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

A001333 Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).

Original entry on oeis.org

1, 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, 22619537, 54608393, 131836323, 318281039, 768398401, 1855077841, 4478554083, 10812186007, 26102926097, 63018038201, 152139002499, 367296043199
Offset: 0

Views

Author

N. J. A. Sloane and R. K. Guy

Keywords

Comments

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].
Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011
Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012
Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman
a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.
In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017
Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002
a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.
a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
Binomial transform of A077957. - Paul Barry, Feb 25 2003
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004
For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004
Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004
Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]
Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008
The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008
Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008
Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008
From Charlie Marion, Jan 07 2009: (Start)
In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then b(1,n)=a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
Cf. A000129, A142238-A142239, A153313-A153318.
(End)
This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010
Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(8,2) = (0 0 1 0)
(0 1 0 1)
(1 0 2 0)
(0 2 0 1).
Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)
(End)
For n >= 1, row sums of triangle
m/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....2
.2..|..1.....2.....4
.3..|..1.....4.....4.....8
.4..|..1.....4....12.....8....16
.5..|..1.....6....12....32....16....32
.6..|..1.....6....24....32....80....32....64
.7..|..1.....8....24....80....80...192....64...128
which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012
The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012
Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012
This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012
a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016
a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018
For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020
For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021
a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022
From Greg Dresden, May 08 2023: (Start)
For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.
_
|| _
|||_|||
|_|
(End)
12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024

Examples

			Convergents are 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.
The 15 3 X 2 crossword grids, with white squares represented by an o:
  ooo ooo ooo ooo ooo ooo ooo oo. o.o .oo o.. .o. ..o oo. .oo
  ooo oo. o.o .oo o.. .o. ..o ooo ooo ooo ooo ooo ooo .oo oo.
G.f. = 1 + x + 3*x^2 + 7*x^3 + 17*x^4 + 41*x^5 + 99*x^6 + 239*x^7 + 577*x^8 + ...

References

  • M. R. Bacon and C. K. Cook, Some properties of Oresme numbers and convolutions ..., Fib. Q., 62:3 (2024), 233-240.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
  • J. Devillet, J.-L. Marichal, and B. Teheux, Classifications of quasitrivial semigroups, Semigroup Forum, 100 (2020), 743-764.
  • Maribel Díaz Noguera [Maribel Del Carmen Díaz Noguera], Rigoberto Flores, Jose L. Ramirez, and Martha Romero Rojas, Catalan identities for generalized Fibonacci polynomials, Fib. Q., 62:2 (2024), 100-111.
  • Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
  • R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 288.
  • A. F. Horadam, R. P. Loh, and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979.
  • Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
  • Kin Y. Li, Math Problem Book I, 2001, p. 24, Problem 159.
  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 102, Problem 10.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Volume 1 (1986), p. 203, Example 4.1.2.
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.
  • R. C. Tilley et al., The cell growth problem for filaments, Proc. Louisiana Conf. Combinatorics, ed. R. C. Mullin et al., Baton Rouge, 1970, 310-339.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.

Links

Crossrefs

For denominators see A000129.
See A040000 for the continued fraction expansion of sqrt(2).
See also A078057 which is the same sequence without the initial 1.
Cf. also A002203, A152113.
Row sums of unsigned Chebyshev T-triangle A053120. a(n)= A054458(n, 0) (first column of convolution triangle).
Row sums of A140750, A160756, A135837.
Equals A034182(n-1) + 2 and A084128(n)/2^n. First differences of A052937. Partial sums of A052542. Pairwise sums of A048624. Bisection of A002965.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Second row of the array in A135597.
Cf. A048211, A153588, A174283, A174284, A174285, A174286, A176499, A176500, A176501, A176502.
Cf. A055099.
Cf. A028859, A001906 / A088305, A033303, A000225, A095263, A003945, A006356, A002478, A214260, A001911 and A000217 for other restricted ternary words.
Cf. Triangle A106513 (alternating row sums).
Equals A293004 + 1.
Cf. A033539, A332602, A086395 (subseq. of primes).

Programs

  • Haskell
    a001333 n = a001333_list !! n
a001333_list = 1 : 1 : zipWith (+)
                       a001333_list (map (* 2) $ tail a001333_list)
-- Reinhard Zumkeller, Jul 08 2012
  • Magma
    [n le 2 select 1 else 2*Self(n-1)+Self(n-2): n in [1..35]]; // Vincenzo Librandi, Nov 10 2018
  • Maple
    A001333 := proc(n) option remember; if n=0 then 1 elif n=1 then 1 else 2*procname(n-1)+procname(n-2) fi end;
Digits := 50; A001333 := n-> round((1/2)*(1+sqrt(2))^n);
with(numtheory): cf := cfrac (sqrt(2),1000): [seq(nthnumer(cf,i), i=0..50)];
a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^n):
seq(a(n), n=0..33);  # Alois P. Heinz, Aug 01 2008
A001333List := proc(m) local A, P, n; A := [1,1]; P := [1,1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-2]]);
A := [op(A), P[-1]] od; A end: A001333List(32); # Peter Luschny, Mar 26 2022
  • Mathematica
    Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 08 2006 *)
Table[((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2, {n, 0, 29}] // Simplify (* Robert G. Wilson v, May 02 2006 *)
a[0] = 1; a[1] = 1; a[n_] := a[n] = 2a[n - 1] + a[n - 2]; Table[a@n, {n, 0, 29}] (* Robert G. Wilson v, May 02 2006 *)
Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, May 02 2006 *)
a=c=0;t={b=1}; Do[c=a+b+c; AppendTo[t,c]; a=b;b=c,{n,40}]; t (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
LinearRecurrence[{2, 1}, {1, 1}, 40] (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
Join[{1}, Numerator[Convergents[Sqrt[2], 30]]] (* Harvey P. Dale, Aug 22 2011 *)
Table[(-I)^n ChebyshevT[n, I], {n, 10}] (* Eric W. Weisstein, Apr 04 2017 *)
CoefficientList[Series[(-1 + x)/(-1 + 2 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
Table[Sqrt[(ChebyshevT[n, 3] + (-1)^n)/2], {n, 0, 20}] (* Eric W. Weisstein, Apr 17 2018 *)
  • PARI
    {a(n) = if( n<0, (-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [1, 1]}; /* Michael Somos, Sep 02 2012 */
  • PARI
    {a(n) = polchebyshev(n, 1, I) / I^n}; /* Michael Somos, Sep 02 2012 */
  • PARI
    a(n) = real((1 + quadgen(8))^n); \\ Michel Marcus, Mar 16 2021
  • PARI
    { for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[1, 1]; if (a > 10^(10^3 - 6), break); write("b001333.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 12 2009
  • Python
    from functools import cache
@cache
def a(n): return 1 if n < 2 else 2*a(n-1) + a(n-2)
print([a(n) for n in range(32)]) # Michael S. Branicky, Nov 13 2022
  • Sage
    from sage.combinat.sloane_functions import recur_gen2
it = recur_gen2(1,1,2,1)
[next(it) for i in range(30)] ## Zerinvary Lajos, Jun 24 2008
  • Sage
    [lucas_number2(n,2,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009

Formula

a(n) = A055642(A125058(n)). - Reinhard Zumkeller, Feb 02 2007
a(n) = 2a(n-1) + a(n-2);
a(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2.
a(n)+a(n+1) = 2 A000129(n+1). 2*a(n) = A002203(n).
G.f.: (1 - x) / (1 - 2*x - x^2) = 1 / (1 - x / (1 - 2*x / (1 + x))). - Simon Plouffe in his 1992 dissertation.
A000129(2n) = 2*A000129(n)*a(n). - John McNamara, Oct 30 2002
a(n) = (-i)^n * T(n, i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
a(n) = a(n-1) + A052542(n-1), n>1. a(n)/A052542(n) converges to sqrt(1/2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
E.g.f.: exp(x)cosh(x*sqrt(2)). - Paul Barry, May 08 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)2^k. - Paul Barry, May 13 2003
For n > 0, a(n)^2 - (1 + (-1)^(n))/2 = Sum_{k=0..n-1} ((2k+1)*A001653(n-1-k)); e.g., 17^2 - 1 = 288 = 1*169 + 3*29 + 5*5 + 7*1; 7^2 = 49 = 1*29 + 3*5 + 5*1. - Charlie Marion, Jul 18 2003
a(n+2) = A078343(n+1) + A048654(n). - Creighton Dement, Jan 19 2005
a(n) = A000129(n) + A000129(n-1) = A001109(n)/A000129(n) = sqrt(A001110(n)/A000129(n)^2) = ceiling(sqrt(A001108(n))). - Henry Bottomley, Apr 18 2000
Also the first differences of A000129 (the Pell numbers) because A052937(n) = A000129(n+1) + 1. - Graeme McRae, Aug 03 2006
a(n) = Sum_{k=0..n} A122542(n,k). - Philippe Deléham, Oct 08 2006
For another recurrence see A000129.
a(n) = Sum_{k=0..n} A098158(n,k)*2^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1,1; 2,1]^n. - Gary W. Adamson, Mar 12 2008
If p[1]=1, and p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
For n>=2, a(n)=F_n(2)+F_(n+1)(2), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(-n) = (-1)^n * a(n). - Michael Somos, Sep 02 2012
Dirichlet g.f.: (PolyLog(s,1-sqrt(2)) + PolyLog(s,1+sqrt(2)))/2. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A000129(n) - A000129(n-1), where A000129(n) is the n-th Pell Number. Hence the continued fraction is of the form 1-(A000129(n-1)/A000129(n)). - Gregory L. Simay, Nov 09 2018
a(n) = (A000129(n+3) + A000129(n-3))/10, n>=3. - Paul Curtz, Jun 16 2021
a(n) = (A000129(n+6) - A000129(n-6))/140, n>=6. - Paul Curtz, Jun 20 2021
a(n) = round((1/2)*sqrt(Product_{k=1..n} 4*(1 + sin(k*Pi/n)^2))), for n>=1. - Greg Dresden, Dec 28 2021
a(n)^2 + a(n+1)^2 = A075870(n+1) = 2*(b(n)^2 + b(n+1)^2) for all n in Z where b(n) := A000129(n). - Michael Somos, Apr 02 2022
a(n) = 2*A048739(n-2)+1. - R. J. Mathar, Feb 01 2024
Sum_{n>=1} 1/a(n) = 1.5766479516393275911191017828913332473... - R. J. Mathar, Feb 05 2024
From Peter Bala, Jul 06 2025: (Start)
G.f.: Sum_{n >= 1} (-1)^(n+1) * x^(n-1) * Product_{k = 1..n} (1 - k*x)/(1 - 3*x + k*x^2).
The following series telescope:
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 1/a(2*n)) = 1/4, since 1/(a(2*n) + 1/a(2*n)) = 1/A077445(n) + 1/A077445(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) - 1/a(2*n+1)) = 1/8, since. 1/(a(2*n+1) - 1/a(2*n+1)) = 1/(4*Pell(2*n)) + 1/(4*Pell(2*n+2)), where Pell(n) = A000129(n).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) + 9/a(2*n+1)) = 1/10, since 1/(a(2*n+1) + 9/a(2*n+1)) = b(n) + b(n+1), where b(n) = A001109(n)/(2*Pell(2*n-1)*Pell(2*n+1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 1 - sqrt(2)/2 = A268682, since (-1)^(n+1)/(a(n)*a(n+1)) = Pell(n)/a(n) - Pell(n+1)/a(n+1). (End)

Extensions

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, 1, 11, 45, 84, 70, 21, 1, 1, 12, 55, 120, 126, 56, 7, 1, 13, 66, 165, 210, 126, 28, 1, 1, 14, 78, 220, 330, 252, 84, 8, 1, 15, 91, 286, 495, 462
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

T(n,k) is the number of subsets of {1,2,...,n-1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of k-matchings of the path graph P_n. - Emeric Deutsch, Dec 10 2003
T(n,k) = number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2). - Emeric Deutsch, Apr 09 2005
Given any recurrence sequence S(k) = x*a(k-1) + a(k-2), starting (1, x, x^2+1, ...); the (k+1)-th term of the series = f(x) in the k-th degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ...). Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1). - Gary W. Adamson, Apr 16 2008
Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6 - 5x^4 + 6x^2 - 1. - David Callan, Jul 22 2008
T(n,k) is the number of nodes at level k in the Fibonacci tree f(k-1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(-1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k-1), taken as the root of f(k), we attach with a rightmost edge the tree f(k-2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1. - Emeric Deutsch, Jun 21 2011
Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
Riordan array (1/(1-x),x^2/(1-x). - Philippe Deléham, Dec 12 2011
This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number. - Mohammad K. Azarian, Mar 08 2012
If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n-1;x), then we obtain the sequence of Vieta-Fibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (-i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n-1) + b*f(n-2). Then we deduce the relation: f(n) = b^((n-1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n-1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (-1)^k * f(n-k) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n-1;a) are the Vieta-Lucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m-1;a)*F(n-1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n-1;a). Further we have L(n;a) = 2*(-i)^n * T(n;i*x/2), where T(n;x) denotes the n-th Chebyshev polynomial of the first kind. For the proofs, other relations and facts - see Witula-Slota's papers. - Roman Witula, Oct 12 2012
The diagonal sums of this triangle are A000930. - John Molokach, Jul 04 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Nov 29 2014
A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials. - Tom Copeland, Dec 06 2015
For n >= 3, the n-th row gives the coefficients of the independence polynomial of the (n-2)-path graph P_{n-2}. - Eric W. Weisstein, Apr 07 2017
For n >= 2, the n-th row gives the coefficients of the matching-generating polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426. - Tom Copeland, Jul 02 2018
T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
T(n, k) = number of compositions of n+1 into n+1-2*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5. - Michael Somos, Sep 19 2019
From Gary W. Adamson, Apr 25 2022: (Start)
Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all -1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,-1,0; -1,2,-1; 0,-1,2) is x^3 - 5x^2 + 6x - 1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).
Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even-indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and -1's as the sub- and superdiagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,-1,0; -1,2,-1; 0,-1,2) is 3.414... = B(8,1) = a root to x^3 - 6x^2 + 10x - 4. (End)
T(n,k) is the number of edge covers of P_(n+2) with (n-k) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two non-consecutive edges among 2-6. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x) = xE(P_(n-1),x)+xE(P_(n-2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi). - Feryal Alayont, Jun 03 2022
T(n,k) is the number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using k dominoes and n-2*k squares. - Michael A. Allen, Dec 28 2022
T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n-2k)) such that s(i) < s(i+1), s(1) is odd, s(n-2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0. - Molly W. Dunkum, Jun 27 2023

Examples

			The first few Fibonacci polynomials (defined here by F(0,x) = 0, F(1,x) = 1; F(n+1, x) = F(n, x) + x*F(n-1, x)) are:
0: 0
1: 1
2: 1
3: 1 + x
4: 1 + 2*x
5: 1 + 3*x + x^2
6: (1 + x)*(1 + 3*x)
7: 1 + 5*x + 6*x^2 + x^3
8: (1 + 2*x)*(1 + 4*x + 2*x^2)
9: (1 + x)*(1 + 6*x + 9*x^2 + x^3)
10: (1 + 3*x + x^2 )*(1 + 5*x + 5*x^2)
11: 1 + 9*x + 28*x^2 + 35*x^3 + 15*x^4 + x^5
From _Roger L. Bagula_, Feb 20 2009: (Start)
  1
  1
  1   1
  1   2
  1   3   1
  1   4   3
  1   5   6   1
  1   6  10   4
  1   7  15  10   1
  1   8  21  20   5
  1   9  28  35  15   1
  1  10  36  56  35   6
  1  11  45  84  70  21   1
  1  12  55 120 126  56   7 (End)
For n=9 and k=4, T(9,4) = C(5,4) = 5 since there are exactly five size-4 subsets of {1,2,...,8} that contain no consecutive integers, namely, {1,3,5,7}, {1,3,5,8}, {1,3,6,8}, {1,4,6,8}, and {2,4,6,8}. - _Dennis P. Walsh_, Mar 31 2011
When the rows of the triangle are displayed as centered text, the falling diagonal sums are A005314. The first few terms are row1 = 1 = 1; row2 = 1+1 = 2; row3 = 2+1 = 3; row4 = 1+3+1 = 5; row5 = 1+3+4+1 = 9; row6 = 4+6+5+1 = 16; row7 = 1+10+10+6+1 = 28; row8 = 1+5+20+15+7+1 = 49; row9 = 6+15+35+21+8+1 = 86; row10 = 1+21+35+56+28+9+1 = 151. - _John Molokach_, Jul 08 2013
In the example, you can see that the n-th row of Pascal's triangle is given by T(n, 0), T(n+1, 1), ..., T(2n-1, n-1), T(2n, n). - _Daniel Forgues_, Jul 07 2018

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 141ff.
  • C. D. Godsil, Algebraic Combinatorics, Chapman and Hall, New York, 1993.
  • I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. See p. 117.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 182-183.

Links

Crossrefs

Row sums = A000045(n+1) (Fibonacci numbers). - Michael Somos, Apr 02 1999
Cf. A000129, A054123, A052553, A000930, A030528, A053122.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.
Cf. A007318, A025581, A055087.
Cf. A000108, A049310, A055151, A098474, A124644, A134264.
Cf. A054142, A053123, A332602.

Programs

  • Haskell
    a011973 n k = a011973_tabf !! n !! k
a011973_row n = a011973_tabf !! n
a011973_tabf = zipWith (zipWith a007318) a025581_tabl a055087_tabf
-- Reinhard Zumkeller, Jul 14 2015
  • Maple
    a := proc(n) local k; [ seq(binomial(n-k,k),k=0..floor(n/2)) ]; end;
T := proc(n, k): if k<0 or k>floor(n/2) then return(0) fi: binomial(n-k, k) end: seq(seq(T(n,k), k=0..floor(n/2)), n=0..15); # Johannes W. Meijer, Aug 26 2013
  • Mathematica
    (* first: sum method *) Table[CoefficientList[Sum[Binomial[n - m + 1, m]*x^m, {m, 0, Floor[(n + 1)/2]}], x], {n, 0, 12}] (* Roger L. Bagula, Feb 20 2009 *)
(* second: polynomial recursion method *) Clear[L, p, x, n, m]; L[x, 0] = 1; L[x, 1] = 1 + x; L[x_, n_] := L[x, n - 1] + x*L[x, n - 2]; Table[ExpandAll[L[x, n]], {n, 0, 10}]; Table[CoefficientList[ExpandAll[L[x, n]], x], {n, 0, 12}]; Flatten[%] (* Roger L. Bagula, Feb 20 2009 *)
(* Center option shows falling diagonals are A224838 *) Column[Table[Binomial[n - m, m], {n, 0, 25}, {m, 0, Floor[n/2]}], Center] (* John Molokach, Jul 26 2013 *)
Table[ Select[ CoefficientList[ Fibonacci[n, x], x], Positive] // Reverse, {n, 1, 18} ] // Flatten (* Jean-François Alcover, Oct 21 2013 *)
CoefficientList[LinearRecurrence[{1, x}, {1 + x, 1 + 2 x}, {-1, 10}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
CoefficientList[Table[x^((n - 1)/2) Fibonacci[n, 1/Sqrt[x]], {n, 15}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
  • PARI
    {T(n, k) = if( k<0 || 2*k>n, 0, binomial(n-k, k))};
  • Sage
    # Prints the table; cf. A145574.
for n in (2..20): [Compositions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012

Formula

Let F(n, x) be the n-th Fibonacci polynomial in x; the g.f. for F(n, x) is Sum_{n>=0} F(n, x)*y^n = (1 + x*y)/(1 - y - x*y^2). - Paul D. Hanna
T(m, n) = 0 for n != 0 and m <= 1 T(0, 0) = T(1, 0) = 1 T(m, n) = T(m - 1, n) + T(m-2, n-1) for m >= 2 (i.e., like the recurrence for Pascal's triangle A007318, but going up one row as well as left one column for the second summand). E.g., T(7, 2) = 10 = T(6, 2) + T(5, 1) = 6 + 4. - Rob Arthan, Sep 22 2003
G.f. for k-th column: x^(2*k-1)/(1-x)^(k+1).
Identities for the Fibonacci polynomials F(n, x):
F(m+n+1, x) = F(m+1, x)*F(n+1, x) + x*F(m, x)F(n, x).
F(n, x)^2-F(n-1, x)*F(n+1, x) = (-x)^(n-1).
The degree of F(n, x) is floor((n-1)/2) and F(2p, x) = F(p, x) times a polynomial of equal degree which is 1 mod p.
From Roger L. Bagula, Feb 20 2009: (Start)
p(x,n) = Sum_{m=0..floor((n+1)/2)} binomial(n-m+1, m)*x^m;
p(x,n) = p(x, n - 1) + x*p(x, n - 2). (End)
T(n, k) = A102541(2*n+2, 2*k+1) + A102541(2*n+1, 2*k) - A102541(2*n+3, 2*k+1), n >= 0 and 0 <= k <= floor(n/2). - Johannes W. Meijer, Aug 26 2013
G.f.: 1/(1-x-y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ x*y)*x/((2*k+2+ x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
O.g.f. G(x,t) = x/(1-x-tx^2) = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... has the inverse Ginv(x,t) = -[1+x-sqrt[(1+x)^2 + 4tx^2]]/(2tx) = x - x^2 + (1-t) x^3 + (-1+3t) x^4 + ..., an o.g.f. for the signed Motzkin polynomials of A055151, consistent with A134264 with h_0 = 1, h_1 = -1, h_2 = -t, and h_n = 0 otherwise. - Tom Copeland, Jan 21 2016
O.g.f. H(x,t) = x (1+tx)/ [1-x(1+tx)] = x + (1+t) x^2 + (1+2t) x^3 + ... = -L[Cinv(-tx)/t], where L(x) = x/(1+x) with inverse Linv(x) = x/(1-x) and Cinv(x) = x (1-x) is the inverse of C(x) = (1-sqrt(1-4x))/2, the o.g.f. of the shifted Catalan numbers A000108. Then Hinv(x,t) = -C[t Linv(-x)]/t = [-1 + sqrt(1+4tx/(1+x))]/2t = x - (1+t) x^2 + (1+2t+2t^2) x^3 - (1+3t+6t^2+5t^3) x^4 + ..., which is signed A098474, reverse of A124644. - Tom Copeland, Jan 25 2016
T(n, k) = GegenbauerC(k, (n+1)/2-k, 1). - Peter Luschny, May 10 2016

A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 15, 7, 1, 1, 15, 35, 28, 9, 1, 1, 21, 70, 84, 45, 11, 1, 1, 28, 126, 210, 165, 66, 13, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 14 2003

Keywords

Comments

Coefficient array for Morgan-Voyce polynomial b(n,x). A053122 (unsigned) is the coefficient array for B(n,x). Reversal of A054142. - Paul Barry, Jan 19 2004
This triangle is formed from even-numbered rows of triangle A011973 read in reverse order. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k+1 peaks. T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k peaks at height >= 2. T(n,k) is the number of directed column-convex polyominoes of area n+1, having k+1 columns. - Emeric Deutsch, May 31 2004
Riordan array (1/(1-x), x/(1-x)^2). - Paul Barry, May 09 2005
The triangular matrix a(n,k) = (-1)^(n+k)*T(n,k) is the matrix inverse of A039599. - Philippe Deléham, May 26 2005
The n-th row gives absolute values of coefficients of reciprocal of g.f. of bottom-line of n-wave sequence. - Floor van Lamoen (fvlamoen(AT)planet.nl), Sep 24 2006
Unsigned version of A129818. - Philippe Deléham, Oct 25 2007
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k >=1 (height(alpha) = |Im(alpha)|) and of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
A085478 is jointly generated with A078812 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x)+x*v(n-1)x and v(n,x) = u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Per Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle given by (0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 26 2012
T(n,k) is also the number of compositions (ordered partitions) of 2*n+1 into 2*k+1 parts which are all odd. Proof: The o.g.f. of column k, x^k/(1-x)^(2*k+1) for k >= 0, is the o.g.f. of the odd-indexed members of the sequence with o.g.f. (x/(1-x^2))^(2*k+1) (bisection, odd part). Thus T(n,k) is obtained from the sum of the multinomial numbers A048996 for the partitions of 2*n+1 into 2*k+1 parts, all of which are odd. E.g., T(3,1) = 3 + 3 from the numbers for the partitions [1,1,5] and [1,3,3], namely 3!/(2!*1!) and 3!/(1!*2!), respectively. The number triangle with the number of these partitions as entries is A152157. - Wolfdieter Lang, Jul 09 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*A039599(n,k). - R. J. Mathar, Mar 12 2013
T(n,k) = A258993(n+1,k) for k = 0..n-1. - Reinhard Zumkeller, Jun 22 2015
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the algebraic function F(x)*G(x)^n about 0, where F(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) and G(x) = ((1 + sqrt(1 + 4*x))/2)^2. For example, for n = 4, (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) * ((1 + sqrt(1 + 4*x))/2)^8 = (x^4 + 10*x^3 + 15*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 23 2018
Row n also gives the coefficients of the characteristc polynomial of the tridiagonal n X n matrix M_n given in A332602: Phi(n, x) := Det(M_n - x*1_n) = Sum_{k=0..n} T(n, k)*(-x)^k, for n >= 0, with Phi(0, x) := 1. - Wolfdieter Lang, Mar 25 2020
It appears that the largest root of the n-th degree polynomial is equal to the sum of the distinct diagonals of a (2*n+1)-gon including the edge, 1. The largest root of x^3 - 6*x^2 + 5*x - 1 is 5.048917... = the sum of (1 + 1.80193... + 2.24697...). Alternatively, the largest root of the n-th degree polynomial is equal to the square of sigma(2*n+1). Check: 5.048917... is the square of sigma(7), 2.24697.... Given N = 2*n+1, sigma(N) (N odd) can be defined as 1/(2*sin(Pi/(2*N))). Relating to the 9-gon, the largest root of x^4 - 10*x^3 + 15*x^2 - 7*x + 1 is 8.290859..., = the sum of (1 + 1.879385... + 2.532088... + 2.879385...), and is the square of sigma(9), 2.879385... Refer to A231187 for a further clarification of sigma(7). - Gary W. Adamson, Jun 28 2022
For n >=1, the n-th row is given by the coefficients of the minimal polynomial of -4*sin(Pi/(4*n + 2))^2. - Eric W. Weisstein, Jul 12 2023
Denoting this lower triangular array by L, then L * diag(binomial(2*k,k)^2) * transpose(L) is the LDU factorization of A143007, the square array of crystal ball sequences for the A_n X A_n lattices. - Peter Bala, Feb 06 2024
T(n, k) is the number of occurrences of the periodic substring (01)^k in the periodic string (01)^n (see Proposition 4.7 at page 7 in Fang). - Stefano Spezia, Jun 09 2024

Examples

			Triangle begins as:
  1;
  1    1;
  1    3    1;
  1    6    5    1;
  1   10   15    7    1;
  1   15   35   28    9    1;
  1   21   70   84   45   11    1;
  1   28  126  210  165   66   13    1;
  1   36  210  462  495  286   91   15    1;
  1   45  330  924 1287 1001  455  120   17    1;
  1   55  495 1716 3003 3003 1820  680  153   19    1;
...
From _Philippe Deléham_, Mar 26 2012: (Start)
(0, 1, 0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) begins:
  1
  0, 1
  0, 1,  1
  0, 1,  3,   1
  0, 1,  6,   5,   1
  0, 1, 10,  15,   7,   1
  0, 1, 15,  35,  28,   9,  1
  0, 1, 21,  70,  84,  45, 11,  1
  0, 1, 28, 126, 210, 165, 66, 13, 1. (End)

Links

Crossrefs

Row sums: A001519. Signed versions: A123970, A129818.
Cf. A000108, A001006, A007318, A098158, A183160, A078812, A091042.
Cf. A258993, A025581, A051162, A054142, A332602.
Cf. A231187, A143007.

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k, 2*k) ))); # G. C. Greubel, Aug 01 2019
  • Haskell
    a085478 n k = a085478_tabl !! n !! k
a085478_row n = a085478_tabl !! n
a085478_tabl = zipWith (zipWith a007318) a051162_tabl a025581_tabl
-- Reinhard Zumkeller, Jun 22 2015
  • Magma
    [Binomial(n+k, 2*k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019
  • Maple
    T := (n,k) -> binomial(n+k,2*k): seq(seq(T(n,k), k=0..n), n=0..11);
  • Mathematica
    (* First program *)
u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A085478 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A078812 *) (*Clark Kimberling, Feb 25 2012 *)
(* Second program *)
Table[Binomial[n + k, 2 k], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 01 2019 *)
CoefficientList[Table[Fibonacci[2 n + 1, Sqrt[x]], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Jul 03 2023 *)
Join[{{1}}, CoefficientList[Table[MinimalPolynomial[-4 Sin[Pi/(4 n + 2)]^2, x], {n, 20}], x]] (* Eric W. Weisstein, Jul 12 2023 *)
  • PARI
    T(n,k) = binomial(n+k,n-k)
  • Sage
    [[binomial(n+k,2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

Formula

T(n, k) = (n+k)!/((n-k)!*(2*k)!).
G.f.: (1-z)/((1-z)^2-tz). - Emeric Deutsch, May 31 2004
Row sums are A001519 (Fibonacci(2n+1)). Diagonal sums are A011782. Binomial transform of A026729 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
T(n, 0) = 1, T(n, k) = 0 if n=0} T(n-1-j, k-1)*(j+1). T(0, 0) = 1, T(0, k) = 0 if k>0; T(n, k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j). For the column k, g.f.: Sum_{n>=0} T(n, k)*x^n = (x^k) / (1-x)^(2*k+1). - Philippe Deléham, Feb 15 2004
Sum_{k=0..n} T(n,k)*x^(2*k) = A000012(n), A001519(n+1), A001653(n), A078922(n+1), A007805(n), A097835(n), A097315(n), A097838(n), A078988(n), A097841(n), A097727(n), A097843(n), A097730(n), A098244(n), A097733(n), A098247(n), A097736(n), A098250(n), A097739(n), A098253(n), A097742(n), A098256(n), A097767(n), A098259(n), A097770(n), A098262(n), A097773(n), A098292(n), A097776(n) for x=0,1,2,...,27,28 respectively. - Philippe Deléham, Dec 31 2007
T(2*n,n) = A005809(n). - Philippe Deléham, Sep 17 2009
A183160(n) = Sum_{k=0..n} T(n,k)*T(n,n-k). - Paul D. Hanna, Dec 27 2010
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Feb 06 2012
O.g.f. for column k: x^k/(1-x)^(2*k+1), k >= 0. [See the o.g.f. of the triangle above, and a comment on compositions. - Wolfdieter Lang, Jul 09 2012]
E.g.f.: (2/sqrt(x + 4))*sinh((1/2)*t*sqrt(x + 4))*cosh((1/2)*t*sqrt(x)) = t + (1 + x)*t^3/3! + (1 + 3*x + x^2)*t^5/5! + (1 + 6*x + 5*x^2 + x^3)*t^7/7! + .... Cf. A091042. - Peter Bala, Jul 29 2013
T(n, k) = A065941(n+3*k, 4*k) = A108299(n+3*k, 4*k) = A194005(n+3*k, 4*k). - Johannes W. Meijer, Sep 05 2013
Sum_{k=0..n} (-1)^k*T(n,k)*A000108(k) = A000007(n) for n >= 0. - Werner Schulte, Jul 12 2017
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A001006(n) for n >= 0. - Werner Schulte, Jul 12 2017
From Peter Bala, Jun 26 2025: (Start)
The n-th row polynomial b(n, x) = (-1)^n * U(2*n, (i/2)*sqrt(x)), where U(n,x) is the n-th Chebyshev polynomial of the second kind.
b(n, x) = (-1)^n * Dir(n, -1 - x/2), where Dir(n, x) is the n-th row polynomial of the triangle A244419.
b(n, -1 - x) is the n-th row polynomial of A098493. (End)

A005021 Random walks (binomial transform of A006054).

Original entry on oeis.org

1, 5, 19, 66, 221, 728, 2380, 7753, 25213, 81927, 266110, 864201, 2806272, 9112264, 29587889, 96072133, 311945595, 1012883066, 3288813893, 10678716664, 34673583028, 112584429049, 365559363741, 1186963827439, 3854047383798, 12514013318097, 40632746115136
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of walks of length 2n+5 in the path graph P_6 from one end to the other one. Example: a(1)=5 because in the path ABCDEF we have ABABCDEF, ABCBCDEF, ABCDCDEF, ABCDEDEF and ABCDEFEF. - Emeric Deutsch, Apr 02 2004
Since a(n) is the binomial transform of A006054 from formula (3.63) in the Witula-Slota-Warzynski paper, it follows that a(n)=A(n;1)*(B(n;-1)-C(n;-1))-B(n;1)*B(n;-1)+C(n;1)*(A(n;-1)-B(n;-1)+C(n;-1)), where A(n;1)=A077998(n), B(n;1)=A006054(n+1), C(n;1)=A006054(n), A(n;-1)=A121449(n), B(n+1;-1)=-A085810(n+1), C(n;-1)=A215404(n) and A(n;d), B(n;d), C(n;d), n in N, d in C, denote the quasi-Fibonacci numbers defined and discussed in comments in A121449 and in the cited paper. - Roman Witula, Aug 09 2012
From Wolfdieter Lang, Mar 30 2020: (Start)
With offset -4 this sequence 6, 1, 0, 0, 1, 5, ... appears in the formula for the n-th power of the 3 X 3 tridiagonal Matrix M_3 = Matrix([1,1,0], [1,2,1], [0,1,2]) from A332602: (M_3)^n = a(n-2)*(M_3)^2 - (6*a(n-3) - a(n-4))*M_3 + a(n-3)*1_3, with the 3 X 3 unit matrix 1_3, for n >= 0. Proof from Cayley-Hamilton: (M_3)^n = 5*(M_3)^3 - 6*M_3 + 1_3 (see A332602 for the characteristic polynomial Phi(3, x)), and the recurrence (M_3)^n = M_3*(M_3)^(n-1). For (M_3)^n[1,1] = 2*a(n-2) - 5*a(n-3) + a(n-4), for n >= 0, see A080937(n).
The formula for a(n) in terms of r = rho(7) = A160389 given below shows that a(n)/a(n-1) converges to rho(7)^2 = A116425 = 3.2469796... for n -> infinity. This is because r - 2/r = 0.692..., and r - 1 - 1/r = 0.137... .
(End)

References

  • W. Feller, An Introduction to Probability Theory and its Applications, 3rd ed, Wiley, New York, 1968, p. 96.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Double partial sums of A060557. Bisection of A052547.
Cf. A094789, A094790, A080937, A160389, A116425, A322602.

Programs

  • Magma
    I:=[1,5,19]; [n le 3 select I[n] else 5*Self(n-1)-6*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015
  • Maple
    a:=k->sum(binomial(5+2*k,7*j+k-2),j=ceil((2-k)/7)..floor((7+k)/7))-sum(binomial(5+2*k,7*j+k-1),j=ceil((1-k)/7)..floor((6+k)/7)): seq(a(k),k=0..25);
A005021:=-(z-1)*(z-5)/(-1+5*z-6*z**2+z**3); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence apart from the initial 1
  • Mathematica
    LinearRecurrence[{5,-6,1}, {1,5,19}, 50] (* Roman Witula, Aug 09 2012 *)
CoefficientList[Series[1/(1 - 5 x + 6 x^2 - x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 18 2015 *)
  • PARI
    x='x+O('x^30); Vec(1/(1-5*x+6*x^2-x^3)) \\ G. C. Greubel, Apr 19 2018

Formula

G.f.: 1/(1-5x+6x^2-x^3). - Emeric Deutsch, Apr 02 2004
a(n) = 5*a(n-1) -6*a(n-2) +a(n-3). - Emeric Deutsch, Apr 02 2004
a(n) = Sum_{j=-infinity..infinity} (binomial(5+2*k, 7*j+k-2) - binomial(5+2*k, 7*j+k-1)) (a finite sum).
a(n-2) = 2^n*C(n;1/2)=(1/7)*((c(2)-c(4))*(c(4))^(2n) + (c(4)-c(1))*(c(1))^(2n) + (c(1)-c(2))*(c(2))^(2n)), where a(-2)=a(-1):=0, c(j):=2*cos(2Pi*j/7). This formula follows from the Binet formula for C(n;d)--one of the quasi-Fibonacci numbers (see comments in A121449 and the formula (3.17) in the Witula-Slota-Warzynski paper). - Roman Witula, Aug 09 2012
In terms of the algebraic number r = rho(7) = 2*cos(Pi/7) = A160389 of degree 3 the preceding formula gives a(n) = r^(2*(n+2))*(A1(r) + A2(r)*(r - 2/r)^(2*(n+1)) = A3(r)*(r - 1 - 1/r)^(2*(n+1)))/7, for n >= -4 (see a comment above for this offset), with A1(r) = -r^2 + 2*r + 1, A2(r) = -r^2 - r + 2, and A3(r) = 2*r^2 - r - 3. - Wolfdieter Lang, Mar 30 2020

Extensions

a(25)-a(26) from Vincenzo Librandi, Sep 18 2015

A080937 Number of Catalan paths (nonnegative, starting and ending at 0, step +/-1) of 2*n steps with all values <= 5.

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 131, 417, 1341, 4334, 14041, 45542, 147798, 479779, 1557649, 5057369, 16420730, 53317085, 173118414, 562110290, 1825158051, 5926246929, 19242396629, 62479659622, 202870165265, 658715265222, 2138834994142, 6944753544643, 22549473023585
Offset: 0

Views

Author

Henry Bottomley, Feb 25 2003

Keywords

Comments

With interpolated zeros (1,0,1,0,2,...), counts closed walks of length n at start or end node of P_6. The sequence (0,1,0,2,...) counts walks of length n between the start and second node. - Paul Barry, Jan 26 2005
HANKEL transform of sequence and the sequence omitting a(0) is the sequence A130716. This is the unique sequence with that property. - Michael Somos, May 04 2012
From Wolfdieter Lang, Mar 30 2020: (Start)
a(n) is also the upper left entry of the n-th power of the 3 X 3 tridiagonal matrix M_3 = Matrix([1,1,0], [1,2,1], [0,1,2]) from A332602: a(n) = ((M_3)^n)[1,1].
Proof: (M_3)^n = b(n-2)*(M_3)^2 - (6*b(n-3) - b(n-4))*M_3 + b(n-3)*1_3, for n >= 0, with b(n) = A005021(n), for n >= -4. For the proof of this see a comment in A005021. Hence (M_3)^n[1,1] = 2*b(n-2) - 5*b(n-3) + b(n-4), for n >= 0. This proves the 3 X 3 part of the conjecture in A332602 by Gary W. Adamson.
The formula for a(n) given below in terms of r = rho(7) = A160389 proves that a(n)/a(n-1) converges to rho(7)^2 = A116425 = 3.2469796..., because r - 2/r = 0.6920... < 1, and r^2 - 3 = 0.2469... < 1. This limit was conjectured in A332602 by Gary W. Adamson.
(End)

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 131*x^6 + 417*x^7 + 1341*x^8 + ...

Links

Crossrefs

Cf. A000007, A000012, A001519, A007051, A011782, A024175, A080937, A080938.
Cf. A033191 which essentially provide the same sequence for different limits and tend to A000108.
Cf. A005021, A094790, A094789.
Cf. A211216, A005021, A160389, A116425, A332602.

Programs

  • Magma
    I:=[1,1,2]; [n le 3 select I[n] else 5*Self(n-1)-6*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jan 09 2016
  • Maple
    a:= n-> (<<0|1|0>, <0|0|1>, <1|-6|5>>^n. <<1, 1, 2>>)[1, 1]:
seq(a(n), n=0..35);  # Alois P. Heinz, Nov 09 2012
  • Mathematica
    nn=56;Select[CoefficientList[Series[(1-4x^2+3x^4)/(1-5x^2+6x^4-x^6), {x,0,nn}], x],#>0 &] (* Geoffrey Critzer, Jan 26 2014 *)
LinearRecurrence[{5,-6,1},{1,1,2},30] (* Jean-François Alcover, Jan 09 2016 *)
  • PARI
    a=vector(99); a[1]=1; a[2]=2;a[3]=5; for(n=4,#a,a[n]=5*a[n-1]-6*a[n-2] +a[n-3]); a \\ Charles R Greathouse IV, Jun 10 2011
  • PARI
    {a(n) = if( n<0, n = -n; polcoeff( (1 - 3*x + x^2) / (1 - 6*x + 5*x^2 - x^3) + x * O(x^n), n), polcoeff( (1 - 4*x + 3*x^2) / (1 - 5*x + 6*x^2 - x^3) + x * O(x^n), n))} /* Michael Somos, May 04 2012 */

Formula

a(n) = A080934(n,5).
G.f.: (1-4*x+3*x^2)/(1-5*x+6*x^2-x^3). - Ralf Stephan, May 13 2003
a(n) = 5*a(n-1) - 6*a(n-2) + a(n-3). - Herbert Kociemba, Jun 11 2004
a(n) = A096976(2*n). - Floor van Lamoen, Nov 02 2005
a(n) = (4/7-4/7*cos(1/7*Pi)^2)*(4*(cos(Pi/7))^2)^n + (1/7-2/7*cos(1/7*Pi) + 4/7*cos(1/7*Pi)^2)*(4*(cos(2*Pi/7))^2)^n + (2/7+2/7*cos(1/7*Pi))*(4*(cos(3*Pi/7))^2)^n for n>=0. - Richard Choulet, Apr 19 2010
G.f.: 1 / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x))))). - Michael Somos, May 04 2012
a(-n) = A038213(n). a(n + 2) * a(n) - a(n + 1)^2 = a(1 - n). Convolution inverse is A123183 with A123183(0)=1. - Michael Somos, May 04 2012
From Wolfdieter Lang, Mar 30 2020: (Start)
In terms of the algebraic number r = rho(7) = A160389 of degree 3 the formula given by Richard Choulet becomes a(n) = (1/7)*(r)^(2*n)*(C1(r) + C2(r)*(r - 2/r)^(2*n) + C3(r)*(r^2 - 3)^(2*n)), with C1(r) = 4 - r^2, C2(r) = 1 - r + r^2, and C3 = 2 + r.
a(n) = ((M_3)^n)[1,1] = 2*b(n-2) - 5*b(n-3) + b(n-4), for n >= 0, with the 3 X 3 tridiagonal matrix M_3 = Matrix([1,1,0], [1,2,1], [0,1,2]) from A332602, and b(n) = A005021(n) (with offset n >= -4). (End)

A094256 Expansion of x / ( (x-1)*(x^3 - 9*x^2 + 6*x - 1) ).

Original entry on oeis.org

1, 7, 34, 143, 560, 2108, 7752, 28101, 100947, 360526, 1282735, 4552624, 16131656, 57099056, 201962057, 714012495, 2523515514, 8916942687, 31504028992, 111295205284, 393151913464, 1388758662221, 4905479957435, 17327203698086, 61202661233823, 216176614077600
Offset: 1

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Author

Gary W. Adamson, Apr 25 2004

Keywords

Comments

Previous name was: Let M = the 4 X 4 matrix [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / -1 10 -15 7]. Perform M^n * [1 0 0 0] = [p q r s]. Then a(n-3), a(n-2), a(n-1), a(n) = -p, -q, -r, -s respectively.
a(n)/a(n-1) tends to 3.53208888624... = 4*cos^2(Pi/9), which is an eigenvalue of the matrix and a root of the polynomial x^4 - 6x^3 + 15x^2 -10x + 1 = 0 (having roots 4*cos^2(r*Pi/9), with r = 1,2,3,4).
Number of (s(0), s(1), ..., s(2n+4)) such that 0 < s(i) < 9 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+4, s(0) = 1, s(2n+4) = 7. - Herbert Kociemba, Jun 13 2004
From Wolfdieter Lang, Mar 27 2020: (Start)
This sequence, with offset -5, starting with -85, -10, -1, 0, 0, 0, 1, 7, ... appears in the formula for the n-th power of the 4 X 4 tridiagonal matrix given in A332602 as M_4 = matrix([1,1,0,0], [1,2,1,0], [0,1,2,1], [0,0,1,2]): (M_4)^n = a(n-2)*(M_4)^3 + b(n)*(M_4)^2 + c(n)*M_4 - a(n-3)*1_4, for n >= 0, with the 4 X 4 unit Matrix 1_4, b(n) = -15*a(n-3) + 10*a(n-4) - a(n-5), and c(n) = 10*a(n-3) - a(n-4). Proof from the characteristc polynomial of M_4 (see a comment in A332602) and the Cayley-Hamilton theorem.
From the proof that A094829(n+3)/A094829(n+2) -> rho(9)^2 = A332438 for n-> infinitiy, with rho(9) = 2*cos(Pi/9) = A332437 (see a comment in A094829), and a formula given below the same limit is obtained for a(n+1)/a(n) for n -> infinity, as stated in a comment above. (End)

Examples

			a(2), a(3), a(4), a(5) = 7, 34, 143, 560, since M^5 * [1 0 0 0] = [ -7 -34 -143 -560].
Cayley-Hamilton: (M_4)^5 = a(3)*(M_4)^3 + b(5)*(M_4)^2 + c(5)*M_4 - a(2)*1_4 = 34*(M_4)^3 - 95*(M_4)^2 + 69*M_4 - 7*1_4. - _Wolfdieter Lang_, Mar 27 2020

References

  • C. V. Durell and A. Robson, "Advanced Trigonometry", Dover 2003, p. 216.

Links

Crossrefs

a(n) = A005023(n-1), n > 1. - R. J. Mathar, Sep 05 2008
Cf. A080938, A094829, A332602, A332437, A332438.

Programs

  • Magma
    I:=[1,7,34,143]; [n le 4 select I[n] else 7*Self(n-1) - 15*Self(n-2) + 10*Self(n-3) - Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015
  • Mathematica
    Table[ (MatrixPower[{{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-1, 10, -15, 7}}, n].{-1, 0, 0, 0})[[4]], {n, 24}] (* Robert G. Wilson v, Apr 28 2004 *)
LinearRecurrence[{7, -15, 10, -1}, {1, 7, 34, 143}, 40] (* Vincenzo Librandi, Jul 25 2015 *)
  • PARI
    Vec(x / ( (x-1)*(x^3-9*x^2+6*x-1) ) + O(x^30)) \\ Michel Marcus, Jul 25 2015

Formula

From Herbert Kociemba, Jun 13 2004: (Start)
a(n) = (2/9)*Sum_{r=1..8} sin(r*Pi/9)*sin(7*r*Pi/9)*(2*cos(r*Pi/9))^(2n+4).
a(n) = 7*a(n-1) - 15*a(n-2) + 10*a(n-3) - a(n-4).
G.f.: x / ( (x-1)*(x^3 - 9*x^2 + 6*x - 1) ). (End)
3*a(n) = 1 - A094829(n+2) + 8*A094829(n+1) - A094829(n). - R. J. Mathar, Jun 29 2012 [offset corrected, and A094829(1) = 0. - Wolfdieter Lang, Mar 27 2020]
a(n) = (1/3)*(1 + 2*A094829(n+1) + 8*A094829(n) - A094829(n-1)), for n >= 1, with A094829(1) and A094829(0) = 0. - Wolfdieter Lang, Mar 27 2020

Extensions

More terms from Robert G. Wilson v, Apr 28 2004
a(25)-a(26) from Vincenzo Librandi, Jul 25 2015
New name (using g.f. from Herbert Kociemba) from Joerg Arndt, Jul 25 2015

A080938 Number of Catalan paths (nonnegative, starting and ending at 0, step +-1) of 2*n steps with all values less than or equal to 7.

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 132, 429, 1429, 4846, 16645, 57686, 201158, 704420, 2473785, 8704089, 30664890, 108126325, 381478030, 1346396146, 4753200932, 16783118309, 59266297613, 209302921830, 739203970773, 2610763825782, 9221050139566, 32568630376132
Offset: 0

Views

Author

Henry Bottomley, Feb 25 2003

Keywords

Comments

From Wolfdieter Lang, Mar 27 2020: (Start)
a(n) also gives the upper left entry of the n-th power of the 4 X 4 tridiagonal matrix M_4, given in A332602: M_4 = Matrix([1,1,0,0], [1,2,1,0], [0,1,2,1], [0,0,1,2]): a(n) = (M_4)^n[1,1]. Proof from the formula for (M_4)^n, given in a comment in A094256, derived from the Cayley-Hamilton theorem, which leads to the recurrence. The formula for a(n) in terms of A094256 is given below.
For A094256(n+1)/A094256(n), like for A094829(n+1)/A094829(n), the limit for n -> infinity is rho(9)^2 = A332438 = 3.53208888..., with rho(9) = 2*cos(Pi/9) = A332437. Therefore the formula of a(n) in terms of A094256 shows that the same limit is reached for a(n+1)/a(n). See this conjecture by Gary W. Adamson in A332602.
(End)

Examples

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

Links

Crossrefs

Cf. A000007, A000012, A011782, A001519, A007051, A080937, A024175, A033191 which essentially provide the same sequence for different limits and tend to A000108.
Cf. A211216, A094826 (first differences), A094829, A094829, A332602, A332437, A332438.

Programs

  • Magma
    I:=[1,1,2,5]; [n le 4 select I[n] else 7*Self(n-1)-15*Self(n-2)+10*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Nov 30 2018
  • Mathematica
    CoefficientList[Series[(1 - 2 x) (2 x^2 - 4 x + 1) / ((x - 1) (x^3 - 9 x^2 + 6 x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 30 2018 *)
LinearRecurrence[{7, -15, 10, -1}, {1, 1, 2, 5}, 30] (* Jean-François Alcover, Jan 07 2019 *)
  • PARI
    {a(n) = local(A); A = 1; for( i=1, 7, A = 1 / (1 - x*A)); polcoeff( A + x * O(x^n), n)} /* Michael Somos, May 12 2012 */

Formula

a(n) = A080934(n,7).
G.f.: -(2*x - 1)*(2*x^2 - 4*x + 1) / ( (x - 1)*(x^3 - 9*x^2 + 6*x - 1) ). - Ralf Stephan, May 13 2003
a(n) = 7*a(n-1) - 15*a(n-2) + 10*a(n-3) - a(n-4). - Herbert Kociemba, Jun 13 2004
G.f.: 1 / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x))))))). - Michael Somos, May 12 2012
a(n) = 5*b(n-2) - 21*b(n-3) + 19*b(n-4) - 2*b(n-5), for n >= 0, with b(n) = A094256(n), for n >= -5. See a comment in A094256 for this offset, and the above comment. - Wolfdieter Lang, Mar 28 2020
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