A046161 -id:A046161 - cp's OEIS frontend

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

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

N. J. A. Sloane

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

			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)

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.
Fürlinger, J.; Hofbauer, J., q-Catalan numbers. J. Combin. Theory Ser. A 40 (1985), no. 2, 248-264. MR0814413 (87e:05017)
M. Gardner, Time Travel and Other Mathematical Bewilderments, Chap. 20 pp. 253-266, W. H. Freeman NY 1988.
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Index entries for "core" sequences
Index entries for sequences related to necklaces
Index entries for sequences related to parenthesizing
Index entries for sequences related to rooted trees
Index entries for sequences related to Benford's law

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}.

A000108:=List([0..30],n->Binomial(2*n,n)/(n+1)); # Muniru A Asiru, Feb 17 2018

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

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

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

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

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 *)

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

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

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

PARI
```
a(n) = (2*n)! / n! / (n+1)!
```

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));

{a(n) = if( n<1, n==0, polcoeff( serreverse( x / (1 + x)^2 + x * O(x^n)), n))}; /* Michael Somos */

(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

x='x+O('x^40); Vec((1-sqrt(1-4*x))/(2*x)) \\ Altug Alkan, Oct 13 2015

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

# 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

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

# 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

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

A001511 The ruler function: exponent of the highest power of 2 dividing 2n. Equivalently, the 2-adic valuation of 2n.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 7, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 6, 1, 2, 1, 3, 1, 2, 1, 4, 1

Offset: 1

N. J. A. Sloane

Number of 2's dividing 2*n.
a(n) is equivalently the exponent of the smallest power of 2 which does not divide n. - David James Sycamore, Oct 02 2023
a(n) - 1 is the number of trailing zeros in the binary expansion of n.
If you are counting in binary and the least significant bit is numbered 1, the next bit is 2, etc., a(n) is the bit that is incremented when increasing from n-1 to n. - Jud McCranie, Apr 26 2004
Number of steps to reach an integer starting with (n+1)/2 and using the map x -> x*ceiling(x) (cf. A073524).
a(n) is the number of the disk to be moved at the n-th step of the optimal solution to Towers of Hanoi problem (comment from Andreas M. Hinz).
Shows which bit to flip when creating the binary reflected Gray code (bits are numbered from the right, offset is 1). This is essentially equivalent to Hinz's comment. - Adam Kertesz, Jul 28 2001
a(n) is the Hamming distance between n and n-1 (in binary). This is equivalent to Kertesz's comments above. - Tak-Shing Chan (chan12(AT)alumni.usc.edu), Feb 25 2003
Let S(0) = {1}, S(n) = {S(n-1), S(n-1)-{x}, x+1} where x = last term of S(n-1); sequence gives S(infinity). - Benoit Cloitre, Jun 14 2003
The sum of all terms up to and including the first occurrence of m is 2^m-1. - Donald Sampson (marsquo(AT)hotmail.com), Dec 01 2003
m appears every 2^m terms starting with the 2^(m-1)th term. - Donald Sampson (marsquo(AT)hotmail.com), Dec 08 2003
Sequence read mod 4 gives A092412. - Philippe Deléham, Mar 28 2004
If q = 2n/2^A001511(n) and if b(m) is defined by b(0)=q-1 and b(m)=2*b(m-1)+1, then 2n = b(A001511(n)) + 1. - Gerald McGarvey, Dec 18 2004
Repeating pattern ABACABADABACABAE ... - Jeremy Gardiner, Jan 16 2005
Relation to C(n) = Collatz function iteration using only odd steps: a(n) is the number of right bits set in binary representation of A004767(n) (numbers of the form 4*m+3). So for m=A004767(n) it follows that there are exactly a(n) recursive steps where mBetween every two instances of any positive integer m there are exactly m distinct values (1 through m-1 and one value greater than m). - Franklin T. Adams-Watters, Sep 18 2006
Number of divisors of n of the form 2^k. - Giovanni Teofilatto, Jul 25 2007
Every prefix up to (but not including) the first occurrence of some k >= 2 is a palindrome. - Gary W. Adamson, Sep 24 2008
1 interleaved with (2 interleaved with (3 interleaved with ( ... ))). - Eric D. Burgess (ericdb(AT)gmail.com), Oct 17 2009
A054525 (Möbius transform) * A001511 = A036987 = A047999^(-1) * A001511. - Gary W. Adamson, Oct 26 2009
Equals A051731 * A036987, (inverse Möbius transform of the Fredholm-Rueppel sequence) = A047999 * A036987. - Gary W. Adamson, Oct 26 2009
Cf. A173238, showing links between generalized ruler functions and A000041. - Gary W. Adamson, Feb 14 2010
Given A000041, P(x) = A(x)/A(x^2) with P(x) = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), A(x) = (1 + x + 3x^2 + 4x^3 + 10x^4 + 13x^5 + ...), A(x^2) = (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...), where A092119 = (1, 1, 3, 4, 10, ...) = Euler transform of the ruler sequence, A001511. - Gary W. Adamson, Feb 11 2010
Subtracting 1 from every term and deleting any 0's yields the same sequence, A001511. - Ben Branman, Dec 28 2011
In the listing of the compositions of n as lists in lexicographic order, a(k) is the last part of composition(k) for all k <= 2^(n-1) and all n, see example. - Joerg Arndt, Nov 12 2012
According to Hinz, et al. (see links), this sequence was studied by Louis Gros in his 1872 pamphlet "Théorie du Baguenodier" and has therefore been called the Gros sequence.
First n terms comprise least squarefree word of length n using positive integers, where "squarefree" means that the word contains no consecutive identical subwords; e.g., 1 contains no square; 11 contains a square but 12 does not; 121 contains no square; both 1211 and 1212 have squares but 1213 does not; etc. - Clark Kimberling, Sep 05 2013
Length of 0-run starting from 2 (10, 100, 110, 1000, 1010, ...), or length of 1-run starting from 1 (1, 11, 101, 111, 1001, 1011, ...) of every second number, from right to left in binary representation. - Armands Strazds, Apr 13 2017
a(n) is also the frequency of the largest part in the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 24 2017
As A000005(n) equals the number of even divisors of 2n and A001227(n) = A001227(2n), the formula A001511(n) = A000005(n)/A001227(n) might be read as "The number of even divisors of 2n is always divisible by the number of odd divisors of 2n" (where number of divisors means sum of zeroth powers of divisors). Conjecture: For any nonnegative integer k, the sum of the k-th powers of even divisors of n is always divisible by the sum of the k-th powers of odd divisors of n. - Ivan N. Ianakiev, Jul 06 2019
From Benoit Cloitre, Jul 14 2022: (Start)
To construct the sequence, start from 1's separated by a place 1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,,1,...
Then put the 2's in every other remaining place
  1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,,1,2,1,...
Then the 3's in every other remaining place
  1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,,1,2,1,...
Then the 4's in every other remaining place
  1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,,1,2,1,3,1,2,1,4,1,2,1,...
By iterating this process, we get the ruler function 1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,... (End)
a(n) is the least positive integer k for which there does not exist i+j=n and a(i)=a(j)=k (cf. A322523). - Rémy Sigrist and Jianing Song, Aug 23 2022
a(n) is the smallest positive integer that does not occur in the coincidences of the sequence so far a(1..n-1) and its reverse. - Neal Gersh Tolunsky, Jan 18 2023
The geometric mean of this sequence approaches the Somos constant (A112302). - Jwalin Bhatt, Jan 31 2025

			For example, 2^1|2, 2^2|4, 2^1|6, 2^3|8, 2^1|10, 2^2|12, ... giving the initial terms 1, 2, 1, 3, 1, 2, ...
From _Omar E. Pol_, Jun 12 2009: (Start)
Triangle begins:
1;
2,1;
3,1,2,1;
4,1,2,1,3,1,2,1;
5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
6,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1;
7,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,3,...
(End)
S(0) = {} S(1) = 1 S(2) = 1, 2, 1 S(3) = 1, 2, 1, 3, 1, 2, 1 S(4) = 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
From _Joerg Arndt_, Nov 12 2012: (Start)
The 16 compositions of 5 as lists in lexicographic order:
[ n]  a(n)  composition
[ 1]  [ 1]  [ 1 1 1 1 1 ]
[ 2]  [ 2]  [ 1 1 1 2 ]
[ 3]  [ 1]  [ 1 1 2 1 ]
[ 4]  [ 3]  [ 1 1 3 ]
[ 5]  [ 1]  [ 1 2 1 1 ]
[ 6]  [ 2]  [ 1 2 2 ]
[ 7]  [ 1]  [ 1 3 1 ]
[ 8]  [ 4]  [ 1 4 ]
[ 9]  [ 1]  [ 2 1 1 1 ]
[10]  [ 2]  [ 2 1 2 ]
[11]  [ 1]  [ 2 2 1 ]
[12]  [ 3]  [ 2 3 ]
[13]  [ 1]  [ 3 1 1 ]
[14]  [ 2]  [ 3 2 ]
[15]  [ 1]  [ 4 1 ]
[16]  [ 5]  [ 5 ]
a(n) is the last part in each list.
(End)
From _Omar E. Pol_, Aug 20 2013: (Start)
Also written as a triangle in which the right border gives A000027 and row lengths give A011782 and row sums give A000079 the sequence begins:
1;
2;
1,3;
1,2,1,4;
1,2,1,3,1,2,1,5;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6;
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,7;
(End)
G.f. = x + 2*x^2 + x^3 + 3*x^4 + x^5 + 2*x^6 + x^7 + 4*x^8 + x^9 + 2*x^10 + ...

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003.
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L. Gros, Théorie du Baguenodier, Aimé Vingtrinier, Lyon, 1872.
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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).

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
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Antonin Callard, Léo Paviet Salomon, and Pascal Vanier, Computability of extender sets in multidimensional subshifts, arXiv:2401.07549 [cs.DM], 2024.
Imanuel Chen and Michael Z. Spivey, Integral Generalized Binomial Coefficients of Multiplicative Functions, Preprint 2015; Summer Research Paper 238, Univ. Puget Sound.
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Eric Weisstein's World of Mathematics, Binary Carry Sequence, Ruler Function, and Towers of Hanoi
Index entries for "core" sequences
Index entries for sequences related to binary expansion of n
Index entries for sequences that are fixed points of mappings
Index entries for sequences related to Towers of Hanoi

Column 1 of table A050600.
Sequence read mod 2 gives A035263.
Sequence is bisection of A007814, A050603, A050604, A067029, A089309.
This is Guy Steele's sequence GS(4, 2) (see A135416).
Cf. A005187 (partial sums), A085058 (bisection), A112302 (geometric mean), A171977 (2^a(n)).
Cf. A000005, A000041, A000079, A003188, A003278, A003602, A007949, A018238, A047999, A051731, A054525, A054852, A065176, A082850, A089080, A092119, A117303, A129360, A130093, A173238, A181988, A220466.
Cf. A287896, A002487, A209229 (Mobius trans.), A092673 (Dirichlet inv.).
Cf. generalized ruler functions for k=3,4,5: A051064, A115362, A055457.

a001511 n = length $ takeWhile ((== 0) . (mod n)) a000079_list
-- Reinhard Zumkeller, Sep 27 2011

a001511 n | odd n = 1 | otherwise = 1 + a001511 (n `div` 2)
-- Walt Rorie-Baety, Mar 22 2013

nmax=5;r=1;for n=2:nmax;r=[r n r];end % Adriano Caroli, Feb 26 2016

[Valuation(2*n,2): n in [1..105]]; // Bruno Berselli, Nov 23 2015

A001511 := n->2-wt(n)+wt(n-1); # where wt is defined in A000120
# This is the binary logarithm of the denominator of (256^n-1)B_{8n}/n, in Maple parlance a := n -> log[2](denom((256^n-1)*bernoulli(8*n)/n)). - Peter Luschny, May 31 2009
A001511 := n -> padic[ordp](2*n,2): seq(A001511(n), n=1..105);  # Peter Luschny, Nov 26 2010
a:= n-> ilog2((Bits[Xor](2*n, 2*n-1)+1)/2): seq(a(n), n=1..50);  # Gary Detlefs, Dec 13 2018

Array[ If[ Mod[ #, 2] == 0, FactorInteger[ # ][[1, 2]], 0] &, 105] + 1 (* or *)
Nest[ Flatten[ # /. a_Integer -> {1, a + 1}] &, {1}, 7] (* Robert G. Wilson v, Mar 04 2005 *)
IntegerExponent[2*n, 2] (* Alexander R. Povolotsky, Aug 19 2011 *)
myHammingDistance[n_, m_] := Module[{g = Max[m, n], h = Min[m, n]}, b1 = IntegerDigits[g, 2]; b2 = IntegerDigits[h, 2, Length[b1]]; HammingDistance[b1, b2]] (* Vladimir Shevelev A206853 *) Table[ myHammingDistance[n, n - 1], {n, 111}] (* Robert G. Wilson v, Apr 05 2012 *)
Table[Position[Reverse[IntegerDigits[n,2]],1,1,1],{n,110}]//Flatten (* Harvey P. Dale, Aug 18 2017 *)

a(n) = sum(k=0,floor(log(n)/log(2)),floor(n/2^k)-floor((n-1)/2^k)) /* Ralf Stephan */

a(n)=if(n%2,1,factor(n)[1,2]+1) /* Jon Perry, Jun 06 2004 */

{a(n) = if( n, valuation(n, 2) + 1, 0)}; /* Michael Somos, Sep 30 2006 */

{a(n)=if(n==1,1,polcoeff(x-sum(k=1, n-1, a(k)*x^k*(1-x^k)*(1-x+x*O(x^n))), n))} /* Paul D. Hanna, Jun 22 2007 */

def a(n): return bin(n)[2:][::-1].index("1") + 1 # Indranil Ghosh, May 11 2017

A001511 = lambda n: (n&-n).bit_length() # M. F. Hasler, Apr 09 2020

def A001511(n): return (~n & n-1).bit_length()+1 # Chai Wah Wu, Jul 01 2022

[valuation(2*n,2) for n in (1..105)]  # Bruno Berselli, Nov 23 2015

(define (A001511 n) (let loop ((n n) (e 1)) (if (odd? n) e (loop (/ n 2) (+ 1 e))))) ;; Antti Karttunen, Oct 06 2017

a(n) = A007814(n) + 1 = A007814(2*n).
a(2*n+1) = 1; a(2*n) = 1 + a(n). - Philippe Deléham, Dec 08 2003
a(n) = 2 - A000120(n) + A000120(n-1), n >= 1. - Daniele Parisse
a(n) = 1 + log_2(abs(A003188(n) - A003188(n-1))).
Multiplicative with a(p^e) = e+1 if p = 2; 1 if p > 2. - David W. Wilson, Aug 01 2001
For any real x > 1/2: lim_{N->infinity} (1/N)*Sum_{n=1..N} x^(-a(n)) = 1/(2*x-1); also lim_{N->infinity} (1/N)*Sum_{n=1..N} 1/a(n) = log(2). - Benoit Cloitre, Nov 16 2001
s(n) = Sum_{k=1..n} a(k) is asymptotic to 2*n since s(n) = 2*n - A000120(n). - Benoit Cloitre, Aug 31 2002
For any n >= 0, for any m >= 1, a(2^m*n + 2^(m-1)) = m. - Benoit Cloitre, Nov 24 2002
a(n) = Sum_{d divides n and d is odd} mu(d)*tau(n/d). - Vladeta Jovovic, Dec 04 2002
G.f.: A(x) = Sum_{k>=0} x^(2^k)/(1-x^(2^k)). - Ralf Stephan, Dec 24 2002
a(1) = 1; for n > 1, a(n) = a(n-1) + (-1)^n*a(floor(n/2)). - Vladeta Jovovic, Apr 25 2003
A fixed point of the mapping 1->12; 2->13; 3->14; 4->15; 5->16; ... . - Philippe Deléham, Dec 13 2003
Product_{k>0} (1+x^k)^a(k) is g.f. for A000041(). - Vladeta Jovovic, Mar 26 2004
G.f. A(x) satisfies A(x) = A(x^2) + x/(1-x). - Franklin T. Adams-Watters, Feb 09 2006
a(A118413(n,k)) = A002260(n,k); = a(A118416(n,k)) = A002024(n,k); a(A014480(n)) = A003602(A014480(n)). - Reinhard Zumkeller, Apr 27 2006
Ordinal transform of A003602. - Franklin T. Adams-Watters, Aug 28 2006 (The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.)
Could be extended to n <= 0 using a(-n) = a(n), a(0) = 0, a(2*n) = a(n)+1 unless n=0. - Michael Somos, Sep 30 2006
A094267(2*n) = A050603(2*n) = A050603(2*n + 1) = a(n). - Michael Somos, Sep 30 2006
Sequence = A129360 * A000005 = M*V, where M = an infinite lower triangular matrix and V = d(n) as a vector: [1, 2, 2, 3, 2, 4, ...]. - Gary W. Adamson, Apr 15 2007
Row sums of triangle A130093. - Gary W. Adamson, May 13 2007
Dirichlet g.f.: zeta(s)*2^s/(2^s-1). - Ralf Stephan, Jun 17 2007
a(n) = -Sum_{d divides n} mu(2*d)*tau(n/d). - Benoit Cloitre, Jun 21 2007
G.f.: x/(1-x) = Sum_{n>=1} a(n)*x^n*( 1 - x^n ). - Paul D. Hanna, Jun 22 2007
2*n = 2^a(n)* A000265(n). - Eric Desbiaux, May 14 2009 [corrected by Alejandro Erickson, Apr 17 2012]
Multiplicative with a(2^k) = k + 1, a(p^k) = 1 for any odd prime p. - Franklin T. Adams-Watters, Jun 09 2009
With S(n): 2^n - 1 first elements of the sequence then S(0) = {} (empty list) and if n > 0, S(n) = S(n-1), n, S(n-1). - Yann David (yann_david(AT)hotmail.com), Mar 21 2010
a(n) = log_2(A046161(n)/A046161(n-1)). - Johannes W. Meijer, Nov 04 2012
a((2*n-1)*2^p) = p+1, p >= 0 and n >= 1. - Johannes W. Meijer, Feb 05 2013
a(n+1) = 1 + Sum_{j=0..ceiling(log_2(n+1))} (j * (1 - abs(sign((n mod 2^(j + 1)) - 2^j + 1)))). - Enrico Borba, Oct 01 2015
Conjecture: a(n) = A181988(n)/A003602(n). - L. Edson Jeffery, Nov 21 2015
a(n) = log_2(A006519(n)) + 1. - Doug Bell, Jun 02 2017
Inverse Moebius transform of A209229. - Andrew Howroyd, Aug 04 2018
a(n) = 1 + (A183063(n)/A001227(n)). - Omar E. Pol, Nov 06 2018 (after Franklin T. Adams-Watters)
a(n) = log_2((Xor(2*n,2*n-1)+1)/2). - Gary Detlefs, Dec 13 2018
(2^(a(n)-1)-1)*(n mod 4) = 2*floor(((n+1) mod 4)/3). - Gary Detlefs, Dec 14 2018
a(n) = A000005(n)/A001227(n). - Ivan N. Ianakiev, Jul 05 2019
a(n) = Sum_{j=1..r} (j/2^j)*(Product_{k=1..j} (1 - (-1)^floor( (n+2^(j-1))/2^(k-1) ))), for n < a predefined 2^r. - Adriano Caroli, Sep 30 2019

Name edited following suggestion by David James Sycamore, Oct 05 2023

A005187 a(n) = a(floor(n/2)) + n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.

Original entry on oeis.org

0, 1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, 22, 23, 25, 26, 31, 32, 34, 35, 38, 39, 41, 42, 46, 47, 49, 50, 53, 54, 56, 57, 63, 64, 66, 67, 70, 71, 73, 74, 78, 79, 81, 82, 85, 86, 88, 89, 94, 95, 97, 98, 101, 102, 104, 105, 109, 110, 112, 113, 116, 117, 119, 120, 127, 128

Offset: 0

N. J. A. Sloane, May 20 1991; Allan Wilks, Dec 11 1999

Also exponent of the largest power of 2 dividing (2n)! (A010050) and (2n)!! (A000165).
Write n in binary: 1ab..yz, then a(n) = 1ab..yz + ... + 1ab + 1a + 1. - Ralf Stephan, Aug 27 2003
Also numbers having a partition into distinct Mersenne numbers > 0; A079559(a(n))=1; complement of A055938. - Reinhard Zumkeller, Mar 18 2009
Wikipedia's article on the "Whitney Immersion theorem" mentions that the a(n)-dimensional sphere arises in the Immersion Conjecture proved by Ralph Cohen in 1985. - Jonathan Vos Post, Jan 25 2010
For n > 0, denominators for consecutive pairs of integral numerator polynomials L(n+1,x) for the Legendre polynomials with o.g.f. 1 / sqrt(1-tx+x^2). - Tom Copeland, Feb 04 2016
a(n) is the total number of pointers in the first n elements of a perfect skip list. - Alois P. Heinz, Dec 14 2017
a(n) is the position of the n-th a (indexing from 0) in the fixed point of the morphism a -> aab, b -> b. - Jeffrey Shallit, Dec 24 2020
Numbers that can be expressed as the sum of distinct numbers of the form 2^k - 1 (lenient Mersenne numbers, A000225). This follows from the 2N - Hamming weight definition. A corollary is that these are the numbers with no 2 in their skew-binary representation (cf. A169683). - Allan C. Wechsler, Feb 25 2025

			G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 8*x^5 + 10*x^6 + 11*x^7 + 15*x^8 + ...

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

N. J. A. Sloane, Table of n, a(n) for n = 0..20000 (first 1000 terms from T. D. Noe)
Jean-Paul Allouche, Jean Bétréma, and Jeffrey Shallit, Sur des points fixes de morphismes d'un monoïde libre, RAIRO-Theor. Inf. Appl. 23 (1989), 235-249.
Laurent Alonso, Edward M. Reingold, and René Schott, Determining the majority, Inform. Process. Lett. 47 (1993), no. 5, 253-255.
Laurent Alonso, Edward M. Reingold, and René Schott, The average-case complexity of determining the majority, SIAM J. Comput. 26 (1997), no. 1, 1-14.
Barry Brent, On the Constant Terms of Certain Laurent Series, Preprints (2023) 2023061164.
Sung-Hyuk Cha, On Integer Sequences Derived from Balanced k-ary Trees, Applied Mathematics in Electrical and Computer Engineering, 2012.
Sung-Hyuk Cha, On Complete and Size Balanced k-ary Tree Integer Sequences, International Journal of Applied Mathematics and Informatics, Issue 2, Volume 6, 2012, pp. 67-75. - From _N. J. A. Sloane_, Dec 24 2012
Ralph L. Cohen, The Immersion Conjecture for Differentiable Manifolds, The Annals of Mathematics, 1985: 237-328.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, preprint, 2016.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Keith Johnson and Kira Scheibelhut, Rational Polynomials That Take Integer Values at the Fibonacci Numbers, American Mathematical Monthly 123.4 (2016): 338-346. See p. 340.
Tanya Khovanova, There are no coincidences, arXiv preprint 1410.2193 [math.CO], 2014.
Anunay Kulshrestha, On Hamming Distance between base-n representations of whole numbers, arXiv:1203.4547 [cs.DM], 2012.
Michael E. Saks and Michael Werman, On computing majority by comparisons, Combinatorica 11 (1991), no. 4, 383-387.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Wikipedia, Whitney Immersion Theorem.
Allan Wilks, Email to N. J. A. Sloane, Jul 7 1988.

Cf. A001790, A011371, A067080, A098844, A132027, A004128, A054899, A046161.
Cf. A001511 (first differences), A122247 (partial sums), A055938 (complement).
Cf. A004134, A010050, A000165.
Cf. A000120, A079559, A085354, A030308, A000225.

a005187 n = a005187_list !! n
a005187_list = 0 : zipWith (+) [1..] (map (a005187 . (`div` 2)) [1..])
-- Reinhard Zumkeller, Nov 07 2011, Oct 05 2011

[n + Valuation(Factorial(n), 2): n in [0..70]]; // Vincenzo Librandi, Jun 11 2019

A005187 := n -> 2*n - add(i, i=convert(n, base, 2)):
seq(A005187(n), n=0..65); # Peter Luschny, Apr 08 2014

a[0] = 0; a[n_] := a[n] = a[Floor[n/2]] + n; Table[ a[n], {n, 0, 50}] (* or *)
Table[IntegerExponent[(2n)!, 2], {n, 0, 65}] (* Robert G. Wilson v, Apr 19 2006 *)
Table[2n-DigitCount[2n,2,1],{n,0,70}] (* Harvey P. Dale, May 24 2014 *)

{a(n) = if( n<0, 0, valuation((2*n)!, 2))}; /* Michael Somos, Oct 24 2002 */

{a(n) = if( n<0, 0, sum(k=1, n, (2*n)\2^k))}; /* Michael Somos, Oct 24 2002 */

{a(n) = if( n<0, 0, 2*n - subst( Pol( binary( n ) ), x, 1) ) }; /* Michael Somos, Aug 28 2007 */

a(n)=my(s=n);while(n>>=1,s+=n);s \\ Charles R Greathouse IV, Apr 07 2012

a(n)=2*n-hammingweight(n) \\ Charles R Greathouse IV, Jan 07 2013

def A005187(n): return 2*n-bin(n).count('1') # Chai Wah Wu, Jun 03 2021

@CachedFunction
def A005187(n): return A005187(n//2) + n if n > 0 else 0
[A005187(n) for n in range(66)]  # Peter Luschny, Dec 13 2012

a(n) = A011371(2n+1) = A011371(n) + n, n >= 0.
A046161(n) = 2^a(n).
For m>0, let q = floor(log_2(m)); a(2m+1) = 2^q + 3m + Sum_{k>=1} floor((m-2^q)/2^k); a(2m) = a(2m+1) - 1. - Len Smiley
a(n) = Sum_{k >= 0} floor(n/2^k) = n + A011371(n). - Henry Bottomley, Jul 03 2001
G.f.: A(x) = Sum_{k>=0} x^(2^k)/((1-x)*(1-x^(2^k))). - Ralf Stephan, Dec 24 2002
a(n) = Sum_{k=1..n} A001511(k), sum of binary Hamming distances between consecutive integers up to n. - Gary W. Adamson, Jun 15 2003
Conjecture: a(n) = 2n + O(log(n)). - Benoit Cloitre, Oct 07 2003 [true as a(n) = 2*n - hamming_weight(2*n). Joerg Arndt, Jun 10 2019]
Sum_{n=2^k..2^(k+1)-1} a(n) = 3*4^k - (k+4)*2^(k-1) = A085354(k). - Philippe Deléham, Feb 19 2004
From Hieronymus Fischer, Aug 14 2007: (Start)
Recurrence: a(n) = n + a(floor(n/2)); a(2n) = 2n + a(n); a(n*2^m) = 2*n*(2^m-1) + a(n).
  a(2^m) = 2^(m+1) - 1, m >= 0.
Asymptotic behavior: a(n) = 2n + O(log(n)), a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= 2n-1; equality holds for powers of 2.
a(n) >= 2n-1-floor(log_2(n)); equality holds for n = 2^m-1, m > 0.
lim inf (2n - a(n)) = 1, for n-->oo.
lim sup (2n - log_2(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_2(n)) = 1, for n-->oo. (End)
a(n) = 2n - A000120(n). - Paul Barry, Oct 26 2007
PURRS demo results: Bounds for a(n) = n + a(n/2) with initial conditions a(1) = 1: a(n) >= -2 + 2*n - log(n)*log(2)^(-1), a(n) <= 1 + 2*n for each n >= 1. - Alexander R. Povolotsky, Apr 06 2008
If n = 2^a_1 + 2^a_2 + ... + 2^a_k, then a(n) = n-k. This can be used to prove that 2^n maximally divides (2n!)/n!. - Jon Perry, Jul 16 2009
a(n) = Sum_{k>=0} A030308(n,k)*A000225(k+1). - Philippe Deléham, Oct 16 2011
a(n) = log_2(denominator(binomial(-1/2,n))). - Peter Luschny, Nov 25 2011
a(2n+1) = a(2n) + 1. - M. F. Hasler, Jan 24 2015
a(n) = A004134(n) - n. - Cyril Damamme, Aug 04 2015
G.f.: (1/(1 - x))*Sum_{k>=0} (2^(k+1) - 1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017

A001790 Numerators in expansion of 1/sqrt(1-x).

Original entry on oeis.org

1, 1, 3, 5, 35, 63, 231, 429, 6435, 12155, 46189, 88179, 676039, 1300075, 5014575, 9694845, 300540195, 583401555, 2268783825, 4418157975, 34461632205, 67282234305, 263012370465, 514589420475, 8061900920775, 15801325804719

Offset: 0

N. J. A. Sloane

Also numerator of e(n-1,n-1) (see Maple line).
Leading coefficient of normalized Legendre polynomial.
Common denominator of expansions of powers of x in terms of Legendre polynomials P_n(x).
Also the numerator of binomial(2*n,n)/2^n. - T. D. Noe, Nov 29 2005
This sequence gives the numerators of the Maclaurin series of the Lorentz factor (see Wikipedia link) of 1/sqrt(1-b^2) = dt/dtau where b=u/c is the velocity in terms of the speed of light c, u is the velocity as observed in the reference frame where time t is measured and tau is the proper time. - Stephen Crowley, Apr 03 2007
Truncations of rational expressions like those given by the numerator operator are artifacts in integer formulas and have many disadvantages. A pure integer formula follows. Let n$ denote the swinging factorial and sigma(n) = number of '1's in the base-2 representation of floor(n/2). Then a(n) = (2*n)$ / sigma(2*n) = A056040(2*n) / A060632(2*n+1). Simply said: this sequence is the odd part of the swinging factorial at even indices. - Peter Luschny, Aug 01 2009
It appears that a(n) = A060818(n)*A001147(n)/A000142(n). - James R. Buddenhagen, Jan 20 2010
The convolution of sequence binomial(2*n,n)/4^n with itself is the constant sequence with all terms = 1.
a(n) equals the denominator of Hypergeometric2F1[1/2, n, 1 + n, -1] (see Mathematica code below). - John M. Campbell, Jul 04 2011
a(n) = numerator of (1/Pi)*Integral_{x=-oo..+oo} 1/(x^2-2*x+2)^n dx. - Leonid Bedratyuk, Nov 17 2012
a(n) = numerator of the mean value of cos(x)^(2*n) from x = 0 to 2*Pi. - Jean-François Alcover, Mar 21 2013
Constant terms for normalized Legendre polynomials. - Tom Copeland, Feb 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums:
a(n)/A060818(n) = (-2)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} a(k)/A060818(k) = -i.
Sum_{k>=0} (-1)^k*a(k)/A060818(k) = 1/sqrt(3).
Sum_{k>=0} (-1)^(k+1)*a(k)/A060818(k) = -1/sqrt(3).
a(n)/A046161(n) = (-1)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} (-1)^k*a(k)/A046161(k) = 1/sqrt(2).
Sum_{k>=0} (-1)^(k+1)*a(k)/A046161(k) = -1/sqrt(2). (End)
a(n) = numerator of (1/Pi)*Integral_{x=-oo..+oo} 1/(x^2+1)^n dx. (n=1 is the Cauchy distribution.) - Harry Garst, May 26 2017
Let R(n, d) = (Product_{j prime to d} Pochhammer(j / d, n)) / n!. Then the numerators of R(n, 2) give this sequence and the denominators are A046161. For d = 3 see A273194/A344402. - Peter Luschny, May 20 2021
Using WolframAlpha, it appears a(n) gives the numerator in the residues of f(z) = 2z choose z at odd negative half integers. E.g., the residues of f(z) at z = -1/2, -3/2, -5/2 are 1/(2*Pi), 1/(16*Pi), and 3/(256*Pi) respectively. - Nicholas Juricic, Mar 31 2022
a(n) is the numerator of (1/Pi) * Integral_{x=-oo..+oo} sech(x)^(2*n+1) dx. The corresponding denominator is A046161. - Mohammed Yaseen, Jul 29 2023
a(n) is the numerator of (1/Pi) * Integral_{x=0..Pi/2} sin(x)^(2*n) dx. The corresponding denominator is A101926(n). - Mohammed Yaseen, Sep 19 2023

			1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, 12155/128, 46189/256, ...
binomial(2*n,n)/4^n => 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, 6435/32768, ...

P. J. Davis, Interpolation and Approximation, Dover Publications, 1975, p. 372.
W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 2nd ed. New York: Wiley, 1968; Chap. III, Eq. 4.1.
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).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equation 6:14:6 at page 51.
J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 102.

Robert G. Wilson v, Table of n, a(n) for n = 0..1666 (first 201 terms from T. D. Noe)
Horst Alzer and Bent Fuglede, Normalized binomial mid-coefficients and power means, Journal of Number Theory, Volume 115, Issue 2, December 2005, Pages 284-294.
C. M. Bender and K. A. Milton, Continued fraction as a discrete nonlinear transform, arXiv:hep-th/9304062, 1993 (see V_n with N=1).
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table [Annotated scanned copy].
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table, Phil. Mag., 33 (1942), 1-12 (plus tables).
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
S. Łukaszyk and W. Bieniawski, Assembly Theory of Binary Messages, Mathematics, 12(10) (2024), 1600.
Peter Luschny, Die schwingende Fakultät und Orbitalsysteme, August 2011.
V. H. Moll, The evaluation of integrals: a personal story, Notices Amer. Math. Soc., 49 (No. 3, March 2002), 311-317.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
H. E. Salzer, Coefficients for expressing the first twenty-four powers in terms of the Legendre polynomials, Math. Comp., 3 (1948), 16-18.
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages).
Eric Weisstein's World of Mathematics, Binomial Series.
Eric Weisstein's World of Mathematics, Legendre Polynomial.
Wikipedia, Lorentz Factor.

Cf. A000142, A000984, A001147, A001316, A001800, A001801, A001803.
Cf. A005187, A007814, A008316, A046161, A056040, A086117, A094638.
Cf. A101926, A123854, A273194, A344402.
Cf. A060818 (denominator of binomial(2*n,n)/2^n), A061549 (denominators).
Cf. A123854 (denominators).
Cf. A161198 (triangle of coefficients for (1-x)^((-1-2*n)/2)).
Cf. A163590 (odd part of the swinging factorial).
Cf. A001405.
First column and diagonal 1 of triangle A100258.
Bisection of A036069.
Bisections give A061548 and A063079.
Inverse Moebius transform of A180403/A046161.
Numerators of [x^n]( (1-x)^(p/2) ): A161202 (p=5), A161200 (p=3), A002596 (p=1), this sequence (p=-1), A001803 (p=-3), A161199 (p=-5), A161201 (p=-7).

A001790:= func< n | Numerator((n+1)*Catalan(n)/4^n) >;
[A001790(n): n in [0..40]]; // G. C. Greubel, Sep 23 2024

e := proc(l,m) local k; add(2^(k-2*m)*binomial(2*m-2*k,m-k)*binomial(m+k,m)*binomial(k,l),k=l..m); end;
# From Peter Luschny, Aug 01 2009: (Start)
swing := proc(n) option remember; if n = 0 then 1 elif irem(n, 2) = 1 then swing(n-1)*n else 4*swing(n-1)/n fi end:
sigma := n -> 2^(add(i,i=convert(iquo(n,2),base,2))):
a := n -> swing(2*n)/sigma(2*n); # (End)
A001790 := proc(n) binomial(2*n, n)/4^n ; numer(%) ; end proc : # R. J. Mathar, Jan 18 2013

Numerator[ CoefficientList[ Series[1/Sqrt[(1 - x)], {x, 0, 25}], x]]
Table[Denominator[Hypergeometric2F1[1/2, n, 1 + n, -1]], {n, 0, 34}]   (* John M. Campbell, Jul 04 2011 *)
Numerator[Table[(-2)^n*Sqrt[Pi]/(Gamma[1/2 - n]*Gamma[1 + n]),{n,0,20}]] (* Ralf Steiner, Apr 07 2017 *)
Numerator[Table[Binomial[2n,n]/2^n, {n, 0, 25}]] (* Vaclav Kotesovec, Apr 07 2017 *)
Table[Numerator@LegendreP[2 n, 0]*(-1)^n, {n, 0, 25}] (* Andres Cicuttin, Jan 22 2018 *)
A = {1}; Do[A = Append[A, 2^IntegerExponent[n, 2]*(2*n - 1)*A[[n]]/n], {n, 1, 25}]; Print[A] (* John Lawrence, Jul 17 2020 *)

{a(n) = if( n<0, 0, polcoeff( pollegendre(n), n) * 2^valuation((n\2*2)!, 2))};

a(n)=binomial(2*n,n)>>hammingweight(n); \\ Gleb Koloskov, Sep 26 2021

# uses[A000120]
@CachedFunction
def swing(n):
    if n == 0: return 1
    return swing(n-1)*n if is_odd(n) else 4*swing(n-1)/n
A001790 = lambda n: swing(2*n)/2^A000120(2*n)
[A001790(n) for n in (0..25)]  # Peter Luschny, Nov 19 2012

a(n) = numerator( binomial(2*n,n)/4^n ) (cf. A046161).
a(n) = A000984(n)/A001316(n) where A001316(n) is the highest power of 2 dividing C(2*n, n) = A000984(n). - Benoit Cloitre, Jan 27 2002
a(n) = denominator of (2^n/binomial(2*n,n)). - Artur Jasinski, Nov 26 2011
a(n) = numerator(L(n)), with rational L(n):=binomial(2*n,n)/2^n. L(n) is the leading coefficient of the Legendre polynomial P_n(x).
L(n) = (2*n-1)!!/n! with the double factorials (2*n-1)!! = A001147(n), n >= 0.
Numerator in (1-2t)^(-1/2) = 1 + t + (3/2)t^2 + (5/2)t^3 + (35/8)t^4 + (63/8)t^5 + (231/16)t^6 + (429/16)t^7 + ... = 1 + t + 3*t^2/2! + 15*t^3/3! + 105*t^4/4! + 945*t^5/5! + ... = e.g.f. for double factorials A001147 (cf. A094638). - Tom Copeland, Dec 04 2013
From Ralf Steiner, Apr 08 2017: (Start)
a(n)/A061549(n) = (-1/4)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} a(k)/A061549(k) = 2/sqrt(3).
Sum_{k>=0} (-1)^k*a(k)/A061549(k) = 2/sqrt(5).
Sum_{k>=0} (-1)^(k+1)*a(k)/A061549(k) = -2/sqrt(5).
a(n)/A123854(n) = (-1/2)^n*sqrt(Pi)/(gamma(1/2 - n)*gamma(1 + n)).
Sum_{k>=0} a(k)/A123854(k) = sqrt(2).
Sum_{k>=0} (-1)^k*a(k)/A123854(k) = sqrt(2/3).
Sum_{k>=0} (-1)^(k+1)*a(k)/A123854(k) = -sqrt(2/3). (End)
a(n) = 2^A007814(n)*(2*n-1)*a(n-1)/n. - John Lawrence, Jul 17 2020
Sum_{k>=0} A086117(k+3)/a(k+2) = Pi. - Antonio Graciá Llorente, Aug 31 2024
a(n) = A001803(n)/(2*n+1). - G. C. Greubel, Sep 23 2024

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A180403 Numerator in Moebius transform of A001790/A046161.

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A206771 0 followed by the numerators of the reduced (A001803(n) + A001790(n)) / (2*A046161(n)).

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A280443 a(n) = A280442(n)/A223067(n) = A067624(n)*A046161(n)/A223068(n).

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Maple

Sage

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A127943 a(n) = 2^binomial(n+1,2)/A046161(n).

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GAP

Magma

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PARI

PARI

Sage

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A173396 a(n) = A046161(n) + A001803(n).

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GAP

Magma

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A242735 Array read by antidiagonals: form difference table of the sequence of rationals 0, 0 followed by A001803(n)/A046161(n), then extract numerators.

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Mathematica

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

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