cp's OEIS Frontend

Number of lattice paths from (0,0) to the line x=n-1 that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and three types of steps H=(1,0) (left factors of 3-Motzkin steps). Example: a(3)=17 because we have UD, UU, 9 HH paths, 3 HU paths and 3 UH paths. - Emeric Deutsch, Jan 22 2004

Also a(n) = number of integer strings s(0), ..., s(n) counted by array U in A026386 that have s(n)=1; a(n) = U(2n-1, n-1).

The Hankel transform of [1,1,4,17,75,339,1558,...] is [1,3,8,21,55,144,377,...] (see A001906). - Philippe Deléham, Apr 13 2007

Number of peaks in all skew Dyck paths of semilength n. A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps. Example: a(2)=4 because in the 3 (=A002212(2)) skew Dyck paths (UD)(UD), U(UD)D and U(UD)L we have altogether 4 peaks (shown between parentheses). - Emeric Deutsch, Jul 25 2007

Hankel transform of this sequence gives A000012 = [1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007

5th binomial transform of (-1)^n*A000108. - Paul Barry, Jan 13 2009

From Gary W. Adamson, May 17 2009: (Start)

Convolved with A007317, (1, 2, 5, 15, 51, ...) = A026376: (1, 6, 30, 144, ...)

Equals A026375, (1, 3, 11, 45, 195, ...) convolved with A002212 prefaced with

a 1: (1, 1, 3, 10, 36, 137, ...). (End)

From Tom Copeland, Nov 09 2014: (Start)

The 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 o.g.f. [1-sqrt(1-4x/(1+(1-t)x))]/2 and inverse x(1-x)/[1+(t-1)x(1-x)]. See A091867 for more info on this family. Here the interpolation is t=-4 (mod signs in the results).

Let C(x) = [1 - sqrt(1-4x)]/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).

O.g.f: G(x) = [-1 + sqrt(1 + 4*x/(1-5x))]/2 = -C[P(-x,5)].

Inverse O.g.f: Ginv(x) = x*(1+x)/[1 + 5x*(1+x)] = -P(Cinv(-x),-5) (signed A039717). (End)

A free pure symmetric multifunction (PSM) in one symbol x is either (case 1) the symbol x, or (case 2) an expression of the form h[g_1,...,g_k] where h is a PSM in x, each of the g_i for i=1..(k>0) is a PSM in x, and for i < j we have g_i <= g_j under a canonical total ordering such as the Mathematica ordering. The number of positions in a PSM is the number of brackets [...] plus the number of x's.

Mirror image of triangle A133336. - Philippe Deléham, Dec 10 2008

From Tom Copeland, Oct 09 2011: (Start)

With polynomials

P(0,t) = 0

P(1,t) = 1

P(2,t) = t

P(3,t) = t + 2 t^2

P(4,t) = t + 5 t^2 + 5 t^3

P(5,t) = t + 9 t^2 + 21 t^3 + 14 t^4

The o.g.f. A(x,t) = {1+x-sqrt[(1-x)^2-4xt]}/[2(1+t)] (see Drake et al.).

B(x,t)= x-t x^2/(1-x)= x-t(x^2+x^3+x^4+...) is the comp. inverse in x.

Let h(x,t) = 1/(dB/dx) = (1-x)^2/(1+(1+t)*x*(x-2)) = 1/(1-t(2x+3x^2+4x^3+...)), an o.g.f. in x for row polynomials in t of A181289. Then P(n,t) is given by (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A = exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t). These results are a special case of A133437 with u(x,t) = B(x,t), i.e., u_1=1 and (u_n)=-t for n > 1. See A001003 for t = 1. (End)

Let U(x,t) = [A(x,t)-x]/t, then U(x,0) = -dB(x,t)/dt and U satisfies dU/dt = UdU/dx, the inviscid Burgers' equation (Wikipedia), also called the Hopf equation (see Buchstaber et al.). Also U(x,t) = U(A(x,t),0) = U(x+tU,0) since U(x,0) = [x-B(x,t)]/t. - Tom Copeland, Mar 12 2012

Diagonals of A132081 are essentially rows of this sequence. - Tom Copeland, May 08 2012

T(r, s) is the number of [0,r]-covering hierarchies with s segments (see Kreweras). - Michel Marcus, Nov 22 2014

From Yu Hin Au, Dec 07 2019: (Start)

T(n,k) is the number of small Schröder n-paths (lattice paths from (0,0) to (2n,0) using steps U=(1,1), F=(2,0), D=(1,-1) with no F step on the x-axis) that has exactly k U steps.

T(n,k) is the number of Schröder trees (plane rooted tree where each internal node has at least two children) with exactly n+1 leaves and k internal nodes. (End)

Number of ordered trees with n edges having k leaves at odd height. Row sums are the Catalan numbers (A000108). T(n,0)=A005043(n). Sum_{k=0..n} k*T(n,k) = binomial(2n-2,n-1).

T(n,k)=number of Dyck paths of semilength n and having k ascents of length 1 (an ascent is a maximal string of consecutive up steps). Example: T(4,2)=6 because we have UdUduud, UduuddUd, uuddUdUd, uudUdUdd, UduudUdd and uudUddUd (the ascents of length 1 are indicated by U instead of u).

T(n,k) is the number of Łukasiewicz paths of length n having k level steps (i.e., (1,0)). 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(4,1)=4 because we have HU(2)DD, U(2)HDD, U(2)DHD and U(2)DDH, where H=(1,0), U(1,1), U(2)=(1,2) and D=(1,-1). - Emeric Deutsch, Jan 06 2005

T(n,k) = number of noncrossing partitions of [n] containing k singleton blocks. Also, T(n,k) = number of noncrossing partitions of [n] containing k adjacencies. An adjacency is an occurrence of 2 consecutive integers in the same block (here 1 and n are considered consecutive). In fact, the statistics # singletons and # adjacencies have a symmetric joint distribution.

Exponential Riordan array [e^x*(Bessel_I(0,2x)-Bessel_I(1,2x)),x]. - Paul Barry, Mar 03 2011

T(n,k) is the number of ordered trees having n edges and exactly k nodes with one child. - Geoffrey Critzer, Feb 25 2013

From Tom Copeland, Nov 04 2014: (Start)

Summing the coeff. of the partitions in A134264 for a Lagrange inversion formula (see also A249548) containing (h_1)^k = (1')^k gives this triangle, so this array's o.g.f. H(x,t) = x + t * x^2 + (1 + t^2) * x^3 ... is the inverse of the o.g.f. of A104597 with a sign change, i.e., H^(-1)(x,t) = (x-x^2) / [1 + (t-1)(x-x^2)] = Cinv(x)/[1 + (t-1)Cinv(x)] = P[Cinv(x),t-1] where Cinv(x)= x * (1-x) is the inverse of C(x) = [1-sqrt(1-4*x)]/2, an o.g.f. for the Catalan numbers A000108, and P(x,t) = x/(1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x/(1-t*x). Therefore,

O.g.f.: H(x,t) = C[Pinv(x,t-1)] = C[P(x,1-t)] = C[x/(1-(t-1)x)] = {1-sqrt[1-4*x/(1-(t-1)x)]}/2 (for A091867). Reprising,

Inverse O.g.f.: H^(-1)(x,t) = x*(1-x) / [1 + (t-1)x(1-x)] = P[Cinv(x),t-1].

From general arguments in A134264, the row polynomials are an Appell sequence with lowering operator d/dt, having the umbral property (p(.,t)+a)^n=p(n,t+a) with e.g.f. = e^(x*t)/w(x), where 1/w(x)= e.g.f. of first column for the Motzkin numbers in A005043. (Mislabeled argument corrected on Jan 31 2016.)

Cf. A124644 (t-shifted polynomials), A026378 (t=-4), A001700 (t=-3), A005773 (t=-2), A126930 (t=-1) and A210736 (t=-1, a(0)=0, unsigned), A005043 (t=0), A000108 (t=1), A007317 (t=2), A064613 (t=3), A104455 (t=4), A030528 (for inverses).

(End)

The sequence of binomial transforms A126930, A005043, A000108, ... in the above comment appears in A126930 and the link therein to a paper by F. Fite et al. on page 42. - Tom Copeland, Jul 23 2016

We define the n-th standard ordered rooted tree to be obtained by taking the (n-1)-th composition in standard order (graded reverse-lexicographic, A066099) as root and replacing each part with its own standard ordered rooted tree. This ranking is an ordered variation of Matula-Goebel numbers, giving a bijective correspondence between positive integers and unlabeled ordered rooted trees.

We define the n-th standard ordered rooted tree to be obtained by taking the (n-1)-th composition in standard order (graded reverse-lexicographic, A066099) as root and replacing each part with its own standard ordered rooted tree. This ranking is an ordered variation of Matula-Goebel numbers, giving a bijective correspondence between positive integers and unlabeled ordered rooted trees.

Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1-M)^n = A057681(n)-a(n)*M+z(n)*M^2, where z(0)=z(1)=0 and, apparently, z(n+2)=A057083(n). - Stanislav Sykora, Jun 10 2012

From Tom Copeland, Nov 09 2014: (Start)

This 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 interp. (here t=-2) o.g.f. G(x,t) = x(1-x)/[1+(t-1)x(1-x)] and inverse o.g.f. Ginv(x,t) = [1-sqrt(1-4x/(1+(1-t)x))]/2 (Cf. A005773 and A091867 and A030528 for more info on this family). (End)

{A057681, A057682, A*}, where A* is A057083 prefixed by two 0's, is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x)} of order 3. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Jul 31 2017

We define the standard ordered rooted tree (SORT)-number of an unlabeled ordered rooted tree to be one plus the standard composition number (A066099) of the SORT-numbers of the branches, or 1 if there are no branches. This ranking is an ordered variation of Matula-Goebel numbers, giving a bijective correspondence between positive integers and unlabeled ordered rooted trees.

We define the n-th standard ordered rooted tree to be obtained by taking the (n-1)-th composition in standard order (graded reverse-lexicographic, A066099) as root and replacing each part with its own standard ordered rooted tree. This ranking is an ordered variation of Matula-Goebel numbers, giving a bijective correspondence between positive integers and unlabeled ordered rooted trees.

A generalized Bethe tree is an unlabeled rooted tree where all branches directly under the same root are equal.

T(n,k) = number of Dyck n-paths that start UU, contain no DUDU and no subpath of the form UUPDD with P a nonempty Dyck path and whose terminal descent has length n-k+2. For example, T(5,4)=2 counts UUDUUDUDDD, UUUDDUUDDD (each ending with exactly n-k+2=3 Ds). - David Callan, Sep 25 2006