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.

Showing 1-10 of 10 results.

A244316 a(0) = 0, after which, if A176137(n) = 1, a(n) = A001511(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).

Original entry on oeis.org

0, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 5, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 5, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 3, 4, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 6
Offset: 0

Views

Author

Antti Karttunen, Jun 25 2014

Keywords

Comments

For n >= 1, a(n) tells the one-based position of the digit (from the right) where the iteration stopped at, when constructing a Semigreedy Catalan representation of n as described in A244159.
Algorithm for constructing the sequence: Find the largest Catalan number which is less than or equal to n (this is A081290(n) = A000108(k), where k = A244160(n), that is, the corresponding index of that Catalan number), and subtract that from n. Then check whether the previous Catalan number, C(m) = A000108(m), where m = k-1, exceeds the remaining n, and if it does not, then subtract that also from n, and keep on doing the same for lesser and lesser Catalan numbers, comparing and also subtracting them (whenever it is possible without going less than zero) from n, until either n becomes zero, or after subtracting C(1) = 1 from n, it still has not reached zero. In the latter case, find again the largest Catalan number which is less than or equal to remaining n, and start the process again. However, when at some point n finally reaches zero, then the index k of the last Catalan number, A000108(k) which was subtracted from n before it reached zero, is our result, a(n) = k. [Here n = the original value of n, from which we started subtracting initially from].
If n is one of the terms of A197433, meaning that if it can be represented as a sum of distinct Catalan numbers as n = C(i) + C(j) + ... + C(k) (which representation then necessarily is unique), then a(n) = min(i,j,...,k).

Links

Crossrefs

One more than A244315.
Cf. A000108, A001511, A176137, A197433, A244230, A244159, A244160, A081290.

Formula

a(0) = 0, and for n >= 1, if A176137(n) = 1, a(n) = A001511(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).
For n >= 1, a(n) = A244315(n)+1.
For n >= 1, a(A000108(n)) = n and a(A014138(n)) = a(A014143(n)) = 1.

A244315 a(0) = 0, after which, if A176137(n) = 1, a(n) = A007814(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).

Original entry on oeis.org

0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 3, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 3, 0, 1, 0, 0, 2, 4, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 3, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 3, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 3, 0, 1, 0, 0, 2, 4, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 2, 3, 0, 1, 0, 0, 2, 0, 1, 0, 0, 1, 0, 5
Offset: 0

Views

Author

Antti Karttunen, Jun 25 2014

Keywords

Comments

For n >= 1, a(n) tells the zero-based position of the digit (from the right) where the iteration stopped at, when constructing a Semigreedy Catalan representation of n as described in A244159.

Links

Crossrefs

One less than A244316.
Cf. A000108, A007814, A176137, A197433, A244230, A244159.

Formula

a(0) = 0, and for n >= 1, if A176137(n) = 1, a(n) = A007814(A244230(n)), otherwise a(n) = a(n-A197433(A244230(n)-1)).

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

Examples

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

References

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Links

Crossrefs

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

Programs

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

Formula

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

A014138 Partial sums of (Catalan numbers starting 1, 2, 5, ...).

Original entry on oeis.org

0, 1, 3, 8, 22, 64, 196, 625, 2055, 6917, 23713, 82499, 290511, 1033411, 3707851, 13402696, 48760366, 178405156, 656043856, 2423307046, 8987427466, 33453694486, 124936258126, 467995871776, 1757900019100
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of paths starting from the root in all ordered trees with n+1 edges (a path is a nonempty tree with no vertices of outdegree greater than 1). Example: a(2)=8 because the five trees with three edges have altogether 1+0+2+2+3=8 paths hanging from the roots. - Emeric Deutsch, Oct 20 2002
a(n) is the sum of the mean maximal pyramid size over all Dyck (n+1)-paths. Also, a(n) = sum of the mean maximal sawtooth size over all Dyck (n+1)-paths. A pyramid (resp. sawtooth) in a Dyck path is a subpath of the form U^k D^k (resp. (UD)^k) with k>=1 and k is its size. For example, the maximal pyramids in the Dyck path uUUDD|UD|UDdUUDD are indicated by uppercase letters (and separated by a vertical bar). Their sizes are 2,1,1,2 left to right and the mean maximal pyramid size of the path is 6/4 = 3/2. Also, the mean maximal sawtooth size of this path is (1+2+1)/3 = 4/3. - David Callan, Jun 07 2006
p^2 divides a(p-1) for prime p of form p=6k+1 (A002476(k)). - Alexander Adamchuk, Jul 03 2006
p^2 divides a(p^2-1) for prime p>3. p^2 divides a(p^3-1) for prime p=7,13,19,... prime p in the form p=6k+1. - Alexander Adamchuk, Jul 03 2006
Row sums of triangle A137614. - Gary W. Adamson, Jan 30 2008
Equals INVERTi transform of A095930: (1, 4, 15, 57, 220, 859, ...). - Gary W. Adamson, May 15 2009
a(n) < A000108(n+1), therefore A176137(n) <= 1. - Reinhard Zumkeller, Apr 10 2010
a(n) is also the sum of the numbers in Catalan's triangle (A009766) from row 0 to row n. - Patrick Labarque, Jul 27 2010
Equals the Catalan sequence starting (1, 1, 2, ...) convolved with A014137 starting (1, 2, 4, 9, ...). - Gary W. Adamson, May 20 2013
p divides a((p-3)/2) for primes {11,23,47,59,...} = A068231 primes congruent to 11 mod 12. - Alexander Adamchuk, Dec 27 2013
a(n) is the number of parking functions of size n avoiding the patterns 132, 213, and 231. - Lara Pudwell, Apr 10 2023

Links

Crossrefs

Cf. A000108, A002476, A005554, A068231, A095930, A137614, A155587.

Programs

  • Haskell
    a014138 n = a014138_list !! n
a014138_list = scanl1 (+) a000108_list  -- Reinhard Zumkeller, Mar 01 2013
  • Maple
    a:=n->sum((binomial(2*j,j)/(j+1)),j=1..n): seq(a(n), n=0..24); # Zerinvary Lajos, Dec 01 2006
# Second program:
A014138 := series(exp(2*x)*(BesselI(0, 2*x)/2 - BesselI(1, 2*x)) + exp(x)*(3/2*int(BesselI(0, 2*x)*exp(x), x) - 1/2), x = 0, 26):
seq(n!*coeff(A014138, x, n), n = 0 .. 24); # Mélika Tebni, Aug 31 2024
  • Mathematica
    Table[Sum[(2k)!/k!/(k+1)!,{k,1,n}],{n,1,70}] (* Alexander Adamchuk, Jul 03 2006 *)
Join[{0},Accumulate[CatalanNumber[Range[30]]]] (* Harvey P. Dale, Jan 25 2013 *)
CoefficientList[Series[(1 - 2 x - (1 - 4 x)^(1/2))/(2 x (1 - x)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 21 2015 *)
a[0] := 0; a[n_] := Sum[CatalanNumber[k], {k, 1, n}]; Table[a[n], {n,0,50}] (* G. C. Greubel, Jan 14 2017 *)
  • PARI
    Vec((1-2*x-(1-4*x)^(1/2))/(2*x*(1-x))) \\ Charles R Greathouse IV, Feb 11 2011
  • Python
    from _future_ import division
A014138_list, b, s = [0], 1, 0
for n in range(1,10**2):
    s += b
    A014138_list.append(s)
    b = b*(4*n+2)//(n+2) # Chai Wah Wu, Jan 28 2016

Formula

a(n) = A014137(n)-1.
G.f.: (1-2*x-sqrt(1-4x))/(2x(1-x)) = (C(x)-1)/(1-x) where C(x) is the generating function for the Catalan numbers. - Rocio Blanco, Apr 02 2007
a(n) = Sum_{k=1..n} A000108(k). - Alexander Adamchuk, Jul 03 2006
Binomial transform of A005554: (1, 2, 3, 6, 13, 30, 72, ...). - Gary W. Adamson, Nov 23 2007
D-finite with recurrence: (n+1)*a(n) + (1-5n)*a(n-1) + 2*(2n-1)*a(n-2) = 0. - R. J. Mathar, Dec 14 2011
Equals the Catalan sequence starting (1, 1, 2, ...) convolved with A014137 starting (1, 2, 4, 9, ...). - Gary W. Adamson, May 20 2013
G.f.: 1/x - G(0)/(1-x)/x, where G(k) = 1 - x/(1 - x/(1 - x/(1 - x/G(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
G.f.: 1/x - T(0)/(2*x*(1-x)), where T(k) = 2*x*(2*k+1)+ k+2 - 2*x*(k+2)*(2*k+3)/T(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 27 2013
a(n) ~ 2^(2*n+2)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 10 2013
a(n) = Sum_{i+jA000108. - Yuchun Ji, Jan 10 2019
E.g.f.: exp(2*x)*(BesselI(0, 2*x)/2 - BesselI(1, 2*x)) + exp(x)*(3/2*Integral_{x=-oo..oo} BesselI(0,2*x)*exp(x) dx - 1/2). - Mélika Tebni, Aug 31 2024

Extensions

Edited by Max Alekseyev, Sep 13 2009 (including adding an initial 0)
Definition edited by N. J. A. Sloane, Oct 03 2009

A014418 Representation of n in base of Catalan numbers (a classic greedy version).

Original entry on oeis.org

0, 1, 10, 11, 20, 100, 101, 110, 111, 120, 200, 201, 210, 211, 1000, 1001, 1010, 1011, 1020, 1100, 1101, 1110, 1111, 1120, 1200, 1201, 1210, 1211, 2000, 2001, 2010, 2011, 2020, 2100, 2101, 2110, 2111, 2120, 2200, 2201, 2210, 2211, 10000
Offset: 0

Views

Author

Olivier Gérard

Keywords

Comments

From Antti Karttunen, Jun 22 2014: (Start)
Also called "Greedy Catalan Base" for short.
Note: unlike A239903, this is a true base system, thus A244158(a(n)) = n holds for all n. See also A244159 for another, "less greedy" Catalan Base number system.
No digits larger than 3 will ever appear, because C(n+1)/C(n) approaches 4 from below, but never reaches it. [Where C(n) is the n-th Catalan number, A000108(n)].
3-digits cannot appear earlier than at the fifth digit-position from the right, the first example being a(126) = 30000.
The last digit is always either 0 or 1. (Cf. the sequences A244222 and A244223 which give the corresponding k for "even" and "odd" representations). No term ends as ...21.
No two "odd" terms (ending with 1) may occur consecutively.
A244217 gives the k for which a(k) starts with the digit 1, while A244216 gives the k for which a(k) starts with the digit 2 or 3.
A000108(n+1) gives the position of numeral where 1 is followed by n zeros.
A014138 gives the positions of repunits.
A197433 gives such k that a(k) = A239903(k). [Actually, such k, that the underlying strings of digits/numbers are same].
For the explanations, see the attached notes.
(End)

Examples

			A simple weighted sum of Sum_{k} digit(k)*C(k) [where C(k) = A000108(k), and digit(1) is the rightmost digit] recovers the natural number n (which the given numeral a(n) represents) as follows:
a(11) = 201, and indeed 2*C(3) + 0*C(2) + 1*C(1) = 2*5 + 0*2 + 1*1 = 11.
a(126) = 30000, and indeed, 3*C(5) = 3*42 = 126.

Links

Crossrefs

Cf. A014420 (gives the sum of digits), A244221 (same sequence reduced modulo 2, or equally, the last digit of a(n)), A244216, A244217, A244222, A244223, A000108, A007623, A197433, A239903, A244155, A244158, A244320, A244318, A244159 (a variant), A244161 (in base-4), A014417 (analogous sequence for Fibonacci numbers).
Cf. also A033552, A176137, A161227, A161228.

Programs

  • Mathematica
    CatalanBaseIntDs[n_] := Module[{m, i, len, dList, currDigit}, i = 1; While[n > CatalanNumber[i], i++]; m = n; len = i; dList = Table[0, {len}]; Do[currDigit = 0; While[m >= CatalanNumber[j], m = m - CatalanNumber[j]; currDigit++]; dList[[len - j + 1]] = currDigit, {j, i, 1, -1}]; If[dList[[1]] == 0, dList = Drop[dList, 1]]; FromDigits@ dList]; Array [CatalanBaseIntDs, 50, 0] (* Robert G. Wilson v, Jul 02 2014 *)
  • Python
    from sympy import catalan
def a244160(n):
    if n==0: return 0
    i=1
    while True:
        if catalan(i)>n: break
        else: i+=1
    return i - 1
def a(n):
    if n==0: return 0
    x=a244160(n)
    return 10**(x - 1) + a(n - catalan(x))
print([a(n) for n in range(51)]) # Indranil Ghosh, Jun 08 2017

Formula

From Antti Karttunen, Jun 23 2014: (Start)
a(0) = 0, a(n) = 10^(A244160(n)-1) + a(n-A000108(A244160(n))). [Here A244160 gives the index of the largest Catalan number that still fits into the sum].
a(n) = A007090(A244161(n)).
For all n, A000035(a(n)) = A000035(A244161(n)) = A244221(n).
(End)

Extensions

Description clarified by Antti Karttunen, Jun 22 2014

A244159 Semigreedy Catalan Representation of nonnegative integers.

Original entry on oeis.org

0, 1, 10, 11, 12, 100, 101, 110, 111, 112, 121, 122, 123, 211, 1000, 1001, 1010, 1011, 1012, 1100, 1101, 1110, 1111, 1112, 1121, 1122, 1123, 1211, 1212, 1221, 1222, 1223, 1232, 1233, 1234, 1322, 2111, 2112, 2121, 2122, 2123, 2211, 10000
Offset: 0

Views

Author

Antti Karttunen, Jun 23 2014

Keywords

Comments

Algorithm for constructing the sequence: Define a(0) as 0, and for larger values of n, find first the largest Catalan number which is less than or equal to n [which is A081290(n)], and the index k = A244160(n), of that Catalan number. Initialize a vector of k zeros, [0, 0, ..., 0]. Set n_remaining = n - A000108(k) and add 1 to the leftmost element of vector, so that it will become [1, 0, ..., 0]. Then check whether the previous Catalan number, C(m) = A000108(m), where m = k-1, exceeds the n_remaining, and provided that C(m) <= n_remaining, then set n_remaining = n_remaining - C(m) and increment by one the m-th element of the vector (where the 1st element is the rightmost), otherwise just decrement m by one and keep on doing the same with lesser and lesser Catalan numbers, and whenever it is possible to subtract them from n_remaining (without going less than zero), do so and increment the corresponding m-th element of the vector, as long as either n_remaining becomes zero, or after subtracting C(1) = 1 from n_remaining, it still has not reached zero. In the latter case, find again the largest Catalan number which is less than or equal to n_remaining, and start the process again. However, after a finite number of such iterations, n_remaining will finally reach zero, and the result of a(n) is then the vector of numbers constructed, concatenated together and represented as a decimal number.
This shares with "Greedy Catalan Base" (A014418) the property that a simple weighted sum of Sum_{k=1..} digit(k)*C(k) recovers the natural number n, which the given numeral string like A014418(n) or here, a(n), represents. (Here C(k) = the k-th Catalan number, A000108(k), and digit(1) = the digit in the rightmost, least significant digit position.)
In this case, A244158(a(n)) = n holds for only up to 33603, after which comes the first representation containing a "digit" larger than nine, at a(33604), where the underlying string of numbers is [1,2,3,4,5,6,7,8,9,10] but the decimal system used here can no more unambiguously represent them.
On the other hand, with the given Scheme-functions, we always get n back with: (CatBaseSumVec (A244159raw n)).
For n >= 1, A014138(n) gives the positions of repunits: 1, 11, 111, 1111, ...
The "rep-2's": 22222, 222222, 2222222, 22222222, 222222222, ..., etc., occur in positions 128, 392, 1250, 4110, 13834, ... i.e. 2*A014138(n) for n >= 5.

Examples

			For n = 18, the largest Catalan number <= 18 is C(4) = 14.
Thus we initialize a vector of four zeros [0, 0, 0, 0] and increment the first element to 1: [1, 0, 0, 0] and subtract 14 from 18 to get the remainder 4.
We see that the next smaller Catalan number, C(3) = 5 is greater than 4, so we cannot subtract it without going negative, so the second leftmost element of the vector stays as zero.
We next check C(2) = 2, which is less than 4, thus we increment the zero at that point to 1, and subtract 4 - 2 to get 2.
We compare 2 to C(1) = 1, and as 1 <= 2, it is subtracted 2-1 = 1, and the corresponding element in the vector incremented, thus after the first round, the vector is now [1, 0, 1, 1], and n remaining is 1.
So we start the second round because n has not yet reached the zero, and look for the largest Catalan number <= 1, which in this case is C(1) = 1, so we subtract it from remaining n, and increment the element in the position 1, after which n has reached zero, and the vector is now [1, 0, 1, 2], whose concatenation as decimal numbers thus yields a(18) = 1012.

Links

Crossrefs

Cf. A014418 (a classical greedy variant), A244231 (maximum "digit value"), A244232 (sum of digits), A244233 (product of digits), A244314 (positive terms which have at least one zero digit), A244316 (the one-based position of digit incremented last in the described process).
Cf. also A007088, A014138, A176137, A197433, A244230, A244158, A244160.
Differs from A239903 for the first time at n=10, where a(10) = 121, while A239903(10) = 120.

Formula

If A176137(n) = 1, a(n) = A007088(A244230(n)), otherwise a(n) = A007088(A244230(n)-1) + a(n-A197433(A244230(n)-1)).
For all n, a(A197433(n)) = A007088(n).
For all n >= 1, a(A000108(n)) = 10^(n-1).
Each a(A014143(n)) has a "triangular" representation [1, 2, 3, ..., n, n+1].

A197433 Sum of distinct Catalan numbers: a(n) = Sum_{k>=0} A030308(n,k)*C(k+1) where C(n) is the n-th Catalan number (A000108). (C(0) and C(1) not treated as distinct.)

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 8, 14, 15, 16, 17, 19, 20, 21, 22, 42, 43, 44, 45, 47, 48, 49, 50, 56, 57, 58, 59, 61, 62, 63, 64, 132, 133, 134, 135, 137, 138, 139, 140, 146, 147, 148, 149, 151, 152, 153, 154, 174, 175, 176, 177, 179, 180, 181, 182, 188, 189, 190, 191, 193, 194, 195, 196
Offset: 0

Views

Author

Philippe Deléham, Oct 15 2011

Keywords

Comments

Replace 2^k with A000108(k+1) in binary expansion of n.
From Antti Karttunen, Jun 22 2014: (Start)
On the other hand, A244158 is similar, but replaces 10^k with A000108(k+1) in decimal expansion of n.
This sequence gives all k such that A014418(k) = A239903(k), which are precisely all nonnegative integers k whose representations in those two number systems contain no digits larger than 1. From this also follows that this is a subsequence of A244155.
(End)

Links

Crossrefs

Characteristic function: A176137.
Subsequence of A244155.
Cf. A000108, A030308, A197432, A014418, A239903, A244158, A244159, A244230, A244231, A244232, A244315, A244316.
Cf. also A060112.
Other sequences that are built by replacing 2^k in binary representation with other numbers: A022290 (Fibonacci), A029931 (natural numbers), A059590 (factorials), A089625 (primes), A197354 (odd numbers).

Programs

  • Mathematica
    nmax = 63;
a[n_] := If[n == 0, 0, SeriesCoefficient[(1/(1-x))*Sum[CatalanNumber[k+1]* x^(2^k)/(1 + x^(2^k)), {k, 0, Log[2, n] // Ceiling}], {x, 0, n}]];
Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Nov 18 2021, after Ilya Gutkovskiy *)

Formula

For all n, A244230(a(n)) = n. - Antti Karttunen, Jul 18 2014
G.f.: (1/(1 - x))*Sum_{k>=0} Catalan number(k+1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017

Extensions

Name clarified by Antti Karttunen, Jul 18 2014

A244230 a(n) is the least k such that A197433(k) >= n.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 5, 6, 7, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 12, 13, 14, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 18, 19, 20, 20, 21, 22, 23, 24, 24, 24, 24, 24, 24, 25, 26, 27, 28, 28, 29, 30, 31, 32, 32, 32, 32
Offset: 0

Views

Author

Antti Karttunen, Jun 25 2014

Keywords

Comments

For n >= 1, a(n) is the total number of ways the natural numbers in range 1 .. n can be represented as sums of distinct Catalan numbers (A000108). Note that for any one number, number of such solutions may be at most one. In other words, this sequence is one less than the partial sums of A176137 (number of partitions of n into distinct Catalan numbers).

Links

Crossrefs

The first differences give A176137 from its term a(1) onward.
Cf. A000108, A197433.

Programs

  • Mathematica
    nmax = 68;
A197433[n_] := If[n == 0, 0, SeriesCoefficient[(1/(1-x))*Sum[ CatalanNumber[k+1]*x^(2^k)/(1+x^(2^k)), {k, 0, Log[2, n] // Ceiling}], {x, 0, n}]];
a[n_] := For[k = 0, True, k++, If[A197433[k] >= n, Return[k]]];
Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Nov 18 2021, after Ilya Gutkovskiy in A197433 *)

Formula

For all n >= 0, a(A197433(n)) = n. [This works as an inverse function for the injection A197433].

A244232 Sum of "digit values" in Semigreedy Catalan Representation of n, A244159.

Original entry on oeis.org

0, 1, 1, 2, 3, 1, 2, 2, 3, 4, 4, 5, 6, 4, 1, 2, 2, 3, 4, 2, 3, 3, 4, 5, 5, 6, 7, 5, 6, 6, 7, 8, 8, 9, 10, 8, 5, 6, 6, 7, 8, 6, 1, 2, 2, 3, 4, 2, 3, 3, 4, 5, 5, 6, 7, 5, 2, 3, 3, 4, 5, 3, 4, 4, 5, 6, 6, 7, 8, 6, 7, 7, 8, 9, 9, 10, 11, 9, 6, 7, 7, 8, 9, 7, 8, 8, 9, 10, 10, 11, 12, 10, 11, 11, 12, 13, 13, 14, 15, 13, 10, 11, 11, 12, 13, 11, 6, 7, 7, 8, 9, 7, 8, 8, 9, 10, 10, 11, 12, 10, 7, 8, 8, 9, 10, 8, 9, 9, 10, 11, 11, 12, 1
Offset: 0

Views

Author

Antti Karttunen, Jun 25 2014

Keywords

Comments

Note that a(33604) = A000217(10) = 55 because the sum is computed from the underlying list (vector) of numbers, and thus is not subject to any corruption by decimal representation as A244159 itself is.
Equivalent description: partition n "greedily" as terms of A197433, i.e. n = A197433(i) + A197433(j) + ... + A197433(k), always using the largest term of A197433 that still "fits in" (i.e. is <= n remaining). Then a(n) = A000120(i) + A000120(j) + ... + A000120(k).

Examples

			For n=18, using the alternative description, we see that it is partitioned  into the terms of A197433 as a greedy sum A197433(11) + A197433(1) = 17 + 1. Thus a(18) = A000120(11) + A000120(1) = 3+1 = 4.
For n=128, we see that is likewise represented as A197433(31) + A197433(31) = 64 + 64. Thus a(128) = 2*A000120(31) = 10.

Links

Crossrefs

Cf. A000108, A000120, A176137, A197433, A244230, A244159, A244231, A244233, A244234, A014420, A236855.

Formula

If A176137(n) = 1, a(n) = A000120(A244230(n)), otherwise a(n) = A000120(A244230(n)-1) + a(n-A197433(A244230(n)-1)).
For all n, a(A000108(n)) = 1. [And moreover, Catalan numbers, A000108, give all such k that a(k) = 1].
For all n, a(A014138(n)) = n and a(A014143(n)) = A000217(n+1).

A033552 Number of partitions into Catalan numbers.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 5, 6, 7, 8, 10, 11, 13, 14, 17, 19, 22, 24, 27, 30, 34, 37, 41, 44, 49, 53, 58, 62, 68, 73, 80, 85, 92, 98, 106, 113, 121, 128, 137, 145, 155, 163, 175, 184, 197, 207, 220, 232, 246, 259, 274, 287, 304, 318, 336, 351, 371, 388, 409, 427, 449, 469
Offset: 0

Views

Author

Marc LeBrun

Keywords

Examples

			n=4 has 3 partitions: 2+2, 2+1+1, 1+1+1+1.
n=5 has 4 partitions: 5, 2+2+1, 2+1+1+1, 1+1+1+1+1.

Links

Crossrefs

Cf. A000108.
Cf. A176137. [Reinhard Zumkeller, Apr 09 2010]

Formula

G.f.: Product_{n>=1} 1/(1 - x^(binomial(2*n, n)/(n+1))).
a(n) = f(n,1,1) with f(m,k,c) = if c > m then 0^m else f(m-c,k,c) + f(m,k+1,2*c*(2*k+1)/(k+2)). [Reinhard Zumkeller, Apr 09 2010]
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