cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152, 69533550916004, 263747951750360, 1002242216651368, 3814986502092304
Offset: 0

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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

  • 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.
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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

A000984 Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.

Original entry on oeis.org

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720, 8233430727600, 32247603683100, 126410606437752, 495918532948104, 1946939425648112
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).
Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.
Number of possible interleavings of a program with n atomic instructions when executed by two processes. - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001
Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002
Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of directed, convex polyominoes having semiperimeter n+2.
Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002
The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its g.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003
Number of possible values of a 2n-bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003
Ordered partitions of n with zeros to n+1, e.g., for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003
Number of nondecreasing sequences of n integers from 0 to n: a(n) = Sum_{i_1=0..n} Sum_{i_2=i_1..n}...Sum_{i_n=i_{n-1}..n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003
Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003
a(n-1) = number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. E.g., n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004
The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.
Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004
Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2) * (sqrt(n)) * zeta(1/2). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. - Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009. - N. J. A. Sloane, Oct 25 2015]
The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 70-80.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73-107) who showed that it holds for all n>4. - Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]
p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006
The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006
Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007
So this is the 2-dimensional analog of A008793. - William Entriken, Aug 06 2013
The number of walks of length 2n on an infinite linear lattice that begins and ends at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Integral representation: C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e., C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008
Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 2008
Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11." - Jonathan Vos Post, Nov 14 2008
Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333, ...). - Gary W. Adamson, May 15 2009
Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009
Number of n X n binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.
Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011
Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012
From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1 + a2*b2 + ... + a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1) + 4*(5) + 6*(10) + 4*(10) + 1*(5)) = 252, the sixth term in this sequence.
Take from Pascal's triangle row(n) with terms b1, b2, ..., b(n+1) and row(n+2) with terms c1, c2, ..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1). Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012
a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p. 4.) - Gary Detlefs, Feb 16 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. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = Sum_{k=0..n} (m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
This comment generalizes the comment dated Oct 31 2012 and the second of the sequence's original comments. For j = 1 to n, a(n) = Sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*Sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). - Charlie Marion, Jun 07 2013
The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013
Number of distinct strings of length 2n using n letters A and n letters B. - Hans Havermann, May 07 2014
From Fung Lam, May 19 2014: (Start)
Expansion of G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=-1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=-1 recovers the g.f. of A000108 as xA^2-A+1=0. The present sequence A000984 refers to q=-2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q-4*q-4)*n + 2*(q*q-q-1))*a(n+1) - 2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=-q. Asymptotics: a(n) ~ ((q+2)/(q+1))*(q^2/(-q-1))^n, q<=-3, a(n) ~ (-1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ -Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=-3), A076035 (q=-4), A076036 (q=-5), A127628 (q=-6), A126694 (q=-7), A115970 (q=-8). (End)
a(n)*(2^n)^(j-2) equals S(n), where S(n) is the n-th number in the self-convolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(4-2) = 252*1024 = 258048. The self-convolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing). - Bob Selcoe, Jul 16 2014
The variance of the n-th difference of a sequence of pairwise uncorrelated random variables each with variance 1. - Liam Patrick Roche, Jun 04 2015
Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(1-4z). - Emeric Deutsch, Jun 17 2015
Number of monomials of degree at most n in n variables. - Ran Pan, Sep 26 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r, then V(n, 2^n) / Pi = V(n-1, 2^n) * a(n/2) for all even n. - Peter Luschny, Oct 12 2015
a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0). - Anton Zakharov, Jul 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)/(-2)^k = 1/sqrt(3).
Sum_{k>=0} a(k)/(-1)^k = 1/sqrt(5).
Sum_{k>=0} a(k)/(-1/2)^k = 1/3.
Sum_{k>=0} a(k)/(1/2)^k = -1/sqrt(7)i.
Sum_{k>=0} a(k)/(1)^k = -1/sqrt(3)i.
Sum_{k>=0} a(k)/2^k = -i. (End)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18] - Eric M. Schmidt, Jul 17 2017
The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1 - (x + y)), 1/(1 - (x + y*z)), 1/(1 - (x + x*y + y*z)) or 1/(1 - (x + y + y*z)). - Peter Bala, Jan 30 2018
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.
a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)
All binary self-dual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together. - Nathan J. Russell, Nov 25 2018
From Isaac Saffold, Dec 28 2018: (Start)
Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,
[(m+n)/p] == [n/p]*Sum_{k=0..(p-1)/2} (-m/(4*n))^k * a(k) (mod p).
Evaluating this identity for m = -1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p-1)/2} (1/4)^k * a(k) is divisible by p. (End)
Number of vertices of the subgraph of the (2n-1)-dimensional hypercube induced by all bitstrings with n-1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle. - Torsten Muetze, Feb 11 2019
a(n) is the number of walks of length 2n from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {3>1, 1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element but smaller than the third elements. - Sergey Kitaev, Dec 08 2020
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 1 through a(2) = 6 compositions are:
(1) (2,1) (3,2)
(1,1,1) (1,2,2)
(2,2,1)
(1,1,2,1)
(2,1,1,1)
(1,1,1,1,1)
The following relate to these compositions:
- The unordered version is A000070.
- The alternating sum -1 version is counted by A001791, ranked by A345910/A345912.
- The alternating sum 0 version is counted by A088218, ranked by A344619.
- Including even indices gives A126869.
- The complement is counted by A202736.
- Ranked by A345909 (reverse: A345911).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 1 than 0's. For example, the a(2) = 6 binary numbers are: 10011, 10101, 10110, 11001, 11010, 11100.
(End)
From Michael Wallner, Jan 25 2022: (Start)
a(n) is the number of nx2 Young tableaux with a single horizontal wall between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
Example for a(2)=6:
3 4 2 4 3 4 3|4 4|3 2|4
1|2, 1|3, 2|1, 1 2, 1 2, 1 3
a(n) is also the number of nx2 Young tableaux with n "walls" between the first and second column.
Example for a(2)=6:
3|4 2|4 4|3 3|4 4|3 4|2
1|2, 1|3, 1|2, 2|1, 2|1, 3|1 (End)
From Shel Kaphan, Jan 12 2023: (Start)
a(n)/4^n is the probability that a fair coin tossed 2n times will come up heads exactly n times and tails exactly n times, or that a random walk with steps of +-1 will return to the starting point after 2n steps (not necessarily for the first time). As n becomes large, this number asymptotically approaches 1/sqrt(n*Pi), using Stirling's approximation for n!.
a(n)/(4^n*(2n-1)) is the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. The absolute value of the n-th term of A144704 is denominator of this fraction.
Considering all possible random walks of exactly 2n steps with steps of +-1, a(n)/(2n-1) is the number of such walks that return to the starting point for the first time after 2n steps. See the absolute values of A002420 or A284016 for these numbers. For comparison, as mentioned by Stefan Hollos, Dec 10 2007, a(n) is the number of such walks that return to the starting point after 2n steps, but not necessarily for the first time. (End)
p divides a((p-1)/2) + 1 for primes p of the form 4*k+3 (A002145). - Jules Beauchamp, Feb 11 2023
Also the size of the shuffle product of two words of length n, such that the union of the two words consist of 2n distinct elements. - Robert C. Lyons, Mar 15 2023
a(n) is the number of vertices of the n-dimensional cyclohedron W_{n+1}. - Jose Bastidas, Mar 25 2025
Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of increasing sizes, followed by a series of reversals of decreasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025

Examples

			G.f.: 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
For n=2, a(2) = 4!/(2!)^2 = 24/4 = 6, and this is the middle coefficient of the binomial expansion (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. - _Michael B. Porter_, Jul 06 2016

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 160.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line -3, with a=b=n.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 101.
  • Emeric Deutsch and Louis W. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31 (2001), 31-38.
  • Henry W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
  • Ronald. L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
  • Martin Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
  • Leonard Lipshitz and A. van der Poorten, "Rational functions, diagonals, automata and arithmetic", in Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), 339-358.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • 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).

Links

Crossrefs

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A081696, A182400. Differs from A071976 at 10th term.
Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.
Cf. A226078, A051924 (first differences).
Row sums of A059481, A008459, A152229, A158815, A205946.
Cf. A258290 (arithmetic derivative). Cf. A098616, A214377.
See A261009 for a conjecture about this sequence.
Cf. A046521 (first column).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Cf. A000346, A001700, A001791, A008549, A032443, A073016, A097805, A126869.
Cf. A003418, A056040, A001316, A261130, A356637.

Programs

  • GAP
    List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018
  • Haskell
    a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011
  • Magma
    a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];
  • Maple
    A000984 := n-> binomial(2*n,n); seq(A000984(n), n=0..30);
with(combstruct); [seq(count([S,{S=Prod(Set(Z,card=i),Set(Z,card=i))}, labeled], size=(2*i)), i=0..20)];
with(combstruct); [seq(count([S,{S=Sequence(Union(Arch,Arch)), Arch=Prod(Epsilon, Sequence(Arch),Z)},unlabeled],size=i), i=0..25)];
with(combstruct):bin := {B=Union(Z,Prod(B,B))}: seq (count([B,bin,unlabeled],size=n)*n, n=1..25); # Zerinvary Lajos, Dec 05 2007
A000984List := proc(m) local A, P, n; A := [1,2]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
A := [op(A), 2*P[-1]] od; A end: A000984List(28); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)
CoefficientList[Series[1/Sqrt[1-4x],{x,0,25}],x]  (* Harvey P. Dale, Mar 14 2011 *)
  • Maxima
    A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */
  • PARI
    A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=prodeuler(p=2,2*n,p^(fv(2*n,p)-2*fv(n,p))) \\ Charles R Greathouse IV, Aug 21 2013
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=my(s=1);forprime(p=2,2*n,s*=p^(fv(2*n,p)-2*fv(n,p)));s \\ Charles R Greathouse IV, Aug 21 2013
  • Python
    from _future_ import division
A000984_list, b = [1], 1
for n in range(10**3):
    b = b*(4*n+2)//(n+1)
    A000984_list.append(b) # Chai Wah Wu, Mar 04 2016

Formula

a(n)/(n+1) = A000108(n), the Catalan numbers.
G.f.: A(x) = (1 - 4*x)^(-1/2) = 1F0(1/2;;4x).
a(n+1) = 2*A001700(n) = A030662(n) + 1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n) =A099976.
D-finite with recurrence: n*a(n) + 2*(1-2*n)*a(n-1)=0.
a(n) = 2^n/n! * Product_{k=0..n-1} (2*k+1).
a(n) = a(n-1)*(4-2/n) = Product_{k=1..n} (4-2/k) = 4*a(n-1) + A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = -A005408(n-1)*A002420(n). - Henry Bottomley, Nov 10 2000
Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 4^n / sqrt(Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
Integral representation as n-th moment of a positive function on the interval [0, 4]: a(n) = Integral_{x=0..4}(x^n*((x*(4-x))^(-1/2))/Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Sep 17 2001
Sum_{n>=1} 1/a(n) = (2*Pi*sqrt(3) + 9)/27. [Lehmer 1985, eq. (15)] - Benoit Cloitre, May 01 2002 (= A073016. - Bernard Schott, Jul 20 2022)
a(n) = Max_{ (i+j)!/(i!j!) | 0<=i,j<=n }. - Benoit Cloitre, May 30 2002
a(n) = Sum_{k=0..n} binomial(n+k-1,k), row sums of A059481. - Vladeta Jovovic, Aug 28 2002
E.g.f.: exp(2*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 08 2002
E.g.f.: I_0(2*x) = Sum a(n)*x^(2*n)/(2*n)!, where I_0 is Bessel function. - Michael Somos, Sep 09 2002
a(n) = Sum_{k=0..n} binomial(n, k)^2. - Benoit Cloitre, Jan 31 2003
Determinant of n X n matrix M(i, j) = binomial(n+i, j). - Benoit Cloitre, Aug 28 2003
Given m = C(2*n, n), let f be the inverse function, so that f(m) = n. Letting q denote -log(log(16)/(m^2*Pi)), we have f(m) = ceiling( (q + log(q)) / log(16) ). - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Oct 30 2003
a(n) = 2*Sum_{k=0..(n-1)} a(k)*a(n-k+1)/(k+1). - Philippe Deléham, Jan 01 2004
a(n+1) = Sum_{j=n..n*2+1} binomial(j, n). E.g., a(4) = C(7,3) + C(6,3) + C(5,3) + C(4,3) + C(3,3) = 35 + 20 + 10 + 4 + 1 = 70. - Jon Perry, Jan 20 2004
a(n) = (-1)^(n)*Sum_{j=0..(2*n)} (-1)^j*binomial(2*n, j)^2. - Helena Verrill (verrill(AT)math.lsu.edu), Jul 12 2004
a(n) = Sum_{k=0..n} binomial(2n+1, k)*sin((2n-2k+1)*Pi/2). - Paul Barry, Nov 02 2004
a(n-1) = (1/2)*(-1)^n*Sum_{0<=i, j<=n}(-1)^(i+j)*binomial(2n, i+j). - Benoit Cloitre, Jun 18 2005
a(n) = C(2n, n-1) + C(n) = A001791(n) + A000108(n). - Lekraj Beedassy, Aug 02 2005
G.f.: c(x)^2/(2*c(x)-c(x)^2) where c(x) is the g.f. of A000108. - Paul Barry, Feb 03 2006
a(n) = A006480(n) / A005809(n). - Zerinvary Lajos, Jun 28 2007
a(n) = Sum_{k=0..n} A106566(n,k)*2^k. - Philippe Deléham, Aug 25 2007
a(n) = Sum_{k>=0} A039599(n, k). a(n) = Sum_{k>=0} A050165(n, k). a(n) = Sum_{k>=0} A059365(n, k)*2^k, n>0. a(n+1) = Sum_{k>=0} A009766(n, k)*2^(n-k+1). - Philippe Deléham, Jan 01 2004
a(n) = 4^n*Sum_{k=0..n} C(n,k)(-4)^(-k)*A000108(n+k). - Paul Barry, Oct 18 2007
a(n) = Sum_{k=0..n} A039598(n,k)*A059841(k). - Philippe Deléham, Nov 12 2008
A007814(a(n)) = A000120(n). - Vladimir Shevelev, Jul 20 2009
From Paul Barry, Aug 05 2009: (Start)
G.f.: 1/(1-2x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction);
G.f.: 1/(1-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
If n>=3 is prime, then a(n) == 2 (mod 2*n). - Vladimir Shevelev, Sep 05 2010
Let A(x) be the g.f. and B(x) = A(-x), then B(x) = sqrt(1-4*x*B(x)^2). - Vladimir Kruchinin, Jan 16 2011
a(n) = (-4)^n*sqrt(Pi)/(gamma((1/2-n))*gamma(1+n)). - Gerry Martens, May 03 2011
a(n) = upper left term in M^n, M = the infinite square production matrix:
2, 2, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
1, 1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, 1, .... - Gary W. Adamson, Jul 14 2011
a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011
E.g.f.: hypergeometric([1/2],[1],4*x). - Wolfdieter Lang, Jan 13 2012
a(n) = 2*Sum_{k=0..n-1} a(k)*A000108(n-k-1). - Alzhekeyev Ascar M, Mar 09 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 28 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2*H(k)/(2*H(n)-H(2*n)), n>0, where H(n) is the n-th harmonic number. - Gary Detlefs, Mar 19 2013
G.f.: Q(0)*(1-4*x), where Q(k) = 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,0,-1/2-n,-1). - Karol A. Penson, Jul 27 2013
a(n) = 2^(4*n)/((2*n+1)*Sum_{k=0..n} (-1)^k*C(2*n+1,n-k)/(2*k+1)). - Mircea Merca, Nov 12 2013
a(n) = C(2*n-1,n-1)*C(4*n^2,2)/(3*n*C(2*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
Sum_{n>=0} a(n)/n! = A234846. - Richard R. Forberg, Feb 10 2014
0 = a(n)*(16*a(n+1) - 6*a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 17 2014
a(n+1) = 4*a(n) - 2*A000108(n). Also a(n) = 4^n*Product_{k=1..n}(1-1/(2*k)). - Stanislav Sykora, Aug 09 2014
G.f.: Sum_{n>=0} x^n/(1-x)^(2*n+1) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n) = (-4)^n*binomial(-1/2,n). - Jean-François Alcover, Feb 10 2015
a(n) = 4^n*hypergeom([-n,1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
a(n) ~ 4^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)/sqrt((4*n+1) *Pi). - Peter Luschny, Oct 14 2015
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). Compare with the o.g.f. B(x) of A098616, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
a(n) = GegenbauerC(n,-n,-1). - Peter Luschny, May 07 2016
a(n) = gamma(1+2*n)/gamma(1+n)^2. - Andres Cicuttin, May 30 2016
Sum_{n>=0} (-1)^n/a(n) = 4*(5 - sqrt(5)*log(phi))/25 = 0.6278364236143983844442267..., where phi is the golden ratio. - Ilya Gutkovskiy, Jul 04 2016
From Peter Bala, Jul 22 2016: (Start)
This sequence occurs as the closed-form expression for several binomial sums:
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(2*n + 1,k).
a(n) = 2*Sum_{k = 0..2*n-1} (-1)^(n+k)*binomial(2*n - 1,k)*binomial(2*n,k) for n >= 1.
a(n) = 2*Sum_{k = 0..n-1} binomial(n - 1,k)*binomial(n,k) for n >= 1.
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x - k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... both Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y + k,n) and Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x - k,n)*binomial(y - k,n) appear to equal Kronecker's delta(n,0).
a(n) = (-1)^n*Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y - k,n) appears to equal Kronecker's delta(n,0).
a(n) = Sum_{k = 0..2n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)* binomial(n + k,n)^2. (Gould, Vol. 7, 5.23).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n,n + k)*binomial(n + k,n)^2. (End)
From Ralf Steiner, Apr 07 2017: (Start)
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for q in N, p in Z/{-4q< (some p) <-2}.
...
Sum_{k>=0} a(k)/(-4)^k = 1/sqrt(2).
Sum_{k>=0} a(k)/(17/4)^k = sqrt(17).
Sum_{k>=0} a(k)/(18/4)^k = 3.
Sum_{k>=0} a(k)/5^k = sqrt(5).
Sum_{k>=0} a(k)/6^k = sqrt(3).
Sum_{k>=0} a(k)/8^k = sqrt(2).
...
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for p>4q.(End)
Boas-Buck recurrence: a(n) = (2/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n, 0). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(2*n+1, k) for n in N. - Rene Adad, Sep 30 2017
a(n) = A034870(n,n). - Franck Maminirina Ramaharo, Nov 26 2018
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*(sqrt(4-x) + sqrt(x)*arcsin(sqrt(x)/2)) / (4-x)^(3/2).
E.g.f. for {1/a(n)}: 1 + exp(x/4)*sqrt(Pi*x)*erf(sqrt(x)/2)/2.
Sum_{n>=0} (-1)^n/a(n) = 4*(1/5 - arcsinh(1/2)/(5*sqrt(5))). (End)
From Peter Luschny, Sep 08 2022: (Start)
a(n) = 2^(2*n)*Product_{k=1..2*n} k^((-1)^(k+1)) = A056040(2*n).
a(n) = A001316(n) * A356637(n) * A261130(n) for n >= 2. (End)
a(n) = 4^n*binomial(n-1/2,-1/2) = 4^n*GegenbauerC(n,1/4,1). - Gerry Martens, Oct 19 2022
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^2 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^2 * binomial(2*n, n+k)^3 = x*(x + 2*n)*a(n) (x arbitrary). Compare with the identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^2 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Mar 31 2024: (Start)
4^n*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
16^n = Sum_{k = 0..2*n} a(k)*a(2*n-k). (End)
From Gary Detlefs, May 28 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k)*binomial(2*k,k)*2^(n-2*k). (H. W. Gould) - Gary Detlefs, May 28 2024
a(n) = Sum_{k=0..2*n} (-1)^k*binomial(2n,k)*binomial(2*n+2*k,n+k)*3^(2*n-k). (H. W. Gould) (End)
a(n) = Product_{k>=n+1} k^2/(k^2 - n^2). - Antonio Graciá Llorente, Sep 08 2024
a(n) = Product_{k=1..n} A003418(floor(2*n/k))^((-1)^(k+1)) (Golomb, 2003). - Amiram Eldar, Aug 08 2025

A000302 Powers of 4: a(n) = 4^n.

Original entry on oeis.org

1, 4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 1048576, 4194304, 16777216, 67108864, 268435456, 1073741824, 4294967296, 17179869184, 68719476736, 274877906944, 1099511627776, 4398046511104, 17592186044416, 70368744177664, 281474976710656
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Same as Pisot sequences E(1, 4), L(1, 4), P(1, 4), T(1, 4). Essentially same as Pisot sequences E(4, 16), L(4, 16), P(4, 16), T(4, 16). See A008776 for definitions of Pisot sequences.
The convolution square root of this sequence is A000984, the central binomial coefficients: C(2n,n). - T. D. Noe, Jun 11 2002
With P(n) being the number of integer partitions of n, p(i) as the number of parts of the i-th partition of n, d(i) as the number of different parts of the i-th partition of n, m(i, j) the multiplicity of the j-th part of the i-th partition of n, one has a(n) = Sum_{i = 1..P(n)} p(i)!/(Product_{j = 1..d(i)} m(i, j)!) * 2^(n-1). - Thomas Wieder, May 18 2005
Sums of rows of the triangle in A122366. - Reinhard Zumkeller, Aug 30 2006
Hankel transform of A076035. - Philippe Deléham, Feb 28 2009
Equals the Catalan sequence: (1, 1, 2, 5, 14, ...), convolved with A032443: (1, 3, 11, 42, ...). - Gary W. Adamson, May 15 2009
Sum of coefficients of expansion of (1 + x + x^2 + x^3)^n.
a(n) is number of compositions of natural numbers into n parts less than 4. For example, a(2) = 16 since there are 16 compositions of natural numbers into 2 parts less than 4.
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 1, a(n) equals the number of 4-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Squares in A002984. - Reinhard Zumkeller, Dec 28 2011
Row sums of Pascal's triangle using the rule that going left increases the value by a factor of k = 3. For example, the first three rows are {1}, {3, 1}, and {9, 6, 1}. Using this rule gives row sums as (k+1)^n. - Jon Perry, Oct 11 2012
First differences of A002450. - Omar E. Pol, Feb 20 2013
Sum of all peak heights in Dyck paths of semilength n+1. - David Scambler, Apr 22 2013
Powers of 4 exceed powers of 2 by A020522 which is the m-th oblong number A002378(m), m being the n-th Mersenne number A000225(n); hence, we may write, a(n) = A000079(n) + A002378(A000225(n)). - Lekraj Beedassy, Jan 17 2014
a(n) is equal to 1 plus the sum for 0 < k < 2^n of the numerators and denominators of the reduced fractions k/2^n. - J. M. Bergot, Jul 13 2015
Binomial transform of A000244. - Tony Foster III, Oct 01 2016
From Ilya Gutkovskiy, Oct 01 2016: (Start)
Number of nodes at level n regular 4-ary tree.
Partial sums of A002001. (End)
Satisfies Benford's law [Berger-Hill, 2011]. - N. J. A. Sloane, Feb 08 2017
Also the number of connected dominating sets in the (n+1)-barbell graph. - Eric W. Weisstein, Jun 29 2017
Side length of the cells at level n in a pyramid scheme where a square grid is decomposed into overlapping 2 X 2 blocks (cf. Kropatsch, 1985). - Felix Fröhlich, Jul 04 2019
a(n-1) is the number of 3-compositions of n; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020

References

  • H. W. Gould, Combinatorial Identities, 1972, eq. (1.93), p. 12.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, eq. (5.39), p. 187.
  • D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.
  • 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).
  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.

Links

Crossrefs

Cf. A024036, A052539, A032443, A000351 (Binomial transform).
Cf. A249307.
Cf. A083420.

Programs

Formula

a(n) = 4^n.
a(0) = 1; a(n) = 4*a(n-1).
G.f.: 1/(1-4*x).
E.g.f.: exp(4*x).
a(n) = Sum_{k = 0..n} binomial(2k, k) * binomial(2(n - k), n - k). - Benoit Cloitre, Jan 26 2003 [See Graham et al., eq. (5.39), p. 187. - Wolfdieter Lang, Aug 16 2019]
1 = Sum_{n >= 1} 3/a(n) = 3/4 + 3/16 + 3/64 + 3/256 + 3/1024, ...; with partial sums: 3/4, 15/16, 63/64, 255/256, 1023/1024, ... - Gary W. Adamson, Jun 16 2003
a(n) = A001045(2*n) + A001045(2*n+1). - Paul Barry, Apr 27 2004
A000005(a(n)) = A005408(n+1). - Reinhard Zumkeller, Mar 04 2007
a(n) = Sum_{j = 0..n} 2^(n - j)*binomial(n + j, j). - Peter C. Heinig (algorithms(AT)gmx.de), Apr 06 2007
Hankel transform of A115967. - Philippe Deléham, Jun 22 2007
a(n) = 6*Stirling2(n+1, 4) + 6*Stirling2(n+1, 3) + 3*Stirling2(n+1, 2) + 1 = 2*Stirling2(2^n, 2^n - 1) + Stirling2(n+1, 2) + 1. - Ross La Haye, Jun 26 2008
a(n) = A159991(n)/A001024(n) = A047653(n) + A181765(n). A160700(a(n)) = A010685(n). - Reinhard Zumkeller, May 02 2009
a(n) = A188915(A006127(n)). - Reinhard Zumkeller, Apr 14 2011
a(n) = Sum_{k = 0..n} binomial(2*n+1, k). - Mircea Merca, Jun 25 2011
Sum_{n >= 1} Mobius(n)/a(n) = 0.1710822479183... - R. J. Mathar, Aug 12 2012
a(n) = Sum_{k = 0..n} binomial(2*k + x, k)*binomial(2*(n - k) - x, n - k) for every real number x. - Rui Duarte and António Guedes de Oliveira, Feb 16 2013
a(n) = 5*a(n - 1) - 4*a(n - 2). - Jean-Bernard François, Sep 12 2013
a(n) = (2*n+1) * binomial(2*n,n) * Sum_{j=0..n} (-1)^j/(2*j+1)*binomial(n,j). - Vaclav Kotesovec, Sep 15 2013
a(n) = A000217(2^n - 1) + A000217(2^n). - J. M. Bergot, Dec 28 2014
a(n) = (2^n)^2 = A000079(n)^2. - Doug Bell, Jun 23 2015
a(n) = A002063(n)/3 - A004171(n). - Zhandos Mambetaliyev, Nov 19 2016
a(n) = (1/2) * Product_{k = 0..n} (1 + (2*n + 1)/(2*k + 1)). - Peter Bala, Mar 06 2018
a(n) = A001045(n+1)*A001045(n+2) + A001045(n)^2. - Ezhilarasu Velayutham, Aug 30 2019
a(n) = denominator(zeta_star({2}(n + 1))/zeta(2*n + 2)) where zeta_star is the multiple zeta star values and ({2}_n) represents (2, ..., 2) where the multiplicity of 2 is n. - _Roudy El Haddad, Feb 22 2022
a(n) = 1 + 3*Sum_{k=0..n} binomial(2*n, n+k)*(k|9), where (k|9) is the Jacobi symbol. - Greg Dresden, Oct 11 2022
a(n) = Sum_{k = 0..n} binomial(2*n+1, 2*k) = Sum_{k = 0..n} binomial(2*n+1, 2*k+1). - Sela Fried, Mar 23 2023

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A001791 a(n) = binomial coefficient C(2n, n-1).

Original entry on oeis.org

0, 1, 4, 15, 56, 210, 792, 3003, 11440, 43758, 167960, 646646, 2496144, 9657700, 37442160, 145422675, 565722720, 2203961430, 8597496600, 33578000610, 131282408400, 513791607420, 2012616400080, 7890371113950, 30957699535776, 121548660036300, 477551179875952
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of peaks at even level in all Dyck paths of semilength n+1. Example: a(2)=4 because UDUDUD, UDUU*DD, UU*DDUD, UU*DU*DD, UUUDDD, where U=(1,1), D=(1,-1) and the peaks at even level are shown by *. - Emeric Deutsch, Dec 05 2003
Also number of long ascents (i.e., ascents of length at least two) in all Dyck paths of semilength n+1. Example: a(2)=4 because in the five Dyck paths of semilength 3, namely UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and (UUU)DDD, we have four long ascents (shown between parentheses). Here U=(1,1) and D=(1,-1). Also number of branch nodes (i.e., vertices of outdegree at least two) in all ordered trees with n+1 edges. - Emeric Deutsch, Feb 22 2004
Number of lattice paths from (0,0) to (n,n) with steps E=(1,0) and N=(0,1) which touch or cross the line x-y=1. Example: For n=2 these are the paths EENN, ENEN, ENNE and NEEN. - Herbert Kociemba, May 23 2004
Narayana transform (A001263) of [1, 3, 5, 7, 9, ...] = (1, 4, 15, 56, 210, ...). Row sums of triangles A136534 and A136536. - Gary W. Adamson, Jan 04 2008
Starting with offset 1 = the Catalan sequence starting (1, 2, 5, 14, ...) convolved with A000984: (1, 2, 6, 20, ...). - Gary W. Adamson, May 17 2009
Also number of peaks plus number of valleys in all Dyck n-paths. - David Scambler, Oct 08 2012
Apparently counts UDDUD in all Dyck paths of semilength n+2. - David Scambler, Apr 22 2013
Apparently the number of peaks strictly left of the midpoint in all Dyck paths of semilength n+1. - David Scambler, Apr 30 2013
For n>0, a(n) is the number of compositions of n into at most n parts if zeros are allowed as parts (so-called "weak" compositions). - L. Edson Jeffery, Jul 24 2014
Number of paths in the half-plane x >= 0, from (0,0) to (2n,2), and consisting of steps U=(1,1) and D=(1,-1). For example, for n=2, we have the 4 paths: UUUD, UUDU, UDUU, DUUU. - José Luis Ramírez Ramírez, Apr 19 2015
For n>1, 1/a(n) is the probability that when a stick is broken up at n points independently and uniformly chosen at random along its length any triple of pieces of the n+1 pieces can form a triangle. The corresponding probability for the existence of at least one triple is A339392(n)/A339393(n). - Amiram Eldar, Dec 04 2020
a(n) is the number of lattice paths of 2n steps taken from the step set {U=(1,1), D=(1,-1)} that start at the origin, never go below the x-axis, and end strictly above the x-axis; more succinctly, proper left factors of Dyck paths. For example, a(2)=4 counts UUUU, UUUD, UUDU, UDUU. - David Callan and Emeric Deutsch, Jan 25 2021
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum -1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(1) = 1 through a(3) = 15 compositions are:
(1,2) (2,3) (3,4)
(1,3,1) (1,4,2)
(1,1,1,2) (2,4,1)
(1,2,1,1) (1,1,2,3)
(1,2,2,2)
(1,3,2,1)
(2,1,1,3)
(2,2,1,2)
(2,3,1,1)
(1,1,1,3,1)
(1,2,1,2,1)
(1,3,1,1,1)
(1,1,1,1,1,2)
(1,1,1,2,1,1)
(1,2,1,1,1,1)
The following relate to these compositions.
- The unordered version is A000070.
- Allowing any negative alternating sum gives A000346.
- The opposite (positive 1) version is A000984.
- The version for reverse-alternating sum is also A001791 (this sequence).
- Taking alternating sum -2 instead of -1 gives A002054.
- The shifted version for alternating sum 0 is counted by A088218 and ranked by A344619.
- Ranked by A345910 (reverse: A345912).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 0 than 1's. For example, the a(2) = 4 binary numbers are: 10001, 10010, 10100, 11000.
(End)
The diagonal of a square n X n matrix where cells of the first row are the nonnegative integers and cells of subsequent rows are sums of cells of the previous row up to and including n. - Torlach Rush, Apr 24 2024
For n>=1, a(n) is the independence number of the odd graph O_{n+1}. - Miquel A. Fiol, Jul 07 2024

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Cornelius Lanczos, Applied Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 517.
  • R. C. Mullin, E. Nemeth and P. J. Schellenberg, The enumeration of almost cubic maps, pp. 281-295 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 1, edited R. C. Mullin et al., 1970.
  • 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).

Links

Crossrefs

Diagonal 3 of triangle A100257.
First differences are in A076540.
Cf. A000108, A000984, A002378, A002390, A025566, A132812.
A345197 counts compositions by length and alternating sum.
Cf. A000070, A000302, A000346, A002054, A008549, A032443, A088218, A097805, A163493, A202736, A345910.

Programs

  • GAP
    List([0..30],n->Binomial(2*n,n-1)); # Muniru A Asiru, Aug 09 2018
  • Magma
    [Binomial(2*n, n-1): n in [0..30]]; // Vincenzo Librandi, Apr 20 2015
  • Mathematica
    Table[Binomial[2n,n-1],{n,0,30}] (* Harvey P. Dale, Jul 12 2012 *)
CoefficientList[ Series[(1 - 2x - Sqrt[1 - 4x])/(2x*Sqrt[1 - 4x]), {x, 0, 26}], x] (* Robert G. Wilson v, Aug 10 2018 *)
  • Maxima
    A001791(n):=binomial(2*n,n-1)$
makelist(A001791(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    a(n)=if(n<1,0,(2*n)!/(n+1)!/(n-1)!)

Formula

a(n) = n*A000108(n).
G.f.: x*(d/dx)c(x) where c(x) = Catalan g.f. - Wolfdieter Lang
Convolution of A001700 (central binomial of odd order) and A000108 (Catalan): a(n+1) = Sum_{k=0..n} C(k)*binomial(2*(n-k)+1, n-k), C(k): Catalan. - Wolfdieter Lang
E.g.f.: exp(2x) * I_1(2x), where I_1 is Bessel function. - Michael Somos, Sep 08 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n, k+1). - Paul Barry, May 15 2003
a(n) = Sum_{i=1..n} binomial(i+n-1, n).
G.f.: (1-2x-sqrt(1-4x))/(2x*sqrt(1-4x)). - Emeric Deutsch, Dec 05 2003
a(n) = A092956/(n!). - Amarnath Murthy, Jun 16 2004
a(n) = binomial(2n,n) - A000108(n). - Paul Barry, Apr 21 2005
a(n) = (1/(2*Pi))*Integral_{x=0..4} (x^n*(x-2)/sqrt(x(4-x))) is the moment sequence representation. - Paul Barry, Jan 11 2007
Row sums of triangle A132812 starting (1, 4, 15, 56, 210, ...). - Gary W. Adamson, Sep 01 2007
Starting (1, 4, 15, 56, 210, ...) gives the binomial transform of A025566 starting (1, 3, 8, 22, 61, 171, ...). - Gary W. Adamson, Sep 01 2007
For n >= 1, a(2^n) = 2^(n+1)*A001795(2^(n-1)). - Vladimir Shevelev, Sep 05 2010
D-finite with recurrence: (n-1)*(n+1)*a(n) = 2*n*(2n-1)*a(n-1). - R. J. Mathar, Dec 17 2011
From Sergei N. Gladkovskii, Jul 07 2012: (Start)
G.f.: -1/(2*x) - G(0) where G(k) = 1 - 1/(2*x - 8*x^3*(2*k+1)/(4*x^2*(2*k+1)- (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step);
E.g.f.: BesselI(1,2*x)*exp(2*x) = x*G(0) where G(k) = 1 + 2*x*(4*k+3)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)*(4*k+5)/(x*(4*k+5) + 2*(k+1)*(k+2)/G(k+1))); (continued fraction, 3rd kind, 3-step).
(End)
G.f.: c(x)^3/(2-c(x)) where c(x) is the g.f. for A000108. - Cheyne Homberger, May 05 2014
G.f.: z*C(z)^2/(1-2*z*C(z)), where C(z) is the g.f. of Catalan numbers. - José Luis Ramírez Ramírez, Apr 19 2015
G.f.: x*2F1(3/2,2;3;4x). - R. J. Mathar, Aug 09 2015
a(n) = Sum_{i=1..n} binomial(2*i-2,i-1)*binomial(2*(n-i+1),n-i+2)/(n-i+1). - Vladimir Kruchinin, Sep 07 2015
L.g.f.: 1/(1 - x/(1 - x/(1 - x/(1 - x/(1 - x/(1 - ...)))))) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 10 2017
Sum_{n>=1} 1/a(n) = 1/3 + 5*Pi/(9*sqrt(3)). - Amiram Eldar, Dec 04 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 1/5 + 14*sqrt(5)*log(phi)/25, where log(phi) = A002390. - Amiram Eldar, Feb 20 2021
a(n) = Product_{i=1..(n - 1)} (((4*i + 6)*i + 2)/((i + 2)*i)), for n>=1. - Neven Sajko, Oct 10 2021
a(n) = 2^(2*n)*gamma(n + 1/2)/(sqrt(Pi)*gamma(n)*(n+1)) for n > 0, and a(0) = lim_{n->0} a(n). - Karol A. Penson, Apr 24 2025

A000346 a(n) = 2^(2*n+1) - binomial(2*n+1, n+1).

Original entry on oeis.org

1, 5, 22, 93, 386, 1586, 6476, 26333, 106762, 431910, 1744436, 7036530, 28354132, 114159428, 459312152, 1846943453, 7423131482, 29822170718, 119766321572, 480832549478, 1929894318332, 7744043540348, 31067656725032, 124613686513778, 499744650202436
Offset: 0

Views

Author

N. J. A. Sloane, Wolfdieter Lang

Keywords

Comments

Also a(n) = 2nd elementary symmetric function of binomial(n,0), binomial(n,1), ..., binomial(n,n).
Also a(n) = one half the sum of the heights, over all Dyck (n+2)-paths, of the vertices that are at even height and terminate an upstep. For example with n=1, these vertices are indicated by asterisks in the 5 Dyck 3-paths: UU*UDDD, UU*DU*DD, UDUU*DD, UDUDUD, UU*DDUD, yielding a(1)=(2+4+2+0+2)/2=5. - David Callan, Jul 14 2006
Hankel transform is (-1)^n*(2n+1); the Hankel transform of sum(k=0..n, C(2*n,k) ) - C(2n,n) is (-1)^n*n. - Paul Barry, Jan 21 2007
Row sums of the Riordan matrix (1/(1-4x),(1-sqrt(1-4x))/2) (A187926). - Emanuele Munarini, Mar 16 2011
From Gus Wiseman, Jul 19 2021: (Start)
For n > 0, a(n-1) is also the number of integer compositions of 2n with nonzero alternating sum, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A053754 /\ A345921. For example, the a(3-1) = 22 compositions of 6 are:
(6) (1,5) (1,1,4) (1,1,1,3) (1,1,1,1,2)
(2,4) (1,2,3) (1,1,3,1) (1,1,2,1,1)
(4,2) (1,4,1) (1,2,1,2) (2,1,1,1,1)
(5,1) (2,1,3) (1,3,1,1)
(2,2,2) (2,1,2,1)
(3,1,2) (3,1,1,1)
(3,2,1)
(4,1,1)
(End)

Examples

			G.f. = 1 + 5*x + 22*x^2 + 93*x^3 + 386*x^4 + 1586*x^5 + 6476*x^6 + ...

References

  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.
  • D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.
  • 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).

Links

Crossrefs

Cf. A000108, A014137, A014318. A column of A058893. Subdiagonal of A053979.
Bisection of A058622 and (possibly) A007008.
Cf. A045621, A068551.
Cf. A033504, A038806, A055217, A069731.
Even bisection of A294175 (without the first two terms).
The following relate to compositions of 2n with alternating sum k.
- The k > 0 case is counted by A000302.
- The k <= 0 case is counted by A000302.
- The k != 0 case is counted by A000346 (this sequence).
- The k = 0 case is counted by A001700 or A088218.
- The k < 0 case is counted by A008549.
- The k >= 0 case is counted by A114121.
A011782 counts compositions.
A086543 counts partitions with nonzero alternating sum (bisection: A182616).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A345197 counts compositions by length and alternating sum.
Cf. A000070, A001791, A007318, A025047, A027306, A032443, A053754, A120452, A163493, A239830, A344611, A345921.

Programs

  • Magma
    [2^(2*n+1) - Binomial(2*n+1,n+1): n in [0..30]]; // Vincenzo Librandi, Jun 07 2011
  • Maple
    seq(2^(2*n+1)-binomial(2*n,n)*(2*n+1)/(n+1), n=0..12); # Emanuele Munarini, Mar 16 2011
  • Mathematica
    Table[2^(2n+1)-Binomial[2n,n](2n+1)/(n+1),{n,0,20}] (* Emanuele Munarini, Mar 16 2011 *)
a[ n_] := If[ n<-4, 0, (4^(n + 1) - Binomial[2 n + 2, n + 1]) / 2]; (* Michael Somos, Jan 25 2014 *)
  • Maxima
    makelist(2^(2*n+1)-binomial(2*n,n)*(2*n+1)/(n+1),n,0,12); /* Emanuele Munarini, Mar 16 2011 */
  • PARI
    {a(n) = if( n<-4, 0, n++; (2^(2*n) - binomial(2*n, n)) / 2)}; /* Michael Somos, Jan 25 2014 */

Formula

G.f.: c(x)/(1-4x), c(x) = g.f. of Catalan numbers.
Convolution of Catalan numbers and powers of 4.
Also one half of convolution of central binomial coeffs. A000984(n), n=0, 1, 2, ... with shifted central binomial coeffs. A000984(n), n=1, 2, 3, ...
a(n) = A045621(2n+1) = (1/2)*A068551(n+1).
a(n) = Sum_{k=0..n} A000984(k)*A001700(n-k). - Philippe Deléham, Jan 22 2004
a(n) = Sum_{k=0..n+1} binomial(n+k, k-1)2^(n-k+1). - Paul Barry, Nov 13 2004
a(n) = Sum_{i=0..n} binomial(2n+2, i). See A008949. - Ed Catmur (ed(AT)catmur.co.uk), Dec 09 2006
a(n) = Sum_{k=0..n+1, C(2n+2,k)} - C(2n+2,n+1). - Paul Barry, Jan 21 2007
Logarithm g.f. log(1/(2-C(x)))=sum(n>0, a(n)/n*x^n), C(x)=(1-sqrt(1-4*x))/2x (A000108). - Vladimir Kruchinin, Aug 10 2010
D-finite with recurrence: (n+3) a(n+2) - 2(4n+9) a(n+1) + 8(2n+3) a(n) = 0. - Emanuele Munarini, Mar 16 2011
E.g.f.:exp(2*x)*(2*exp(2*x) - BesselI(0,2*x) - BesselI(1,2*x)).
This is the first derivative of exp(2*x)*(exp(2*x) - BesselI(0,2*x))/2. See the e.g.f. of A032443 (which has a plus sign) and the remarks given there. - Wolfdieter Lang, Jan 16 2012
a(n) = Sum_{0<=iMircea Merca, Apr 05 2012
A000346 = A004171 - A001700. See A032443 for the sum. - M. F. Hasler, Jan 02 2014
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n>-5. - Michael Somos, Jan 25 2014
REVERT transform is [1,-5,28,-168,1056,...] = alternating signed version of A069731. - Michael Somos, Jan 25 2014
Convolution square is A038806. - Michael Somos, Jan 25 2014
BINOMIAL transform of A055217(n-1) is a(n-1). - Michael Somos, Jan 25 2014
(n+1) * a(n) = A033504(n). - Michael Somos, Jan 25 2014
Recurrence: (n+1)*a(n) = 512*(2*n-7)*a(n-5) + 256*(13-5*n)*a(n-4) + 64*(10*n-17)*a(n-3) + 32*(4-5*n)*a(n-2) + 2*(10*n+1)*a(n-1), n>=5. - Fung Lam, Mar 21 2014
Asymptotic approximation: a(n) ~ 2^(2n+1)*(1-1/sqrt(n*Pi)). - Fung Lam, Mar 21 2014
a(n) = Sum_{m = n+2..2*(n+1)} binomial(2*(n+1), m), n >= 0. - Wolfdieter Lang, May 22 2015
a(n) = A000302(n) + A008549(n). - Gus Wiseman, Jul 19 2021
a(n) = Sum_{j=1..n+1} Sum_{k=1..j} 2^(j-k)*binomial(n+k-1, n). - Fabio Visonà, May 04 2022
a(n) = (1/2)*(-1)^n*binomial(-(n+1), n+2)*hypergeom([1, 2*n + 3], [n + 3], 1/2). - Peter Luschny, Nov 29 2023

Extensions

Corrected by Christian G. Bower

A027306 a(n) = 2^(n-1) + ((1 + (-1)^n)/4)*binomial(n, n/2).

Original entry on oeis.org

1, 1, 3, 4, 11, 16, 42, 64, 163, 256, 638, 1024, 2510, 4096, 9908, 16384, 39203, 65536, 155382, 262144, 616666, 1048576, 2449868, 4194304, 9740686, 16777216, 38754732, 67108864, 154276028, 268435456, 614429672, 1073741824, 2448023843
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

Inverse binomial transform of A027914. Hankel transform (see A001906 for definition) is {1, 2, 3, 4, ..., n, ...}. - Philippe Deléham, Jul 21 2005
Number of walks of length n on a line that starts at the origin and ends at or above 0. - Benjamin Phillabaum, Mar 05 2011
Number of binary integers (i.e., with a leading 1 bit) of length n+1 which have a majority of 1-bits. E.g., for n+1=4: (1011, 1101, 1110, 1111) a(3)=4. - Toby Gottfried, Dec 11 2011
Number of distinct symmetric staircase walks connecting opposite corners of a square grid of side n > 1. - Christian Barrientos, Nov 25 2018
From Gus Wiseman, Aug 20 2021: (Start)
Also the number of integer compositions of n + 1 with alternating sum > 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A345917. For example, the a(0) = 1 through a(4) = 11 compositions are:
(1) (2) (3) (4) (5)
(21) (31) (32)
(111) (112) (41)
(211) (113)
(122)
(212)
(221)
(311)
(1121)
(2111)
(11111)
The following relate to these compositions:
- The unordered version is A027193.
- The complement is counted by A058622.
- The reverse unordered version is A086543.
- The version for alternating sum >= 0 is A116406.
- The version for alternating sum < 0 is A294175.
- Ranked by A345917. (End)
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 07 2022

Examples

			From _Gus Wiseman_, Aug 20 2021: (Start)
The a(0) = 1 through a(4) = 11 binary numbers with a majority of 1-bits (Gottfried's comment) are:
  1   11   101   1011   10011
           110   1101   10101
           111   1110   10110
                 1111   10111
                        11001
                        11010
                        11011
                        11100
                        11101
                        11110
                        11111
The version allowing an initial zero is A058622.
(End)

References

  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992, Eq. (4.2.1.6)

Links

Crossrefs

a(n) = Sum{(k+1)T(n, m-k)}, 0<=k<=[ (n+1)/2 ], T given by A008315.
Column k=2 of A226873. - Alois P. Heinz, Jun 21 2013
Cf. A008949, A248574, A001405, A246437.
The even bisection is A000302.
The odd bisection appears to be A032443.
Cf. A000984, A001700, A001791, A008549, A011782, A088218, A097805, A163493, A182616, A345197.

Programs

  • GAP
    List([0..35],n->Sum([0..Int(n/2)],k->Binomial(n,k))); # Muniru A Asiru, Nov 27 2018
  • Haskell
    a027306 n = a008949 n (n `div` 2)  -- Reinhard Zumkeller, Nov 14 2014
  • Magma
    [2^(n-1)+(1+(-1)^n)/4*Binomial(n, n div 2): n in [0..40]]; // Vincenzo Librandi, Jun 19 2016
  • Maple
    a:= proc(n) add(binomial(n, j), j=0..n/2) end:
seq(a(n), n=0..32); # Zerinvary Lajos, Mar 29 2009
  • Mathematica
    Table[Sum[Binomial[n, k], {k, 0, Floor[n/2]}], {n, 1, 35}]
(* Second program: *)
a[0] = a[1] = 1; a[2] = 3; a[n_] := a[n] = (2(n-1)(2a[n-2] + a[n-1]) - 8(n-2) a[n-3])/n; Array[a, 33, 0] (* Jean-François Alcover, Sep 04 2016 *)
  • PARI
    a(n)=if(n<0,0,(2^n+if(n%2,0,binomial(n, n/2)))/2)

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k).
Odd terms are 2^(n-1). Also a(2n) - 2^(2n-1) is given by A001700. a(n) = 2^n + (n mod 2)*binomial(n, (n-1)/2).
E.g.f.: (exp(2x) + I_0(2x))/2.
O.g.f.: 2*x/(1-2*x)/(1+2*x-((1+2*x)*(1-2*x))^(1/2)). - Vladeta Jovovic, Apr 27 2003
a(n) = A008949(n, floor(n/2)); a(n) + a(n-1) = A248574(n), n > 0. - Reinhard Zumkeller, Nov 14 2014
From Peter Bala, Jul 21 2015: (Start)
a(n) = [x^n]( 2*x - 1/(1 - x) )^n.
O.g.f.: (1/2)*( 1/sqrt(1 - 4*x^2) + 1/(1 - 2*x) ).
Inverse binomial transform is (-1)^n*A246437(n).
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + ... is the o.g.f. for A001405. (End)
a(n) = Sum_{k=1..floor((n+1)/2)} binomial(n-1,(2n+1-(-1)^n)/4 -k). - Anthony Browne, Jun 18 2016
D-finite with recurrence: n*a(n) + 2*(-n+1)*a(n-1) + 4*(-n+1)*a(n-2) + 8*(n-2)*a(n-3) = 0. - R. J. Mathar, Aug 09 2017

Extensions

Better description from Robert G. Wilson v, Aug 30 2000 and from Yong Kong (ykong(AT)curagen.com), Dec 28 2000

A008549 Number of ways of choosing at most n-1 items from a set of size 2*n+1.

Original entry on oeis.org

0, 1, 6, 29, 130, 562, 2380, 9949, 41226, 169766, 695860, 2842226, 11576916, 47050564, 190876696, 773201629, 3128164186, 12642301534, 51046844836, 205954642534, 830382690556, 3345997029244, 13475470680616, 54244942336114, 218269673491780, 877940640368572
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Area under Dyck excursions (paths ending in 0): a(n) is the sum of the areas under all Dyck excursions of length 2*n (nonnegative walks beginning and ending in 0 with jumps -1,+1).
Number of inversions in all 321-avoiding permutations of [n+1]. Example: a(2)=6 because the 321-avoiding permutations of [3], namely 123,132,312,213,231, have 0, 1, 2, 1, 2 inversions, respectively. - Emeric Deutsch, Jul 28 2003
Convolution of A001791 and A000984. - Paul Barry, Feb 16 2005
a(n) = total semilength of "longest Dyck subpath" starting at an upstep U taken over all upsteps in all Dyck paths of semilength n. - David Callan, Jul 25 2008
[1,6,29,130,562,2380,...] is convolution of A001700 with itself. - Philippe Deléham, May 19 2009
From Ran Pan, Feb 04 2016: (Start)
a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) bounce off the diagonal y = x to the right. This is related to paired pattern P_2 in Pan and Remmel's link and more details can be found in Section 3.2 in the link.
a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) horizontally cross the diagonal y = x. This is related to paired pattern P_3 in Pan and Remmel's link and more details can be found in Section 3.3 in the link.
2*a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) bounce off the diagonal y = x. This is related to paired pattern P_2 and P_4 in Pan and Remmel's link and more details can be found in Section 4.2 in the link.
2*a(n) is the total number of times that all the North-East lattice paths from (0,0) to (n+1,n+1) cross the diagonal y = x. This is related to paired pattern P_3 and P_4 in Pan and Remmel's link and more details can be found in Section 4.3 in the link. (End)
From Gus Wiseman, Jul 17 2021: (Start)
Also the number of integer compositions of 2*(n+1) with alternating sum < 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(3) = 29 compositions of 8 are:
(1,7) (1,5,2) (1,1,1,5) (1,1,1,4,1) (1,1,1,1,1,3)
(2,6) (1,6,1) (1,1,2,4) (1,2,1,3,1) (1,1,1,2,1,2)
(3,5) (2,5,1) (1,2,1,4) (1,3,1,2,1) (1,1,1,3,1,1)
(1,2,2,3) (1,4,1,1,1) (1,2,1,1,1,2)
(1,3,1,3) (1,2,1,2,1,1)
(1,3,2,2) (1,3,1,1,1,1)
(1,4,1,2)
(1,4,2,1)
(1,5,1,1)
(2,1,1,4)
(2,2,1,3)
(2,3,1,2)
(2,4,1,1)
Also the number of integer compositions of 2*(n+1) with reverse-alternating sum < 0. For a bijection, keep the odd-length compositions and reverse the even-length ones.
Also the number of 2*(n+1)-digit binary numbers with more 0's than 1's. For example, the a(2) = 6 binary numbers are: 100000, 100001, 100010, 100100, 101000, 110000; or in decimal: 32, 33, 34, 36, 40, 48.
(End)

Examples

			a(2) = 6 because there are 6 ways to choose at most 1 item from a set of size 5: You can choose the empty set, or you can choose any of the five one-element sets.
G.f. = x + 6*x^2 + 29*x^3 + 130*x^4 + 562*x^5 + 2380*x^6 + 9949*x^7 + ...

References

  • D. Phulara and L. W. Shapiro, Descendants in ordered trees with a marked vertex, Congressus Numerantium, 205 (2011), 121-128.

Links

Crossrefs

Cf. A038608, A045894, A057571, A109196, A153338.
Odd bisection of A294175 (even is A000346).
For integer compositions of 2*(n+1) with alternating sum k < 0 we have:
- The opposite (k > 0) version is A000302.
- The weak (k <= 0) version is (also) A000302.
- The k = 0 version is A001700 or A088218.
- The reverse-alternating version is also A008549 (this sequence).
- These compositions are ranked by A053754 /\ A345919.
- The complement (k >= 0) is counted by A114121.
- The case of reversed integer partitions is A344743(n+1).
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A345197 counts compositions by length and alternating sum.
Cf. A000070, A001791, A007318, A025047, A027306, A032443, A058622, A120452, A163493, A239830, A344611.

Programs

  • Magma
    [4^n-Binomial(2*n+1, n): n in [0..30]]; // Vincenzo Librandi, Feb 04 2016
  • Maple
    A008549:=n->4^n-binomial(2*n+1,n): seq(A008549(n), n=0..30);
  • Mathematica
    Table[4^n-Binomial[2n+1,n],{n,0,30}] (* Harvey P. Dale, May 11 2011 *)
a[ n_] := If[ n<-4, 0, 4^n - Binomial[2 n + 2, n + 1] / 2] (* Michael Somos, Jan 25 2014 *)
  • PARI
    {a(n)=if(n<0, 0, 4^n - binomial(2*n+1, n))} /* Michael Somos Oct 31 2006 */
  • PARI
    {a(n) = if( n<-4, 0, n++; (4^n / 2 - binomial(2*n, n)) / 2)} /* Michael Somos, Jan 25 2014 */
  • Python
    import math
def C(n,r):
    f=math.factorial
    return f(n)/f(r)/f(n-r)
def A008549(n):
    return int((4**n)-C(2*n+1,n)) # Indranil Ghosh, Feb 18 2017

Formula

a(n) = 4^n - C(2*n+1, n).
a(n) = Sum_{k=1..n} Catalan(k)*4^(n-k): convolution of Catalan numbers and powers of 4.
G.f.: x*c(x)^2/(1 - 4*x), c(x) = g.f. of Catalan numbers. - Wolfdieter Lang
Note Sum_{k=0..2*n+1} binomial(2*n+1, k) = 2^(2n+1). Therefore, by the symmetry of Pascal's triangle, Sum_{k=0..n} binomial(2*n+1, k) = 2^(2*n) = 4^n. This explains why the following two expressions for a(n) are equal: Sum_{k=0..n-1} binomial(2*n+1, k) = 4^n - binomial(2*n+1, n). - Dan Velleman
G.f.: (2*x^2 - 1 + sqrt(1 - 4*x^2))/(2*(1 + 2*x)*(2*x - 1)*x^3).
a(n) = Sum_{k=0..n} C(2*k, k)*C(2*(n-k), n-k-1). - Paul Barry, Feb 16 2005
Second binomial transform of 2^n - C(n, floor(n/2)) = A045621(n). - Paul Barry, Jan 13 2006
a(n) = Sum_{0 < i <= k < n} binomial(n, k+i)*binomial(n, k-i). - Mircea Merca, Apr 05 2012
D-finite with recurrence (n+1)*a(n) + 2*(-4*n-1)*a(n-1) + 8*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012
0 = a(n) * (256*a(n+1) - 224*a(n+2) + 40*a(n+3)) + a(n+1) * (-32*a(n+1) + 56*a(n+2) - 14*a(n+3)) + a(n+2) * (-2*a(n+2) + a(n+3)) if n > -5. - Michael Somos, Jan 25 2014
Convolution square is A045894. - Michael Somos, Jan 25 2014
HANKEL transform is [0, -1, 2, -3, 4, -5, ...]. - Michael Somos, Jan 25 2014
BINOMIAL transform of [0, 0, 1, 3, 11, 35,...] (A109196) is [0, 0, 1, 6, 29, 130, ...]. - Michael Somos, Jan 25 2014
(n+1) * a(n) = A153338(n+1). - Michael Somos, Jan 25 2014
a(n) = Sum_{m = n+2..2*n+1} binomial(2*n+1,m), n >= 0. - Wolfdieter Lang, May 22 2015
E.g.f.: (exp(2*x) - BesselI(0,2*x) - BesselI(1,2*x))*exp(2*x). - Ilya Gutkovskiy, Aug 30 2016

Extensions

Better description from Dan Velleman (djvelleman(AT)amherst.edu), Dec 01 2000

A116406 Expansion of ((1 + x - 2x^2) + (1+x)*sqrt(1-4x^2))/(2(1-4x^2)).

Original entry on oeis.org

1, 1, 2, 3, 7, 11, 26, 42, 99, 163, 382, 638, 1486, 2510, 5812, 9908, 22819, 39203, 89846, 155382, 354522, 616666, 1401292, 2449868, 5546382, 9740686, 21977516, 38754732, 87167164, 154276028, 345994216, 614429672, 1374282019, 2448023843
Offset: 0

Views

Author

Paul Barry, Feb 13 2006

Keywords

Comments

Interleaving of A114121 and A032443. Row sums of A116405. Binomial transform is A116409.
Appears to be the number of n-digit binary numbers not having more zeros than ones; equivalently, the number of unrestricted Dyck paths of length n not going below the axis. - Ralf Stephan, Mar 25 2008
From Gus Wiseman, Jun 20 2021: (Start)
Also the number compositions of n with alternating sum >= 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. The a(0) = 1 through a(5) = 11 compositions are:
() (1) (2) (3) (4) (5)
(11) (21) (22) (32)
(111) (31) (41)
(112) (113)
(121) (122)
(211) (212)
(1111) (221)
(311)
(1121)
(2111)
(11111)
(End)
From J. Stauduhar, Jan 14 2022: (Start)
Also, for n >= 2, first differences of partial row sums of Pascal's triangle. The first ceiling(n/2)+1 elements of rows n=0 to n=4 in Pascal's triangle are:
1
1 1
1 2
1 3 3
1 4 6
...
The cumulative sums of these partial rows form the sequence 1,3,6,13,24,..., and its first differences are a(2),a(3),a(4),... in this sequence.
(End)

Crossrefs

The alternating sum = 0 case is A001700 or A088218.
The alternating sum > 0 case appears to be A027306.
The bisections are A032443 (odd) and A114121 (even).
The alternating sum <= 0 version is A058622.
The alternating sum < 0 version is A294175.
The restriction to reversed partitions is A344607.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A124754 gives the alternating sum of standard compositions.
A344610 counts partitions by sum and positive reverse-alternating sum.
A344616 lists the alternating sums of partitions by Heinz number.
Cf. A000041, A000070, A000097, A003242, A006330, A028260, A058696, A119899, A239830, A344605, A344611, A344650, A344739.

Programs

  • Mathematica
    CoefficientList[Series[((1+x-2x^2)+(1+x)Sqrt[1-4x^2])/(2(1-4x^2)),{x,0,40}],x] (* Harvey P. Dale, Aug 16 2012 *)
ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],ats[#]>=0&]],{n,0,15}] (* Gus Wiseman, Jun 20 2021 *)

Formula

a(n) = A114121(n/2)*(1+(-1)^n)/2 + A032443((n-1)/2)*(1-(-1)^n)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-1,k). - Paul Barry, Oct 06 2007
Conjecture: n*(n-3)*a(n) +2*(-n^2+4*n-2)*a(n-1) -4*(n-2)^2*a(n-2) +8*(n-2)*(n-3)*a(n-3)=0. - R. J. Mathar, Nov 28 2014
a(n) ~ 2^(n-2) * (1 + (3+(-1)^n)/sqrt(2*Pi*n)). - Vaclav Kotesovec, May 30 2016
a(n) = 2^(n-1) - A294175(n). - Gus Wiseman, Jun 27 2021

A345197 Concatenation of square matrices A(n), each read by rows, where A(n)(k,i) is the number of compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Jul 03 2021

Keywords

Comments

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

Examples

			The matrices for n = 1..7:
  1   0 1   0 0 1   0 0 0 1   0 0 0 0 1   0 0 0 0 0 1   0 0 0 0 0 0 1
      1 0   1 1 0   1 1 1 0   1 1 1 1 0   1 1 1 1 1 0   1 1 1 1 1 1 0
            0 1 0   0 1 2 0   0 1 2 3 0   0 1 2 3 4 0   0 1 2 3 4 5 0
                    0 1 0 0   0 2 2 0 0   0 3 4 3 0 0   0 4 6 6 4 0 0
                              0 0 1 0 0   0 0 2 3 0 0   0 0 3 6 6 0 0
                                          0 0 1 0 0 0   0 0 3 3 0 0 0
                                                        0 0 0 1 0 0 0
Matrix n = 5 counts the following compositions:
           i=-3:        i=-1:          i=1:            i=3:        i=5:
        -----------------------------------------------------------------
   k=1: |    0            0             0               0          (5)
   k=2: |   (14)         (23)          (32)            (41)         0
   k=3: |    0          (131)       (221)(122)   (311)(113)(212)    0
   k=4: |    0       (1211)(1112)  (2111)(1121)         0           0
   k=5: |    0            0          (11111)            0           0

Links

Crossrefs

The number of nonzero terms in each matrix appears to be A000096.
The number of zeros in each matrix appears to be A000124.
Row sums and column sums both appear to be A007318 (Pascal's triangle).
The matrix sums are A131577.
Antidiagonal sums appear to be A163493.
The reverse-alternating version is also A345197 (this sequence).
Antidiagonals are A345907.
Traces are A345908.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Other tetrangles: A318393, A318816, A320808, A321912.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000097, A000346, A007318, A008549, A025047, A032443, A034871, A114121, A120452, A238279, A239830, A344604.

Programs

  • Mathematica
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==k&&ats[#]==i&]],{n,0,6},{k,1,n},{i,-n+2,n,2}]

A008949 Triangle read by rows of partial sums of binomial coefficients: T(n,k) = Sum_{i=0..k} binomial(n,i) (0 <= k <= n); also dimensions of Reed-Muller codes.

Original entry on oeis.org

1, 1, 2, 1, 3, 4, 1, 4, 7, 8, 1, 5, 11, 15, 16, 1, 6, 16, 26, 31, 32, 1, 7, 22, 42, 57, 63, 64, 1, 8, 29, 64, 99, 120, 127, 128, 1, 9, 37, 93, 163, 219, 247, 255, 256, 1, 10, 46, 130, 256, 382, 466, 502, 511, 512, 1, 11, 56, 176, 386, 638, 848, 968, 1013, 1023, 1024, 1, 12, 67, 232, 562, 1024, 1486, 1816, 1981, 2036, 2047, 2048
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The second-left-from-middle column is A000346: T(2n+2, n) = A000346(n). - Ed Catmur (ed(AT)catmur.co.uk), Dec 09 2006
T(n,k) is the maximal number of regions into which n hyperplanes of co-dimension 1 divide R^k (the Cake-Without-Icing numbers). - Rob Johnson, Jul 27 2008
T(n,k) gives the number of vertices within distance k (measured along the edges) of an n-dimensional unit cube, (i.e., the number of vertices on the hypercube graph Q_n whose distance from a reference vertex is <= k). - Robert Munafo, Oct 26 2010
A triangle formed like Pascal's triangle, but with 2^n for n >= 0 on the right border instead of 1. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
Consider each "1" as an apex of two sequences: the first is the set of terms in the same row as the "1", but the rightmost term in the row repeats infinitely. Example: the row (1, 4, 7, 8) becomes (1, 4, 7, 8, 8, 8, ...). The second sequence begins with the same "1" but is the diagonal going down and to the right, thus: (1, 5, 16, 42, 99, 219, 466, ...). It appears that for all such sequence pairs, the binomial transform of the first, (1, 4, 7, 8, 8, 8, ...) in this case; is equal to the second: (1, 5, 16, 42, 99, ...). - Gary W. Adamson, Aug 19 2015
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let q(n) be the sum of polynomials in the n-th generation of T*. For n >= 0, row n of A008949 gives the coefficients of q(n+1); e.g., (row 3) = (1, 4, 7, 8) matches x^3 + 4*x^2 + 7*x + 9, which is the sum of the 8 polynomials in the 4th generation of T*. - Clark Kimberling, Jun 16 2016
T(n,k) is the number of subsets of [n]={1,...,n} of at most size k. Equivalently, T(n,k) is the number of subsets of [n] of at least size n-k. Counting the subsets of at least size (n-k) by conditioning on the largest element m of the smallest (n-k) elements of such a subset provides the formula T(n,k) = Sum_{m=n-k..n} C(m-1,n-k-1)*2^(n-m), and, by letting j=m-n+k, we obtain T(n,k) = Sum_{j=0..k} C(n+j-k-1,j)*2^(k-j). - Dennis P. Walsh, Sep 25 2017
If the interval of integers 1..n is shifted up or down by k, making the new interval 1+k..n+k or 1-k..n-k, then T(n-1,n-1-k) (= 2^(n-1)-T(n-1,k-1)) is the number of subsets of the new interval that contain their own cardinal number as an element. - David Pasino, Nov 01 2018

Examples

			Triangle begins:
  1;
  1,  2;
  1,  3,  4;
  1,  4,  7,   8;
  1,  5, 11,  15,  16;
  1,  6, 16,  26,  31,  32;
  1,  7, 22,  42,  57,  63,  64;
  1,  8, 29,  64,  99, 120, 127, 128;
  1,  9, 37,  93, 163, 219, 247, 255,  256;
  1, 10, 46, 130, 256, 382, 466, 502,  511,  512;
  1, 11, 56, 176, 386, 638, 848, 968, 1013, 1023, 1024;
  ...

References

  • F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier-North Holland, 1978, p. 376.

Links

Crossrefs

Diagonals are given by A000079, A000225, A000295, A002662, A002663, A002664, A035038-A035042.
Columns are given by A000012, A000027, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863. - Ken Shirriff, Jun 28 2011
Row sums sequence is A001792.
T(n, m)= A055248(n, n-m).
Cf. A110555, A007318, A000346, A171886, A228196, A228576.
Cf. also A027306, A249111, A163866, A261363.
Cf. A000071, A001924

Programs

  • GAP
    T:=Flat(List([0..11],n->List([0..n],k->Sum([0..k],j->Binomial(n+j-k-1,j)*2^(k-j))))); # Muniru A Asiru, Nov 25 2018
  • Haskell
    a008949 n k = a008949_tabl !! n !! k
a008949_row n = a008949_tabl !! n
a008949_tabl = map (scanl1 (+)) a007318_tabl
-- Reinhard Zumkeller, Nov 23 2012
  • Magma
    [[(&+[Binomial(n,j): j in [0..k]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Nov 25 2018
  • Maple
    A008949 := proc(n,k) local i; add(binomial(n,i),i=0..k) end; # Typo corrected by R. J. Mathar, Oct 26 2010
  • Mathematica
    Table[Length[Select[Subsets[n], (Length[ # ] <= k) &]], {n, 0, 12}, {k, 0, n}] // Grid (* Geoffrey Critzer, May 13 2009 *)
Flatten[Accumulate/@Table[Binomial[n,i],{n,0,20},{i,0,n}]] (* Harvey P. Dale, Aug 08 2015 *)
T[ n_, k_] := If[ n < 0 || k > n, 0, Binomial[n, k] Hypergeometric2F1[1, -k, n + 1 - k, -1]]; (* Michael Somos, Aug 05 2017 *)
  • PARI
    A008949(n)=T8949(t=sqrtint(2*n-sqrtint(2*n)),n-t*(t+1)/2)
T8949(r,c)={ 2*c > r || return(sum(i=0,c,binomial(r,i))); 1<M. F. Hasler, May 30 2010
  • PARI
    {T(n, k) = if(k>n, 0, sum(i=0, k, binomial(n, i)))}; /* Michael Somos, Aug 05 2017 */
  • PARI
    row(n) = my(v=vector(n+1, k, binomial(n,k-1))); vector(#v, k, sum(i=1, k, v[i])); \\ Michel Marcus, Apr 13 2025
  • Sage
    [[sum(binomial(n,j) for j in range(k+1)) for k in range(n+1)] for n in range(12)] # G. C. Greubel, Nov 25 2018

Formula

From partial sums across rows of Pascal triangle A007318.
T(n, 0) = 1, T(n, n) = 2^n, T(n, k) = T(n-1, k-1) + T(n-1, k), 0 < k < n.
G.f.: (1 - x*y)/((1 - y - x*y)*(1 - 2*x*y)). - Antonio Gonzalez (gonfer00(AT)gmail.com), Sep 08 2009
T(2n,n) = A032443(n). - Philippe Deléham, Sep 16 2009
T(n,k) = 2 T(n-1,k-1) + binomial(n-1,k) = 2 T(n-1,k) - binomial(n-1,k). - M. F. Hasler, May 30 2010
T(n,k) = binomial(n,n-k)* 2F1(1, -k; n+1-k; -1). - Olivier Gérard, Aug 02 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 18 2013
T(n,floor(n/2)) = A027306(n). - Reinhard Zumkeller, Nov 14 2014
T(n,n) = 2^n, otherwise for 0 <= k <= n-1, T(n,k) = 2^n - T(n,n-k-1). - Bob Selcoe, Mar 30 2017
For fixed j >= 0, lim_{n -> oo} T(n+1,n-j+1)/T(n,n-j) = 2. - Bob Selcoe, Apr 03 2017
T(n,k) = Sum_{j=0..k} C(n+j-k-1,j)*2^(k-j). - Dennis P. Walsh, Sep 25 2017

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Mar 23 2000
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