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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A339470 Decimal expansion of log(phi)^2, where phi is the golden ratio (A002390^2).

Original entry on oeis.org

2, 3, 1, 5, 6, 4, 8, 2, 0, 5, 7, 7, 1, 9, 4, 3, 9, 2, 4, 9, 6, 9, 2, 9, 0, 7, 1, 2, 3, 1, 5, 3, 2, 7, 6, 0, 0, 1, 6, 4, 0, 6, 3, 5, 0, 0, 4, 9, 2, 9, 8, 8, 7, 0, 8, 1, 5, 3, 0, 1, 2, 2, 8, 6, 8, 9, 7, 9, 5, 3, 4, 5, 5, 6, 6, 9, 6, 1, 8, 1, 2, 9, 8, 5, 0, 5, 4
Offset: 0

Views

Author

Robert Bilinski, Dec 06 2020

Keywords

Examples

			0.2315648205771943924969290712315327600164063500492988708153012286...

Links

Crossrefs

Cf. A002390, A001622, A086467.

Programs

  • Mathematica
    RealDigits[Log[GoldenRatio]^2, 10, 100][[1]] (* Amiram Eldar, Dec 06 2020 *)
  • PARI
    asinh(1/2)^2 \\ Michel Marcus, Dec 06 2020

Formula

Equals arcsinh(1/2)^2 = A002390^2.
Equals (1/2)*Sum_{k>=1} ((k!)^2*(-1)^(k+1))/((2*k)!*k^2) = A086467/2.
Equals (1/3)*(zeta(2) - Sum_{k>=1} ((k!)^2*(-1)^k)/((2*k)!*(2*k+1)^2)).
Equals (1/2)*Sum_{k>=1} (-1)^(k+1)/A002736(k).

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

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

Examples

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

References

  • The large number of references and links demonstrates the ubiquity of the Catalan numbers.
  • R. Alter, Some remarks and results on Catalan numbers, pp. 109-132 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 2, edited R. C. Mullin et al., 1971.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, many references.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 53.
  • J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, 1995, ch. 4, pp. 96-106.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 183, 196, etc.).
  • Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, Non-contiguous pattern avoidance in binary trees. Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227.
  • E. Deutsch, Dyck path enumeration, Discrete Math., 204, 167-202, 1999.
  • E. Deutsch and L. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31, 31-38, 2001.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, 207-208.
  • Tomislav Doslic and Darko Veljan, Logarithmic behavior of some combinatorial sequences. Discrete Math. 308 (2008), no. 11, 2182-2212. MR2404544 (2009j:05019)
  • S. Dulucq and J.-G. Penaud, Cordes, arbres et permutations. Discrete Math. 117 (1993), no. 1-3, 89-105.
  • A. Errera, Analysis situs - Un problème d'énumération, Mémoires Acad. Bruxelles, Classe des sciences, Série 2, Vol. XI, Fasc. 6, No. 1421 (1931), 26 pp.
  • Ehrenfeucht, Andrzej; Haemer, Jeffrey; Haussler, David. Quasimonotonic sequences: theory, algorithms and applications. SIAM J. Algebraic Discrete Methods 8 (1987), no. 3, 410-429. MR0897739 (88h:06026)
  • I. M. H. Etherington, Non-associate powers and a functional equation. The Mathematical Gazette, 21 (1937): 36-39; addendum 21 (1937), 153.
  • I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
  • I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
  • K. Fan, Structure of a Hecke algebra quotient, J. Amer. Math. Soc., 10 (1997), 139-167.
  • Susanna Fishel, Myrto Kallipoliti and Eleni Tzanaki, Facets of the Generalized Cluster Complex and Regions in the Extended Catalan Arrangement of Type A, The electronic Journal of Combinatorics 20(4) (2013), #P7.
  • D. Foata and D. Zeilberger, A classic proof of a recurrence for a very classical sequence, J. Comb Thy A 80 380-384 1997.
  • H. G. Forder, Some problems in combinatorics, Math. Gazette, vol. 45, 1961, 199-201.
  • Fürlinger, J.; Hofbauer, J., q-Catalan numbers. J. Combin. Theory Ser. A 40 (1985), no. 2, 248-264. MR0814413 (87e:05017)
  • M. Gardner, Time Travel and Other Mathematical Bewilderments, Chap. 20 pp. 253-266, W. H. Freeman NY 1988.
  • James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
  • M. C. Golumbic and A. N. Trenk, Tolerance graphs, Vol. 89, Cambridge University Press, 2004, pp. 32.
  • S Goodenough, C Lavault, Overview on Heisenberg—Weyl Algebra and Subsets of Riordan Subgroups, The Electronic Journal of Combinatorics, 22(4) (2015), #P4.16,
  • H. W. Gould, Research bibliography of two special number sequences, Mathematica Monongaliae, Vol. 12, 1971.
  • D. Gouyou-Beauchamps, Chemins sous-diagonaux et tableau de Young, pp. 112-125 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986.
  • M. Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 53-63 and 85-93.
  • J. L. Gross and J. Yellen, eds., Handbook of Graph Theory, CRC Press, 2004; p. 530.
  • N. S. S. Gu, N. Y. Li and T. Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
  • R. K. Guy, Dissecting a polygon into triangles, Research Paper #9, Math. Dept., Univ. Calgary, 1967.
  • R. K. Guy and J. L. Selfridge, The nesting and roosting habits of the laddered parenthesis. Amer. Math. Monthly 80 (1973), 868-876.
  • Peter Hajnal and Gabor V. Nagy, A bijective proof of Shapiro's Catalan convolution, Elect. J. Combin., 21 (2014), #P2.42.
  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 67, (3.3.23).
  • F. Harary, G. Prins, and W. T. Tutte, The number of plane trees. Indag. Math. 26, 319-327, 1964.
  • J. Harris, Algebraic Geometry: A First Course (GTM 133), Springer-Verlag, 1992, pages 245-247.
  • S. Heubach, N. Y. Li and T. Mansour, Staircase tilings and k-Catalan structures, Discrete Math., 308 (2008), 5954-5964.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • Higgins, Peter M. Combinatorial results for semigroups of order-preserving mappings. Math. Proc. Camb. Phil. Soc. (1993), 113: 281-296.
  • B. D. Hughes, Random Walks and Random Environments, Oxford 1995, vol. 1, p. 513, Eq. (7.282).
  • F. Hurtado, M. Noy, Ears of triangulations and Catalan numbers, Discrete Mathematics, Volume 149, Issues 1-3, Feb 22 1996, Pages 319-324.
  • M. Janjic, Determinants and Recurrence Sequences, Journal of Integer Sequences, 2012, Article 12.3.5.
  • R. H. Jeurissen, Raney and Catalan, Discrete Math., 308 (2008), 6298-6307.
  • M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 36.
  • Kim, Ki Hang; Rogers, Douglas G.; Roush, Fred W. Similarity relations and semiorders. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 577-594, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561081 (81i:05013)
  • Klarner, D. A. A Correspondence Between Sets of Trees. Indag. Math. 31, 292-296, 1969.
  • M. Klazar, On numbers of Davenport-Schinzel sequences, Discr. Math., 185 (1998), 77-87.
  • D. E. Knuth, The Art of Computer Programming, 2nd Edition, Vol. 1, Addison-Wesley, 1973, pp. 238.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.6 (p. 450).
  • Thomas Koshy and Mohammad Salmassi, "Parity and Primality of Catalan Numbers", College Mathematics Journal, Vol. 37, No. 1 (Jan 2006), pp. 52-53.
  • M. Kosters, A theory of hexaflexagons, Nieuw Archief Wisk., 17 (1999), 349-362.
  • E. Krasko, A. Omelchenko, Brown's Theorem and its Application for Enumeration of Dissections and Planar Trees, The Electronic Journal of Combinatorics, 22 (2015), #P1.17.
  • C. Krishnamachary and M. Bheemasena Rao, Determinants whose elements are Eulerian, prepared Bernoullian and other numbers, J. Indian Math. Soc., 14 (1922), 55-62, 122-138 and 143-146.
  • P. Lafar and C. T. Long, A combinatorial problem, Amer. Math. Mnthly, 69 (1962), 876-883.
  • Laradji, A. and Umar, A. On certain finite semigroups of order-decreasing transformations I, Semigroup Forum 69 (2004), 184-200.
  • P. J. Larcombe, On pre-Catalan Catalan numbers: Kotelnikow (1766), Mathematics Today, 35 (1999), p. 25.
  • P. J. Larcombe, On the history of the Catalan numbers: a first record in China, Mathematics Today, 35 (1999), p. 89.
  • P. J. Larcombe, The 18th century Chinese discovery of the Catalan numbers, Math. Spectrum, 32 (1999/2000), 5-7.
  • P. J. Larcombe and P. D. C. Wilson, On the trail of the Catalan sequence, Mathematics Today, 34 (1998), 114-117.
  • P. J. Larcombe and P. D. C. Wilson, On the generating function of the Catalan sequence: a historical perspective, Congress. Numer., 149 (2001), 97-108.
  • G. S. Lueker, Some techniques for solving recurrences, Computing Surveys, 12 (1980), 419-436.
  • J. J. Luo, Antu Ming, the first inventor of Catalan numbers in the world [in Chinese], Neimenggu Daxue Xuebao, 19 (1998), 239-245.
  • C. L. Mallows, R. J. Vanderbei, Which Young Tableaux Can Represent an Outer Sum?, Journal of Integer Sequences, Vol. 18, 2015, #15.9.1.
  • Toufik Mansour, Matthias Schork, and Mark Shattuck, Catalan numbers and pattern restricted set partitions. Discrete Math. 312(2012), no. 20, 2979-2991. MR2956089
  • Toufik Mansour and Simone Severini, Enumeration of (k,2)-noncrossing partitions, Discrete Math., 308 (2008), 4570-4577.
  • M. E. Mays and Jerzy Wojciechowski, A determinant property of Catalan numbers. Discrete Math. 211, No. 1-3, 125-133 (2000). Zbl 0945.05037
  • D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344.
  • A. Milicevic and N. Trinajstic, "Combinatorial Enumeration in Chemistry", Chem. Modell., Vol. 4, (2006), pp. 405-469.
  • Miller, Steven J., ed. Benford's Law: Theory and Applications. Princeton University Press, 2015.
  • David Molnar, "Wiggly Games and Burnside's Lemma", Chapter 8, The Mathematics of Various Entertaining Subjects: Volume 3 (2019), Jennifer Beineke & Jason Rosenhouse, eds. Princeton University Press, Princeton and Oxford, p. 102.
  • C. O. Oakley and R. J. Wisner, Flexagons, Amer. Math. Monthly, 64 (1957), 143-154.
  • A. Panholzer and H. Prodinger, Bijections for ternary trees and non-crossing trees, Discrete Math., 250 (2002), 181-195 (see Eq. 4).
  • Papoulis, Athanasios. "A new method of inversion of the Laplace transform."Quart. Appl. Math 14.405-414 (1957): 124.
  • S. G. Penrice, Stacks, bracketings and CG-arrangements, Math. Mag., 72 (1999), 321-324.
  • C. A. Pickover, Wonders of Numbers, Chap. 71, Oxford Univ. Press NY 2000.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • G. Pólya, On the number of certain lattice polygons. J. Combinatorial Theory 6 1969 102-105. MR0236031 (38 #4329)
  • C. Pomerance, Divisors of the middle binomial coefficient, Amer. Math. Monthly, 112 (2015), 636-644.
  • Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
  • Ronald C. Read, "The Graph Theorists who Count -- and What They Count", in 'The Mathematical Gardner', in D. A. Klarner, Ed., pp. 331-334, Wadsworth CA 1989.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 101.
  • J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222.
  • T. Santiago Costa Oliveira, "Catalan traffic" and integrals on the Grassmannian of lines, Discr. Math., 308 (2007), 148-152.
  • A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
  • E. Schröder, Vier combinatorische Probleme, Z. f. Math. Phys., 15 (1870), 361-376.
  • Shapiro, Louis W. Catalan numbers and "total information" numbers. Proceedings of the Sixth Southeastern Conference on Combinatorics, Graph Theory, and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1975), pp. 531-539. Congressus Numerantium, No. XIV, Utilitas Math., Winnipeg, Man., 1975. MR0398853 (53 #2704).
  • L. W. Shapiro, A short proof of an identity of Touchard's concerning Catalan numbers, J. Combin. Theory, A 20 (1976), 375-376.
  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
  • L. W. Shapiro, W.-J. Woan and S. Getu, The Catalan numbers via the World Series, Math. Mag., 66 (1993), 20-22.
  • D. M. Silberger, Occurrences of the integer (2n-2)!/n!(n-1)!, Roczniki Polskiego Towarzystwa Math. 13 (1969): 91-96.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • S. Snover and S. Troyer, Multidimensional Catalan numbers, Abstracts 848-05-94 and 848-05-95, 848th Meeting, Amer. Math. Soc., Worcester Mass., March 15-16, 1989.
  • R. P. Stanley, Enumerative Combinatorics, Wadsworth, Vol. 1, 1986, Vol. 2, 1999; see especially Chapter 6.
  • R. P. Stanley, Recent Progress in Algebraic Combinatorics, Bull. Amer. Math. Soc., 40 (2003), 55-68.
  • Richard P. Stanley, "Catalan Numbers", Cambridge University Press, 2015.
  • J. J. Sylvester, On reducible cyclodes, Coll. Math. Papers, Vol. 2, see especially page 670, where Catalan numbers appear.
  • Thiel, Marko. "A new cyclic sieving phenomenon for Catalan objects." Discrete Mathematics 340.3 (2017): 426-429.
  • I. Vun and P. Belcher, Catalan numbers, Mathematical Spectrum, 30 (1997/1998), 3-5.
  • D. Wells, Penguin Dictionary of Curious and Interesting Numbers, Entry 42 p 121, Penguin Books, 1987.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 41.
  • J. Wuttke, The zig-zag walk with scattering and absorption on the real half line and in a lattice model, J. Phys. A 47 (2014), 215203, 1-9.

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

A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

Original entry on oeis.org

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803
Offset: 0

Views

Author

N. J. A. Sloane, May 24 1994

Keywords

Comments

Cf. A000204 for Lucas numbers beginning with 1.
Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014
Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017
This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003
For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.
a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).
Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007
From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)
The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.
The first six multiplication formulas are:
L(2n) = L(n)^2 - 2*(-1)^n;
L(3n) = L(n)^3 - 3*(-1)^n*L(n);
L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;
L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);
L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.
Generally, L(n) | L(mn) if and only if m is odd.
In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:
For a >= b and odd b, L(a + b) + L(a - b) = 5*F(a)*F(b).
For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).
For a >= b and odd b, L(a + b) - L(a - b) = L(a)*L(b).
For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).
A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:
L(n) - L(n - 3) = 2*L(n - 2);
L(n) - L(n - 4) = 5*F(n - 2);
L(n) - L(n - 6) = 4*L(n - 3);
L(n) - L(n - 12) = 40*F(n - 6);
L(n) - L(n - 60) = 4160200*F(n - 30).
These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).
The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)
Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009
A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011
The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014
Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014
If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015
A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015
A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017
Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017
For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
From Klaus Purath, Apr 19 2019: (Start)
While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:
First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.
Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.
Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)
L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019
If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019
From Kai Wang, Dec 18 2019: (Start)
L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).
Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)
From Peter Bala, Dec 23 2021: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)
For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024
For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

Examples

			G.f. = 2 + x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + ...

References

  • P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 69.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 32,50.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 499.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 46.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 112, 202-203.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, pp. 287-288.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 148.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.
  • Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, 2001.
  • C. N. Menhinick, The Fibonacci Resonance and other new Golden Ratio discoveries, Onperson, (2015), pages 200-206.
  • Paulo Ribenboim, My Numbers, My Friends: Popular Lectures on Number Theory, Springer-Verlag, NY, 2000, p. 3.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 45-46, 59.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See pp. 83-84.

Links

Crossrefs

Cf. A000204. A000045(n) = (2*L(n + 1) - L(n))/5.
First row of array A103324.
a(n) = A101220(2, 0, n), for n > 0.
a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k - 1), for k > 0.
Cf. A131774, A001622, A002878 (L(2n+1)), A005248 (L(2n)), A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods), A057854 (complement), A354265 (generalized Lucas numbers).
Cf. sequences with formula Fibonacci(n+k)+Fibonacci(n-k) listed in A280154.
Subsequence of A047201.

Programs

  • Haskell
    a000032 n = a000032_list !! n
a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)
-- Reinhard Zumkeller, Aug 20 2011
  • Magma
    [Lucas(n): n in [0..120]];
  • Maple
    with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)
  • Mathematica
    a[0] := 2; a[n] := Nest[{Last[#], First[#] + Last[#]} &, {2, 1}, n] // Last
Array[2 Fibonacci[# + 1] - Fibonacci[#] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
Table[LucasL[n], {n, 0, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{1, 1}, {2, 1}, 40] (* Harvey P. Dale, Sep 07 2013 *)
LucasL[Range[0, 20]] (* Eric W. Weisstein, Aug 07 2017 *)
CoefficientList[Series[(-2 + x)/(-1 + x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
  • PARI
    {a(n) = if(n<0, (-1)^n * a(-n), if( n<2, 2-n, a(n-1) + a(n-2)))};
  • PARI
    {a(n) = if(n<0, (-1)^n * a(-n), polsym(x^2 - x - 1, n)[n+1])};
  • PARI
    {a(n) = real((2 + quadgen(5)) * quadgen(5)^n)};
  • PARI
    a(n)=fibonacci(n+1)+fibonacci(n-1) \\ Charles R Greathouse IV, Jun 11 2011
  • PARI
    polsym(1+x-x^2, 50) \\ Charles R Greathouse IV, Jun 11 2011
  • Python
    def A000032_gen(): # generator of terms
    a, b = 2, 1
    while True:
        yield a
        a, b = b, a+b
it = A000032_gen()
A000032_list = [next(it) for  in range(50)] # _Cole Dykstra, Aug 02 2022
  • Python
    from sympy import lucas
def A000032(n): return lucas(n) # Chai Wah Wu, Sep 23 2023
  • Python
    [(i:=3)+(j:=-1)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 02 2025
  • Sage
    [lucas_number2(n,1,-1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

Formula

G.f.: (2 - x)/(1 - x - x^2).
L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n = phi^n + (1-phi)^n.
L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).
L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - Jeff Burch, Dec 11 1999
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001
L(n) = F(n) + 2*F(n - 1) = F(n + 1) + F(n - 1). - Henry Bottomley, Apr 12 2000
a(n) = sqrt(F(n)^2 + 4*F(n + 1)*F(n - 1)). - Benoit Cloitre, Jan 06 2003 [Corrected by Gary Detlefs, Jan 21 2011]
a(n) = 2^(1 - n)*Sum_{k=0..floor(n/2)} C(n, 2k)*5^k. a(n) = 2T(n, i/2)( - i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = - 1. - Paul Barry, Nov 15 2003
L(n) = 2*F(n + 1) - F(n). - Paul Barry, Mar 22 2004
a(n) = (phi)^n + ( - phi)^( - n). - Paul Barry, Mar 12 2005
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:
L(n + m) + (-1)^m*L(n - m) = L(n)*L(m).
L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m).
L(n + m + k) + (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = L(n)*L(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*F(n)*L(m)*F(k).
L(n + m + k) + (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = 5*F(n)*F(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*L(n)*F(m)*F(k). (End)
Inverse: floor(log_phi(a(n)) + 1/2) = n, for n>1. Also for n >= 0, floor((1/2)*log_phi(a(n)*a(n+1))) = n. Extension valid for all integers n: floor((1/2)*sign(a(n)*a(n+1))*log_phi|a(n)*a(n+1)|) = n {where sign(x) = sign of x}. - Hieronymus Fischer, May 02 2007
Let f(n) = phi^n + phi^(-n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1) - 2*Sum_{k>=0} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide Colazingari, Jul 01 2016
Starting (1, 3, 4, 7, 11, ...) = row sums of triangle A131774. - Gary W. Adamson, Jul 14 2007
a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n. - Gary W. Adamson, Mar 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
For odd n: a(n) = floor(1/(fract(phi^n))); for even n>0: a(n) = ceiling(1/(1 - fract(phi^n))). This follows from the basic property of the golden ratio phi, which is phi - phi^(-1) = 1 (see general formula described in A001622).
a(n) = round(1/min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)
E.g.f.: exp(phi*x) + exp(-x/phi) with phi: = (1 + sqrt(5))/2 (golden section). 1/phi = phi - 1. See another form given in the Smiley e.g.f. comment. - Wolfdieter Lang, May 15 2010
L(n)/L(n - 1) -> A001622. - Vincenzo Librandi, Jul 17 2010
a(n) = 2*a(n-2) + a(n-3), n>2. - Gary Detlefs, Sep 09 2010
L(n) = floor(1/fract(Fibonacci(n)*phi)), for n odd. - Hieronymus Fischer, Oct 20 2010
L(n) = ceiling(1/(1 - fract(Fibonacci(n)*phi))), for n even. - Hieronymus Fischer, Oct 20 2010
L(n) = 2^n * (cos(Pi/5)^n + cos(3*Pi/5)^n). - Gary Detlefs, Nov 29 2010
L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)), n != 0. - Gary Detlefs, Dec 13 2010
L(n) = sqrt(A001254(n)) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n+1)). - Gary Detlefs, Dec 26 2010
L(n) = floor(phi^n) + ((-1)^n + 1)/2 = A014217(n) +((-1)^n+1)/2, where phi = A001622. - Gary Detlefs, Jan 20 2011
L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2), n>2. - Gary Detlefs, May 19 2011
For n >= 2, a(n) = round(phi^n) where phi is the golden ratio. - Arkadiusz Wesolowski, Jul 20 2012
a(p*k) == a(k) (mod p) for primes p. a(2^s*n) == a(n)^(2^s) (mod 2) for s = 0,1,2.. a(2^k) == - 1 (mod 2^k). a(p^2*k) == a(k) (mod p) for primes p and s = 0,1,2,3.. [Hoggatt and Bicknell]. - R. J. Mathar, Jul 24 2012
From Gary Detlefs, Dec 21 2012: (Start)
L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n.
L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k.
where phi = (1 + sqrt(5))/2, F(n) = A000045(n).
(End)
L(n) = n * Sum_{k=0..floor(n/2)} binomial(n - k,k)/(n - k), n>0 [H. W. Gould]. - Gary Detlefs, Jan 20 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(5*k-1))/((x*(5*k+4)) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
L(n) = F(n) + F(n-1) + F(n-2) + F(n-3). - Bob Selcoe, Jun 17 2013
L(n) = round(sqrt(L(2n-1) + L(2n-2))). - Richard R. Forberg, Jun 24 2014
L(n) = (F(n+1)^2 - F(n-1)^2)/F(n) for n>0. - Richard R. Forberg, Nov 17 2014
L(n+2) = 1 + A001610(n+1) = 1 + Sum_{k=0..n} L(k). - Tom Edgar, Apr 15 2015
L(i+j+1) = L(i)*F(j) + L(i+1)*F(j+1) with F(n)=A000045(n). - J. M. Bergot, Feb 12 2016
a(n) = (L(n+1)^2 + 5*(-1)^n)/L(n+2). - J. M. Bergot, Apr 06 2016
Dirichlet g.f.: PolyLog(s,-1/phi) + PolyLog(s,phi), where phi is the golden ratio. - Ilya Gutkovskiy, Jul 01 2016
L(n) = F(n+2) - F(n-2). - Yuchun Ji, Feb 14 2016
L(n+1) = A087131(n+1)/2^(n+1) = 2^(-n)*Sum_{k=0..n} binomial(n,k)*5^floor((k+1)/2). - Tony Foster III, Oct 14 2017
L(2*n) = (F(k+2*n) + F(k-2*n))/F(k); n >= 1, k >= 2*n. - David James Sycamore, May 04 2018
From Greg Dresden and Shaoxiong Yuan, Jul 16 2019: (Start)
L(3n + 4)/L(3n + 1) has continued fraction: n 4's followed by a single 7.
L(3n + 3)/L(3n) has continued fraction: n 4's followed by a single 2.
L(3n + 2)/L(3n - 1) has continued fraction: n 4's followed by a single -3. (End)
From Klaus Purath, Sep 15 2019: (Start)
All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).
L(n) = (2*L(n+2) - L(n-3))/5.
L(n) = (2*L(n-2) + L(n+3))/5.
L(n) = F(n-3) + 2*F(n).
L(n) = 2*F(n+2) - 3*F(n).
L(n) = (3*F(n-1) + F(n+2))/2.
L(n) = 3*F(n-3) + 4*F(n-2).
L(n) = 4*F(n+1) - F(n+3).
L(n) = (F(n-k) + F(n+k))/F(k) with odd k>0.
L(n) = (F(n+k) - F(n-k))/F(k) with even k>0.
L(n) = A001060(n-1) - F(n+1).
L(n) = (A022121(n-1) - F(n+1))/2.
L(n) = (A022131(n-1) - F(n+1))/3.
L(n) = (A022139(n-1) - F(n+1))/4.
L(n) = (A166025(n-1) - F(n+1))/5.
The following two formulas apply for all sequences of the Fibonacci type.
(a(n-2*k) + a(n+2*k))/a(n) = L(2*k).
(a(n+2*k+1) - a(n-2*k-1))/a(n) = L(2*k+1). (End)
L(n) = F(n-k)*L(k+1) + F(n-k-1)*L(k), for all k >= 0, where F(n) = A000045(n). - Michael Tulskikh, Dec 06 2019
F(n+2*m) = L(m)*F(n+m) + (-1)^(m-1)*F(n) for all n >= 0 and m >= 0. - Alexander Burstein, Mar 31 2022
a(n) = i^(n-1)*cos(n*c)/cos(c) = i^(n-1)*cos(c*n)*sec(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022
From Yike Li and Greg Dresden, Aug 25 2022: (Start)
L(2*n) = 5*binomial(2*n-1,n) - 2^(2*n-1) + 5*Sum_{j=1..n/5} binomial(2*n,n+5*j) for n>0.
L(2*n+1) = 2^(2n) - 5*Sum_{j=0..n/5} binomial(2*n+1,n+5*j+3). (End)
From Andrea Pinos, Jul 04 2023: (Start)
L(n) ~ Gamma(1/phi^n) + gamma.
L(n) = Re(phi^n + e^(i*Pi*n)/phi^n). (End)
L(n) = ((Sum_{i=0..n-1} L(i)^2) - 2)/L(n-1). - Jules Beauchamp, May 03 2025
From Peter Bala, Jul 09 2025: (Start)
The following series telescope:
For k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

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

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Cornelius Lanczos, Applied Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 517.
  • R. C. Mullin, E. Nemeth and P. J. Schellenberg, The enumeration of almost cubic maps, pp. 281-295 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 1, edited R. C. Mullin et al., 1970.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

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

Programs

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

Formula

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

A079586 Decimal expansion of Sum_{k>=1} 1/F(k) where F(k) is the k-th Fibonacci number A000045(k).

Original entry on oeis.org

3, 3, 5, 9, 8, 8, 5, 6, 6, 6, 2, 4, 3, 1, 7, 7, 5, 5, 3, 1, 7, 2, 0, 1, 1, 3, 0, 2, 9, 1, 8, 9, 2, 7, 1, 7, 9, 6, 8, 8, 9, 0, 5, 1, 3, 3, 7, 3, 1, 9, 6, 8, 4, 8, 6, 4, 9, 5, 5, 5, 3, 8, 1, 5, 3, 2, 5, 1, 3, 0, 3, 1, 8, 9, 9, 6, 6, 8, 3, 3, 8, 3, 6, 1, 5, 4, 1, 6, 2, 1, 6, 4, 5, 6, 7, 9, 0, 0, 8, 7, 2, 9, 7, 0, 4
Offset: 1

Views

Author

Benoit Cloitre, Jan 26 2003

Keywords

Comments

André-Jeannin proved that this constant is irrational.
This constant does not belong to the quadratic number field Q(sqrt(5)) (Bundschuh and Väänänen, 1994). - Amiram Eldar, Oct 30 2020

Examples

			3.35988566624317755317201130291892717968890513373...

References

  • Daniel Duverney, Number Theory, World Scientific, 2010, 5.22, pp.75-76.
  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 358.

Links

Crossrefs

Cf. A000045, A000796, A001622, A002163, A002390, A084119, A093540, A016628, A105199, A153386, A153387.
Cf. A350901, A350902, A350903, A350904.

Programs

  • Maple
    Digits := 120: c := Pi/2 + I*arccsch(2):
Jeannin := n -> sqrt(5/4)*add(I^(1-j)/sin(j*c), j = 1..n):
evalf(Jeannin(1000)); # Peter Luschny, Nov 15 2023
  • Mathematica
    digits = 105; Sqrt[5]*NSum[(-1)^n/(GoldenRatio^(2*n + 1) - (-1)^n), {n, 0, Infinity}, WorkingPrecision -> digits, NSumTerms -> digits] // RealDigits[#, 10, digits] & // First (* Jean-François Alcover, Apr 09 2013 *)
First@RealDigits[Sqrt[5]/4 ((Log[5] + 2 QPolyGamma[1, 1/GoldenRatio^4] - 4 QPolyGamma[1, 1/GoldenRatio^2])/(2 Log[GoldenRatio]) + EllipticTheta[2, 0, 1/GoldenRatio^2]^2), 10, 105] (* Vladimir Reshetnikov, Nov 18 2015 *)
  • PARI
    /* Fast computation without splitting into even and odd indices, see the Arndt reference */
lambert2(x, a, S)=
{
/* Return G(x,a) = Sum_{n>=1} a*x^n/(1-a*x^n) (generalized Lambert series)
   computed as Sum_{n=1..S} x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) )
   As series in x correct up to order S^2.
   We also have G(x,a) = Sum_{n>=1} a^n*x^n/(1-x^n) */
    return( sum(n=1,S, x^(n^2)*a^n*( 1/(1-x^n) + a*x^n/(1-a*x^n) ) ) );
}
inv_fib_sum(p=1, q=1, S)=
{
/* Return Sum_{n>=1} 1/f(n) where f(0)=0, f(1)=1, f(n) = p*f(n-1) + q*f(n-1)
   computed using generalized Lambert series.
   Must have p^2+4*q > 0 */
    my(al,be);
    \\ Note: the q here is -q in the Horadam paper.
    \\ The following numerical examples are for p=q=1:
    al=1/2*(p+sqrt(p^2+4*q));  \\ == +1.6180339887498...
    be=1/2*(p-sqrt(p^2+4*q));  \\ == -0.6180339887498...
    return( (al-be)*( 1/(al-1) + lambert2(be/al, 1/al, S) ) ); \\ == 3.3598856...
}
default(realprecision,100);
S = 1000; /* (be/al)^S == -0.381966^S == -1.05856*10^418 << 10^-100 */
inv_fib_sum(1,1,S) /* 3.3598856... */ /* Joerg Arndt, Jan 30 2011 */
  • PARI
    suminf(k=1, 1/(fibonacci(k))) \\ Michel Marcus, Feb 19 2019
  • Sage
    m=120; numerical_approx(sum(1/fibonacci(k) for k in (1..10*m)), digits=m) # G. C. Greubel, Feb 20 2019

Formula

Alternating series representation: 3 + Sum_{k >= 1} (-1)^(k+1)/(F(k)*F(k+1)*F(k+2)). - Peter Bala, Nov 30 2013
From Amiram Eldar, Oct 04 2020: (Start)
Equals sqrt(5) * Sum_{k>=0} (1/(phi^(2*k+1) - 1) - 2*phi^(2*k+1)/(phi^(4*(2*k+1)) - 1)), where phi is the golden ratio (A001622) (Greig, 1977).
Equals sqrt(5) * Sum_{k>=0} (-1)^k/(phi^(2*k+1) - (-1)^k) (Griffin, 1992).
Equals A153386 + A153387. (End)
From Gleb Koloskov, Sep 14 2021: (Start)
Equals 1 + c1*(c2 + 32*Integral_{x=0..infinity} f(x) dx),
where c1 = sqrt(5)/(8*log(phi)) = A002163/(8*A002390),
c2 = 2*arctan(2)+log(5) = 2*A105199+A016628,
phi = (1+sqrt(5))/2 = A001622,
f(x) = sin(x)*(4+cos(2*x))/((exp(Pi*x/log(phi))-1)*(2*cos(2*x)+3)*(7-2*cos(2*x))) (End)
From Amiram Eldar, Jan 27 2022: (Start)
Equals 3 + 2 * Sum_{k>=1} 1/(F(2*k-1)*F(2*k+1)*F(2*k+2)) (Bruckman, 1987).
Equals 2 + Sum_{k>=1} 1/A350901(k) (André-Jeannin, Problem H-450, 1991).
Equals lim_{n->oo} A350903(n)/(A350904(n)*A350902(n)) (André-Jeannin, 1991). (End)
Equals sqrt(5/4)*Sum_{j>=1} i^(1-j)/sin(j*c) where c = Pi/2 + i*arccsch(2). - Peter Luschny, Nov 15 2023
Equals lim_{n->oo} A203006(n)/A003266(n) (Z.W. M. Trzaska, 1996). - Raul Prisacariu, Sep 04 2024

A130233 a(n) is the maximal k such that Fibonacci(k) <= n (the "lower" Fibonacci Inverse).

Original entry on oeis.org

0, 2, 3, 4, 4, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
Offset: 0

Views

Author

Hieronymus Fischer, May 17 2007

Keywords

Comments

Inverse of the Fibonacci sequence (A000045), nearly, since a(Fibonacci(n)) = n except for n = 1 (see A130234 for another version). a(n) + 1 is equal to the partial sum of the Fibonacci indicator sequence (see A104162).

Examples

			a(10) = 6, since Fibonacci(6) = 8 <= 10 but Fibonacci(7) = 13 > 10.

Links

Crossrefs

Cf. A130235 (partial sums), A104162 (first differences).
Other related sequences: A000045, A130234, A130237, A130239, A130255, A130259, A108852. Lucas inverse: A130241.
Cf. A001622 (golden ratio), A002390 (its log).

Programs

  • Mathematica
    fibLLog[0] := 0; fibLLog[1] := 2; fibLLog[n_Integer] := fibLLog[n] = If[n < Fibonacci[fibLLog[n - 1] + 1], fibLLog[n - 1], fibLLog[n - 1] + 1]; Table[fibLLog[n], {n, 0, 88}] (* Alonso del Arte, Sep 01 2013 *)
  • PARI
    a(n)=log(sqrt(5)*n+1.5)\log((1+sqrt(5))/2) \\ Charles R Greathouse IV, Mar 21 2012

Formula

a(n) = floor(log_phi((sqrt(5)*n + sqrt(5*n^2+4))/2)) where phi = (1+sqrt(5))/2 = A001622.
a(n) = floor(arcsinh(sqrt(5)*n/2) / log(phi)), with log(phi) = A002390.
a(n) = A130234(n+1) - 1.
G.f.: g(x) = 1/(1-x) * Sum_{k>=1} x^Fibonacci(k).
a(n) = floor(log_phi(sqrt(5)*n+1)), n >= 0, where phi is the golden ratio. - Hieronymus Fischer, Jul 02 2007

A004146 Alternate Lucas numbers - 2.

Original entry on oeis.org

0, 1, 5, 16, 45, 121, 320, 841, 2205, 5776, 15125, 39601, 103680, 271441, 710645, 1860496, 4870845, 12752041, 33385280, 87403801, 228826125, 599074576, 1568397605, 4106118241, 10749957120, 28143753121, 73681302245, 192900153616, 505019158605, 1322157322201
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

This is the r=5 member in the r-family of sequences S_r(n) defined in A092184 where more information can be found.
Number of spanning trees of the wheel W_n on n+1 vertices. - Emeric Deutsch, Mar 27 2005
Also number of spanning trees of the n-helm graph. - Eric W. Weisstein, Jul 16 2011
a(n) is the smallest number requiring n terms when expressed as a sum of Lucas numbers (A000204). - David W. Wilson, Jan 10 2006
This sequence has a primitive prime divisor for all terms beyond the twelfth. - Anthony Flatters (Anthony.Flatters(AT)uea.ac.uk), Aug 17 2007
From Giorgio Balzarotti, Mar 11 2009: (Start)
Determinant of power series of gamma matrix with determinant 1:
a(n) = Determinant(A + A^2 + A^3 + A^4 + A^5 + ... + A^n)
where A is the submatrix A(1..2,1..2) of the matrix with factorial determinant
A = [[1,1,1,1,1,1,...],[1,2,1,2,1,2,...],[1,2,3,1,2,3,...],[1,2,3,4,1,2,...],
[1,2,3,4,5,1,...],[1,2,3,4,5,6,...],...]. Note: Determinant A(1..n,1..n)= (n-1)!.
See A158039, A158040, A158041, A158042, A158043, A158044, for sequences of matrix 2!,3!,... (End)
The previous comment could be rephrased as: a(n) = -det(A^n - I) where I is the 2 X 2 identity matrix and A = [1, 1; 1, 2]. - Peter Bala, Mar 20 2015
a(n) is also the number of points of Arnold's "cat map" that are on orbits of period n-1. This is a map of the two-torus T^2 into itself. If we regard T^2 as R^2 / Z^2, the action of this map on a two vector in R^2 is multiplication by the unit-determinant matrix A = [2, 1;1, 1], with the vector components taken modulo one. As such, an explicit formula for the n-th entry of this sequence is -det(I-A^n). - Bruce Boghosian, Apr 26 2009
7*a(n) gives the total number of vertices in a heptagonal hyperbolic lattice {7,3} with n total levels, in which an open heptagon is centered at the origin. - Robert M. Ziff, Apr 10 2011
The sequence is the case P1 = 5, P2 = 6, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014
Determinants of the spiral knots S(3,k,(1,-1)). a(k) = det(S(3,k,(1,-1))). These knots are also the weaving knots W(k,3) and the Turk's Head Links THK(3,k). - Ryan Stees, Dec 14 2014
Even-indexed Fibonacci numbers (1, 3, 8, 21, ...) convolved with (1, 2, 2, 2, 2, ...). - Gary W. Adamson, Aug 09 2016
a(n) is the number of ways to tile a bracelet of length n with 1-color square, 2-color dominos, 3-color trominos, etc. - Yu Xiao, May 23 2020
a(n) is the number of face-labeled unfoldings of a pyramid whose base is a simple n-gon. Cf. A103536. - Rick Mabry, Apr 17 2023

Examples

			For k=3, b(3) = sqrt(5)*b(2) - b(1) = 5 - 1 = 4, so det(S(3,3,(1,-1))) = 4^2 = 16.
G.f. = x + 5*x^2 + 16*x^3 + 45*x^4 + 121*x^5 + 320*x^4 + 841*x^5 + ... - _Michael Somos_, Feb 10 2023

References

  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (p. 193, Problem 3.3.40 (a)).
  • N. Hartsfield and G. Ringel, Pearls in Graph Theory, p. 102. Academic Press: 1990.
  • B. Hasselblatt and A. Katok, "Introduction to the Modern Theory of Dynamical Systems," Cambridge University Press, 1997.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

This is the r=5 member of the family S_r(n) defined in A092184.
Cf. A005248. Partial sums of A002878. Pairwise sums of A027941. Bisection of A074392.
Sequence A032170, the Möbius transform of this sequence, is then the number of prime periodic orbits of Arnold's cat map. - Bruce Boghosian, Apr 26 2009
Cf. A000032, A100047, A001350.
Cf. also A158039, A158040, A158041, A158042, A158043, A158044.
Cf. A103536 for the number of geometrically distinct edge-unfoldings of the regular pyramid. - Rick Mabry, Apr 17 2023

Programs

  • Magma
    [Lucas(n)-2: n in [0..60 by 2]]; // Vincenzo Librandi, Mar 20 2015
  • Mathematica
    Table[LucasL[2*n] - 2, {n, 0, 20}]
(* Second program: *)
LinearRecurrence[{4, -4, 1}, {0, 1, 5}, 30] (* Jean-François Alcover, Jan 08 2019 *)
  • PARI
    a(n) = { we = quadgen(5);((1+we)^n) + ((2-we)^n) - 2;} /* Michel Marcus, Aug 18 2012 */

Formula

a(n) = A005248(n) - 2 = A000032(2*n) - 2.
a(n+1) = 3*a(n) - a(n-1) + 2.
G.f.: x*(1+x)/(1-4*x+4*x^2-x^3) = x*(1+x)/((1-x)*(1-3*x+x^2)).
a(n) = 2*(T(n, 3/2)-1) with Chebyshev's polynomials T(n, x) of the first kind. See their coefficient triangle A053120.
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3), n>=3, a(0)=0, a(1)=1, a(2)=5.
a(n) = 2*T(n, 3/2) - 2, with twice the Chebyshev polynomials of the first kind, 2*T(n, x=3/2) = A005248(n).
a(n) = b(n) + b(n-1), n>=1, with b(n):=A027941(n-1), n>=1, b(-1):=0, the partial sums of S(n, 3) = U(n, 3/2) = A001906(n+1), with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind.
a(2n) = A000204(2n)^2 - 4 = 5*A000045(2n)^2; a(2n+1) = A000204(2n+1)^2. - David W. Wilson, Jan 10 2006
a(n) = ((3+sqrt(5))/2)^n + ((3-sqrt(5))/2)^n - 2. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jun 09 2001
a(n) = b(n-1) + b(n-2), n>=1, with b(n):=A027941(n), b(-1):=0, partial sums of S(n, 3) = U(n, 3/2) = A001906(n+1), Chebyshev's polynomials of the second kind.
a(n) = n*Sum_{k=1..n} binomial(n+k-1,2*k-1)/k, n > 0. - Vladimir Kruchinin, Sep 03 2010
a(n) = floor(tau^(2*n)*(tau^(2*n) - floor(tau^(2*n)))), where tau = (1+sqrt(5))/2. - L. Edson Jeffery, Aug 26 2013
From Peter Bala, Apr 03 2014: (Start)
a(n) = U(n-1,sqrt(5)/2)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -3/2; 1, 5/2] and T(n,x) denotes the Chebyshev polynomial of the first kind.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
a(k) = det(S(3,k,(1,-1))) = b(k)^2, where b(1)=1, b(2)=sqrt(5), b(k)=sqrt(5)*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + Sum_{n >= 1} Fibonacci(2*n)*x^n. Cf. A001350. - Peter Bala, Mar 19 2015
E.g.f.: exp(phi^2*x) + exp(x/phi^2) - 2*exp(x), where phi = (1 + sqrt(5))/2. - G. C. Greubel, Aug 24 2015
a(n) = a(-n) for all n in Z. - Michael Somos, Aug 27 2015
From Peter Bala, Jun 03 2016: (Start)
a(n) = Lucas(2*n) - Lucas(0*n);
a(n)^2 = Lucas(4*n) - 3*Lucas(2*n) + 3*Lucas(0*n) - Lucas(-2*n);
a(n)^3 = Lucas(6*n) - 5*Lucas(4*n) + 10*Lucas(2*n) - 10*Lucas(0*n) + 5*Lucas(-2*n) - Lucas(-4*n) and so on (follows from Binet's formula for Lucas(2*n) and the algebraic identity (x + 1/x - 2)^m = f(x) + f(1/x) where f(x) = (x - 1)^(2*m - 1)/x^(m-1) ). (End)
Limit_{n->infinity} a(n+1)/a(n) = (3 + sqrt(5))/2 = A104457. - Ilya Gutkovskiy, Jun 03 2016
a(n) = (phi^n - phi^(-n))^2, where phi = A001622 = (1 + sqrt(5))/2. - Diego Rattaggi, Jun 10 2020
a(n) = 4*sinh(n*A002390)^2, where A002390 = arcsinh(1/2). - Gleb Koloskov, Sep 18 2021
a(n) = 5*F(n)^2 = L(n)^2 - 4 if n even and a(n) = L(n)^2 = 5*F(n)^2 - 4 if n odd. - Michael Somos, Feb 10 2023
a(n) = n^2 + Sum_{i=1..n-1} i*a(n-i). - Fern Gossow, Dec 03 2024

Extensions

Correction to formula from Nephi Noble (nephi(AT)math.byu.edu), Apr 09 2002
Chebyshev comments from Wolfdieter Lang, Sep 10 2004

A139339 Decimal expansion of the square root of the golden ratio.

Original entry on oeis.org

1, 2, 7, 2, 0, 1, 9, 6, 4, 9, 5, 1, 4, 0, 6, 8, 9, 6, 4, 2, 5, 2, 4, 2, 2, 4, 6, 1, 7, 3, 7, 4, 9, 1, 4, 9, 1, 7, 1, 5, 6, 0, 8, 0, 4, 1, 8, 4, 0, 0, 9, 6, 2, 4, 8, 6, 1, 6, 6, 4, 0, 3, 8, 2, 5, 3, 9, 2, 9, 7, 5, 7, 5, 5, 3, 6, 0, 6, 8, 0, 1, 1, 8, 3, 0, 3, 8, 4, 2, 1, 4, 9, 8, 8, 4, 6, 0, 2, 5, 8, 5, 3, 8, 5, 1
Offset: 1

Views

Author

Mohammad K. Azarian, Apr 14 2008

Keywords

Comments

The hyperbolas x^2 - y^2 = 1 and xy = 1 meet at (c, 1/c) and (-c, -1/c), where c = sqrt(golden ratio); see the Mathematica program for a graph. - Clark Kimberling, Oct 19 2011
An algebraic integer of degree 4. Minimal polynomial: x^4 - x^2 - 1. - Charles R Greathouse IV, Jan 07 2013
Also the limiting value of the ratio of the slopes of the tangents drawn to the function y=sqrt(x) from the abscissa F(n) points (where F(n)=A000045(n) are the Fibonacci numbers and n > 0). - Burak Muslu, Apr 04 2021
The length of the base of the isosceles triangle of smallest perimeter which circumscribes a unit-diameter semicircle (DeTemple, 1992). - Amiram Eldar, Jan 22 2022
The unique real solution to arcsec(x) = arccot(x). - Wolfe Padawer, Apr 14 2023

Examples

			1.2720196495140689642524224617374914917156080418400...

References

  • B. Muslu, Sayılar ve Bağlantılar 2, Luna, 2021, pages 45-48.

Links

Crossrefs

Cf. A000045, A001622, A094214, A104457, A098317, A002390; A197762 (related intersection of hyperbolas).

Programs

  • Maple
    Digits:=100: evalf(sqrt((1+sqrt(5))/2)); # Muniru A Asiru, Sep 11 2018
  • Mathematica
    N[Sqrt[GoldenRatio], 100]
FindRoot[x*Sqrt[-1 + x^2] == 1, {x, 1.2, 1.3}, WorkingPrecision -> 110]
Plot[{Sqrt[-1 + x^2], 1/x}, {x, 0, 3}] (* Clark Kimberling, Oct 19 2011 *)
  • PARI
    sqrt((1+sqrt(5))/2) \\ Charles R Greathouse IV, Jan 07 2013
  • PARI
    a(n) = sqrtint(10^(2*n-2)*quadgen(5))%10; \\ Chittaranjan Pardeshi, Aug 24 2024

Formula

Equals sqrt((1 + sqrt(5))/2).
Equals 1/sqrt(A094214). - Burak Muslu, Apr 04 2021
From Amiram Eldar, Feb 07 2022: (Start)
Equals 1/A197762.
Equals tan(arccos(1/phi)).
Equals cot(arcsin(1/phi)). (End)
From Gerry Martens, Jul 30 2023: (Start)
Equals 5^(1/4)*cos(arctan(2)/2).
Equals Re(sqrt(1+2*i)) (the imaginary part is A197762). (End)

A086466 Decimal expansion of 2*sqrt(5)/5 arccsch(2).

Original entry on oeis.org

4, 3, 0, 4, 0, 8, 9, 4, 0, 9, 6, 4, 0, 0, 4, 0, 3, 8, 8, 8, 9, 4, 3, 3, 2, 3, 2, 9, 5, 0, 6, 0, 5, 4, 2, 5, 4, 2, 4, 5, 7, 0, 6, 8, 2, 5, 4, 0, 2, 8, 9, 6, 5, 4, 7, 5, 7, 0, 0, 6, 1, 0, 3, 9, 9, 2, 5, 6, 1, 2, 1, 5, 4, 6, 1, 1, 3, 1, 9, 6, 1, 3, 6, 1, 4, 9, 0, 2, 6, 4, 6, 9, 7, 2, 1, 9, 9, 5, 5, 4, 0, 6
Offset: 0

Views

Author

Eric W. Weisstein, Jul 21 2003

Keywords

Comments

Equals the value of the Dirichlet L-series of the non-principal character modulo 5 (A080891) at s=1. - Jianing Song, Nov 16 2019

Examples

			0.43040894096400403888943323295060542542457...

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.2, p. 7.

Links

Crossrefs

Cf. A086465, A086467, A086468.
Cf. A000032, A000045, A080891, A328717 (s=2), A344041.

Programs

  • Mathematica
    2*Log[GoldenRatio]/Sqrt[5] // RealDigits[#, 10, 102]& // First (* Jean-François Alcover, Apr 18 2014 *)
  • PARI
    2*log((1+sqrt(5))/2)/sqrt(5) \\ Stefano Spezia, Oct 15 2024

Formula

Equals Sum_{k>=1} (-1)^(k-1)/(k*binomial(2*k,k)).
Equals A010532 * A002390 / 10. - R. J. Mathar, Jul 26 2010
Also equals f'(0) = 2*log(phi)/sqrt(5), with f(x) = (phi^x-cos(Pi*x)*phi^-x)/sqrt(5), the real Fibonacci interpolating function. - Jean-François Alcover, Apr 04 2014
Equals Sum_{k>=1} A080891(k)/k = Sum_{k>=1} Kronecker(5,k)/k = 1 - 1/2 - 1/3 + 1/4 + 1/6 - 1/7 - 1/8 + 1/9 + ... - Jianing Song, Nov 16 2019
Equals Sum_{k>=1} F(k)/(k*2^(k+1)), where F(k) is the k-th Fibonacci number (A000045). - Amiram Eldar, Aug 10 2020
Sum_{k>=1} (2*k+1)*Lucas(k)/(k*(k+1)*2^k) = 10*c + 2 = 6.3040894096... where c is this constant (Seiffert, 1994). - Amiram Eldar, Jan 15 2022
Equals Sum_{k>=1} F(k)/(k*3^k), where F(k) is the k-th Fibonacci number (A000045). - Amiram Eldar, Jul 02 2023
Equals 1/Product_{p prime} (1 - Kronecker(5,p)/p), where Kronecker(5,p) = 0 if p = 5, 1 if p == 1 or 4 (mod 5) or -1 if p == 2 or 3 (mod 5). - Amiram Eldar, Dec 17 2023
Equals A344041/2. - Hugo Pfoertner, Oct 16 2024

A121839 Decimal expansion of Sum_{k>=1} 1/C(k), where C(k) is a Catalan Number (A000108).

Original entry on oeis.org

1, 8, 0, 6, 1, 3, 3, 0, 5, 0, 7, 7, 0, 7, 6, 3, 4, 8, 9, 1, 5, 2, 9, 2, 3, 6, 7, 0, 0, 6, 3, 1, 8, 0, 3, 2, 5, 4, 5, 9, 5, 8, 4, 9, 9, 9, 1, 5, 2, 3, 2, 9, 1, 4, 4, 6, 9, 7, 7, 2, 6, 6, 3, 7, 9, 5, 0, 2, 7, 6, 9, 6, 9, 3, 8, 9, 4, 9, 0, 6, 1, 4, 9, 7, 0, 7, 2, 2, 2, 1, 6, 9, 8, 3, 1, 3, 7, 8, 5, 2, 8, 2
Offset: 1

Views

Author

Alexander Adamchuk, Aug 28 2006

Keywords

Examples

			1.806133050770763489152923670063180325459584999152...

Links

Crossrefs

Cf. A000108, A002390, A268813 (essentially the same).

Programs

  • Magma
    SetDefaultRealField(RealField(100)); R:=RealField(); 1 + 4*Sqrt(3)*Pi(R)/27; // G. C. Greubel, Nov 04 2018
  • Maple
    evalf(1 + Sum((-1)^n*(2*n+1)/(9*n*(n+1)/2+1), n=0..infinity), 120); # Vaclav Kotesovec, May 31 2015
  • Mathematica
    RealDigits[N[Sum[n!(n + 1)!/(2n)!, {n, 1, Infinity}], 150]]
RealDigits[N[1+4*Sqrt[3]*Pi/27,100]][[1]]
  • PARI
    default(realprecision,100); 1 + 4*sqrt(3)*Pi/27

Formula

Reciprocal Catalan Constant C = 1 + 4*sqrt(3)*Pi/27.
This number is f(1) where f(x) = -1 + 2*(sqrt(4-x)*(8+x) + 12 * sqrt(x) * arctan(sqrt(x)/sqrt(4-x))) / sqrt((4-x)^5). This form corresponds to a generating function of the reciprocal Catalan numbers in the sense of Sprugnoli. - Juan M. Marquez, Mar 05 2009
Equals -1 + hypergeom([1,2],[1/2],1/4); note hypergeom([1,2],[1/2],x/4) = 1/1 + 1/1*x + 1/2*x^2 + 1/5*x^3 + 1/14*x^4 + 1/42*x^5 + ... is the g.f. for the inverse Catalan numbers (including C(0)). - Joerg Arndt, Apr 06 2013
From Vaclav Kotesovec, May 31 2015: (Start)
Equals 1 + Integral_{x=0..1} Product_{k>=1} (1-x^(9*k))^3 dx.
Equals 1 + Sum_{n>=0} (-1)^n * (2*n+1) / (9*n*(n+1)/2 + 1).
(End)
Equals 1 + Integral_{0..inf} x^3 BesselI_0(x) BesselK_0(x)^2 dx. - Jean-François Alcover, Jun 06 2016
From Amiram Eldar, Jul 05 2020: (Start)
Equals 1 + gamma(4/3)*gamma(5/3).
Equals 1 + Integral_{x=0..oo} dx/(1 + x^3)^2. (End)
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