi, Apr 02 2011

(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

(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

(Maxima) A000108(n):=binomial(2*n, n)/(n+1)$ makelist(A000108(n), n, 0, 30); [Martin Ettl, Oct 24 2012]

(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

CROSSREFS

Cf. A000984, A001453, A002420, A048990, A024492, A000142, A022553, A039599, A003046, A069640, A094216, A094638, A014137, A014138, A094639, A099731, A008549, A008276, A094638, (|A008276|), A094216, A094639, A000984, A000245, A002057, A000344, A003517, A000588, A003518, A003519, A001392, A124926, A098597, A086117, A137697, A000957, A068875, A032443, A179277, A154559, A059288, A129763, A003046, A032357, A014140, A120304, A211611, A119822, A129763, A167892, A167893, A161581, A006480, A001791.

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).

Cf. A038003, A119861, A119908, A120274, A120275 (odd Catalan number).

Cf. A002390 (decimal expansion of natural logarithm of golden ratio).

Sequence in context: A115140 A120588 A168491 * A057413 A126567 A125501

Adjacent sequences:  A000105 A000106 A000107 * A000109 A000110 A000111

KEYWORD

core,nonn,easy,eigen,nice,changed

AUTHOR

N. J. A. Sloane

STATUS

approved

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Last modified October 21 01:22 EDT 2014. Contains 248371 sequences.