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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A001006 Motzkin numbers: number of ways of drawing any number of nonintersecting chords joining n (labeled) points on a circle.

Original entry on oeis.org

1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559, 400763223, 1129760415, 3192727797, 9043402501, 25669818476, 73007772802, 208023278209, 593742784829, 1697385471211
Offset: 0

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Number of 4321-, (3412,2413)-, (3412,3142)- and 3412-avoiding involutions in S_n.
Number of sequences of length n-1 consisting of positive integers such that the first and last elements are 1 or 2 and the absolute difference between any 2 consecutive elements is 0 or 1. - Jon Perry, Sep 04 2003
From David Callan, Jul 15 2004: (Start)
Also number of Motzkin n-paths: paths from (0,0) to (n,0) in an n X n grid using only steps U = (1,1), F = (1,0) and D = (1,-1).
Number of Dyck n-paths with no UUU. (Given such a Dyck n-path, change each UUD to U, then change each remaining UD to F. This is a bijection to Motzkin n-paths. Example with n=5: U U D U D U U D D D -> U F U D D.)
Number of Dyck (n+1)-paths with no UDU. (Given such a Dyck (n+1)-path, mark each U that is followed by a D and each D that is not followed by a U. Then change each unmarked U whose matching D is marked to an F. Lastly, delete all the marked steps. This is a bijection to Motzkin n-paths. Example with n=6 and marked steps in small type: U U u d D U U u d d d D u d -> U U u d D F F u d d d D u d -> U U D F F D.) (End)
a(n) is the number of strings of length 2n+2 from the following recursively defined set: L contains the empty string and, for any strings a and b in L, we also find (ab) in L. The first few elements of L are e, (), (()), ((())), (()()), (((()))), ((()())), ((())()), (()(())) and so on. This proves that a(n) is less than or equal to C(n), the n-th Catalan number. See Orrick link (2024). - Saul Schleimer (saulsch(AT)math.rutgers.edu), Feb 23 2006 (Additional linked comment added by William P. Orrick, Jun 13 2024.)
a(n) = number of Dyck n-paths all of whose valleys have even x-coordinate (when path starts at origin). For example, T(4,2)=3 counts UDUDUUDD, UDUUDDUD, UUDDUDUD. Given such a path, split it into n subpaths of length 2 and transform UU->U, DD->D, UD->F (there will be no DUs for that would entail a valley with odd x-coordinate). This is a bijection to Motzkin n-paths. - David Callan, Jun 07 2006
Also the number of standard Young tableaux of height <= 3. - Mike Zabrocki, Mar 24 2007
a(n) is the number of RNA shapes of size 2n+2. RNA Shapes are essentially Dyck words without "directly nested" motifs of the form A[[B]]C, for A, B and C Dyck words. The first RNA Shapes are []; [][]; [][][], [[][]]; [][][][], [][[][]], [[][][]], [[][]][]; ... - Yann Ponty (ponty(AT)lri.fr), May 30 2007
The sequence is self-generated from top row A going to the left starting (1,1) and bottom row = B, the same sequence but starting (0,1) and going to the right. Take dot product of A and B and add the result to n-th term of A to get the (n+1)-th term of A. Example: a(5) = 21 as follows: Take dot product of A = (9, 4, 2, 1, 1) and (0, 1, 1, 2, 4) = (0, + 4 + 2 + 2 + 4) = 12; which is added to 9 = 21. - Gary W. Adamson, Oct 27 2008
Equals A005773 / A005773 shifted (i.e., (1,2,5,13,35,96,...) / (1,1,2,5,13,35,96,...)). - Gary W. Adamson, Dec 21 2008
Starting with offset 1 = iterates of M * [1,1,0,0,0,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
a(n) is the number of involutions of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(4)=9; indeed, p=3412=(13)(24) is the only involution of {1,2,3,4} with genus > 0. This follows easily from the fact that a permutation p of {1,2,...,n} has genus 0 if and only if the cycle decomposition of p gives a noncrossing partition of {1,2,...,n} and each cycle of p is increasing (see Lemma 2.1 of the Dulucq-Simion reference). [Also, redundantly, for p=3412=(13)(24) we have cp'=2341*3412=4123=(1432) and so g(p)=(1/2)(4+1-2-1)=1.] - Emeric Deutsch, May 29 2010
Let w(i,j,n) denote walks in N^2 which satisfy the multivariate recurrence w(i,j,n) = w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) with boundary conditions w(0,0,0) = 1 and w(i,j,n) = 0 if i or j or n is < 0. Then a(n) = Sum_{i = 0..n, j = 0..n} w(i,j,n) is the number of such walks of length n. - Peter Luschny, May 21 2011
a(n)/a(n-1) tends to 3.0 as N->infinity: (1+2*cos(2*Pi/N)) relating to longest odd N regular polygon diagonals, by way of example, N=7: Using the tridiagonal generator [cf. comment of Jan 07 2009], for polygon N=7, we extract an (N-1)/2 = 3 X 3 matrix, [0,1,0; 1,1,1; 0,1,1] with an e-val of 2.24697...; the longest Heptagon diagonal with edge = 1. As N tends to infinity, the diagonal lengths tend to 3.0, the convergent of the sequence. - Gary W. Adamson, Jun 08 2011
Number of (n+1)-length permutations avoiding the pattern 132 and the dotted pattern 23\dot{1}. - Jean-Luc Baril, Mar 07 2012
Number of n-length words w over alphabet {a,b,c} such that for every prefix z of w we have #(z,a) >= #(z,b) >= #(z,c), where #(z,x) counts the letters x in word z. The a(4) = 9 words are: aaaa, aaab, aaba, abaa, aabb, abab, aabc, abac, abca. - Alois P. Heinz, May 26 2012
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=1, r(k)<=k, and r(k)!=r(k-1); for example, the 9 RGS for n=4 are 1010, 1012, 1201, 1210, 1212, 1230, 1231, 1232, 1234. - Joerg Arndt, Apr 16 2013
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=0, r(k)<=k and r(k)-r(k-1) != 1; for example, the 9 RGS for n=4 are 0000, 0002, 0003, 0004, 0022, 0024, 0033, 0222, 0224. - Joerg Arndt, Apr 17 2013
Number of (4231,5276143)-avoiding involutions in S_n. - Alexander Burstein, Mar 05 2014
a(n) is the number of increasing unary-binary trees with n nodes that have an associated permutation that avoids 132. For more information about unary-binary trees with associated permutations, see A245888. - Manda Riehl, Aug 07 2014
a(n) is the number of involutions on [n] avoiding the single pattern p, where p is any one of the 8 (classical) patterns 1234, 1243, 1432, 2134, 2143, 3214, 3412, 4321. Also, number of (3412,2413)-, (3412,3142)-, (3412,2413,3142)-avoiding involutions on [n] because each of these 3 sets actually coincides with the 3412-avoiding involutions on [n]. This is a complete list of the 8 singles, 2 pairs, and 1 triple of 4-letter classical patterns whose involution avoiders are counted by the Motzkin numbers. (See Barnabei et al. 2011 reference.) - David Callan, Aug 27 2014
From Tony Foster III, Jul 28 2016: (Start)
A series created using 2*a(n) + a(n+1) has Hankel transform of F(2n), offset 3, F being the Fibonacci bisection, A001906 (empirical observation).
A series created using 2*a(n) + 3*a(n+1) + a(n+2) gives the Hankel transform of Sum_{k=0..n} k*Fibonacci(2*k), offset 3, A197649 (empirical observation). (End)
Conjecture: (2/n)*Sum_{k=1..n} (2k+1)*a(k)^2 is an integer for each positive integer n. - Zhi-Wei Sun, Nov 16 2017
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n." - Eric M. Schmidt, Dec 16 2017
Number of U_{k}-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are P-equivalent iff the positions of pattern P are identical in these paths. - Sergey Kirgizov, Apr 08 2018
If tau_1 and tau_2 are two distinct permutation patterns chosen from the set {132,231,312}, then a(n) is the number of valid hook configurations of permutations of [n+1] that avoid the patterns tau_1 and tau_2. - Colin Defant, Apr 28 2019
Number of permutations of length n that are sorted to the identity by a consecutive-321-avoiding stack followed by a classical-21-avoiding stack. - Colin Defant, Aug 29 2020
From Helmut Prodinger, Dec 13 2020: (Start)
a(n) is the number of paths in the first quadrant starting at (0,0) and consisting of n steps from the infinite set {(1,1), (1,-1), (1,-2), (1,-3), ...}.
For example, denoting U=(1,1), D=(1,-1), D_ j=(1,-j) for j >= 2, a(4) counts UUUU, UUUD, UUUD_2, UUUD_3, UUDU, UUDD, UUD_2U, UDUU, UDUD.
This step set is inspired by {(1,1), (1,-1), (1,-3), (1,-5), ...}, suggested by Emeric Deutsch around 2000.
See Prodinger link that contains a bijection to Motzkin paths. (End)
Named by Donaghey (1977) after the Israeli-American mathematician Theodore Motzkin (1908-1970). In Sloane's "A Handbook of Integer Sequences" (1973) they were called "generalized ballot numbers". - Amiram Eldar, Apr 15 2021
Number of Motzkin n-paths a(n) is split into A107587(n), number of even Motzkin n-paths, and A343386(n), number of odd Motzkin n-paths. The value A107587(n) - A343386(n) can be called the "shadow" of a(n) (see A343773). - Gennady Eremin, May 17 2021
Conjecture: If p is a prime of the form 6m+1 (A002476), then a(p-2) is divisible by p. Currently, no counterexample exists for p < 10^7. Personal communication from Robert Gerbicz: mod such p this is equivalent to A066796 with comment: "Every A066796(n) from A066796((p-1)/2) to A066796(p-1) is divisible by prime p of form 6m+1". - Serge Batalov, Feb 08 2022
From Rob Burns, Nov 11 2024: (Start)
The conjecture is proved in the 2017 paper by Rob Burns in the Links below. The result is contained in Tables 4 and 5 of the paper, which show that a(p-2) == 0 (mod p) when p == 1 (mod 6) and a(p-2) == -1 (mod p) when p == -1 (mod 6).
In fact, the 2017 paper by Burns establishes more general congruences for a(p^k - 2) where k >= 1.
If p == 1 (mod 6) then a(p^k - 2) == 0 (mod p) for k >= 1.
If p == -1 (mod 6) then a(p^k - 2) == -1 (mod p) when k is odd and a(p^k - 2) == 0 (mod p) when k is even.
These are consequences of the transitions provided in Tables 4, 5 and 6 of the paper.
The 2024 paper by Nadav Kohen also proves the conjecture. Proposition 6 of the paper states that a prime p divides a(p-2) if and only if p = (1 mod 3). (End)
From Peter Bala, Feb 10 2022: (Start)
Conjectures:
(1) For prime p == 1 (mod 6) and n, r >= 1, a(n*p^r - 2) == -A005717(n-1) (mod p), where we take A005717(0) = 0 to match Batalov's conjecture above.
(2) For prime p == 5 (mod 6) and n >= 1, a(n*p - 2) == -A005773(n) (mod p).
(3) For prime p >= 3 and k >= 1, a(n + p^k) == a(n) (mod p) for 0 <= n <= (p^k - 3).
(4) For prime p >= 5 and k >= 2, a(n + p^k) == a(n) (mod p^2) for 0 <= n <= (p^(k-1) - 3). (End)
The Hankel transform of this sequence with a(0) omitted gives the period-6 sequence [1, 0, -1, -1, 0, 1, ...] which is A010892 with its first term omitted, while the Hankel transform of the current sequence is the all-ones sequence A000012, and also it is the unique sequence with this property which is similar to the unique Hankel transform property of the Catalan numbers. - Michael Somos, Apr 17 2022
The number of terms in which the exponent of any variable x_i is not greater than 2 in the expansion of Product_{j=1..n} Sum_{i=1..j} x_i. E.g.: a(4) = 9: 3*x1^2*x2^2, 4*x1^2*x2*x3, 2*x1^2*x2*x4, x1^2*x3^2, x1^2*x3*x4, 2*x1*x2^2*x3, x1*x2^2*x4, x1*x2*x3^2, x1*x2*x3*x4. - Elif Baser, Dec 20 2024

Examples

			G.f.: 1 + x + 2*x^2 + 4*x^3 + 9*x^4 + 21*x^5 + 51*x^6 + 127*x^7 + 323*x^8 + ...
.
The 21 Motzkin-paths of length 5: UUDDF, UUDFD, UUFDD, UDUDF, UDUFD, UDFUD, UDFFF, UFUDD, UFDUD, UFDFF, UFFDF, UFFFD, FUUDD, FUDUD, FUDFF, FUFDF, FUFFD, FFUDF, FFUFD, FFFUD, FFFFF.
		

References

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  • F. R. Bernhart, Catalan, Motzkin, and Riordan numbers, Discr. Math., 204 (1999) 73-112.
  • R. Bojicic and M. D. Petkovic, Orthogonal Polynomials Approach to the Hankel Transform of Sequences Based on Motzkin Numbers, Bulletin of the Malaysian Mathematical Sciences, 2015, doi:10.1007/s40840-015-0249-3.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pp. 24, 298, 618, 912.
  • A. J. Bu, Automated counting of restricted Motzkin paths, Enumerative Combinatorics and Applications, ECA 1:2 (2021) Article S2R12.
  • Naiomi Cameron, JE McLeod, Returns and Hills on Generalized Dyck Paths, Journal of Integer Sequences, Vol. 19, 2016, #16.6.1.
  • L. Carlitz, Solution of certain recurrences, SIAM J. Appl. Math., 17 (1969), 251-259.
  • 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.
  • D. E. Davenport, L. W. Shapiro, and L. C. Woodson, The Double Riordan Group, The Electronic Journal of Combinatorics, 18(2) (2012), #P33.
  • E. Deutsch and L. Shapiro, A survey of the Fine numbers, Discrete Math., 241 (2001), 241-265.
  • T. Doslic, D. Svrtan, and D. Veljan, Enumerative aspects of secondary structures, Discr. Math., 285 (2004), 67-82.
  • 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 R. Simion, Combinatorial statistics on alternating permutations, J. Algebraic Combinatorics, 8, 1998, 169-191.
  • M. Dziemianczuk, "Enumerations of plane trees with multiple edges and Raney lattice paths." Discrete Mathematics 337 (2014): 9-24.
  • Wenjie Fang, A partial order on Motzkin paths, Discrete Math., 343 (2020), #111802.
  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (5.2.10).
  • N. S. S. Gu, N. Y. Li, and T. Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
  • Kris Hatch, Presentation of the Motzkin Monoid, Senior Thesis, Univ. Cal. Santa Barbara, 2012; http://ccs.math.ucsb.edu/senior-thesis/Kris-Hatch.pdf.
  • V. Jelinek, Toufik Mansour, and M. Shattuck, On multiple pattern avoiding set partitions, Advances in Applied Mathematics Volume 50, Issue 2, February 2013, pp. 292-326.
  • Hana Kim and R. P. Stanley, A refined enumeration of hex trees and related polynomials, http://www-math.mit.edu/~rstan/papers/hextrees.pdf, Preprint 2015.
  • S. Kitaev, Patterns in Permutations and Words, Springer-Verlag, 2011. See p. 399 Table A.7.
  • A. Kuznetsov et al., Trees associated with the Motzkin numbers, J. Combin. Theory, A 76 (1996), 145-147.
  • T. Lengyel, On divisibility properties of some differences of Motzkin numbers, Annales Mathematicae et Informaticae, 41 (2013) pp. 121-136.
  • W. A. Lorenz, Y. Ponty, and P. Clote, Asymptotics of RNA Shapes, Journal of Computational Biology. 2008, 15(1): 31-63. doi:10.1089/cmb.2006.0153.
  • Piera Manara and Claudio Perelli Cippo, The fine structure of 4321 avoiding involutions and 321 avoiding involutions, PU. M. A. Vol. 22 (2011), 227-238; http://www.mat.unisi.it/newsito/puma/public_html/22_2/manara_perelli-cippo.pdf.
  • Toufik Mansour, Restricted 1-3-2 permutations and generalized patterns, Annals of Combin., 6 (2002), 65-76.
  • Toufik Mansour, Matthias Schork, and Mark Shattuck, Catalan numbers and pattern restricted set partitions. Discrete Math. 312(2012), no. 20, 2979-2991. MR2956089.
  • T. S. Motzkin, Relations between hypersurface cross ratios and a combinatorial formula for partitions of a polygon, for permanent preponderance and for non-associative products, Bull. Amer. Math. Soc., 54 (1948), 352-360.
  • Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
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  • A. Sapounakis, I. Tasoulas, and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
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  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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  • Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
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  • Wen-Jin Woan, A combinatorial proof of a recursive relation of the Motzkin sequence by lattice paths. Fibonacci Quart. 40 (2002), no. 1, 3-8.
  • Wen-jin Woan, A Recursive Relation for Weighted Motzkin Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.1.6.
  • F. Yano and H. Yoshida, Some set partition statistics in non-crossing partitions and generating functions, Discr. Math., 307 (2007), 3147-3160.

Crossrefs

Bisections: A026945, A099250.
Sequences related to chords in a circle: A001006, A054726, A006533, A006561, A006600, A007569, A007678. See also entries for chord diagrams in Index file.
a(n) = A005043(n)+A005043(n+1).
A086246 is another version, although this is the main entry. Column k=3 of A182172.
Motzkin numbers A001006 read mod 2,3,4,5,6,7,8,11: A039963, A039964, A299919, A258712, A299920, A258711, A299918, A258710.
Cf. A004148, A004149, A023421, A023422, A023423, A290277 (inv. Euler Transf.).

Programs

  • Haskell
    a001006 n = a001006_list !! n
    a001006_list = zipWith (+) a005043_list $ tail a005043_list
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    # Three different Maple scripts for this sequence:
    A001006 := proc(n)
        add(binomial(n,2*k)*A000108(k),k=0..floor(n/2)) ;
    end proc:
    A001006 := proc(n) option remember; local k; if n <= 1 then 1 else procname(n-1) + add(procname(k)*procname(n-k-2),k=0..n-2); end if; end proc:
    # n -> [a(0),a(1),..,a(n)]
    A001006_list := proc(n) local w, m, j, i; w := proc(i,j,n) option remember;
    if min(i,j,n) < 0 or max(i,j) > n then 0
    elif n = 0 then if i = 0 and j = 0 then 1 else 0 fi else
    w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) fi end:
    [seq( add( add( w(i, j, m), i = 0..m), j = 0..m), m = 0..n)] end:
    A001006_list(29); # Peter Luschny, May 21 2011
  • Mathematica
    a[0] = 1; a[n_Integer] := a[n] = a[n - 1] + Sum[a[k] * a[n - 2 - k], {k, 0, n - 2}]; Array[a, 30]
    (* Second program: *)
    CoefficientList[Series[(1 - x - (1 - 2x - 3x^2)^(1/2))/(2x^2), {x, 0, 29}], x] (* Jean-François Alcover, Nov 29 2011 *)
    Table[Hypergeometric2F1[(1-n)/2, -n/2, 2, 4], {n,0,29}] (* Peter Luschny, May 15 2016 *)
    Table[GegenbauerC[n,-n-1,-1/2]/(n+1),{n,0,100}] (* Emanuele Munarini, Oct 20 2016 *)
    MotzkinNumber = DifferenceRoot[Function[{y, n}, {(-3n-3)*y[n] + (-2n-5)*y[n+1] + (n+4)*y[n+2] == 0, y[0] == 1, y[1] == 1}]];
    Table[MotzkinNumber[n], {n, 0, 29}] (* Jean-François Alcover, Oct 27 2021 *)
  • Maxima
    a[0]:1$
    a[1]:1$
    a[n]:=((2*n+1)*a[n-1]+(3*n-3)*a[n-2])/(n+2)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • Maxima
    M(n) := coeff(expand((1+x+x^2)^(n+1)),x^n)/(n+1);
    makelist(M(n),n,0,60); /* Emanuele Munarini, Apr 04 2012 */
    
  • Maxima
    makelist(ultraspherical(n,-n-1,-1/2)/(n+1),n,0,12); /* Emanuele Munarini, Oct 20 2016 */
    
  • PARI
    {a(n) = polcoeff( ( 1 - x - sqrt((1 - x)^2 - 4 * x^2 + x^3 * O(x^n))) / (2 * x^2), n)}; /* Michael Somos, Sep 25 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( x / (1 + x + x^2) + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp(x + x * O(x^n)) * besseli(1, 2 * x + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */
    
  • Python
    from gmpy2 import divexact
    A001006 = [1, 1]
    for n in range(2, 10**3):
        A001006.append(divexact(A001006[-1]*(2*n+1)+(3*n-3)*A001006[-2],n+2))
    # Chai Wah Wu, Sep 01 2014
    
  • Python
    def mot():
        a, b, n = 0, 1, 1
        while True:
            yield b//n
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
    A001006 = mot()
    print([next(A001006) for n in range(30)]) # Peter Luschny, May 16 2016
    
  • Python
    # A simple generator of Motzkin-paths (see the first comment of David Callan).
    C = str.count
    def aGen(n: int):
        a = [""]
        for w in a:
            if len(w) == n:
                if C(w, "U") == C(w, "D"): yield w
            else:
                for j in "UDF":
                    u = w + j
                    if C(u, "U") >= C(u, "D"): a += [u]
        return a
    for n in range(6):
        MP = [w for w in aGen(n)];
        print(len(MP), ":", MP)  # Peter Luschny, Dec 03 2024

Formula

G.f.: A(x) = ( 1 - x - (1-2*x-3*x^2)^(1/2) ) / (2*x^2).
G.f. A(x) satisfies A(x) = 1 + x*A(x) + x^2*A(x)^2.
G.f.: F(x)/x where F(x) is the reversion of x/(1+x+x^2). - Joerg Arndt, Oct 23 2012
a(n) = (-1/2) Sum_{i+j = n+2, i >= 0, j >= 0} (-3)^i*C(1/2, i)*C(1/2, j).
a(n) = (3/2)^(n+2) * Sum_{k >= 1} 3^(-k) * Catalan(k-1) * binomial(k, n+2-k). [Doslic et al.]
a(n) ~ 3^(n+1)*sqrt(3)*(1 + 1/(16*n))/((2*n+3)*sqrt((n+2)*Pi)). [Barcucci, Pinzani and Sprugnoli]
Limit_{n->infinity} a(n)/a(n-1) = 3. [Aigner]
a(n+2) - a(n+1) = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). [Bernhart]
a(n) = (1/(n+1)) * Sum_{i} (n+1)!/(i!*(i+1)!*(n-2*i)!). [Bernhart]
From Len Smiley: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k+1), inv. Binomial Transform of A000108.
a(n) = (1/(n+1))*Sum_{k=0..ceiling((n+1)/2)} binomial(n+1, k)*binomial(n+1-k, k-1);
D-finite with recurrence: (n+2)*a(n) = (2*n+1)*a(n-1) + (3*n-3)*a(n-2). (End)
a(n) = Sum_{k=0..n} C(n, 2k)*A000108(k). - Paul Barry, Jul 18 2003
E.g.f.: exp(x)*BesselI(1, 2*x)/x. - Vladeta Jovovic, Aug 20 2003
a(n) = A005043(n) + A005043(n+1).
The Hankel transform of this sequence gives A000012 = [1, 1, 1, 1, 1, 1, ...]. E.g., Det([1, 1, 2, 4; 1, 2, 4, 9; 2, 4, 9, 21; 4, 9, 21, 51]) = 1. - Philippe Deléham, Feb 23 2004
a(m+n) = Sum_{k>=0} A064189(m, k)*A064189(n, k). - Philippe Deléham, Mar 05 2004
a(n) = (1/(n+1))*Sum_{j=0..floor(n/3)} (-1)^j*binomial(n+1, j)*binomial(2*n-3*j, n). - Emeric Deutsch, Mar 13 2004
a(n) = A086615(n) - A086615(n-1) (n >= 1). - Emeric Deutsch, Jul 12 2004
G.f.: A(x)=(1-y+y^2)/(1-y)^2 where (1+x)*(y^2-y)+x=0; A(x)=4*(1+x)/(1+x+sqrt(1-2*x-3*x^2))^2; a(n)=(3/4)*(1/2)^n*Sum_(k=0..2*n, 3^(n-k)*C(k)*C(k+1, n+1-k) ) + 0^n/4 [after Doslic et al.]. - Paul Barry, Feb 22 2005
G.f.: c(x^2/(1-x)^2)/(1-x), c(x) the g.f. of A000108. - Paul Barry, May 31 2006
Asymptotic formula: a(n) ~ sqrt(3/4/Pi)*3^(n+1)/n^(3/2). - Benoit Cloitre, Jan 25 2007
a(n) = A007971(n+2)/2. - Zerinvary Lajos, Feb 28 2007
a(n) = (1/(2*Pi))*Integral_{x=-1..3} x^n*sqrt((3-x)*(1+x)) is the moment representation. - Paul Barry, Sep 10 2007
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, Phi([1]) is the Catalan numbers A000108. The present sequence is Phi([0,1,1]), see the 6th formula. - Gary W. Adamson, Oct 27 2008
G.f.: 1/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/.... (continued fraction). - Paul Barry, Dec 06 2008
G.f.: 1/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-.... (continued fraction). - Paul Barry, Feb 08 2009
a(n) = (-3)^(1/2)/(6*(n+2)) * (-1)^n*(3*hypergeom([1/2, n+1],[1],4/3) - hypergeom([1/2, n+2],[1],4/3)). - Mark van Hoeij, Nov 12 2009
G.f.: 1/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-... (continued fraction). - Paul Barry, Mar 02 2010
G.f.: 1/(1-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, Jan 26 2011 [Adds apparently a third '1' in front. - R. J. Mathar, Jan 29 2011]
Let A(x) be the g.f., then B(x)=1+x*A(x) = 1 + 1*x + 1*x^2 + 2*x^3 + 4*x^4 + 9*x^5 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+x) (continued fraction); more generally B(x)=C(x/(1+x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = (2/Pi)*Integral_{x=-1..1} (1+2*x)^n*sqrt(1-x^2). - Peter Luschny, Sep 11 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = 1/2/(x^2)-1/2/x-1/2/(x^2)*G(0); G(k) = 1+(4*k-1)*x*(2+3*x)/(4*k+2-x*(2+3*x)*(4*k+1)*(4*k+2) /(x*(2+3*x)*(4*k+1)+(4*k+4)/G(k+1))), if -1 < x < 1/3; (continued fraction). - Sergei N. Gladkovskii, Dec 01 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = (-1 + 1/G(0))/(2*x); G(k) = 1-2*x/(1+x/(1+x/(1-2*x/(1-x/(2-x/G(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, Dec 11 2011
0 = a(n) * (9*a(n+1) + 15*a(n+2) - 12*a(n+3)) + a(n+1) * ( -3*a(n+1) + 10*a(n+2) - 5*a(n+3)) + a(n+2) * (a(n+2) + a(n+3)) unless n=-2. - Michael Somos, Mar 23 2012
a(n) = (-1)^n*hypergeometric([-n,3/2],[3],4). - Peter Luschny, Aug 15 2012
Representation in terms of special values of Jacobi polynomials P(n,alpha,beta,x), in Maple notation: a(n)= 2*(-1)^n*n!*JacobiP(n,2,-3/2-n,-7)/(n+2)!, n>=0. - Karol A. Penson, Jun 24 2013
G.f.: Q(0)/x - 1/x, where Q(k) = 1 + (4*k+1)*x/((1+x)*(k+1) - x*(1+x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1+x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
Catalan(n+1) = Sum_{k=0..n} binomial(n,k)*a(k). E.g.: 42 = 1*1 + 4*1 + 6*2 + 4*4 + 1*9. - Doron Zeilberger, Mar 12 2015
G.f. A(x) with offset 1 satisfies: A(x)^2 = A( x^2/(1-2*x) ). - Paul D. Hanna, Nov 08 2015
a(n) = GegenbauerPoly(n,-n-1,-1/2)/(n+1). - Emanuele Munarini, Oct 20 2016
a(n) = a(n-1) + A002026(n-1). Number of Motzkin paths that start with an F step plus number of Motzkin paths that start with an U step. - R. J. Mathar, Jul 25 2017
G.f. A(x) satisfies A(x)*A(-x) = F(x^2), where F(x) is the g.f. of A168592. - Alexander Burstein, Oct 04 2017
G.f.: A(x) = exp(int((E(x)-1)/x dx)), where E(x) is the g.f. of A002426. Equivalently, E(x) = 1 + x*A'(x)/A(x). - Alexander Burstein, Oct 05 2017
G.f. A(x) satisfies: A(x) = Sum_{j>=0} x^j * Sum_{k=0..j} binomial(j,k)*x^k*A(x)^k. - Ilya Gutkovskiy, Apr 11 2019
From Gennady Eremin, May 08 2021: (Start)
G.f.: 2/(1 - x + sqrt(1-2*x-3*x^2)).
a(n) = A107587(n) + A343386(n) = 2*A107587(n) - A343773(n) = 2*A343386(n) + A343773(n). (End)
Revert transform of A049347 (after Michael Somos). - Gennady Eremin, Jun 11 2021
Sum_{n>=0} 1/a(n) = 2.941237337631025604300320152921013604885956025483079699366681494505960039781389... - Vaclav Kotesovec, Jun 17 2021
Let a(-1) = (1 - sqrt(-3))/2 and a(n) = a(-3-n)*(-3)^(n+3/2) for all n in Z. Then a(n) satisfies my previous formula relation from Mar 23 2012 now for all n in Z. - Michael Somos, Apr 17 2022
Let b(n) = 1 for n <= 1, otherwise b(n) = Sum_{k=2..n} b(k-1) * b(n-k), then a(n) = b(n+1) (conjecture). - Joerg Arndt, Jan 16 2023
From Peter Bala, Feb 03 2024: (Start)
G.f.: A(x) = 1/(1 + x)*c(x/(1 + x))^2 = 1 + x/(1 + x)*c(x/(1 + x))^3, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
A(x) = 1/(1 - 3*x)*c(-x/(1 -3*x))^2.
a(n+1) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n, k)*A000245(k+1).
a(n) = 3^n * Sum_{k = 0..n} (-3)^(-k)*binomial(n, k)*Catalan(k+1).
a(n) = 3^n * hypergeom([3/2, -n], [3], 4/3). (End)
G.f. A(x) satisfies A(x) = exp( x*A(x) + Integral x*A(x)/(1 - x^2*A(x)) dx ). - Paul D. Hanna, Mar 04 2024
a(n) = hypergeom([-n/2,1/2-n/2],[2],4). - Karol A. Penson, May 18 2025

A002426 Central trinomial coefficients: largest coefficient of (1 + x + x^2)^n.

Original entry on oeis.org

1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, 8953, 25653, 73789, 212941, 616227, 1787607, 5196627, 15134931, 44152809, 128996853, 377379369, 1105350729, 3241135527, 9513228123, 27948336381, 82176836301, 241813226151, 712070156203, 2098240353907, 6186675630819
Offset: 0

Views

Author

Keywords

Comments

Number of ordered trees with n + 1 edges, having root of odd degree and nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 02 2002
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), running from (0,0) to (n,0) (i.e., grand Motzkin paths of length n). For example, a(3) = 7 because we have HHH, HUD, HDU, UDH, DUH, UHD and DHU. - Emeric Deutsch, May 31 2003
Number of lattice paths from (0,0) to (n,n) using steps (2,0), (0,2), (1,1). It appears that 1/sqrt((1 - x)^2 - 4*x^s) is the g.f. for lattice paths from (0,0) to (n,n) using steps (s,0), (0,s), (1,1). - Joerg Arndt, Jul 01 2011
Number of lattice paths from (0,0) to (n,n) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Binomial transform of A000984, with interpolated zeros. - Paul Barry, Jul 01 2003
Number of leaves in all 0-1-2 trees with n edges, n > 0. (A 0-1-2 tree is an ordered tree in which every vertex has at most two children.) - Emeric Deutsch, Nov 30 2003
a(n) is the number of UDU-free paths of n + 1 upsteps (U) and n downsteps (D) that start U. For example, a(2) = 3 counts UUUDD, UUDDU, UDDUU. - David Callan, Aug 18 2004
Diagonal sums of triangle A063007. - Paul Barry, Aug 31 2004
Number of ordered ballots from n voters that result in an equal number of votes for candidates A and B in a three candidate election. Ties are counted even when candidates A and B lose the election. For example, a(3) = 7 because ballots of the form (voter-1 choice, voter-2 choice, voter-3 choice) that result in equal votes for candidates A and B are the following: (A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A) and (C,C,C). - Dennis P. Walsh, Oct 08 2004
a(n) is the number of weakly increasing sequences (a_1,a_2,...,a_n) with each a_i in [n]={1,2,...,n} and no element of [n] occurring more than twice. For n = 3, the sequences are 112, 113, 122, 123, 133, 223, 233. - David Callan, Oct 24 2004
Note that n divides a(n+1) - a(n). In fact, (a(n+1) - a(n))/n = A007971(n+1). - T. D. Noe, Mar 16 2005
Row sums of triangle A105868. - Paul Barry, Apr 23 2005
Number of paths of length n with steps U = (1,1), D = (1,-1) and H = (1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having no H steps on the x-axis. Example: a(3) = 7 because we have UDU, UHD, UHH, UHU, UUD, UUH and UUU. - Emeric Deutsch, Oct 07 2007
Equals right border of triangle A152227; starting with offset 1, the row sums of triangle A152227. - Gary W. Adamson, Nov 29 2008
Starting with offset 1 = iterates of M * [1,1,1,...] where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
a(n) is prime for n = 2, 3 and 4, with no others for n <= 10^5 (E. W. Weisstein, Mar 14 2005). It has apparently not been proved that no [other] prime central trinomials exist. - Jonathan Vos Post, Mar 19 2010
a(n) is not divisible by 3 for n whose base-3 representation contains no 2 (A005836).
a(n) = number of (n-1)-lettered words in the alphabet {1,2,3} with as many occurrences of the substring (consecutive subword) [1,2] as those of [2,1]. See the papers by Ekhad-Zeilberger and Zeilberger. - N. J. A. Sloane, Jul 05 2012
a(n) = coefficient of x^n in (1 + x + x^2)^n. - L. Edson Jeffery, Mar 23 2013
a(n) is the number of ordered pairs (A,B) of subsets of {1,2,...,n} such that (i.) A and B are disjoint and (ii.) A and B contain the same number of elements. For example, a(2) = 3 because we have: ({},{}) ; ({1},{2}) ; ({2},{1}). - Geoffrey Critzer, Sep 04 2013
Also central terms of A082601. - Reinhard Zumkeller, Apr 13 2014
a(n) is the number of n-tuples with entries 0, 1, or 2 and with the sum of entries equal to n. For n=3, the seven 3-tuples are (1,1,1), (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1), and (2,1,0). - Dennis P. Walsh, May 08 2015
The series 2*a(n) + 3*a(n+1) + a(n+2) = 2*A245455(n+3) has Hankel transform of L(2n+1)*2^n, offset n = 1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
The series (2*a(n) + 3*a(n+1) + a(n+2))/2 = A245455(n+3) has Hankel transform of L(2n+1), offset n=1, L being a Lucas number, see A002878 (empirical observation). - Tony Foster III, Sep 05 2016
Conjecture: An integer n > 3 is prime if and only if a(n) == 1 (mod n^2). We have verified this for n up to 8*10^5, and proved that a(p) == 1 (mod p^2) for any prime p > 3 (cf. A277640). - Zhi-Wei Sun, Nov 30 2016
This is the analog for Coxeter type B of Motzkin numbers (A001006) for Coxeter type A. - F. Chapoton, Jul 19 2017
a(n) is also the number of solutions to the equation x(1) + x(2) + ... + x(n) = 0, where x(1), ..., x(n) are in the set {-1,0,1}. Indeed, the terms in (1 + x + x^2)^n that produce x^n are of the form x^i(1)*x^i(2)*...*x^i(n) where i(1), i(2), ..., i(n) are in {0,1,2} and i(1) + i(2) + ... + i(n) = n. By setting j(t) = i(t) - 1 we obtain that j(1), ..., j(n) satisfy j(1) + ... + j(n) =0 and j(t) in {-1,0,1} for all t = 1..n. - Lucien Haddad, Mar 10 2018
If n is a prime greater than 3 then a(n)-1 is divisible by n^2. - Ira M. Gessel, Aug 08 2021
Let f(m) = ceiling((q+log(q))/log(9)), where q = -log(log(27)/(2*m^2*Pi)) then f(a(n)) = n, for n > 0. - Miko Labalan, Oct 07 2024
Diagonal of the rational function 1 / (1 - x^2 - y^2 - x*y). - Ilya Gutkovskiy, Apr 23 2025

Examples

			For n = 2, (x^2 + x + 1)^2 = x^4 + 2*x^3 + 3*x^2 + 2*x + 1, so a(2) = 3. - _Michael B. Porter_, Sep 06 2016
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 78 and 163, #19.
  • L. Euler, Exemplum Memorabile Inductionis Fallacis, Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 15, p. 59.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 575.
  • P. Henrici, Applied and Computational Complex Analysis. Wiley, NY, 3 vols., 1974-1986. (Vol. 1, p. 42.)
  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 579.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 74.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 6.3.8.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 22.
  • Lin Yang and S.-L. Yang, The parametric Pascal rhombus. Fib. Q., 57:4 (2019), 337-346. See p. 341.

Crossrefs

INVERT transform is A007971. Partial sums are A097893. Squares are A168597.
Main column of A027907. Column k=2 of A305161. Column k=0 of A328347. Column 1 of A201552(?).
Cf. A001006, A002878, A005043, A005717, A082758 (bisection), A273055 (bisection), A102445, A113302, A113303, A113304, A113305 (divisibility of central trinomial coefficients), A152227, A277640.

Programs

  • Haskell
    a002426 n = a027907 n n  -- Reinhard Zumkeller, Jan 22 2013
    
  • Magma
    P:=PolynomialRing(Integers()); [Max(Coefficients((1+x+x^2)^n)): n in [0..26]]; // Bruno Berselli, Jul 05 2011
    
  • Maple
    A002426 := proc(n) local k;
        sum(binomial(n, k)*binomial(n-k, k), k=0..floor(n/2));
    end proc: # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    # Alternatively:
    a := n -> simplify(GegenbauerC(n,-n,-1/2)):
    seq(a(n), n=0..29); # Peter Luschny, May 07 2016
  • Mathematica
    Table[ CoefficientList[ Series[(1 + x + x^2)^n, {x, 0, n}], x][[ -1]], {n, 0, 27}] (* Robert G. Wilson v *)
    a=b=1; Join[{a,b}, Table[c=((2n-1)b + 3(n-1)a)/n; a=b; b=c; c, {n,2,100}]]; Table[Sqrt[-3]^n LegendreP[n,1/Sqrt[-3]],{n,0,26}] (* Wouter Meeussen, Feb 16 2013 *)
    a[ n_] := If[ n < 0, 0, 3^n Hypergeometric2F1[ 1/2, -n, 1, 4/3]]; (* Michael Somos, Jul 08 2014 *)
    Table[4^n *JacobiP[n,-n-1/2,-n-1/2,-1/2], {n,0,29}] (* Peter Luschny, May 13 2016 *)
    a[n_] := a[n] = Sum[n!/((n - 2*i)!*(i!)^2), {i, 0, n/2}]; Table[a[n], {n, 0, 29}] (* Shara Lalo and Zagros Lalo, Oct 03 2018 *)
  • Maxima
    trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
    makelist(trinomial(n,n),n,0,12); /* Emanuele Munarini, Mar 15 2011 */
    
  • Maxima
    makelist(ultraspherical(n,-n,-1/2),n,0,12); /* Emanuele Munarini, Dec 20 2016 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, n))};
    
  • PARI
    /* as lattice paths: same as in A092566 but use */
    steps=[[2, 0], [0, 2], [1, 1]];
    /* Joerg Arndt, Jul 01 2011 */
    
  • PARI
    a(n)=polcoeff(sum(m=0, n, (2*m)!/m!^2 * x^(2*m) / (1-x+x*O(x^n))^(2*m+1)), n) \\ Paul D. Hanna, Sep 21 2013
    
  • Python
    from math import comb
    def A002426(n): return sum(comb(n,k)*comb(k,n-k) for k in range(n+1)) # Chai Wah Wu, Nov 15 2022
  • Sage
    A002426 = lambda n: hypergeometric([-n/2, (1-n)/2], [1], 4)
    [simplify(A002426(n)) for n in (0..29)]
    # Peter Luschny, Sep 17 2014
    
  • Sage
    def A():
        a, b, n = 1, 1, 1
        yield a
        while True:
            yield b
            n += 1
            a, b = b, ((3 * (n - 1)) * a + (2 * n - 1) * b) // n
    A002426 = A()
    print([next(A002426) for  in range(30)])  # _Peter Luschny, May 16 2016
    

Formula

G.f.: 1/sqrt(1 - 2*x - 3*x^2).
E.g.f.: exp(x)*I_0(2x), where I_0 is a Bessel function. - Michael Somos, Sep 09 2002
a(n) = 2*A027914(n) - 3^n. - Benoit Cloitre, Sep 28 2002
a(n) is asymptotic to d*3^n/sqrt(n) with d around 0.5.. - Benoit Cloitre, Nov 02 2002, d = sqrt(3/Pi)/2 = 0.4886025119... - Alec Mihailovs (alec(AT)mihailovs.com), Feb 24 2005
D-finite with recurrence: a(n) = ((2*n - 1)*a(n-1) + 3*(n - 1)*a(n-2))/n; a(0) = a(1) = 1; see paper by Barcucci, Pinzani and Sprugnoli.
Inverse binomial transform of A000984. - Vladeta Jovovic, Apr 28 2003
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, k/2)*(1 + (-1)^k)/2; a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*binomial(2*k, k). - Paul Barry, Jul 01 2003
a(n) = Sum_{k>=0} binomial(n, 2*k)*binomial(2*k, k). - Philippe Deléham, Dec 31 2003
a(n) = Sum_{i+j=n, 0<=j<=i<=n} binomial(n, i)*binomial(i, j). - Benoit Cloitre, Jun 06 2004
a(n) = 3*a(n-1) - 2*A005043(n). - Joost Vermeij (joost_vermeij(AT)hotmail.com), Feb 10 2005
a(n) = Sum_{k=0..n} binomial(n, k)*binomial(k, n-k). - Paul Barry, Apr 23 2005
a(n) = (-1/4)^n*Sum_{k=0..n} binomial(2*k, k)*binomial(2*n-2*k, n-k)*(-3)^k. - Philippe Deléham, Aug 17 2005
a(n) = A111808(n,n). - Reinhard Zumkeller, Aug 17 2005
a(n) = Sum_{k=0..n} (((1 + (-1)^k)/2)*Sum_{i=0..floor((n-k)/2)} binomial(n, i)*binomial(n-i, i+k)*((k + 1)/(i + k + 1))). - Paul Barry, Sep 23 2005
a(n) = 3^n*Sum_{j=0..n} (-1/3)^j*C(n, j)*C(2*j, j); follows from (a) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/2)^n*Sum_{j=0..n} 3^j*binomial(n, j)*binomial(2*n-2*j, n) = (3/2)^n*Sum_{j=0..n} (1/3)^j*binomial(n, j)*binomial(2*j, n); follows from (c) in A027907. - Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006
a(n) = (1/Pi)*Integral_{x=-1..3} x^n/sqrt((3 - x)*(1 + x)) is moment representation. - Paul Barry, Sep 10 2007
G.f.: 1/(1 - x - 2x^2/(1 - x - x^2/(1 - x - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 05 2009
a(n) = sqrt(-1/3)*(-1)^n*hypergeometric([1/2, n+1], [1], 4/3). - Mark van Hoeij, Nov 12 2009
a(n) = (1/Pi)*Integral_{x=-1..1} (1 + 2*x)^n/sqrt(1 - x^2) = (1/Pi)*Integral_{t=0..Pi} (1 + 2*cos(t))^n. - Eli Wolfhagen, Feb 01 2011
In general, g.f.: 1/sqrt(1 - 2*a*x + x^2*(a^2 - 4*b)) = 1/(1 - a*x)*(1 - 2*x^2*b/(G(0)*(a*x - 1) + 2*x^2*b)); G(k) = 1 - a*x - x^2*b/G(k+1); for g.f.: 1/sqrt(1 - 2*x - 3*x^2) = 1/(1 - x)*(1 - 2*x^2/(G(0)*(x - 1) + 2*x^2)); G(k) = 1 - x - x^2/G(k+1), a = 1, b = 1; (continued fraction). - Sergei N. Gladkovskii, Dec 08 2011
a(n) = Sum_{k=0..floor(n/3)} (-1)^k*binomial(2*n-3*k-1, n-3*k)*binomial(n, k). - Gopinath A. R., Feb 10 2012
G.f.: A(x) = x*B'(x)/B(x) where B(x) satisfies B(x) = x*(1 + B(x) + B(x)^2). - Vladimir Kruchinin, Feb 03 2013 (B(x) = x*A001006(x) - Michael Somos, Jul 08 2014)
G.f.: G(0), where G(k) = 1 + x*(2 + 3*x)*(4*k + 1)/(4*k + 2 - x*(2 + 3*x)*(4*k + 2)*(4*k + 3)/(x*(2 + 3*x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 29 2013
E.g.f.: exp(x) * Sum_{k>=0} (x^k/k!)^2. - Geoffrey Critzer, Sep 04 2013
G.f.: Sum_{n>=0} (2*n)!/n!^2*(x^(2*n)/(1 - x)^(2*n+1)). - Paul D. Hanna, Sep 21 2013
0 = a(n)*(9*a(n+1) + 9*a(n+2) - 6*a(n+3)) + a(n+1)*(3*a(n+1) + 4*a(n+2) - 3*a(n+3)) + a(n+2)*(-a(n+2) + a(n+3)) for all n in Z. - Michael Somos, Jul 08 2014
a(n) = hypergeometric([-n/2, (1-n)/2], [1], 4). - Peter Luschny, Sep 17 2014
a(n) = A132885(n,0), that is, a(n) = A132885(A002620(n+1)). - Altug Alkan, Nov 29 2015
a(n) = GegenbauerC(n,-n,-1/2). - Peter Luschny, May 07 2016
a(n) = 4^n*JacobiP[n,-n-1/2,-n-1/2,-1/2]. - Peter Luschny, May 13 2016
From Alexander Burstein, Oct 03 2017: (Start)
G.f.: A(4*x) = B(-x)*B(3*x), where B(x) is the g.f. of A000984.
G.f.: A(2*x)*A(-2*x) = B(x^2)*B(9*x^2).
G.f.: A(x) = 1 + x*M'(x)/M(x), where M(x) is the g.f. of A001006. (End)
a(n) = Sum_{i=0..n/2} n!/((n - 2*i)!*(i!)^2). [Cf. Lalo and Lalo link. It is Luschny's terminating hypergeometric sum.] - Shara Lalo and Zagros Lalo, Oct 03 2018
From Peter Bala, Feb 07 2022: (Start)
a(n)^2 = Sum_{k = 0..n} (-3)^(n-k)*binomial(2*k,k)^2*binomial(n+k,n-k) and has g.f. Sum_{n >= 0} binomial(2*n,n)^2*x^n/(1 + 3*x)^(2*n+1). Compare with the g.f. for a(n) given above by Hanna.
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.
Conjecture: The stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all prime p >= 5 and positive integers n and k. (End)
a(n) = A005043(n) + A005717(n) for n >= 1. - Amiram Eldar, May 17 2024
For even n, a(n) = (n-1)!!* 2^{n/2}/ (n/2)!* 2F1(-n/2,-n/2;1/2;1/4). For odd n, a(n) = n!! *2^(n/2-1/2) / (n/2-1/2)! * 2F1(1/2-n/2,1/2-n/2;3/2;1/4). - R. J. Mathar, Mar 19 2025

A005043 Riordan numbers: a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1).

Original entry on oeis.org

1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, 4213, 11298, 30537, 83097, 227475, 625992, 1730787, 4805595, 13393689, 37458330, 105089229, 295673994, 834086421, 2358641376, 6684761125, 18985057351, 54022715451, 154000562758, 439742222071, 1257643249140
Offset: 0

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Comments

Also called Motzkin summands or ring numbers.
The old name was "Motzkin sums", used in certain publications. The sequence has the property that Motzkin(n) = A001006(n) = a(n) + a(n+1), e.g., A001006(4) = 9 = 3 + 6 = a(4) + a(5).
Number of 'Catalan partitions', that is partitions of a set 1,2,3,...,n into parts that are not singletons and whose convex hulls are disjoint when the points are arranged on a circle (so when the parts are all pairs we get Catalan numbers). - Aart Blokhuis (aartb(AT)win.tue.nl), Jul 04 2000
Number of ordered trees with n edges and no vertices of outdegree 1. For n > 1, number of dissections of a convex polygon by nonintersecting diagonals with a total number of n+1 edges. - Emeric Deutsch, Mar 06 2002
Number of Motzkin paths of length n with no horizontal steps at level 0. - Emeric Deutsch, Nov 09 2003
Number of Dyck paths of semilength n with no peaks at odd level. Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUDUDD, where U=(1,1), D=(1,-1). Number of Dyck paths of semilength n with no ascents of length 1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUUDDD. - Emeric Deutsch, Dec 05 2003
Arises in Schubert calculus as follows. Let P = complex projective space of dimension n+1. Take n projective subspaces of codimension 3 in P in general position. Then a(n) is the number of lines of P intersecting all these subspaces. - F. Hirzebruch, Feb 09 2004
Difference between central trinomial coefficient and its predecessor. Example: a(6) = 15 = 141 - 126 and (1 + x + x^2)^6 = ... + 126*x^5 + 141*x^6 + ... (Catalan number A000108(n) is the difference between central binomial coefficient and its predecessor.) - David Callan, Feb 07 2004
a(n) = number of 321-avoiding permutations on [n] in which each left-to-right maximum is a descent (i.e., is followed by a smaller number). For example, a(4) counts 4123, 3142, 2143. - David Callan, Jul 20 2005
The Hankel transform of this sequence give A000012 = [1, 1, 1, 1, 1, 1, 1, ...]; example: Det([1, 0, 1, 1; 0, 1, 1, 3; 1, 1, 3, 6; 1, 3, 6, 15]) = 1. - Philippe Deléham, May 28 2005
The number of projective invariants of degree 2 for n labeled points on the projective line. - Benjamin J. Howard (bhoward(AT)ima.umn.edu), Nov 24 2006
Define a random variable X=trA^2, where A is a 2 X 2 unitary symplectic matrix chosen from USp(2) with Haar measure. The n-th central moment of X is E[(X+1)^n] = a(n). - Andrew V. Sutherland, Dec 02 2007
Let V be the adjoint representation of the complex Lie algebra sl(2). The dimension of the invariant subspace of the n-th tensor power of V is a(n). - Samson Black (sblack1(AT)uoregon.edu), Aug 27 2008
Starting with offset 3 = iterates of M * [1,1,1,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 08 2009
a(n) has the following standard-Young-tableaux (SYT) interpretation: binomial(n+1,k)*binomial(n-k-1,k-1)/(n+1)=f^(k,k,1^{n-2k}) where f^lambda equals the number of SYT of shape lambda. - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 02 2010
a(n) is also the sum of the numbers of standard Young tableaux of shapes (k,k,1^{n-2k}) for all 1 <= k <= floor(n/2). - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 10 2010
a(n) is the number of derangements of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(3)=1 because p=231=(123) is the only derangement of {1,2,3} with genus 0. Indeed, cp'=231*312=123=(1)(2)(3) and so g(p) = (1/2)(3+1-1-3)=0. - Emeric Deutsch, May 29 2010
Apparently: Number of Dyck 2n-paths with all ascents length 2 and no descent length 2. - David Scambler, Apr 17 2012
This is true. Proof: The mapping "insert a peak (UD) after each upstep (U)" is a bijection from all Dyck n-paths to those Dyck (2n)-paths in which each ascent is of length 2. It sends descents of length 1 in the n-path to descents of length 2 in the (2n)-path. But Dyck n-paths with no descents of length 1 are equinumerous with Riordan n-paths (Motzkin n-paths with no flatsteps at ground level) as follows. Given a Dyck n-path with no descents of length 1, split it into consecutive step pairs, then replace UU with U, DD with D, UD with a blue flatstep (F), DU with a red flatstep, and concatenate the new steps to get a colored Motzkin path. Each red F will be (immediately) preceded by a blue F or a D. In the latter case, transfer the red F so that it precedes the matching U of the D. Finally, erase colors to get the required Riordan path. For example, with lowercase f denoting a red flatstep, U^5 D^2 U D^4 U^4 D^3 U D^2 -> (U^2, U^2, UD, DU, D^2, D^2, U^2, U^2 D^2, DU, D^2) -> UUFfDDUUDfD -> UUFFDDUFUDD. - David Callan, Apr 25 2012
From Nolan Wallach, Aug 20 2014: (Start)
Let ch[part1, part2] be the value of the character of the symmetric group on n letters corresponding to the partition part1 of n on the conjucgacy class given by part2. Let A[n] be the set of (n+1) partitions of 2n with parts 1 or 2. Then deleting the first term of the sequence one has a(n) = Sum_{k=1..n+1} binomial(n,k-1)*ch[[n,n], A[n][[k]]])/2^n. This via the Frobenius Character Formula can be interpreted as the dimension of the SL(n,C) invariants in tensor^n (wedge^2 C^n).
Explanation: Let p_j denote sum (x_i)^j the sum in k variables. Then the Frobenius formula says then (p_1)^j_1 (p_2)^j_2 ... (p_r)^j_r is equal to sum(lambda, ch[lambda, 1^j_12^j_2 ... r^j_r] S_lambda) with S_lambda the Schur function corresponding to lambda. This formula implies that the coefficient of S([n,n]) in (((p_1)^1+p_2)/2)^n in its expansion in terms of Schur functions is the right hand side of our formula. If we specialize the number of variables to 2 then S[n,n](x,y)=(xy)^n. Which when restricted to y=x^(-1) is 1. That is it is 1 on SL(2).
On the other hand ((p_1)^2+p_2)/2 is the complete homogeneous symmetric function of degree 2 that is tr(S^2(X)). Thus our formula for a(n) is the same as that of Samson Black above since his V is the same as S^2(C^2) as a representation of SL(2). On the other hand, if we multiply ch(lambda) by sgn you get ch(Transpose(lambda)). So ch([n,n]) becomes ch([2,...,2]) (here there are n 2's). The formula for a(n) is now (1/2^n)*Sum_{j=0..n} ch([2,..,2], 1^(2n-2j) 2^j])*(-1)^j)*binomial(n,j), which calculates the coefficient of S_(2,...,2) in (((p_1)^2-p_2)/2)^n. But ((p_1)^2-p_2)/2 in n variables is the second elementary symmetric function which is the character of wedge^2 C^n and S_(2,...,2) is 1 on SL(n).
(End)
a(n) = number of noncrossing partitions (A000108) of [n] that contain no singletons, also number of nonnesting partitions (A000108) of [n] that contain no singletons. - David Callan, Aug 27 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Various relations among the o.g.f.s may be easily constructed, such as Fib[-Mot(-x)] = -P[P(-x)] = x/(1-2*x) a generating fct for 2^n.
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)
Consistent with David Callan's comment above, A249548, provides a refinement of the Motzkin sums into the individual numbers for the non-crossing partitions he describes. - Tom Copeland, Nov 09 2014
The number of lattice paths from (0,0) to (n,0) that do not cross below the x-axis and use up-step=(1,1) and down-steps=(1,-k) where k is a positive integer. For example, a(4) = 3: [(1,1)(1,1)(1,-1)(1,-1)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,1)(1,-3)]. - Nicholas Ham, Aug 19 2015
A series created using 2*(a(n) + a(n+1)) + (a(n+1) + a(n+2)) has Hankel transform of F(2n), offset 3, F being a Fibonacci number, A001906 (Empirical observation). - Tony Foster III, Jul 30 2016
The series a(n) + A001006(n) has Hankel transform F(2n+1), offset n=1, F being the Fibonacci bisection A001519 (empirical observation). - Tony Foster III, Sep 05 2016
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n. Moreover, the number of such algebras having the double centraliser property with respect to a minimal faithful projective-injective module equals the number of Riordan paths, that is, Motzkin paths without level-steps at height zero, of length n." - Eric M. Schmidt, Dec 16 2017
A connection to the Thue-Morse sequence: (-1)^a(n) = (-1)^A010060(n) * (-1)^A010060(n+1) = A106400(n) * A106400(n+1). - Vladimir Reshetnikov, Jul 21 2019
Named by Bernhart (1999) after the American mathematician John Riordan (1903-1988). - Amiram Eldar, Apr 15 2021

Examples

			a(5)=6 because the only dissections of a polygon with a total number of 6 edges are: five pentagons with one of the five diagonals and the hexagon with no diagonals.
G.f. = 1 + x^2 + x^3 + 3*x^4 + 6*x^5 + 15*x^6 + 36*x^7 + 91*x^8 + 232*x^9 + ...
From _Gus Wiseman_, Nov 15 2022: (Start)
The a(0) = 1 through a(6) = 15 lone-child-avoiding (no vertices of outdegree 1) ordered rooted trees with n + 1 vertices (ranked by A358376):
  o  .  (oo)  (ooo)  (oooo)   (ooooo)   (oooooo)
                     ((oo)o)  ((oo)oo)  ((oo)ooo)
                     (o(oo))  ((ooo)o)  ((ooo)oo)
                              (o(oo)o)  ((oooo)o)
                              (o(ooo))  (o(oo)oo)
                              (oo(oo))  (o(ooo)o)
                                        (o(oooo))
                                        (oo(oo)o)
                                        (oo(ooo))
                                        (ooo(oo))
                                        (((oo)o)o)
                                        ((o(oo))o)
                                        ((oo)(oo))
                                        (o((oo)o))
                                        (o(o(oo)))
(End)
		

References

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

Crossrefs

Row sums of triangle A020474, first differences of A082395.
First diagonal of triangular array in A059346.
Binomial transform of A126930. - Philippe Deléham, Nov 26 2009
The Hankel transform of a(n+1) is A128834. The Hankel transform of a(n+2) is floor((2*n+4)/3) = A004523(n+2). - Paul Barry, Mar 08 2011
The Kn11 triangle sums of triangle A175136 lead to A005043(n+2), while the Kn12(n) = A005043(n+4)-2^(n+1), Kn13(n) = A005043(n+6)-(n^2+9*n+56)*2^(n-2) and the Kn4(n) = A005043(2*n+2) = A099251(n+1) triangle sums are related to the sequence given above. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Cf. A187306 (self-convolution), A348210 (column 1).
Bisections: A099251, A099252.

Programs

  • Haskell
    a005043 n = a005043_list !! n
    a005043_list = 1 : 0 : zipWith div
       (zipWith (*) [1..] (zipWith (+)
           (map (* 2) $ tail a005043_list) (map (* 3) a005043_list))) [3..]
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    A005043 := proc(n) option remember; if n <= 1 then 1-n else (n-1)*(2*A005043(n-1)+3*A005043(n-2))/(n+1); fi; end;
    Order := 20: solve(series((x-x^2)/(1-x+x^2),x)=y,x); # outputs g.f.
  • Mathematica
    a[0]=1; a[1]=0; a[n_]:= a[n] = (n-1)*(2*a[n-1] + 3*a[n-2])/(n+1); Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jun 14 2005 *)
    Table[(-3)^(1/2)/6 * (-1)^n*(3*Hypergeometric2F1[1/2,n+1,1,4/3]+ Hypergeometric2F1[1/2,n+2,1,4/3]), {n,0,32}] (* cf. Mark van Hoeij in A001006 *) (* Wouter Meeussen, Jan 23 2010 *)
    RecurrenceTable[{a[0]==1,a[1]==0,a[n]==(n-1) (2a[n-1]+3a[n-2])/(n+1)},a,{n,30}] (* Harvey P. Dale, Sep 27 2013 *)
    a[ n_]:= SeriesCoefficient[2/(1+x +Sqrt[1-2x-3x^2]), {x, 0, n}]; (* Michael Somos, Aug 21 2014 *)
    a[ n_]:= If[n<0, 0, 3^(n+3/2) Hypergeometric2F1[3/2, n+2, 2, 4]/I]; (* Michael Somos, Aug 21 2014 *)
    Table[3^(n+3/2) CatalanNumber[n] (4(5+2n)Hypergeometric2F1[3/2, 3/2, 1/2-n, 1/4] -9 Hypergeometric2F1[3/2, 5/2, 1/2 -n, 1/4])/(4^(n+3) (n+1)), {n, 0, 31}] (* Vladimir Reshetnikov, Jul 21 2019 *)
    Table[Sqrt[27]/8 (3/4)^n CatalanNumber[n] Hypergeometric2F1[1/2, 3/2, 1/2 - n, 1/4], {n, 0, 31}] (* Jan Mangaldan, Sep 12 2021 *)
  • Maxima
    a[0]:1$
    a[1]:0$
    a[n]:=(n-1)*(2*a[n-1]+3*a[n-2])/(n+1)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( (x - x^3) / (1 + x^3) + x * O(x^n)), n))}; /* Michael Somos, May 31 2005 */
    
  • PARI
    my(N=66); Vec(serreverse(x/(1+x*sum(k=1,N,x^k))+O(x^N))) \\ Joerg Arndt, Aug 19 2012
    
  • Python
    from functools import cache
    @cache
    def A005043(n: int) -> int:
        if n <= 1: return 1 - n
        return (n - 1) * (2 * A005043(n - 1) + 3 * A005043(n - 2)) // (n + 1)
    print([A005043(n) for n in range(32)]) # Peter Luschny, Nov 20 2022
  • Sage
    A005043 = lambda n: (-1)^n*jacobi_P(n,1,-n-3/2,-7)/(n+1)
    [simplify(A005043(n)) for n in (0..29)]
    # Peter Luschny, Sep 23 2014
    
  • Sage
    def ms():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        while True:
            yield -b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)/((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)/n
    A005043 = ms()
    print([next(A005043) for  in range(32)]) # _Peter Luschny, May 16 2016
    

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k). a(n) = (1/(n+1)) * Sum_{k=0..ceiling(n/2)} binomial(n+1, k)*binomial(n-k-1, k-1), for n > 1. - Len Smiley. [Comment from Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 02 2010: the latter sum should be over the range k=1..floor(n/2).]
G.f.: (1 + x - sqrt(1-2*x-3*x^2))/(2*x*(1+x)).
G.f.: 2/(1+x+sqrt(1-2*x-3*x^2)). - Paul Peart (ppeart(AT)fac.howard.edu), May 27 2000
a(n+1) + (-1)^n = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). - Bernhart
a(n) = (1/(n+1)) * Sum_{i} (-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i). - Bernhart
G.f. A(x) satisfies A = 1/(1+x) + x*A^2.
E.g.f.: exp(x)*(BesselI(0, 2*x) - BesselI(1, 2*x)). - Vladeta Jovovic, Apr 28 2003
a(n) = A001006(n-1) - a(n-1).
a(n+1) = Sum_{k=0..n} (-1)^k*A026300(n, k), where A026300 is the Motzkin triangle.
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
a(n) = Sum_{k>=0} A086810(n-k, k). - Philippe Deléham, May 30 2005
a(n+2) = Sum_{k>=0} A064189(n-k, k). - Philippe Deléham, May 31 2005
Moment representation: a(n) = (1/(2*Pi))*Int(x^n*sqrt((1+x)(3-x))/(1+x),x,-1,3). - Paul Barry, Jul 09 2006
Inverse binomial transform of A000108 (Catalan numbers). - Philippe Deléham, Oct 20 2006
a(n) = (2/Pi)* Integral_{x=0..Pi} (4*cos(x)^2-1)^n*sin(x)^2 dx. - Andrew V. Sutherland, Dec 02 2007
G.f.: 1/(1-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 22 2009
G.f.: 1/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: 1/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-... (continued fraction). - Paul Barry, Mar 02 2010
a(n) = -(-1)^n * hypergeom([1/2, n+2],[2],4/3) / sqrt(-3). - Mark van Hoeij, Jul 02 2010
a(n) = (-1)^n*hypergeometric([-n,1/2],[2],4). - Peter Luschny, Aug 15 2012
Let A(x) be the g.f., then x*A(x) is the reversion of x/(1 + x^2*Sum_{k>=0} x^k); see A215340 for the correspondence to Dyck paths without length-1 ascents. - Joerg Arndt, Aug 19 2012 and Apr 16 2013
a(n) ~ 3^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 02 2012
G.f.: 2/(1+x+1/G(0)), where G(k) = 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 05 2013
D-finite (an alternative): (n+1)*a(n) = 3*(n-2)*a(n-3) + (5*n-7)*a(n-2) + (n-2)*a(n-1), n >= 3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) = (3^(n+2)/(sqrt(3*n*Pi)*(8*n)))*(1-21/(16*n) + O(1/n^2)) (with contribution by Vaclav Kotesovec). - Fung Lam, Mar 22 2014
a(n) = T(2*n-1,n)/n, where T(n,k) = triangle of A180177. - Vladimir Kruchinin, Sep 23 2014
a(n) = (-1)^n*JacobiP(n,1,-n-3/2,-7)/(n+1). - Peter Luschny, Sep 23 2014
a(n) = Sum_{k=0..n} C(n,k)*(C(k,n-k)-C(k,n-k-1)). - Peter Luschny, Oct 01 2014
Conjecture: a(n) = A002426(n) - A005717(n), n > 0. - Mikhail Kurkov, Feb 24 2019 [The conjecture is true. - Amiram Eldar, May 17 2024]
a(n) = A309303(n) + A309303(n+1). - Vladimir Reshetnikov, Jul 22 2019
From Peter Bala, Feb 11 2022: (Start)
a(n) = A005773(n+1) - 2*A005717(n) for n >= 1.
Conjectures: for n >= 1, n divides a(2*n+1) and 2*n-1 divides a(2*n). (End)

Extensions

Thanks to Laura L. M. Yang (yanglm(AT)hotmail.com) for a correction, Aug 29 2004
Name changed to Riordan numbers following a suggestion from Ira M. Gessel. - N. J. A. Sloane, Jul 24 2020

A027907 Triangle of trinomial coefficients T(n,k) (n >= 0, 0 <= k <= 2*n), read by rows: n-th row is obtained by expanding (1 + x + x^2)^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 10, 16, 19, 16, 10, 4, 1, 1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1, 1, 6, 21, 50, 90, 126, 141, 126, 90, 50, 21, 6, 1, 1, 7, 28, 77, 161, 266, 357, 393, 357, 266, 161, 77, 28, 7, 1, 1, 8, 36, 112, 266
Offset: 0

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Keywords

Comments

When the rows are centered about their midpoints, each term is the sum of the three terms directly above it (assuming the undefined terms in the previous row are zeros). - N. J. A. Sloane, Dec 23 2021
T(n,k) = number of integer strings s(0),...,s(n) such that s(0)=0, s(n)=k, s(i) = s(i-1) + c, where c is 0, 1 or 2. Columns of T include A002426, A005717 and A014531.
Also number of ordered trees having n+1 leaves, all at level three and n+k+3 edges. Example: T(3,5)=3 because we have three ordered trees with 4 leaves, all at level three and 11 edges: the root r has three children; from one of these children two paths of length two are hanging (i.e., 3 possibilities) while from each of the other two children one path of length two is hanging. Diagonal sums are the tribonacci numbers; more precisely: Sum_{i=0..floor(2*n/3)} T(n-i,i) = A000073(n+2). - Emeric Deutsch, Jan 03 2004
T(n,k) = A111808(n,k) for 0 <= k <= n and T(n, 2*n-k) = A111808(n,k) for 0 <= k < n. - Reinhard Zumkeller, Aug 17 2005
The trinomial coefficients, T(n,i), are the absolute value of the coefficients of the chromatic polynomial of P_2 X P_n factored with x*(x-1)^i terms. Example: The chromatic polynomial of P_2 X P_2 is: x*(x-1) - 2*x*(x-1)^2 + x*(x-1)^3 and so T(1,0)=1, T(1,1)=2 and T(1,1) = 1. - Thomas J. Pfaff (tpfaff(AT)ithaca.edu), Oct 02 2006
T(n,k) is the number of distinct ways in which k unlabeled objects can be distributed in n labeled urns allowing at most 2 objects to fall into each urn. - N-E. Fahssi, Mar 16 2008
T(n,k) is the number of compositions of k into n parts p, each part 0 <= p <= 2. Adding 1 to each part, as a corollary, T(n,k) is the number of compositions of n+k into n parts p where 1 <= p <= 3. E.g., T(2,3)=2 since 5 = 3+2 = 2+3. - Steffen Eger, Jun 10 2011
Number of lattice paths from (0,0) to (n,k) using steps (1,0), (1,1), (1,2). - Joerg Arndt, Jul 05 2011
Number of lattice paths from (0,0) to (2*n-k,k) using steps (2,0), (1,1), (0,2). - Werner Schulte, Jan 25 2017
T(n,k) is number of distinct ways to sum the integers -1, 0 , and 1 n times to obtain n-k, where T(n,0) = T(n,2*n+1) = 1. - William Boyles, Apr 23 2017
T(n-1,k-1) is the number of 2-compositions of n with 0's having k parts; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 15 2020
T(n,k) is the number of ways to obtain a sum of n+k when throwing a 3-sided die n times. Follows from the "T(n,k) is the number of compositions of n+k into n parts p where 1 <= p <= 3" comment above. - Feryal Alayont, Dec 30 2024

Examples

			The triangle T(n, k) begins:
  n\k 0   1   2   3   4   5   6   7   8   9 10 11 12
  0:  1
  1:  1   1   1
  2:  1   2   3   2   1
  3:  1   3   6   7   6   3   1
  4:  1   4  10  16  19  16  10   4   1
  5:  1   5  15  30  45  51  45  30  15   5  1
  6:  1   6  21  50  90 126 141 126  90  50 21  6  1
Concatenated rows:
G.f. = 1 + (x^2+x+1)*x + (x^2+x+1)^2*x^4 + (x^2+x+1)^3*x^9 + ...
     = 1 + (x + x^2 + x^3) + (x^4 + 2*x^5 + 3*x^6 + 2*x^7 + x^8) +
  (x^9 + 3*x^10 + 6*x^11 + 7*x^12 + 6*x^13 + 3*x^14 + x^15) + ... .
As a centered triangle, this begins:
           1
        1  1  1
     1  2  3  2  1
  1  3  6  7  6  3  1
		

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.
  • D. C. Fielder and C. O. Alford, Pascal's triangle: top gun or just one of the gang?, in G E Bergum et al., eds., Applications of Fibonacci Numbers Vol. 4 1991 pp. 77-90 (Kluwer).
  • L. Kleinrock, Uniform permutation of sequences, JPL Space Programs Summary, Vol. 37-64-III, Apr 30, 1970, pp. 32-43.

Crossrefs

Columns of T include A002426, A005717, A014531, A005581, A005712, etc. See also A035000, A008287.
First differences are in A025177. Pairwise sums are in A025564.

Programs

  • Haskell
    a027907 n k = a027907_tabf !! n !! k
    a027907_row n = a027907_tabf !! n
    a027907_tabf = [1] : iterate f [1, 1, 1] where
       f row = zipWith3 (((+) .) . (+))
                        (row ++ [0, 0]) ([0] ++ row ++ [0]) ([0, 0] ++ row)
    a027907_list = concat a027907_tabf
    -- Reinhard Zumkeller, Jul 06 2014, Jan 22 2013, Apr 02 2011
  • Maple
    A027907 := proc(n,k) expand((1+x+x^2)^n) ; coeftayl(%,x=0,k) ; end proc:
    seq(seq(A027907(n,k),k=0..2*n),n=0..5) ; # R. J. Mathar, Jun 13 2011
    T := (n,k) -> simplify(GegenbauerC(`if`(kPeter Luschny, May 08 2016
  • Mathematica
    Table[CoefficientList[Series[(Sum[x^i, {i, 0, 2}])^n, {x, 0, 2 n}], x], {n, 0, 10}] // Grid (* Geoffrey Critzer, Mar 31 2010 *)
    Table[Sum[Binomial[n, i]Binomial[n - i, k - 2i], {i, 0, n}], {n, 0, 10}, {k, 0, 2n}] (* Adi Dani, May 07 2011 *)
    T[ n_, k_] := If[ n < 0, 0, Coefficient[ (1 + x + x^2)^n, x, k]]; (* Michael Somos, Nov 08 2016 *)
    Flatten[DeleteCases[#,0]&/@CellularAutomaton[{Total[#] &, {}, 1}, {{1}, 0}, 8] ] (* Giorgos Kalogeropoulos, Nov 09 2021 *)
  • Maxima
    trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
    create_list(trinomial(n,k),n,0,8,k,0,2*n); /* Emanuele Munarini, Mar 15 2011 */
    
  • Maxima
    create_list(ultraspherical(k,-n,-1/2),n,0,6,k,0,2*n); /* Emanuele Munarini, Oct 18 2016 */
    
  • PARI
    {T(n, k) = if( n<0, 0, polcoeff( (1 + x + x^2)^n, k))}; /* Michael Somos, Jun 27 2003 */
    

Formula

G.f.: 1/(1-z*(1+w+w^2)).
T(n,k) = Sum_{r=0..floor(k/3)} (-1)^r*binomial(n, r)*binomial(k-3*r+n-1, n-1).
Recurrence: T(0,0) = 1; T(n,k) = T(n-1,k-2) + T(n-1,k-1) + T(n-1,k-0), with T(n,k) = 0 if k < 0 or k > 2*n:
T(i,0) = T(i, 2*i) = 1 for i >= 0, T(i, 1) = T(i, 2*i-1) = i for i >= 1 and for i >= 2 and 2 <= j <= i-2, T(i, j) = T(i-1, j-2) + T(i-1, j-1) + T(i-1, j).
The row sums are powers of 3 (A000244). - Gerald McGarvey, Aug 14 2004
T(n,k) = Sum_{i=0..floor(k/2)} binomial(n, 2*i+n-k) * binomial(2*i+n-k, i). - Ralf Stephan, Jan 26 2005
T(n,k) = Sum_{j=0..n} binomial(n, j) * binomial(j, k-j). - Paul Barry, May 21 2005
T(n,k) = Sum_{j=0..n} binomial(k-j, j) * binomial(n, k-j). - Paul Barry, Nov 04 2005
From Loic Turban (turban(AT)lpm.u-nancy.fr), Aug 31 2006: (Start)
T(n,k) = Sum_{j=0..n} (-1)^j * binomial(n,j) * binomial(2*n-2*j, k-j); (G. E. Andrews (1990)) obtained by expanding ((1+x)^2 - x)^n.
T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(n-j, k-2*j); obtained by expanding ((1+x) + x^2)^n.
T(n,k) = (-1)^k*Sum_{j=0..n} (-3)^j * binomial(n,j) * binomial(2*n-2*j, k-j); obtained by expanding ((1-x)^2 + 3*x)^n.
T(n,k) = (1/2)^k * Sum_{j=0..n} 3^j * binomial(n,j) * binomial(2*n-2*j, k-2*j); obtained by expanding ((1+x/2)^2 + (3/4)*x^2)^n.
T(n,k) = (2^k/4^n) * Sum_{j=0..n} 3^j * binomial(n,j) * binomial(2*n-2*j, k); obtained by expanding ((1/2+x)^2 + 3/4)^n using T(n,k) = T(2*n-k). (End)
From Paul D. Hanna, Apr 18 2012: (Start)
Let A(x) be the g.f. of the flattened sequence, then:
G.f.: A(x) = Sum_{n>=0} x^(n^2) * (1+x+x^2)^n.
G.f.: A(x) = Sum_{n>=0} x^n*(1+x+x^2)^n * Product_{k=1..n} (1 - (1+x+x^2) * x^(4*k-3)) / (1 - (1+x+x^2)*x^(4*k-1)).
G.f.: A(x) = 1/(1 - x*(1+x+x^2)/(1 + x*(1-x^2)*(1+x+x^2)/(1 - x^5*(1+x+x^2)/(1 + x^3*(1-x^4)*(1+x+x^2)/(1 - x^9*(1+x+x^2)/(1 + x^5*(1-x^6)*(1+x+x^2)/(1 - x^13* (1+x+x^2)/(1 + x^7*(1-x^8)*(1+x+x^2)/(1 - ...))))))))), a continued fraction.
(End)
Triangle: G.f. = Sum_{n>=0} (1+x+x^2)^n * x^(n^2) * y^n. - Daniel Forgues, Mar 16 2015
From Peter Luschny, May 08 2016: (Start)
T(n+1,n)/(n+1) = A001006(n) (Motzkin) for n>=0.
T(n,k) = H(n, k) if k < n else H(n, 2*n-k) where H(n,k) = binomial(n,k)*hypergeom([(1-k)/2, -k/2], [n-k+1], 4).
T(n,k) = GegenbauerC(m, -n, -1/2) where m=k if k < n else 2*n-k. (End)
T(n,k) = (-1)^k * C(2*n,k) * hypergeom([-k, -(2*n-k)], [-n+1/2], 3/4), for all k with 0 <= k <= 2n. - Robert S. Maier, Jun 13 2023
T(n,n) = Sum_{k=0..2*n} (-1)^k*(T(n,k))^2 and T(2*n,2*n) = Sum_{k=0..2*n} (T(n,k))^2 for n >= 0. - Werner Schulte, Nov 08 2016
T(n,n) = A002426(n), central trinomial coefficients. - M. F. Hasler, Nov 02 2019
Sum_{k=0..n-1} T(n, 2*k) = (3^n-1)/2. - Tony Foster III, Oct 06 2020

A005773 Number of directed animals of size n (or directed n-ominoes in standard position).

Original entry on oeis.org

1, 1, 2, 5, 13, 35, 96, 267, 750, 2123, 6046, 17303, 49721, 143365, 414584, 1201917, 3492117, 10165779, 29643870, 86574831, 253188111, 741365049, 2173243128, 6377181825, 18730782252, 55062586341, 161995031226, 476941691177, 1405155255055, 4142457992363
Offset: 0

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Author

Keywords

Comments

This sequence, with first term a(0) deleted, appears to be determined by the conditions that the diagonal and first superdiagonal of U are {1,1,1,1,...} and {2,3,4,5,...,n+1,...} respectively, where A=LU is the LU factorization of the Hankel matrix A given by [{a(1),a(2),...}, {a(2),a(3),...}, ..., {a(n),a(n+1),...}, ...]. - John W. Layman, Jul 21 2000
Also the number of base 3 n-digit numbers (not starting with 0) with digit sum n. For the analogous sequence in base 10 see A071976, see example. - John W. Layman, Jun 22 2002
Also number of paths in an n X n grid from (0,0) to the line x=n-1, using only steps U=(1,1), H=(1,0) and D=(1,-1) (i.e., left factors of length n-1 of Motzkin paths, palindromic Motzkin paths of length 2n-2 or 2n-1). Example: a(3)=5, namely, HH, UD, HU, UH and UU. Also number of ordered trees with n edges and having nonroot nodes of outdegree at most 2. - Emeric Deutsch, Aug 01 2002
Number of symmetric Dyck paths of semilength 2n-1 with no peaks at even level. Example: a(3)=5 because we have UDUDUDUDUD, UDUUUDDDUD, UUUUUDDDDD, UUUDUDUDDD and UUUDDUUDDD, where U=(1,1) and D=(1,-1). Also number of symmetric Dyck paths of semilength 2n with no peaks at even level. Example: a(3)=5 because we have UDUDUDUDUDUD, UDUUUDUDDDUD, UUUDUDUDUDDD, UUUUUDUDDDDD and UUUDDDUUUDDD. - Emeric Deutsch, Nov 21 2003
a(n) is the sum of the (n-1)-st central trinomial coefficient and its predecessor. Example: a(4) = 6 + 7 and (1 + x + x^2)^3 = ... + 6*x^2 + 7*x^3 + ... . - David Callan, Feb 07 2004
a(n) is the number of UDU-free paths of n upsteps (U) and n downsteps (D) that start U (n>=1). Example: a(2)=2 counts UUDD, UDDU. - David Callan, Aug 18 2004
a(n) is also the number of Grand-Dyck paths of semilength n starting with an up-step and avoiding the pattern DUD. - David Bevan, Nov 19 2019
Hankel transform of a(n+1) = [1,2,5,13,35,96,...] gives A000012 = [1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
Equals row sums of triangle A136787 starting (1, 2, 5, 13, 35, ...). - Gary W. Adamson, Jan 21 2008
a(n) is the number of permutations on [n] that avoid the patterns 1-23-4 and 1-3-2, where the omission of a dash in a pattern means the permutation entries must be adjacent. Example: a(4) = 13 counts all 14 (Catalan number) (1-3-2)-avoiding permutations on [4] except 1234. - David Callan, Jul 22 2008
a(n) is also the number of involutions of length 2n-2 which are invariant under the reverse-complement map and have no decreasing subsequences of length 4. - Eric S. Egge, Oct 21 2008
Hankel transform is A010892. - Paul Barry, Jan 19 2009
a(n) is the number of Dyck words of semilength n with no DUUU. For example, a(4) = 14-1 = 13 because there is only one Dyck 4-word containing DUUU, namely UDUUUDDD. - Eric Rowland, Apr 21 2009
Inverse binomial transform of A024718. - Philippe Deléham, Dec 13 2009
Let w(i, j, n) denote walks in N^2 which satisfy the multivariate recurrence
w(i, j, n) = w(i - 1, j, n - 1) + w(i, j - 1, n - 1) + w(i + 1, j - 1,n - 1) with boundary conditions w(0,0,0) = 1 and w(i,j,n) = 0 if i or j or n is < 0. Let alpha(n) the number of such walks of length n, alpha(n) = Sum_{i = 0..n, j=0..n} w(i, j, n). Then a(n+1) = alpha(n). - Peter Luschny, May 21 2011
Number of length-n strings [d(0),d(1),d(2),...,d(n-1)] where 0 <= d(k) <= k and abs(d(k) - d(k-1)) <= 1 (smooth factorial numbers, see example). - Joerg Arndt, Nov 10 2012
a(n) is the number of n-multisets of {1,...,n} containing no pair of consecutive integers (e.g., 111, 113, 133, 222, 333 for n=3). - David Bevan, Jun 10 2013
a(n) is also the number of n-multisets of [n] in which no integer except n occurs exactly once (e.g., 111, 113, 222, 223, 333 for n=3). - David Bevan, Nov 19 2019
Number of minimax elements in the affine Weyl group of the Lie algebra so(2n+1) or the Lie algebra sp(2n). See Panyushev 2005. Cf. A245455. - Peter Bala, Jul 22 2014
The shifted, signed array belongs to an interpolated family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the interpolating (here t=-2) o.g.f. G(x,t) = (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse o.g.f. Ginv(x,t) = x(1-x)/(1+(t-1)x(1-x)) (A057682). See A091867 for more info on this family. - Tom Copeland, Nov 09 2014
Alternatively, this sequence corresponds to the number of positive walks with n steps {-1,0,1} starting at the origin, ending at any altitude, and staying strictly above the x-axis. - David Nguyen, Dec 01 2016
Let N be a squarefree number with n prime factors: p_1 < p_2 < ... < p_n. Let D be its set of divisors, E the subset of D X D made of the (d_1, d_2) for which, provided that we know which p_i are in d_1, which p_i are in d_2, d_1 <= d_2 is provable without needing to know the numerical values of the p_i. It appears that a(n+1) is the number of (d_1, d_2) in E such that d_1 and d_2 are coprime. - Luc Rousseau, Aug 21 2017
Number of ordered rooted trees with n non-root nodes and all non-root nodes having outdegrees 1 or 2. - Andrew Howroyd, Dec 04 2017
a(n) is the number of compositions (ordered partitions) of n where there are A001006(k-1) sorts of part k (see formula by Andrew Howroyd, Dec 04 2017). - Joerg Arndt, Jan 26 2024

Examples

			G.f. = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 35*x^5 + 96*x^6 + 267*x^7 + ...
a(3) = 5, a(4) = 13; since the top row of M^3 = (5, 5, 2, 1, ...)
From _Eric Rowland_, Sep 25 2021: (Start)
There are a(4) = 13 directed animals of size 4:
  O
  O    O    O    OO              O         O
  O    O    OO   O    OO   O    OO   OOO   O    O    OO    O
  O    OO   O    O    OO   OOO  O    O    OO   OOO  OO   OOO  OOOO
(End)
From _Joerg Arndt_, Nov 10 2012: (Start)
There are a(4)=13 smooth factorial numbers of length 4 (dots for zeros):
[ 1]   [ . . . . ]
[ 2]   [ . . . 1 ]
[ 3]   [ . . 1 . ]
[ 4]   [ . . 1 1 ]
[ 5]   [ . . 1 2 ]
[ 6]   [ . 1 . . ]
[ 7]   [ . 1 . 1 ]
[ 8]   [ . 1 1 . ]
[ 9]   [ . 1 1 1 ]
[10]   [ . 1 1 2 ]
[11]   [ . 1 2 1 ]
[12]   [ . 1 2 2 ]
[13]   [ . 1 2 3 ]
(End)
From _Joerg Arndt_, Nov 22 2012: (Start)
There are a(4)=13 base 3 4-digit numbers (not starting with 0) with digit sum 4:
[ 1]   [ 2 2 . . ]
[ 2]   [ 2 1 1 . ]
[ 3]   [ 1 2 1 . ]
[ 4]   [ 2 . 2 . ]
[ 5]   [ 1 1 2 . ]
[ 6]   [ 2 1 . 1 ]
[ 7]   [ 1 2 . 1 ]
[ 8]   [ 2 . 1 1 ]
[ 9]   [ 1 1 1 1 ]
[10]   [ 1 . 2 1 ]
[11]   [ 2 . . 2 ]
[12]   [ 1 1 . 2 ]
[13]   [ 1 . 1 2 ]
(End)
		

References

  • J. E. Goodman and J. O'Rourke, editors, Handbook of Discrete and Computational Geometry, CRC Press, 1997, p. 237.
  • T. Mansour, Combinatorics of Set Partitions, Discrete Mathematics and Its Applications, CRC Press, 2013, p. 377.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.46a.
  • R. P. Stanley, Catalan Numbers, Cambridge, 2015, p. 132.

Crossrefs

See also A005775. Inverse of A001006. Also sum of numbers in row n+1 of array T in A026300. Leading column of array in A038622.
The right edge of the triangle A062105.
Column k=3 of A295679.
Interpolates between Motzkin numbers (A001006) and Catalan numbers (A000108). Cf. A054391, A054392, A054393, A055898.
Except for the first term a(0), sequence is the binomial transform of A001405.
a(n) = A002426(n-1) + A005717(n-1) if n > 0. - Emeric Deutsch, Aug 14 2002

Programs

  • Haskell
    a005773 n = a005773_list !! n
    a005773_list = 1 : f a001006_list [] where
       f (x:xs) ys = y : f xs (y : ys) where
         y = x + sum (zipWith (*) a001006_list ys)
    -- Reinhard Zumkeller, Mar 30 2012
    
  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 2*x/(3*x-1+Sqrt(1-2*x-3*x^2)) )); // G. C. Greubel, Apr 05 2019
  • Maple
    seq( sum(binomial(i-1, k)*binomial(i-k, k), k=0..floor(i/2)), i=0..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    A005773:=proc(n::integer)
    local i, j, A, istart, iend, KartProd, Liste, Term, delta;
        A:=0;
        for i from 0 to n do
            Liste[i]:=NULL;
            istart[i]:=0;
            iend[i]:=n-i+1:
            for j from istart[i] to iend[i] do
                Liste[i]:=Liste[i], j;
            end do;
            Liste[i]:=[Liste[i]]:
        end do;
        KartProd:=cartprod([seq(Liste[i], i=1..n)]);
        while not KartProd[finished] do
            Term:=KartProd[nextvalue]();
            delta:=1;
            for i from 1 to n-1 do
                if (op(i, Term) - op(i+1, Term))^2 >= 2 then
                    delta:=0;
                    break;
                end if;
            end do;
            A:=A+delta;
        end do;
    end proc; # Thomas Wieder, Feb 22 2009:
    # n -> [a(0),a(1),..,a(n)]
    A005773_list := proc(n) local W, m, j, i;
    W := proc(i, j, n) option remember;
    if min(i, j, n) < 0 or max(i, j) > n then 0
    elif n = 0 then if i = 0 and j = 0 then 1 else 0 fi
    else W(i-1,j,n-1)+W(i,j-1,n-1)+W(i+1,j-1,n-1) fi end:
    [1,seq(add(add(W(i,j,m),i=0..m),j=0..m),m=0..n-1)] end:
    A005773_list(27); # Peter Luschny, May 21 2011
    A005773 := proc(n)
        option remember;
        if n <= 1 then
            1 ;
        else
            2*n*procname(n-1)+3*(n-2)*procname(n-2) ;
            %/n ;
        end if;
    end proc:
    seq(A005773(n),n=0..10) ; # R. J. Mathar, Jul 25 2017
  • Mathematica
    CoefficientList[Series[(2x)/(3x-1+Sqrt[1-2x-3x^2]), {x,0,40}], x] (* Harvey P. Dale, Apr 03 2011 *)
    a[0]=1; a[n_] := Sum[k/n*Sum[Binomial[n, j]*Binomial[j, 2*j-n-k], {j, 0, n}], {k, 1, n}]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 31 2015, after Vladimir Kruchinin *)
    A005773[n_] := 2 (-1)^(n+1) JacobiP[n - 1, 3, -n -1/2, -7] / (n^2 + n); A005773[0] := 1; Table[A005773[n], {n, 0, 27}] (* Peter Luschny, May 25 2021 *)
  • PARI
    a(n)=if(n<2,n>=0,(2*n*a(n-1)+3*(n-2)*a(n-2))/n)
    
  • PARI
    for(n=0, 27, print1(if(n==0, 1, sum(k=0, n-1, (-1)^(n - 1 + k)*binomial(n - 1, k)*binomial(2*k + 1, k + 1))),", ")) \\ Indranil Ghosh, Mar 14 2017
    
  • PARI
    Vec(1/(1-serreverse(x*(1-x)/(1-x^3) + O(x*x^25)))) \\ Andrew Howroyd, Dec 04 2017
    
  • Sage
    def da():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        yield 1
        while True:
            yield b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)//n
    A005773 = da()
    print([next(A005773) for  in range(28)]) # _Peter Luschny, May 16 2016
    
  • Sage
    (2*x/(3*x-1+sqrt(1-2*x-3*x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Apr 05 2019
    

Formula

G.f.: 2*x/(3*x-1+sqrt(1-2*x-3*x^2)). - Len Smiley
Also a(0)=1, a(n) = Sum_{k=0..n-1} M(k)*a(n-k-1), where M(n) are the Motzkin numbers (A001006).
D-finite with recurrence n*a(n) = 2*n*a(n-1) + 3*(n-2)*a(n-2), a(0)=a(1)=1. - Michael Somos, Feb 02 2002
G.f.: 1/2+(1/2)*((1+x)/(1-3*x))^(1/2). Related to Motzkin numbers A001006 by a(n+1) = 3*a(n) - A001006(n-1) [see Yaqubi Lemma 2.6].
a(n) = Sum_{q=0..n} binomial(q, floor(q/2))*binomial(n-1, q) for n > 0. - Emeric Deutsch, Aug 15 2002
From Paul Barry, Jun 22 2004: (Start)
a(n+1) = Sum_{k=0..n} (-1)^(n+k)*C(n, k)*C(2*k+1, k+1).
a(n) = 0^n + Sum_{k=0..n-1} (-1)^(n+k-1)*C(n-1, k)*C(2*k+1, k+1). (End)
a(n+1) = Sum_{k=0..n} (-1)^k*3^(n-k)*binomial(n, k)*A000108(k). - Paul Barry, Jan 27 2005
Starting (1, 2, 5, 13, ...) gives binomial transform of A001405 and inverse binomial transform of A001700. - Gary W. Adamson, Aug 31 2007
Starting (1, 2, 5, 13, 35, 96, ...) gives row sums of triangle A132814. - Gary W. Adamson, Aug 31 2007
G.f.: 1/(1-x/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 19 2009
G.f.: 1+x/(1-2*x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-.... (continued fraction). - Paul Barry, Jan 19 2009
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))^2 >= 2 for i=1..n-1 and delta(l_1,l_2,..., l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
INVERT transform of offset Motzkin numbers (A001006): (a(n)){n>=1}=(1,1,2,4,9,21,...). - _David Callan, Aug 27 2009
A005773(n) = ((n+3)*A001006(n+1) + (n-3)*A001006(n)) * (n+2)/(18*n) for n > 0. - Mark van Hoeij, Jul 02 2010
a(n) = Sum_{k=1..n} (k/n * Sum_{j=0..n} binomial(n,j)*binomial(j,2*j-n-k)). - Vladimir Kruchinin, Sep 06 2010
a(0) = 1; a(n+1) = Sum_{t=0..n} n!/((n-t)!*ceiling(t/2)!*floor(t/2)!). - Andrew S. Hays, Feb 02 2011
a(n) = leftmost column term of M^n*V, where M = an infinite quadradiagonal matrix with all 1's in the main, super and subdiagonals, [1,0,0,0,...] in the diagonal starting at position (2,0); and rest zeros. V = vector [1,0,0,0,...]. - Gary W. Adamson, Jun 16 2011
From Gary W. Adamson, Jul 29 2011: (Start)
a(n) = upper left term of M^n, a(n+1) = sum of top row terms of M^n; M = an infinite square production matrix in which the main diagonal is (1,1,0,0,0,...) as follows:
1, 1, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 0, 1, 0, 0, ...
1, 1, 1, 0, 1, 0, ...
1, 1, 1, 1, 0, 1, ...
1, 1, 1, 1, 1, 0, ... (End)
Limit_{n->oo} a(n+1)/a(n) = 3.0 = lim_{n->oo} (1 + 2*cos(Pi/n)). - Gary W. Adamson, Feb 10 2012
a(n) = A025565(n+1) / 2 for n > 0. - Reinhard Zumkeller, Mar 30 2012
With first term deleted: E.g.f.: a(n) = n! * [x^n] exp(x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). - Peter Luschny, Aug 25 2012
G.f.: G(0)/2 + 1/2, where G(k) = 1 + 2*x*(4*k+1)/( (2*k+1)*(1+x) - x*(1+x)*(2*k+1)*(4*k+3)/(x*(4*k+3) + (1+x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 24 2013
a(n) ~ 3^(n-1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Jul 30 2013
For n > 0, a(n) = (-1)^(n+1) * hypergeom([3/2, 1-n], [2], 4). - Vladimir Reshetnikov, Apr 25 2016
a(n) = GegenbauerC(n-2,-n+1,-1/2) + GegenbauerC(n-1,-n+1,-1/2) for n >= 1. - Peter Luschny, May 12 2016
0 = a(n)*(+9*a(n+1) + 18*a(n+2) - 9*a(n+3)) + a(n+1)*(-6*a(n+1) + 7*a(n+2) - 2*a(n+3)) + a(n+2)*(-2*a(n+2) + a(n+3)) for n >= 0. - Michael Somos, Dec 01 2016
G.f.: 1/(1-x*G(x)) where G(x) is g.f. of A001006. - Andrew Howroyd, Dec 04 2017
a(n) = (-1)^(n + 1)*2*JacobiP(n - 1, 3, -n - 1/2, -7)/(n^2 + n). - Peter Luschny, May 25 2021
a(n+1) = A005043(n) + 2*A005717(n) for n >= 1. - Peter Bala, Feb 11 2022
a(n) = Sum_{k=0..n-1} A064189(n-1,k) for n >= 1. - Alois P. Heinz, Aug 29 2022

A111808 Left half of trinomial triangle (A027907), triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 7, 1, 4, 10, 16, 19, 1, 5, 15, 30, 45, 51, 1, 6, 21, 50, 90, 126, 141, 1, 7, 28, 77, 161, 266, 357, 393, 1, 8, 36, 112, 266, 504, 784, 1016, 1107, 1, 9, 45, 156, 414, 882, 1554, 2304, 2907, 3139, 1, 10, 55, 210, 615, 1452, 2850, 4740, 6765, 8350
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 17 2005

Keywords

Comments

Consider a doubly infinite chessboard with squares labeled (n,k), ranks or rows n in Z, files or columns k in Z (Z denotes ...,-2,-1,0,1,2,... ); number of king-paths of length n from (0,0) to (n,k), 0 <= k <= n, is T(n,n-k). - Harrie Grondijs, May 27 2005. Cf. A026300, A114929, A114972.
Triangle of numbers C^(2)(n-1,k), n>=1, of combinations with repetitions from elements {1,2,...,n} over k, such that every element i, i=1,...,n, appears in a k-combination either 0 or 1 or 2 times (cf. also A213742-A213745). - Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

References

  • Harrie Grondijs, Neverending Quest of Type C, Volume B - the endgame study-as-struggle.

Crossrefs

Row sums give A027914; central terms give A027908;
T(n, 0) = 0;
T(n, 1) = n for n>1;
T(n, 2) = A000217(n) for n>1;
T(n, 3) = A005581(n) for n>2;
T(n, 4) = A005712(n) for n>3;
T(n, 5) = A000574(n) for n>4;
T(n, 6) = A005714(n) for n>5;
T(n, 7) = A005715(n) for n>6;
T(n, 8) = A005716(n) for n>7;
T(n, 9) = A064054(n-5) for n>8;
T(n, n-5) = A098470(n) for n>4;
T(n, n-4) = A014533(n-3) for n>3;
T(n, n-3) = A014532(n-2) for n>2;
T(n, n-2) = A014531(n-1) for n>1;
T(n, n-1) = A005717(n) for n>0;
T(n, n) = central terms of A027907 = A002426(n).

Programs

  • Maple
    T := (n,k) -> simplify(GegenbauerC(k, -n, -1/2)):
    for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, May 09 2016
  • Mathematica
    Table[GegenbauerC[k, -n, -1/2], {n,0,10}, {k,0,n}] // Flatten (* G. C. Greubel, Feb 28 2017 *)

Formula

(1 + x + x^2)^n = Sum(T(n,k)*x^k: 0<=k<=n) + Sum(T(n,k)*x^(2*n-k): 0<=k
T(n, k) = A027907(n, k) = Sum_{i=0,..,(k/2)} binomial(n, n-k+2*i) * binomial(n-k+2*i, i), 0<=k<=n.
T(n, k) = GegenbauerC(k, -n, -1/2). - Peter Luschny, May 09 2016

Extensions

Corrected and edited by Johannes W. Meijer, Oct 05 2010

A243827 Number A(n,k) of Dyck paths of semilength n having exactly one occurrence of the consecutive step pattern given by the binary expansion of k, where 1=U=(1,1) and 0=D=(1,-1); square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 3, 1, 0, 0, 0, 0, 1, 4, 6, 1, 0, 0, 0, 0, 1, 2, 11, 10, 1, 0, 0, 0, 0, 0, 4, 6, 26, 15, 1, 0, 0, 0, 0, 0, 1, 11, 16, 57, 21, 1, 0, 0, 0, 0, 0, 1, 4, 26, 45, 120, 28, 1, 0, 0, 0, 0, 1, 1, 5, 15, 57, 126, 247, 36, 1, 0, 0
Offset: 0

Author

Alois P. Heinz, Jun 11 2014

Keywords

Examples

			Square array A(n,k) begins:
  0, 0, 0,  0,   0,    0,   0,    0,    0,    0, ...
  1, 1, 1,  0,   0,    0,   0,    0,    0,    0, ...
  0, 0, 1,  1,   1,    1,   1,    0,    0,    0, ...
  0, 0, 1,  3,   4,    2,   4,    1,    1,    1, ...
  0, 0, 1,  6,  11,    6,  11,    4,    5,    5, ...
  0, 0, 1, 10,  26,   16,  26,   15,   21,   17, ...
  0, 0, 1, 15,  57,   45,  57,   50,   78,   54, ...
  0, 0, 1, 21, 120,  126, 120,  161,  274,  177, ...
  0, 0, 1, 28, 247,  357, 247,  504,  927,  594, ...
  0, 0, 1, 36, 502, 1016, 502, 1554, 3061, 1997, ...
		

Crossrefs

Columns k=2-10 give: A000012(n) for n>0, A000217(n-1) for n>0, A000295(n-1) for n>0, A005717(n-1) for n>1, A000295(n-1) for n>0, A014532(n-2) for n>2, A108863, A244235, A244236.
Main diagonal gives A243770 or column k=1 of A243752.

A014531 Form array in which n-th row is obtained by expanding (1+x+x^2)^n and taking the 2nd column from the center.

Original entry on oeis.org

1, 3, 10, 30, 90, 266, 784, 2304, 6765, 19855, 58278, 171106, 502593, 1477035, 4343160, 12778152, 37616427, 110797569, 326527350, 962803170, 2840372304, 8383467708, 24755608584, 73133433800, 216143407675, 639062383401
Offset: 1

Keywords

Comments

Number of "up" steps in all Motzkin paths of length n+1. E.g. a(2)=3 because in the four Motzkin paths of length 3, HHH, HUD, UDH and UHD, where H=(1,0), U=(1,1), D=(1,-1), we have altogether three U steps. - Emeric Deutsch, Dec 26 2003
a(n-1) = A111808(n,n-2) for n>1. - Reinhard Zumkeller, Aug 17 2005
a(n) = number of paths in the half-plane x>=0, from (0,0) to (n+1,2), and consisting of steps U=(1,1), D=(1,-1) and H=(1,0). For example, for n=2, we have the 3 paths: UUH, HUU, UHU. - José Luis Ramírez Ramírez, Apr 19 2015

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 78.

Crossrefs

First differences are in A025180.

Programs

  • Maple
    seq( add(binomial(i+1,k)*binomial(i-k+1,k+2), k=0..floor(i/2)), i=1..30 ); # Detlef Pauly (dettodet(AT)yahoo.de), Nov 09 2001
    a := n -> simplify(GegenbauerC(n-1, -n-1, -1/2)):
    seq(a(n), n=1..26); # Peter Luschny, May 09 2016
  • Mathematica
    Table[Sum[Binomial[i + 1, k]*Binomial[i - k + 1, k + 2], {k, 0, Floor[i/2]}], {i, 30}] (* Michael De Vlieger, Apr 20 2015 *)
    Table[GegenbauerC[n - 1, -n - 1, -1/2], {n,1,50}] (* G. C. Greubel, Feb 28 2017 *)
  • PARI
    for(n=1,25, print1(sum(k=0,n+1, binomial(n+1,k)*binomial(n-k+1,k+2)), ", ")) \\ G. C. Greubel, Feb 28 2017
  • Sage
    a = lambda n: n*(n+1)*hypergeometric([(1-n)/2, 1-n/2], [3], 4)/2
    [simplify(a(n)) for n in (1..26)] # Peter Luschny, Nov 23 2014
    

Formula

a(n) = A002426(n+1)-A001006(n+1) = a(n-1)+A005717(n)+A014532(n-2) - Henry Bottomley, May 15 2001
E.g.f.: exp(x)*(2*x*BesselI(1, 2*x)+(x-2)*BesselI(2, 2*x))/x. - Vladeta Jovovic, Aug 21 2003
G.f.: [1-2z-z^2-(1-z)q]/(2z^3q), where q=sqrt(1-2z-3z^2). - Emeric Deutsch, Dec 26 2003
a(n) = Sum_{k=0..n+1} C(n+1,k)*C(n-k+1,k+2). - Paul Barry, Sep 20 2004
D-finite with recurrence (n+3)*(n-1)*a(n) -(n+1)*(2n+1)*a(n-2)-3*n*(n+1)*a(n-2)=0. - R. J. Mathar, Dec 08 2011
a(n) = n*(n+1)*hypergeom([(1-n)/2, 1-n/2], [3], 4)/2. - Peter Luschny, Nov 23 2014
G.f.: z*M(z)^2/(1-z-2*z^2*M(z)), where M(z) is the g.f. of Motzkin paths. - José Luis Ramírez Ramírez, Apr 19 2015
a(n) = GegenbauerC(n-1, -n-1, -1/2). - Peter Luschny, May 09 2016
a(n) = Sum_{k>0} k * A055151(n+1,k). - Alois P. Heinz, Mar 29 2020

Extensions

More terms from James Sellers, Feb 05 2000

A091869 Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n having k peaks at even height.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 6, 3, 1, 9, 16, 12, 4, 1, 21, 45, 40, 20, 5, 1, 51, 126, 135, 80, 30, 6, 1, 127, 357, 441, 315, 140, 42, 7, 1, 323, 1016, 1428, 1176, 630, 224, 56, 8, 1, 835, 2907, 4572, 4284, 2646, 1134, 336, 72, 9, 1, 2188, 8350, 14535, 15240, 10710, 5292, 1890, 480, 90, 10, 1
Offset: 1

Author

Emeric Deutsch, Mar 10 2004

Keywords

Comments

Number of ordered trees with n edges having k leaves at even height. Row sums are the Catalan numbers (A000108). T(n,0)=A001006(n-1) (the Motzkin numbers). Sum_{k=0..n-1} k*T(n,k) = binomial(2n-2, n-2) = A001791(n-1). Mirror image of A091187.
T(n,k) is the number of Dyck paths of semilength n and having k dud's (here u=(1,1) and d=(1,-1)). Example: T(4,2)=3 because we have uud(du[d)ud], uu(dud)(dud) and uu(du[d)ud]d (the dud's are shown between parentheses).
T(n,k) is the number of Dyck paths of semilength n and containing exactly k double rises whose matching down steps form a doublefall. Example: UUUDUDDD has 2 double rises but only the first has matching Ds - the path's last 2 steps - forming a doublefall. (Travel horizontally east from an up step to encounter its matching down step.) - David Callan, Jul 15 2004
T(n,k) is the number of ordered trees on n edges containing k edges of outdegree 1. (The outdegree of an edge is the outdegree of its child vertex. Thus edges of outdegree 1 correspond to non-root vertices of outdegree 1.) T(3,2)=2 because
/\.../\.
|.....|.
each have one edge of outdegree 1. - David Callan, Oct 25 2004
Exponential Riordan array [exp(x)*Bessel_I(1,2x)/x, x]. - Paul Barry, Mar 09 2010
T(n, k) is the number of Dyck paths of semilength n and having k udu's (here u=(1,1) and d=(1,-1)). Note that reversing a path swaps u and d, thus udu becomes dud and vice versa. - Michael Somos, Feb 26 2020

Examples

			T(4,1)=6 because we have u(ud)dudud, udu(ud)dud, ududu(ud)d, uuudd(ud)d, u(ud)uuddd and uuu(ud)ddd (here u=(1,1), d=(1,-1) and the peaks at even height are shown between parentheses).
Triangle begins:
    1;
    1,    1;
    2,    2,    1;
    4,    6,    3,    1;
    9,   16,   12,    4,    1;
   21,   45,   40,   20,    5,    1;
   51,  126,  135,   80,   30,    6,   1;
  127,  357,  441,  315,  140,   42,   7,  1;
  323, 1016, 1428, 1176,  630,  224,  56,  8, 1;
  835, 2907, 4572, 4284, 2646, 1134, 336, 72, 9, 1;
  ...
		

Crossrefs

Programs

  • Maple
    T := proc(n,k) if k0, b(x-1, y-1, 0)*z^irem(t*y+t, 2), 0)+
          `if`(y (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0$2)):
    seq(T(n), n=1..16);  # Alois P. Heinz, May 12 2017
  • Mathematica
    (* m = MotzkinNumber *) m[0] = 1; m[n_] := m[n] = m[n - 1] + Sum[m[k]*m[n - 2 - k], {k, 0, n - 2}]; t[n_, n_] = 1; t[n_, k_] := m[n - k]*Binomial[n - 1, k - 1]; Table[t[n, k], {n, 1, 11}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 10 2013 *)
  • PARI
    {T(n, k) = my(y, c, w); if( k<0 || k>=n, 0, w = vector(n);   forvec(v=vector(2*n, k, [0, 1]), c=y=0; for(k=1, 2*n, if( 0>(y += (-1)^v[k]), break)); if( y, next); for(i=1, 2*n-2, c += ([0, 1, 0] == v[i..i+2])); w[c+1]++); w[k+1])}; /* Michael Somos, Feb 26 2020 */

Formula

T(n, k) = binomial(n-1, k)*(Sum_{j=0..ceiling((n-k)/2)} binomial(n-k, j)*binomial(n-k-j, j-1))/(n-k) for 0 <= k < n; T(n, k)=0 for k >= n.
G.f.: G = G(t, z) satisfies z*G^2 - (1 + z - t*z)*G + 1 + z - t*z = 0.
T(n, k) = M(n-k-1)*binomial(n-1, k), where M(n) = A001006(n) are the Motzkin numbers.
T(n+1, k+1) = n*T(n, k)/(k+1). - David Callan, Dec 09 2004
G.f.: 1/(1-x-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-x-xy-x^2/(1-... (continued fraction). - Paul Barry, Aug 03 2009
E.g.f.: exp(x+xy)*Bessel_I(1,2x)/x. - Paul Barry, Mar 10 2010

A189912 Extended Motzkin numbers, Sum_{k>=0} C(n,k)*C(k), where C(k) is the extended Catalan number A057977(k).

Original entry on oeis.org

1, 2, 4, 10, 25, 66, 177, 484, 1339, 3742, 10538, 29866, 85087, 243478, 699324, 2015082, 5822619, 16865718, 48958404, 142390542, 414837699, 1210439958, 3536809521, 10347314544, 30306977757, 88861597426, 260798283502, 766092871654, 2252240916665
Offset: 0

Author

Peter Luschny, May 01 2011

Keywords

Comments

a(n) = Sum_{k=0..n} binomial(n,k)*A057977(k). For comparison:
A001006(n) = Sum_{k=0..n} binomial(n,k)*A057977(k)*[k is even],
A005717(n) = Sum_{k=0..n} binomial(n,k)*A057977(k)*[k is odd].
Thus one might simply say: The extended Motzkin numbers are the binomial sum of the extended Catalan numbers. Moreover: The Catalan numbers aerated with 0's at odd positions (A126120) are the inverse binomial transform of the Motzkin numbers (A001006). The complementary Catalan numbers (A001700) aerated with 0's at even positions (A138364) are the inverse binomial transform of the complementary Motzkin numbers (A005717). The extended Catalan numbers (A057977 = A126120 + A138364) are the inverse binomial transform of the extended Motzkin numbers (A189912).
David Scambler observed that [1, a(n-1)] for n >= 1 count the Dyck paths of semilength n which satisfy the condition "number of peaks <= number of returns + number of hills". - Peter Luschny, Oct 22 2012

Programs

  • Maple
    A189912 := proc(n) local k;
    add(n!/(((n-k)!*iquo(k,2)!^2)*(iquo(k,2)+1)),k=0..n) end:
    M := proc(n) option remember; `if`(n<2, 1, (3*(n-1)*M(n-2)+(2*n+1)*M(n-1))/(n+2)) end:
    A189912 := n -> n*M(n-1)+M(n);
    seq(A189912(i), i=0..28); # Peter Luschny, Sep 12 2011
  • Mathematica
    A057977[n_] := n!/(Quotient[n, 2]!^2*(Quotient[n, 2] + 1)); a[n_] := Sum[Binomial[n, k]*A057977[k], {k, 0, n}]; Table[a[n], {n, 0, 28}] (* Jean-François Alcover, May 21 2013, after Peter Luschny *)
    Table[Sum[n!/(((n-k)!*Floor[k/2]!^2)*(Floor[k/2]+1)), {k,0,n}], {n,0,30}] (* G. C. Greubel, Jan 24 2017 *)
    A057977[n_] :=  Sum[n! (n + 1 - 2 k)/((k + 1)! (k!) (n - 2 k)!), {k, 0, n}] (* Per W. Alexandersson, May 28 2020 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n,k)*k!/( (k\2)!^2 * (k\2+1)) );
    vector(30, n, a(n-1)) \\ G. C. Greubel, Jan 24 2017; Mar 28 2020
  • Sage
    @CachedFunction
    def M(n): return (3*(n-1)*M(n-2)+(2*n+1)*M(n-1))/(n+2) if n>1 else 1
    A189912 = lambda n: n*M(n-1) + M(n)
    [A189912(i) for i in (0..28)] # Peter Luschny, Oct 22 2012
    

Formula

a(n) = Sum_{k=0..n} n!/(((n-k)!*floor(k/2)!^2)*(floor(k/2)+1)).
Recurrence: (n+2)*(n^2 + 2*n - 5)*a(n) = (2*n^3 + 7*n^2 - 14*n - 7)*a(n-1) + 3*(n-1)*(n^2 + 4*n - 2)*a(n-2). - Vaclav Kotesovec, Mar 20 2014
a(n) ~ 3^(n+1/2) / (2*sqrt(Pi*n)). - Vaclav Kotesovec, Mar 20 2014
Conjecture: a(n) = Sum_{k=0..floor(n/2)} (n+1-2*k)*A055151(n,k). - Werner Schulte, Oct 23 2016
a(n) = Sum_{k=0..floor(n/2)} (n+1-2*k)*n!/(k!*(k+1)!*(n-2*k)!). - Per W. Alexandersson, May 28 2020
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