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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

Words of length n in language defined by L = 1 + a + (L)L: L(0)=1, L(1)=a, L(2)=(), L(3)=(a)+()a, L(4)=(())+(a)a+()(), ...

Series reversion of x*A(x) is x*A082582(-x). - Michael Somos, Jul 22 2003

a(n) = number of Motzkin n-paths (A001006) in which no flatstep (F) is immediately followed by either an upstep (U) or a flatstep, in other words, each flatstep is either followed by a downstep (D) or ends the path. For example, a(4)=3 counts UDUD, UFDF, UUDD. - David Callan, Jun 07 2006

a(n) = number of Dyck (n+1)-paths (A000108) containing no UDUs and no subpaths of the form UUPDD where P is a nonempty Dyck path. For example, a(4)=3 counts UUUDDUUDDD, UUDDUUDDUD, UUUDDUDDUD. - David Callan, Sep 25 2006

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_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-1}*a_0 for n >= t. For example phi([1]) is the Catalan numbers A000108. The present sequence is (essentially) phi([0,1,1]). - Gary W. Adamson and R. J. Mathar, Oct 27 2008

The Kn21(n) triangle sums of A175136 lead to A007477(n+1), while the Kn22(n) = A007477(n+3)-1, Kn23(n) = A007477(n+5)-(4+n) and Kn3(n) = A007477(2*n+1) triangle sums of A175136 are related to the sequence given above. For the definition of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011

For n>=2, a(n) gives number of possible, ways to parse an English sentence consisting of just n+1 copies of word "buffalo", with one particular "plausible" grammar. See the Wikipedia page and my Python source at OEIS Wiki. Whether these are really intelligible sentences is of course debatable. See A213705 for a more plausible example in the Finnish language. - Antti Karttunen, Sep 14 2012

Number of Dyck paths of semilength n, having k uu's with midpoint at even height. Example: T(4,1) = 6 because we have u(uu)duddd, u(uu)udddd, udu(uu)ddd, u(uu)dddud, u(uu)ddudd and uud(uu)ddd [here u = (1,1), d = (1,-1) and the uu's with midpoint at even height are shown between parentheses]. Row sums are the Catalan numbers (A000108). T(2n+1,n) = A000108(n) (the Catalan numbers). Sum_{k>=0} k*T(n,k) = binomial(2n-2,n-3) = A002694(n-1).

Sometimes called the Touchard distribution (after Touchard's Catalan number identity). T(n,k) = number of full binary trees on 2n edges with k deep interior vertices (deep interior means you have to traverse at least 2 edges to reach a leaf) = number of binary trees on n-1 edges with k vertices having a full complement of 2 children. - David Callan, Jul 19 2004

From David Callan, Oct 25 2004: (Start)

T(n,k) = number of ordered trees on n edges with k prolific edges. A prolific edge is one whose child vertex has at least two children. For example with n=3, drawing ordered trees down from the root, /|\ has no prolific edge and the only tree with one prolific edge has the shape of an inverted Y, so T(3,1)=1.

Proof: Consider the following bijection, recorded by Emeric Deutsch, from ordered trees on n edges to Dyck n-paths. For a given ordered tree, traverse the tree in preorder (walk-around from root order). To each node of outdegree r there correspond r upsteps followed by 1 downstep; nothing corresponds to the last leaf. This bijection sends prolific edges to noninitial ascents of length >=2, that is, to DUU's. Then reverse the resulting Dyck n-path so that prolific edges correspond to DDU's. (End)

T(n,k) is the number of Łukasiewicz paths of length n having k fall steps (1,-1) that start at an even level. A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1) (see R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps). Example: T(3,1) = 1 because we have U(2)(D)D, where U(2) = (1,2), D = (1,-1) and the fall steps that start at an even level are shown between parentheses. Row n contains ceiling(n/2) terms (n >= 1). - Emeric Deutsch, Jan 06 2005

Number of binary trees with n-1 edges and k+1 leaves (a binary tree is a rooted tree in which each vertex has at most two children and each child of a vertex is designated as its left or right child). - Emeric Deutsch, Jul 31 2006

Number of full binary trees with 2n edges and k+1 vertices both children of which are leaves (n >= 1; a full binary tree is a rooted tree in which each vertex has either 0 or two children). - Emeric Deutsch, Dec 26 2006

Number of ordered trees with n edges and k jumps. In the preorder traversal of an ordered tree, any transition from a node at a deeper level to a node on a strictly higher level is called a jump. - Emeric Deutsch, Jan 18 2007

It is remarkable that we can generate the coefficients of the right hand columns of triangle A175136 with the aid of the coefficients in the rows of the triangle given above. See A175136 for more information. - Johannes W. Meijer, May 06 2011

The antidiagonal sums equal A152225. - Johannes W. Meijer, Sep 13 2012

This array also counts 231-avoiding permutations according to the number of peaks, i.e., positions w[i-1] < w[i] > w[i+1]. For example, 123, 213, 312, and 321 have no peaks, while 132 has one peak. Note also T(n,k) = 2^(n - 1 - 2*k)*A055151(n-1,k). - Kyle Petersen, Aug 02 2013

The Kn4 triangle sums of A175136 lead to the sequence given above (n >= 1). For the definition of the Kn4 and other triangle sums see A180662. - Johannes W. Meijer, May 06 2011

Equals the expected value of trace(O)^(2n), where O is a 3 X 3 orthogonal matrix randomly selected according to Haar measure (see MathOverflow link). - Nathaniel Johnston, Sep 05 2014

From Petros Hadjicostas, Jul 23 2020: (Start)

In Smith (1985), we apparently have a(n) = P(2*n), where P(n) is the number of linearly independent three-dimensional n-th order isotropic tensors. In the paper, he refers to Smith (1968) for more details. It is not clear why he does not list the values of P(2*n+1). See also the 1978 letter of D. L. Andrews to N. J. A. Sloane.

Eric Weisstein gives some details on how the material in Smith (1968) about isotropic tensors is related to Motzkin sums. (End)

Partial sums of A147875.

Equals the fourth right hand column of A175136 for n>=1. - Johannes W. Meijer, May 06 2011

a(n) is the number of triples (w,x,y) havingt all terms in {0,...,n} and x+y>w. - Clark Kimberling, Jun 14 2012

a(n) = A166299(n,0).

a(n) is the number of peakless Motzkin paths of length n with no (1,0)-steps at level 0. Example: a(5)=2 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have UHHHD and UUHDD. - Emeric Deutsch, May 03 2011

From Paul Barry, Mar 31 2011: (Start)

The Hankel transform of a(n+3) is A188444(n+1).

a(n+3) gives the diagonal sums of the triangle A100754.

a(n+3) has g.f. 1/(1-x-x^2/(1-2x+3x^2/(1+2x+x^2/(1-2x-(1/3)x^2/(1-x-(2/3)x^2/(1-2x+(5/2)x^2/(1+2x+(3/2)x^2/(1-...)))))))) (continued fraction) where the coefficients of x^2 have denominators A188442 and numerators A188443. (End)

The Ca1 triangle sums of triangle A175136 lead to this sequence (n>=3). For the definitions of the Ca1 and other triangle sums see A180662. - Johannes W. Meijer, May 06 2011

a(n) is the number of closed Deutsch paths of n steps with all peaks at even height. A Deutsch path is a lattice path of up-steps (1,1) and down-steps (1,-j), j>=1, starting at the origin that never goes below the x-axis, and it is closed if it ends on the x-axis. For example a(5) = 2 counts UUUU4, UU1U2, where U denotes an up-step and a down-step is denoted by its length, and a(6) = 5 counts UUUU13, UUUU22, UUUU31, UU1U11, UU2UU2. - David Callan, Dec 08 2021

Equals the fifth right hand column of A175136.

Equals the sixth right hand column of A175136.

The second left hand column of triangle A175136.

The third left hand column of triangle A175136.