A343386 -id:A343386 - cp's OEIS frontend

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

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

N. J. A. Sloane

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

			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.

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Index entries for "core" sequences

Cf. A026300, A005717, A020474, A001850, A004148. First column of A064191, A064189, A000108, A088615, A007971, A001405, A005817, A049401, A007579, A007578, A097862, A005773, A178515, A217275. First row of A064645.
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.).

a001006 n = a001006_list !! n
a001006_list = zipWith (+) a005043_list $ tail a005043_list
-- Reinhard Zumkeller, Jan 31 2012

# 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

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 *)

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 */

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 */

makelist(ultraspherical(n,-n-1,-1/2)/(n+1),n,0,12); /* Emanuele Munarini, Oct 20 2016 */

{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 */

{a(n) = if( n<0, 0, n++; polcoeff( serreverse( x / (1 + x + x^2) + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */

{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 */

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

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

# 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

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

cp's OEIS Frontend

A343773 Excess of the number of even Motzkin n-paths (A107587) over the odd ones (A343386).

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

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A359364 Triangle read by rows. The Motzkin triangle, the coefficients of the Motzkin polynomials. M(n, k) = binomial(n, k) * CatalanNumber(k/2) if k is even, otherwise 0.

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