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Partial Motzkin paths with air pockets of the first kind avoiding peaks, valleys or double rises

Jean-Luc Baril LIB, Université de Bourgogne
B.P. 47 870, 21078 Dijon Cedex France
E-mail: barjl@u-bourgogne.fr
José L. Ramírez Departamento de Matemáticas, Universidad Nacional de Colombia
Bogotá, Colombia
E-mail: jlramirezr@unal.edu.co
Abstract

Motzkin paths with air pockets (MAP) of the first kind are defined as a generalization of Dyck paths with air pockets. They are lattice paths in 2\mathbb{N}^{2} starting at the origin made of steps U=(1,1)U=(1,1), Dk=(1,k)D_{k}=(1,-k), k1k\geqslant 1 and H=(1,0)H=(1,0), where two down-steps cannot be consecutive. We enumerate MAP and their prefixes avoiding peaks (resp. valleys, resp. double rise) according to the length, the type of the last step, and the height of its end-point. We express our results using Riordan arrays. Finally, we provide constructive bijections between these paths and restricted Dyck and Motzkin paths.

1 Introduction

In a recent paper [2], the authors introduce, study, and enumerate special classes of lattice paths, called Dyck paths with air pockets (DAP for short). Such paths are non empty lattice paths in the first quadrant of 2\mathbb{Z}^{2} starting at the origin, and consisting of up-steps U=(1,1)U=(1,1) and down-steps Dk=(1,k)D_{k}=(1,-k), k1k\geqslant 1, where two down-steps cannot be consecutive. These paths can be viewed as ordinary Dyck paths (i.e., paths in 2\mathbb{N}^{2} starting at the origin, ending on the xx-axis and consisting of UU and D1(=D)D_{1}(=D)), where each maximal run of down-steps is condensed into one large down-step. As mentioned in [2], they also correspond to a stack evolution with (partial) reset operations that cannot be consecutive (see for instance [6]). The authors enumerate these paths with respect to the length, the type (up or down) of the last step and the height of the end-point. Whenever the last point is on the xx-axis, they prove that the DAP of length nn are in one-to-one correspondence with the peak-less Motzkin paths of length n1n-1. They also investigate the popularity of many patterns in these paths and they give asymptotic approximations. In a second work [3], the authors make a study for a generalization of these paths by allowing them to go below the xx-axis. They call these paths Grand Dyck paths with air pockets (GDAP), and they also yield enumerative results for these paths according to the length and several restrictions on the height. In a third paper, Baril and Barry [4] study two generalizations of DAP by allowing some horizontal steps H=(1,0)H=(1,0) with some conditions. They call them Motzkin paths with air pockets of the first and second kind.

In this paper we study Motzkin paths with air pockets of the first kind, which are defined as Motzkin paths (lattice paths in 2\mathbb{N}^{2} starting at the origin and made of UU, DD, and HH), where each maximal run of down-steps is condensed into one large down-step. More precisely, we consider lattice paths in 2\mathbb{N}^{2} starting at the origin, consisting of steps UU, HH, and DkD_{k}, k1k\geqslant 1, where two down-steps cannot be consecutive. We denote by 𝒟\mathcal{D} (resp. \mathcal{M}, resp. 𝒫\mathcal{MP}) the set of Dyck paths (resp. Motzkin paths, resp. Motzkin paths with air pockets of the first kind). Moreover, we denote by 𝒫𝒫\mathcal{PMP} the set of partial Motzkin paths with air pockets (PMAP for short). The MAP of the first kind are enumerated by the sequence A114465. This sequence also counts the Dyck paths having no ascents of length 2 that start at an odd level.

Throughout the paper, we will use the following notations. For k0k\geqslant 0, we consider the generating function fk=fk(z)f_{k}=f_{k}(z) (resp. gk=gk(z)g_{k}=g_{k}(z), resp. hk=hk(z)h_{k}=h_{k}(z)), where the coefficient of znz^{n} in the series expansion is the number of partial Motzkin paths with air pockets of length nn ending at height kk with an up-step, (resp. with a down-step, resp. with a horizontal step HH). We introduce the bivariate generating functions

F(u,z)=k0ukfk(z),G(u,z)=k0ukgk(z), and H(u,z)=k0ukhk(z).F(u,z)=\sum\limits_{k\geqslant 0}u^{k}f_{k}(z),\quad G(u,z)=\sum\limits_{k\geqslant 0}u^{k}g_{k}(z),\mbox{ and }H(u,z)=\sum\limits_{k\geqslant 0}u^{k}h_{k}(z).

For short, we also use the notation F(u),G(u)F(u),G(u), and H(u)H(u) for these functions.

A Riordan array is an infinite lower triangular matrix whose kk-th column has generating function g(z)f(z)kg(z)f(z)^{k} for all k0k\geqslant 0, for some formal power series g(z)g(z) and f(z)f(z), with g(0)0g(0)\neq 0, f(0)=0f(0)=0, and f(0)0f^{\prime}(0)\neq 0. Such a Riordan array is denoted by (g(z),f(z))(g(z),f(z)). We refer to [16, 17] for more details on Riordan arrays. Several authors have used Riordan arrays to study lattice paths; see for example [10, 13, 14, 19, 20, 21].

The outline of this paper is the following. We present enumerative results for partial Motzkin paths with air pockets of the first kind avoiding peaks (resp. avoiding valleys, resp. avoiding double rises), knowing that a peak is an occurrence UDkUD_{k} for some k1k\geqslant 1, a valley is an occurrence DkUD_{k}U for some k1k\geqslant 1, and a double rise is an occurrence UUUU. For each avoidance, we provide bivariate generating functions that count the PMAP with respect to the length, the type of the last step (up, down or horizontal step) and the height of the end-point. All these results are obtained algebraically by using the famous kernel method for solving several systems of functional equations. We express our results using Riordan arrays and we deduce closed forms for PMAP of length nn ending at height kk. Finally, we provide constructive bijections between these paths and some restricted Dyck and Motzkin paths.

2 Partial peak-less Motzkin paths with air pockets

In this section, we study partial Motzkin paths with air pockets of the first kind avoiding occurrences of UDiUD_{i} for all i1i\geqslant 1.

2.1 Enumerative results

Let PP be a length nn PMAP ending at height k0k\geqslant 0 and avoiding the occurrences of UDiUD_{i} for i1i\geqslant 1. If the last step of PP is UU, then k1k\geqslant 1 and we have P=QUP=QU, where QQ is a length (n1)(n-1) MAP ending at height k1k-1 and avoiding the peaks (QQ can be the empty path). So, we obtain the first relation fk=zfk1+zgk1+zhk1f_{k}=zf_{k-1}+zg_{k-1}+zh_{k-1} for k1k\geqslant 1, anchored with f0=1f_{0}=1 by considering the empty path. If the last step of PP is a down-step DiD_{i}, i1i\geqslant 1, then we have P=QDiP=QD_{i}, where QQ is a length (n1)(n-1) PMAP ending at height k+1\ell\geqslant k+1 with no up- and down-steps at its end, and with no peaks. So, we obtain the second relation gk=zk+1hg_{k}=z\sum\limits_{\ell\geqslant k+1}h_{\ell}. If the last step of PP is a horizontal step HH, then we have P=QHP=QH, where QQ is a length (n1)(n-1) PMAP ending at height kk with no peaks, which implies that hk=z(fk+gk+hk)h_{k}=z(f_{k}+g_{k}+h_{k}) for k0k\geqslant 0.

Therefore, we have to solve the following system of equations:

{f0=1, and fk=zfk1+zgk1+zhk1,k1,gk=zk+1h,k0,hk=zfk+zgk+zhk,k0.\left\{\begin{array}[]{l}f_{0}=1,\mbox{ and }f_{k}=zf_{k-1}+zg_{k-1}+zh_{k-1},\quad k\geqslant 1,\\ g_{k}=z\sum\limits_{\ell\geqslant k+1}h_{\ell},\quad k\geqslant 0,\\ h_{k}=zf_{k}+zg_{k}+zh_{k},\quad k\geqslant 0.\\ \end{array}\right. (1)

Multiplying by uku^{k} the recursions in (1) and summing over kk, we have:

F(u)\displaystyle F(u) =1+zk1ukfk1+zk1ukgk1+zk1ukhk1\displaystyle=1+z\sum\limits_{k\geqslant 1}u^{k}f_{k-1}+z\sum\limits_{k\geqslant 1}u^{k}g_{k-1}+z\sum\limits_{k\geqslant 1}u^{k}h_{k-1}
=1+zuF(u)+zuG(u)+zuH(u),\displaystyle=1+zuF(u)+zuG(u)+zuH(u),
G(u)\displaystyle G(u) =zk0uk(k+1h)=zk1hk(1+u++uk1)\displaystyle=z\sum\limits_{k\geqslant 0}u^{k}\Bigl{(}\sum\limits_{\ell\geqslant k+1}h_{\ell}\Bigr{)}=z\sum\limits_{k\geqslant 1}h_{k}(1+u+\cdots+u^{k-1})
=zk1uk1u1hk=zu1(H(u)H(1)),\displaystyle=z\sum\limits_{k\geqslant 1}\frac{u^{k}-1}{u-1}h_{k}=\frac{z}{u-1}(H(u)-H(1)),
H(u)\displaystyle H(u) =zF(u)+zG(u)+zH(u).\displaystyle=zF(u)+zG(u)+zH(u).

Notice that we have F(1)H(1)=1F(1)-H(1)=1 by considering the difference of the first and third equations. Now, setting h1:=H(1)h1:=H(1) and solving these functional equations, we obtain

F(u)=h1uz2+zu+z2uz+1u2z+z2uz+1,F(u)={\frac{{\it h1}\,u{z}^{2}+zu+{z}^{2}-u-z+1}{{u}^{2}z+{z}^{2}-u-z+1}},
G(u)=z(h1uz+zh1h1+z)u2z+z2uz+1,H(u)=z(zh1u+1)u2z+z2uz+1.G(u)=-{\frac{z\left({\it h1}\,uz+z{\it h1}-{\it h1}+z\right)}{{u}^{2}z+{z}^{2}-u-z+1}},\quad H(u)={\frac{z\left(z{\it h1}-u+1\right)}{{u}^{2}z+{z}^{2}-u-z+1}}.

In order to compute h1h1, we use the kernel method (see [1, 11]) on H(u)H(u). We can write the denominator (which is a polynomial in uu of degree 2), as z(ur)(us)z(u-r)(u-s) with

r=1+4z3+4z24z+12z and s=14z3+4z24z+12z.r={\frac{1+\sqrt{-4\,{z}^{3}+4\,{z}^{2}-4\,z+1}}{2z}}\mbox{ and }s={\frac{1-\sqrt{-4\,{z}^{3}+4\,{z}^{2}-4\,z+1}}{2z}}.

Plugging u=su=s (which has a Taylor expansion at z=0z=0) in H(u)z(ur)(us)H(u)z(u-r)(u-s), we obtain the equation zh1s+1=0,zh1-s+1=0, which implies that

h1=s1z.h1=\frac{s-1}{z}.

Finally, after simplifying by the factor (us)(u-s) in the numerators and denominators, we obtain

F(u)=rru,G(u)=s1ru, and H(u)=1ru,F(u)=\frac{r}{r-u},\quad G(u)=\frac{s-1}{r-u},\quad\mbox{ and }\quad H(u)=\frac{1}{r-u},

which induces that

fk=[uk]F(u)=1rk,gk=[uk]G(u)=s1rk+1, and hk=[uk]H(u)=1rk+1.f_{k}=[u^{k}]F(u)=\frac{1}{r^{k}},\quad g_{k}=[u^{k}]G(u)=\frac{s-1}{r^{k+1}},\quad\mbox{ and }\quad h_{k}=[u^{k}]H(u)=\frac{1}{r^{k+1}}.
Theorem 1

The bivariate generating function for the total number of peak-less PMAP with respect to the length and the height of the end-point is given by

𝑇𝑜𝑡𝑎𝑙(z,u)=1z(ru),\mathit{Total}(z,u)=\frac{1}{z(r-u)},

and we have

[uk]𝑇𝑜𝑡𝑎𝑙(z,u)=1zrk+1.[u^{k}]\mathit{Total}(z,u)=\frac{1}{zr^{k+1}}.

Finally, setting t(n,k)=[zn][uk]𝑇𝑜𝑡𝑎𝑙(z,u)t(n,k)=[z^{n}][u^{k}]\mathit{Total}(z,u), we have for n2n\geqslant 2 and k1k\geqslant 1,

t(n,k)=t(n,k1)+t(n1,k)t(n1,k2)t(n2,k),t(n,k)=t(n,k-1)+t(n-1,k)-t(n-1,k-2)-t(n-2,k),

and setting tn:=t(n,0)t_{n}:=t(n,0), then we have

tn=tn1+k=0n3tktnk3+k=2n1(tktk1)tnk1.t_{n}=t_{n-1}+\sum\limits_{k=0}^{n-3}t_{k}t_{n-k-3}+\sum\limits_{k=2}^{n-1}\left(t_{k}-t_{k-1}\right)t_{n-k-1}.

Proof. The first three equalities are immediately deduced from the previous results. Now, let us prove the last equality. Any length nn peak-less MAP is of the form (ii) HPHP where PP is a MAP of length n1n-1, or (iiii) UQHDRUQHDR, where Q,RQ,R are some MAP such that the length of QQ lies into [0,n3][0,n-3], or (iiiiii) PQP^{\sharp}Q, where P=UPDiP^{\sharp}=UP^{\prime}D_{i}, i2i\geqslant 2, and PDi1P^{\prime}D_{i-1} is a MAP of length lying into [2,n1][2,n-1]. The number of PDi1P^{\prime}D_{i-1} of a given length kk is the total number of peak-less MAP of length kk minus the number of peak-less MAP of length kk and ending with HH. Taking into account all these cases, we obtain the result. \Box

Corollary 1

The generating function that counts all peak-less PMAP with respect to the length is given by

𝑇𝑜𝑡𝑎𝑙(z,1)=1z(r1).\mathit{Total}(z,1)=\frac{1}{z(r-1)}.

The first few terms of the series expansion of 𝑇𝑜𝑡𝑎𝑙(z,1)\mathit{Total}(z,1) are

1+2z+4z2+9z3+22z4+56z5+146z6+388z7+1048z8+2869z9+O(x10),1+2z+4z^{2}+9z^{3}+22z^{4}+56z^{5}+146z^{6}+388z^{7}+1048z^{8}+2869z^{9}+O(x^{10}),

which corresponds to the sequence A152225 in [18] counting Dyck paths of semilength n+1n+1 with no peaks of height 0 (mod 3) and no valleys of height 2 (mod 3); see [7].

Corollary 2

The generating function that counts the peak-less MAP with respect to the length is given by

𝑇𝑜𝑡𝑎𝑙(z,0)=1zr.\mathit{Total}(z,0)=\frac{1}{zr}.

The first few terms of the series expansion of 𝑇𝑜𝑡𝑎𝑙(z,0)\mathit{Total}(z,0) are

1+z+z2+2z3+5z4+12z5+29z6+73z7+190z8+505z9+O(z10),1+z+z^{2}+2z^{3}+5z^{4}+12z^{5}+29z^{6}+73z^{7}+190z^{8}+505z^{9}+O(z^{10}),

which corresponds to the sequence A152171 in [18] counting Dyck paths of length 2n2n with no peaks of height 22 (mod 3) and no valleys of height 1 (mod 3). In Section 2.2, we will exhibit a constructive bijection between these two classes of paths.

Let 𝒯\mathcal{T} be the infinite matrix 𝒯:=[t(n,k)]n,k0\mathcal{T}:=[t(n,k)]_{n,k\geqslant 0}, where t(n,k)=[zn][uk]𝑇𝑜𝑡𝑎𝑙(z,u)t(n,k)=[z^{n}][u^{k}]\mathit{Total}(z,u). The first few rows of the matrix 𝒯\mathcal{T} are

𝒯=(100000000110000000121000000233100000566410000121513105100029383324156100739687634021710190248229172110622881).\mathcal{T}=\left(\begin{array}[]{ccccccccc}1&0&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0&0\\ 1&2&1&0&0&0&0&0&0\\ 2&3&3&1&0&0&0&0&0\\ 5&6&6&4&1&0&0&0&0\\ 12&15&13&10&5&1&0&0&0\\ 29&38&33&24&15&6&1&0&0\\ 73&96&87&63&40&21&7&1&0\\ 190&248&229&172&110&62&28&8&1\\ \end{array}\right).

In Figure 1 we show the peak-less PMAP counted by t(5,1)=15t(5,1)=15.

Refer to caption
Figure 1: Peak-less PMAP of length 55 ending at height 11.
Corollary 3

The matrix 𝒯=[t(n,k)]n,k0\mathcal{T}=[t(n,k)]_{n,k\geqslant 0} is a Riordan array defined by

(C(z(1z+z2)),zC(z(1z+z2))),\left(C\left(z(1-z+z^{2})\right),zC\left(z(1-z+z^{2})\right)\right),

where C(z)=114z2zC(z)=\frac{1-\sqrt{1-4z}}{2z} is the generating function of the Catalan numbers cn=1n+1(2nn)c_{n}=\frac{1}{n+1}\binom{2n}{n}.

Proof. Indeed, we directly deduce the result from the following.

[uk]𝑇𝑜𝑡𝑎𝑙(z,u)\displaystyle[u^{k}]\mathit{Total}(z,u) =1zrk+1=1zr1rk=C(z(1z+z2))(zC(z(1z+z2)))k.\displaystyle=\frac{1}{zr^{k+1}}=\frac{1}{zr}\cdot\frac{1}{r^{k}}=C\left(z(1-z+z^{2})\right)\cdot\left(zC\left(z(1-z+z^{2})\right)\right)^{k}.

\Box

Corollary 4

We have

t(n,k)=j=0nkk+12(nj)k+1(2(nj)k+1nkj)a(nkj,j),t(n,k)=\sum_{j=0}^{n-k}\frac{k+1}{2(n-j)-k+1}\binom{2(n-j)-k+1}{n-k-j}a(n-k-j,j),

where a(n,k)=(1)ki=0n(ni)(nik2i)a(n,k)=(-1)^{k}\sum_{i=0}^{n}\binom{n}{i}\binom{n-i}{k-2i}.

Proof. From the Lagrange Inversion Formula (cf. [8]), we know that

C(z)k=n0k2n+k(2n+kn)zn.\displaystyle C(z)^{k}=\sum_{n\geqslant 0}\frac{k}{2n+k}\binom{2n+k}{n}z^{n}. (2)

From definition of Riordan arrays and Eq. (2) we have

t(n,k)\displaystyle t(n,k) =[znk](C(z(1z+z2)))k+1\displaystyle=[z^{n-k}]\left(C\left(z(1-z+z^{2})\right)\right)^{k+1}
=[znk]=0k+12+k+1(2+k+1)z(1z+z2)\displaystyle=[z^{n-k}]\sum_{\ell=0}^{\infty}\frac{k+1}{2\ell+k+1}\binom{2\ell+k+1}{\ell}z^{\ell}(1-z+z^{2})^{\ell}
=[znk]=0k+12+k+1(2+k+1)zj=02a(,j)zj\displaystyle=[z^{n-k}]\sum_{\ell=0}^{\infty}\frac{k+1}{2\ell+k+1}\binom{2\ell+k+1}{\ell}z^{\ell}\sum_{j=0}^{2\ell}a(\ell,j)z^{j}
=[znk]j=0=0k+12(+j/2)+k+1(2(+j/2)+k+1+j/2)a(+j/2,j)zj++j/2.\displaystyle=[z^{n-k}]\sum_{j=0}^{\infty}\sum_{\ell=0}^{\infty}\frac{k+1}{2(\ell+\lfloor j/2\rfloor)+k+1}\binom{2(\ell+\lfloor j/2\rfloor)+k+1}{\ell+\lfloor j/2\rfloor}a(\ell+\lfloor j/2\rfloor,j)z^{j+\ell+\lfloor j/2\rfloor}.

If we take s=j++j/2s=j+\ell+\lfloor j/2\rfloor, then

t(n,k)\displaystyle t(n,k) =[znk]j=0s=jk+12(sj)+k+1(2(sj)+k+1sj)a(sj,j)zs\displaystyle=[z^{n-k}]\sum_{j=0}^{\infty}\sum_{s=j}^{\infty}\frac{k+1}{2(s-j)+k+1}\binom{2(s-j)+k+1}{s-j}a(s-j,j)z^{s}
=j=0nkk+12(nj)k+1(2(nj)k+1nkj)a(nkj,j).\displaystyle=\sum_{j=0}^{n-k}\frac{k+1}{2(n-j)-k+1}\binom{2(n-j)-k+1}{n-k-j}a(n-k-j,j).

\Box

In [15], Rogers gave an equivalent characterization of the Riordan arrays. That is, every element not belonging to row 0 or column 0 in a Riordan array can be expressed as a fixed linear combination of the elements in the preceding row. The AA-sequence is defined to be the sequence coefficients of this linear combination. Analogously, Merlini et al. [9] introduced the ZZ-sequence, that characterizes the elements in column 0, except for the top one.

An infinite lower triangular matrix [dn,k]n,k0[d_{n,k}]_{n,k\geqslant 0} is a Riordan array if and only if d0,00d_{0,0}\neq 0 and there exist two sequences (a0,a1,a2,)(a_{0},a_{1},a_{2},\dots), with a00a_{0}\neq 0, and (z0,z1,z2,)(z_{0},z_{1},z_{2},\dots) (called the AA-sequence and the ZZ-sequence, respectively), such that

dn+1,k+1\displaystyle d_{n+1,k+1} =a0dn,k+a1dn,k+1+a2dn,k+2+\displaystyle=a_{0}d_{n,k}+a_{1}d_{n,k+1}+a_{2}d_{n,k+2}+\cdots for n,k0,\displaystyle\text{for }n,k\geqslant 0, (3)
dn+1,0\displaystyle d_{n+1,0} =z0dn,0+z1dn,1+z2dn,2+\displaystyle=z_{0}d_{n,0}+z_{1}d_{n,1}+z_{2}d_{n,2}+\cdots for n0.\displaystyle\text{for }n\geqslant 0. (4)

The product of two Riordan arrays (g(z),f(z))(g(z),f(z)) and (h(z),l(z))(h(z),l(z)) is defined by

(g(z),f(z))(h(z),l(z))=(g(z)h(f(z)),l(f(z))).(g(z),f(z))*(h(z),l(z))=\left(g(z)h(f(z)),l(f(z))\right). (5)

Under this operation, the set of all Riordan arrays is a group [16]. The identity element is I=(1,z)I=(1,z) and the inverse of (g(z),f(z))(g(z),f(z)) is given by

(g(z),f(z))1=(1/(gf<1>)(z),f<1>(z)),(g(z),f(z))^{-1}=\left(1/\left(g\circ f^{<-1>}\right)(z),f^{<-1>}(z)\right), (6)

where f<1>(z)f^{<-1>}(z) denotes the compositional inverse of f(z)f(z).

The generating functions for the AA-sequence and ZZ-sequence of the Riordan array =(g(z),f(z))\mathcal{F}=(g(z),f(z)), with inverse 1=(d(z),h(z))\mathcal{F}^{-1}=(d(z),h(z)), are given by ([9, 5])

A(z)=zh(z)andZ(z)=1h(z)(1d0,0d(z)),\displaystyle A(z)=\frac{z}{h(z)}\quad\text{and}\quad Z(z)=\frac{1}{h(z)}\left(1-d_{0,0}d(z)\right),

respectively.

From the definition of the AA-sequence and ZZ-sequence for the Riordan arrays we can give an additional recurrence relation for the sequence t(n,k)t(n,k).

Corollary 5

We have

t(n+1,k+1)=j0a(j)t(n,k+j),t(n+1,k+1)=\sum_{j\geqslant 0}a(j)t(n,k+j),

where a(n)=(1)n+1k=1nj=0k1k(jnkj)(kj)(nk2k1)a(n)=(-1)^{n+1}\sum_{k=1}^{n}\sum_{j=0}^{k}\frac{1}{k}\binom{j}{n-k-j}\binom{k}{j}\binom{n-k-2}{k-1} for n1n\geqslant 1 and a(0)=1a(0)=1. Moreover,

tn+1=j0a(j+1)t(n,j).t_{n+1}=\sum_{j\geqslant 0}a(j+1)t(n,j).

Proof. By Equation (6), the inverse of the matrix 𝒯=[t(n,k)]n,k0\mathcal{T}=[t(n,k)]_{n,k\geqslant 0} is given by 𝒯1=(g2(z),zg2(z))\mathcal{T}^{-1}=\left(g_{2}(z),zg_{2}(z)\right), where

g2(z)=1+z2+12z2+4z33z42z3.g_{2}(z)=\frac{-1+z^{2}+\sqrt{1-2z^{2}+4z^{3}-3z^{4}}}{2z^{3}}.

Therefore, the AA-sequence and ZZ-sequence of the Riordan array 𝒯\mathcal{T} have generating functions

A(z)=n0a(n)zn=2z31+z2+12z2+4z33z4 and Z(z)=A(z)1z.\displaystyle A(z)=\sum_{n\geqslant 0}a(n)z^{n}=\frac{2z^{3}}{-1+z^{2}+\sqrt{1-2z^{2}+4z^{3}-3z^{4}}}\quad\text{ and }\quad Z(z)=\frac{A(z)-1}{z}.

The generating function A(z)A(z) corresponds with the sequence A247162, where the explicit formula for a(n)a(n) can be found. From (3) we obtain the result. \Box

The first few values of the sequence a(n)a(n) for n0n\geqslant 0 are

1,1,0,1,0,1,1,2,3,6,10,.1,\quad 1,\quad 0,\quad 1,\quad 0,\quad 1,\quad-1,\quad 2,\quad-3,\quad 6,\quad-10,\dots.

2.2 A bijective approach

Corollary 2 proves that the set of peak-less Motzkin paths with air pockets of length nn (ending on the xx-axis) is equinumerous to the set 𝒟n(2,1)\mathcal{D}_{n}(2,1) of Dyck paths of length 2n2n with no peak at height 2 (mod 3) and no valley at height 1 (mod 3).

Any non-empty peak-less Motzkin path with air pockets is either of the form (1) HαH\alpha or (2) Uα1Uα2UαkHDkβU\alpha_{1}U\alpha_{2}\cdots U\alpha_{k}HD_{k}\beta, where k1k\geqslant 1 and α,α1,,αk,β\alpha,\alpha_{1},\ldots,\alpha_{k},\beta are possibly empty peak-less MAP. We refer to the left part of Figure 2 for an illustration of this form.

Remember that 𝒫\mathcal{MP} denotes the set of Motzkin paths with air pockets of the first kind.

Definition 1

We recursively define the map ψ\psi from 𝒫\mathcal{MP} to n0𝒟n(2,1)\cup_{n\geqslant 0}\mathcal{D}_{n}(2,1) as follows. For α𝒫\alpha\in\mathcal{MP}, we set:

ψ(P)={ϵif P=ϵ(i)UDψ(α)if P=Hα with α𝒫(ii)U3ψ(α1)DUψ(α2)DUDUψ(αk)D3ψ(β)if P=Uα1Uα2UαkHDkβ withk1 and α1,,αk,β𝒫.(iii)\psi(P)=\left\{\begin{array}[]{llr}\epsilon&\text{if }P=\epsilon&(i)\\ UD\psi(\alpha)&\text{if }P=H\alpha\text{ with }\alpha\in\mathcal{MP}&(ii)\\ U^{3}\psi(\alpha_{1})DU\psi(\alpha_{2})DU\ldots DU\psi(\alpha_{k})D^{3}\psi(\beta)&\text{if }P=U\alpha_{1}U\alpha_{2}\ldots U\alpha_{k}HD_{k}\beta\text{ with}\\ &k\geqslant 1\mbox{ and }\alpha_{1},\ldots,\alpha_{k},\beta\in\mathcal{MP}.&(iii)\end{array}\right.

We refer to Figure 3 for an illustration of the third case of the definition of ψ\psi.

α1\alpha_{1}α2\alpha_{2}αk\alpha_{k}β\beta

\quad\Longrightarrow ψ(α1)\psi(\alpha_{1})ψ(α2)\psi(\alpha_{2})ψ(αk)\psi(\alpha_{k})ψ(β)\psi(\beta)

Figure 2: Illustration of the map ψ\psi for the more general case (iiiiii) in Definition 1.

Due to the recursive definition, the image of peak-less MAP of length nn under ψ\psi is a Dyck path of length 2n2n. Moreover it is clear that the obtained path is a Dyck path in n0𝒟n(2,1)\cup_{n\geqslant 0}\mathcal{D}_{n}(2,1). For instance (see Figure 3 for an illustration of this example).

ψ(UUHDHUHUHD3HH)\displaystyle\psi(UUHDHUHUHD_{3}HH) =ψ(UUHDHα1UHα2Uϵα3HD3HHβ\displaystyle=\psi(U\cdot\overbrace{UHDH}^{\alpha_{1}}\cdot U\cdot\overbrace{H}^{\alpha_{2}}\cdot U\cdot\overbrace{\epsilon}^{\alpha_{3}}\cdot HD_{3}\cdot\overbrace{HH}^{\beta}
=U3ψ(UHDH)DUψ(H)DUψ(ϵ)D3ψ(HH)\displaystyle=U^{3}\psi(UHDH)\cdot DU\cdot\psi(H)\cdot DU\cdot\psi(\epsilon)\cdot D^{3}\cdot\psi(HH)
=U3(U3D3ψ(H))DUUDDUD3UDUD\displaystyle=U^{3}\cdot(U^{3}D^{3}\psi(H))\cdot DU\cdot UD\cdot DU\cdot D^{3}\cdot UDUD
=U6D3UD2U2D2UD2DUDUD.\displaystyle=U^{6}D^{3}UD^{2}U^{2}D^{2}UD^{2}DUDUD.
Refer to caption
Figure 3: ψ(UUHDHUHUHD3HH)=UUUUUUDDDUDDUUDDUDDDUDUD\psi(UUHDHUHUHD_{3}HH)=UUUUUUDDDUDDUUDDUDDDUDUD.
Theorem 2

For all n0n\geqslant 0, the map ψ\psi induces a bijection between 𝒫n\mathcal{MP}_{n} and 𝒟n(2,1)\mathcal{D}_{n}(2,1).

Proof. Since 𝒫n\mathcal{MP}_{n} and 𝒟n(2,1)\mathcal{D}_{n}(2,1) have the same cardinality (due to Corollary 2 and A152171 in [18]), it suffices to prove that for P,Q𝒫P,Q\in\mathcal{MP}, PQP\neq Q implies ψ(P)ψ(Q)\psi(P)\neq\psi(Q). A simple induction on nn allows to obtain the result. \Box

3 Partial valley-less Motzkin paths with air pockets

Using the same arguments we used for the system of the previous section, we study partial Motzkin paths with air pockets of the first kind avoiding occurrences of DiUD_{i}U for all i1i\geqslant 1.

3.1 Enumerative results

In the same way as we done in Section 2.1, we have to solve the following system of equations:

{f0=1, and fk=zfk1+zhk1,k1,gk=zk+1f+zk+1h,k0,hk=zfk+zgk+zhk,k0.\left\{\begin{array}[]{l}f_{0}=1,\mbox{ and }f_{k}=zf_{k-1}+zh_{k-1},\quad k\geqslant 1,\\ g_{k}=z\sum\limits_{\ell\geqslant k+1}f_{\ell}+z\sum\limits_{\ell\geqslant k+1}h_{\ell},\quad k\geqslant 0,\\ h_{k}=zf_{k}+zg_{k}+zh_{k},\quad k\geqslant 0.\\ \end{array}\right. (7)

Multiplying by uku^{k} the recursions in (7) and summing over kk, we have:

F(u)\displaystyle F(u) =1+zuF(u)+zuH(u),\displaystyle=1+zuF(u)+zuH(u),
G(u)\displaystyle G(u) =zu1(F(u)F(1)+H(u)H(1)),\displaystyle=\frac{z}{u-1}(F(u)-F(1)+H(u)-H(1)),
H(u)\displaystyle H(u) =zF(u)+zG(u)+zH(u).\displaystyle=zF(u)+zG(u)+zH(u).

Notice that we have (1z)F(1)=1+zH(1)(1-z)F(1)=1+zH(1) by considering the first equation. Now, setting f1:=F(1)f1:=F(1) and solving these functional equations, we obtain

F(u)=f1uz2uz2+zu+z2uz+1u2z+z2uz+1,F(u)={\frac{{\it f1}\,u{z}^{2}-u{z}^{2}+zu+{z}^{2}-u-z+1}{{u}^{2}z+{z}^{2}-u-z+1}},
G(u)=f1uz+f1zzuf1+1u2z+z2uz+1,H(u)=z(f1uzzuf1+u+z)u2z+z2uz+1.G(u)=-{\frac{{\it f1}\,uz+{\it f1}\,z-zu-{\it f1}+1}{{u}^{2}z+{z}^{2}-u-z+1}},\quad H(u)=-{\frac{z\left({\it f1}\,uz-zu-{\it f1}+u+z\right)}{{u}^{2}z+{z}^{2}-u-z+1}}.

In order to compute f1f1, we use the kernel method (see [1, 11]) on F(u)F(u). We can write the denominator (which is a polynomial in uu of degree 2), as z(ur)(us)z(u-r)(u-s) with

r=1+4z3+4z24z+12z and s=14z3+4z24z+12z.r={\frac{1+\sqrt{-4\,{z}^{3}+4\,{z}^{2}-4\,z+1}}{2z}}\quad\mbox{ and }\quad s={\frac{1-\sqrt{-4\,{z}^{3}+4\,{z}^{2}-4\,z+1}}{2z}}.

Plugging u=su=s (which has a Taylor expansion at z=0z=0) in F(u)z(ur)(us)F(u)z(u-r)(u-s), we obtain the equation f1sz2sz2+zs+z2sz+1=0,{\it f1}\,s{z}^{2}-s{z}^{2}+zs+{z}^{2}-s-z+1=0, which implies that

f1=1+s1z.f1=1+\frac{s-1}{z}.

Finally, after simplifying by the factor (us)(u-s) in the numerators and denominators, we obtain

F(u)=rru,G(u)=s1z(ru), and H(u)=sru,F(u)=\frac{r}{r-u},\quad G(u)=\frac{s-1}{z(r-u)},\quad\mbox{ and }\quad H(u)=\frac{s}{r-u},

which implies that

fk=[uk]F(u)=1rk,gk=[uk]G(u)=s1zrk+1, and hk=[uk]H(u)=srk+1.f_{k}=[u^{k}]F(u)=\frac{1}{r^{k}},\quad g_{k}=[u^{k}]G(u)=\frac{s-1}{zr^{k+1}},\quad\mbox{ and }\quad h_{k}=[u^{k}]H(u)=\frac{s}{r^{k+1}}.
Theorem 3

The bivariate generating function for the total number of partial valley-less MAP with respect to the length and the height of the end-point is given by

𝑇𝑜𝑡𝑎𝑙(z,u)=sz(ru),\mathit{Total}(z,u)=\frac{s}{z(r-u)},

and we have

[uk]𝑇𝑜𝑡𝑎𝑙(z,u)=szrk+1.[u^{k}]\mathit{Total}(z,u)=\frac{s}{zr^{k+1}}.

Finally, setting t(n,k)=[zn][uk]𝑇𝑜𝑡𝑎𝑙(z,u)t(n,k)=[z^{n}][u^{k}]\mathit{Total}(z,u), we have for n2n\geqslant 2, k1k\geqslant 1,

t(n,k)=t(n,k1)+t(n1,k)t(n1,k2)t(n2,k),t(n,k)=t(n,k-1)+t(n-1,k)-t(n-1,k-2)-t(n-2,k),

and setting tn:=t(n,0)t_{n}:=t(n,0) and t1=1t_{-1}=1, then we have for n2n\geqslant 2,

tn1=tn2+k=0n3tk1tnk4+k=2n1(tk1tk2)tnk2.t_{n-1}=t_{n-2}+\sum\limits_{k=0}^{n-3}t_{k-1}t_{n-k-4}+\sum\limits_{k=2}^{n-1}\left(t_{k-1}-t_{k-2}\right)t_{n-k-2}.

Proof. The first three equalities are immediately deduced from the previous results. For the last equality, the term tnt_{n} satisfies the same recurrence relation as in Theorem 1 (modulo shift of nn) since the two sequences are equal modulo a shift. \Box

Corollary 6

The generating function that counts the partial valley-less MAP with respect to the length is given by

𝑇𝑜𝑡𝑎𝑙(z,1)=sz(r1).\mathit{Total}(z,1)=\frac{s}{z(r-1)}.

The first few terms of the series expansion of 𝑇𝑜𝑡𝑎𝑙(z,1)\mathit{Total}(z,1) are

1+2z+5z2+13z3+34z4+90z5+242z6+660z7+1821z8+5073z9+O(x10),1+2z+5z^{2}+13z^{3}+34z^{4}+90z^{5}+242z^{6}+660z^{7}+1821z^{8}+5073z^{9}+O(x^{10}),

which does not appear in [18].

Corollary 7

The generating function that counts the partial valley-less MAP with respect to the length is given by

𝑇𝑜𝑡𝑎𝑙(z,0)=szr.\mathit{Total}(z,0)=\frac{s}{zr}.

The first few terms of the series expansion of 𝑇𝑜𝑡𝑎𝑙(z,0)\mathit{Total}(z,0) are

1+z+2z2+5z3+12z4+29z5+73z6+190z7+505z8+1363z9+O(z10),1+z+2z^{2}+5z^{3}+12z^{4}+29z^{5}+73z^{6}+190z^{7}+505z^{8}+1363z^{9}+O(z^{10}),

which corresponds to a shift of the sequence A152171 in [18] counting Dyck paths of length 2n2n with no peaks of height 22 (mod 3) and no valleys of height 1 (mod 3) (see also Section 3.1).

Let 𝒯\mathcal{T} be the infinite matrix 𝒯:=[t(n,k)]n,k0\mathcal{T}:=[t(n,k)]_{n,k\geqslant 0}. The first few rows of the matrix 𝒯\mathcal{T} are

𝒯=(10000000011000000022100000054310000012107410000292618115100073674930166100190175133854722710505467361241139702981).\mathcal{T}=\left(\begin{array}[]{ccccccccc}1&0&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0&0\\ 2&2&1&0&0&0&0&0&0\\ 5&4&3&1&0&0&0&0&0\\ 12&10&7&4&1&0&0&0&0\\ 29&26&18&11&5&1&0&0&0\\ 73&67&49&30&16&6&1&0&0\\ 190&175&133&85&47&22&7&1&0\\ 505&467&361&241&139&70&29&8&1\\ \end{array}\right).

In Figure 4 we show the valley-less PMAP counted by t(4,0)=12t(4,0)=12.

Refer to caption
Figure 4: The 12 UUUU-less PMAP of length 44 ending at height 0.
Corollary 8

The matrix 𝒯=[t(n,k)]n,k0\mathcal{T}=[t(n,k)]_{n,k\geqslant 0} is a Riordan array defined by

((z2z+1)C(z(1z+z2))2,zC(z(1z+z2)))\left((z^{2}-z+1)C\left(z(1-z+z^{2})\right)^{2},zC\left(z(1-z+z^{2})\right)\right)

where C(z)=114z2zC(z)=\frac{1-\sqrt{1-4z}}{2z} is the generating function of the Catalan numbers cn=1n+1(2nn)c_{n}=\frac{1}{n+1}\binom{2n}{n}.

Proof. Indeed, we directly deduce the result from the following.

[uk]𝑇𝑜𝑡𝑎𝑙(z,u)\displaystyle[u^{k}]\mathit{Total}(z,u) =szrk+1=szr1rk=(z2z+1)C(z(1z+z2))2(zC(z(1z+z2)))k.\displaystyle=\frac{s}{zr^{k+1}}=\frac{s}{zr}\cdot\frac{1}{r^{k}}=(z^{2}-z+1)C\left(z(1-z+z^{2})\right)^{2}\cdot\left(zC\left(z(1-z+z^{2})\right)\right)^{k}.

\Box

From a similar argument as in Corollary 4 we obtain the following result.

Corollary 9

We have

t(n,k)=j=0nkk+22(nj)k+2(2(nj)k+2nkj)a(nkj+1,j),t(n,k)=\sum_{j=0}^{n-k}\frac{k+2}{2(n-j)-k+2}\binom{2(n-j)-k+2}{n-k-j}a(n-k-j+1,j),

where a(n,k)=(1)ki=0n(ni)(nik2i)a(n,k)=(-1)^{k}\sum_{i=0}^{n}\binom{n}{i}\binom{n-i}{k-2i}.

3.2 A bijective approach

Corollary 7 and Corollary 2 prove that the set of valley-less Motzkin paths with air pockets of length n1n-1 (ending on the xx-axis) is equinumerous to the set of peak-less Motzkin paths with air pockets of length nn, which is in one-to-one correspondence with the set 𝒟n(2,1)\mathcal{D}_{n}(2,1) of Dyck paths of length 2n2n with no peak at height 2 (mod 3) and no valley at height 1 (mod 3) (see a constructive bijection in Section 2.2). Below, we provide a bijection between valley-less MAP of length n1n-1 and peak-less MAP of length nn.

Any valley-less Motzkin path with air pockets is either of the form (ii) ϵ\epsilon, (iiii) αH\alpha H, (iiiiii) UαDU\alpha D, (iviv) βHUαD\beta HU\alpha D, (vv) UγDkU\gamma D_{k}, or (vivi) βHUγDk\beta HU\gamma D_{k}, where α,β\alpha,\beta are valley-less MAP (possibly empty), and γDk1\gamma D_{k-1} is a valley-less MAP. According to all these cases, we define the map ϕ\phi.

Definition 2

We recursively define the map ϕ\phi from valley-less MAP of length n1n-1 to peak-less MAP of length nn. Let PP be a valley-less MAP, we set:

ϕ(P)={Hif P=ϵ,(i)ϕ(α)Hif P=αH,(ii)Uϕ(α)Dif P=UαD,(iii)ϕ(β)Uϕ(α)Dif P=βHUαD,(iv)ϕ(αDk1)if P=UαDk,(v)ϕ(β)ϕ(αDk1)if P=βHUαDk,(vi)\phi(P)=\left\{\begin{array}[]{llr}H&\text{if }P=\epsilon,&(i)\\ \phi(\alpha)H&\text{if }P=\alpha H,&(ii)\\ U\phi(\alpha)D&\text{if }P=U\alpha D,&(iii)\\ \phi(\beta)U\phi(\alpha)D&\text{if }P=\beta HU\alpha D,&(iv)\\ \phi(\alpha D_{k-1})^{\sharp}&\text{if }P=U\alpha D_{k},&(v)\\ \phi(\beta)\phi(\alpha D_{k-1})^{\sharp}&\text{if }P=\beta HU\alpha D_{k},&(vi)\\ \end{array}\right.

where the \sharp-operator maps a peak-less MAP of the form αDk1\alpha D_{k-1} into the peak-less MAP (αDk1)=UαDk(\alpha D_{k-1})^{\sharp}=U\alpha D_{k}.

Due to the recursive definition, the image of valley-less MAP of length n1n-1 under ϕ\phi is a peak-less MAP of length nn. The recursive definition naturally induces that ϕ\phi is a bijection. Using the bijection ψ\psi presented above, we can easily obtain a constructive bijection between valley-less MAP of length n1n-1 and Dyck paths of length 2n2n with no peak at height 2 (mod 3) and no valley at height 1 (mod 3).

4 Partial UUUU-less Motzkin paths with air pockets

In this section, we study partial Motzkin paths with air pockets of the first kind avoiding occurrences of UUUU.

4.1 Enumerative results

In the same way as we done in the previous section, we have to solve the following system of equations:

{f0=1,f1=z+zg0+zh0, and fk=zgk1+zhk1,k2,gk=zk+1f+zk+1h,k0,hk=zfk+zgk+zhk,k0.\left\{\begin{array}[]{l}f_{0}=1,f_{1}=z+zg_{0}+zh_{0},\mbox{ and }f_{k}=zg_{k-1}+zh_{k-1},\quad k\geqslant 2,\\ g_{k}=z\sum\limits_{\ell\geqslant k+1}f_{\ell}+z\sum\limits_{\ell\geqslant k+1}h_{\ell},\quad k\geqslant 0,\\ h_{k}=zf_{k}+zg_{k}+zh_{k},\quad k\geqslant 0.\\ \end{array}\right. (8)

Multiplying by uku^{k} the recursions in (8) and summing over kk, we have:

F(u)\displaystyle F(u) =1+zu+zuG(u)+zuH(u),\displaystyle=1+zu+zuG(u)+zuH(u),
G(u)\displaystyle G(u) =zu1(F(u)F(1)+H(u)H(1)),\displaystyle=\frac{z}{u-1}(F(u)-F(1)+H(u)-H(1)),
H(u)\displaystyle H(u) =zF(u)+zG(u)+zH(u).\displaystyle=zF(u)+zG(u)+zH(u).

Notice that we have H(1)=(1+z)(F(1)1)H(1)=(1+z)(F(1)-1) by considering the first and the last equation. Now, setting f1:=F(1)f1:=F(1) and solving these functional equations, we obtain

F(u)=f1uz3+2f1uz2+u2z2u2z2uz2+2uz+z2uz+1u2z2+uz3+uz+z2uz+1,F(u)=\frac{\mathit{f1}u\,z^{3}+2\mathit{f1}u\,z^{2}+u^{2}z^{2}-u^{2}z-2u\,z^{2}+2uz+z^{2}-u-z+1}{u^{2}z^{2}+u\,z^{3}+uz+z^{2}-u-z+1},
G(u)=z(f1uz3+2f1uz2uz3+f1z2uz2+f1z+uzz22f1+2)u2z2+uz3+uz+z2uz+1,G(u)=-\frac{z\left(\mathit{f1}u\,z^{3}+2\mathit{f1}u\,z^{2}-u\,z^{3}+\mathit{f1}\,z^{2}-u\,z^{2}+\mathit{f1}z+uz-z^{2}-2\mathit{f1}+2\right)}{u^{2}z^{2}+u\,z^{3}+uz+z^{2}-u-z+1},
H(u)=z(f1uz3+2f1uz2uz3+f1z2u2z2uz2+2f1z+uzz2u2z+1)u2z2+uz3+uz+z2uz+1.H(u)=\frac{z\left(\mathit{f1}u\,z^{3}+2\mathit{f1}u\,z^{2}-u\,z^{3}+\mathit{f1}\,z^{2}-u^{2}z-2u\,z^{2}+2\mathit{f1}z+uz-z^{2}-u-2z+1\right)}{u^{2}z^{2}+u\,z^{3}+uz+z^{2}-u-z+1}.

In order to compute f1f1, we use the kernel method (see [1, 11]) on F(u)F(u). We can write the denominator (which is a polynomial in uu of degree 2), as z2(ur)(us)z^{2}(u-r)(u-s), with

r=z3z+1+z62z4+2z33z22z+12z2 and r={\frac{-{z}^{3}-z+1+\sqrt{{z}^{6}-2\,{z}^{4}+2\,{z}^{3}-3\,{z}^{2}-2\,z+1}}{2{z}^{2}}}\ \mbox{ and }
s=1zz3z62z4+2z33z22z+12z2.s={\frac{1-z-{z}^{3}-\sqrt{{z}^{6}-2\,{z}^{4}+2\,{z}^{3}-3\,{z}^{2}-2\,z+1}}{2{z}^{2}}}.

Plugging u=su=s (which has a Taylor expansion at z=0z=0) in F(u)z2(ur)(us)F(u)z^{2}(u-r)(u-s), we obtain the equation f1sz3+2f1sz2+s2z2s2z2sz2+2sz+z2uz+1=0,\mathit{f1}s\,z^{3}+2\mathit{f1}s\,z^{2}+s^{2}z^{2}-s^{2}z-2s\,z^{2}+2sz+z^{2}-u-z+1=0, which implies that

f1=1+s1z(z+2).f1=1+\frac{s-1}{z(z+2)}.

Finally, after simplifying by the factor (us)(u-s) in the numerators and denominators, we obtain

F(u)=z1z+rz(ru),G(u)=s+zru, and H(u)=zr+1z(ru)1,F(u)=\frac{z-1}{z}+\frac{r}{z(r-u)},\quad G(u)=\frac{s+z}{r-u},\quad\mbox{ and }\quad H(u)=\frac{zr+1}{z(r-u)}-1,

which implies that f0=1f_{0}=1, g0=s+zrg_{0}=\frac{s+z}{r}, h0=1zrh_{0}=\frac{1}{zr} and

fk=[uk]F(u)=1zrk,gk=[uk]G(u)=s+zrk+1, and hk=[uk]H(u)=1+zrzrk+1.f_{k}=[u^{k}]F(u)=\frac{1}{zr^{k}},\quad g_{k}=[u^{k}]G(u)=\frac{s+z}{r^{k+1}},\quad\mbox{ and }\quad h_{k}=[u^{k}]H(u)=\frac{1+zr}{zr^{k+1}}.
Theorem 4

The bivariate generating function for the total number of UUUU-less PMAP with respect to the length and the height of the end-point is given by

𝑇𝑜𝑡𝑎𝑙(z,u)=1+uz(ru)z2,\mathit{Total}(z,u)=\frac{1+uz}{(r-u)z^{2}},

and we have

[u0]𝑇𝑜𝑡𝑎𝑙(z,u)=1z2r,[u^{0}]\mathit{Total}(z,u)=\frac{1}{z^{2}r},

and for k1k\geqslant 1

[uk]𝑇𝑜𝑡𝑎𝑙(z,u)=rz+1z2rk+1.[u^{k}]\mathit{Total}(z,u)=\frac{rz+1}{z^{2}r^{k+1}}.

Finally, setting t(n,k)=[zn][uk]𝑇𝑜𝑡𝑎𝑙(z,u)t(n,k)=[z^{n}][u^{k}]\mathit{Total}(z,u), we have for n2n\geqslant 2, k1k\geqslant 1,

t(n,k)=t(n,k1)+t(n1,k)t(n1,k1)t(n2,k)t(n2,k2)t(n3,k1).t(n,k)=t(n,k-1)+t(n-1,k)-t(n-1,k-1)-t(n-2,k)\\ -t(n-2,k-2)-t(n-3,k-1).

and setting tn:=t(n,0)t_{n}:=t(n,0), then we have

tn=tn1+tn2+tn3+k=2n2tk2tnk2+k=3n1(tk1tk2)tnk1.t_{n}=t_{n-1}+t_{n-2}+t_{n-3}+\sum\limits_{k=2}^{n-2}t_{k-2}t_{n-k-2}+\sum\limits_{k=3}^{n-1}(t_{k-1}-t_{k-2})t_{n-k-1}.

Proof. The first three equalities are immediately deduced from the previous results. Now, let us prove the last equality. Any non-empty length nn UUUU-less MAP is of the form (ii) HPHP where PP is a UUUU-less MAP of length n1n-1, or (iiii) UDQUDQ where QQ is a MAP of length n2n-2 avoiding UUUU, or (iiiiii) UQDRUQDR, where Q,RQ,R are some MAP avoiding UUUU such that the length of QQ lies into [1,n2][1,n-2] and QQ starts and ends with HH, or (iviv) PQPQ, where P=UPDiP=UP^{\prime}D_{i}, i2i\geqslant 2, and PDi1P^{\prime}D_{i-1} is a MAP of length lying into [3,n1][3,n-1] and starting with HH. The number of PDi1P^{\prime}D_{i-1} of a given length kk is the total number of UUUU-less MAP of length k1k-1 minus the total number of UUUU-less MAP of length k2k-2. Taking into account all these cases, we obtain the result.

\Box

Corollary 10

The generating function that counts the partial UUUU-less MAP with respect to the length is given by

𝑇𝑜𝑡𝑎𝑙(z,1)=1+z(r1)z2.\mathit{Total}(z,1)=\frac{1+z}{(r-1)z^{2}}.

The first few terms of the series expansion of 𝑇𝑜𝑡𝑎𝑙(z,1)\mathit{Total}(z,1) are

1+2z+4z2+9z3+21z4+50z5+122z6+302z7+759z8+1928x9+O(x10),1+2z+4z^{2}+9z^{3}+21z^{4}+50z^{5}+122z^{6}+302z^{7}+759z^{8}+1928x^{9}+O(x^{10}),

which does not appear in [18].

Corollary 11

The generating function that counts UUUU-less MAP with respect to the length is given by

𝑇𝑜𝑡𝑎𝑙(z,0)=1z2r.\mathit{Total}(z,0)=\frac{1}{z^{2}r}.

The first few terms of the series expansion of 𝑇𝑜𝑡𝑎𝑙(z,0)\mathit{Total}(z,0) are

1+z+2z2+4z3+9z4+20z5+47z6+112z7+274z8+679x9+O(z10),1+z+2z^{2}+4z^{3}+9z^{4}+20z^{5}+47z^{6}+112z^{7}+274z^{8}+679x^{9}+O(z^{10}),

which corresponds to the sequence A095980 in [18] counting Motzkin paths of length nn with no occurrences of UHUUHU.

Let 𝒯\mathcal{T} be the infinite matrix 𝒯:=[t(n,k)]n,k0\mathcal{T}:=[t(n,k)]_{n,k\geqslant 0}. The first few rows of the matrix 𝒯\mathcal{T} are

𝒯=(100000000110000000220000000441000000993000000202181000004750214000001121215513100002742981433950000).\mathcal{T}=\left(\begin{array}[]{ccccccccc}1&0&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0&0\\ 2&2&0&0&0&0&0&0&0\\ 4&4&1&0&0&0&0&0&0\\ 9&9&3&0&0&0&0&0&0\\ 20&21&8&1&0&0&0&0&0\\ 47&50&21&4&0&0&0&0&0\\ 112&121&55&13&1&0&0&0&0\\ 274&298&143&39&5&0&0&0&0\\ \end{array}\right).

In Figure 5 we show the UUUU-less PMAP counted by t(4,0)=9t(4,0)=9.

Refer to caption
Figure 5: The 9 UUUU-less PMAP of length 44 ending at height 0.

The matrix 𝒢\mathcal{G} is not a (proper) Riordan array. For this reason, we consider the matrix 𝒢:=[g(n,k)]n0,k0\mathcal{G}:=[g(n,k)]_{n\geqslant 0,k\geqslant 0}, where

g(n,k)={1, if n=k=0;t(n+k1,k), if n1.g(n,k)=\begin{cases}1,&\text{ if }n=k=0;\\ t(n+k-1,k),&\text{ if }n\geqslant 1.\end{cases}

The first few rows of the matrix 𝒢\mathcal{G} are

𝒢=(10000000011000000012100000024310000049841000092121135100020505539196100471211431136426710112298372319203973481).\mathcal{G}=\left(\begin{array}[]{ccccccccc}1&0&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0&0\\ 1&2&1&0&0&0&0&0&0\\ 2&4&3&1&0&0&0&0&0\\ 4&9&8&4&1&0&0&0&0\\ 9&21&21&13&5&1&0&0&0\\ 20&50&55&39&19&6&1&0&0\\ 47&121&143&113&64&26&7&1&0\\ 112&298&372&319&203&97&34&8&1\\ \end{array}\right).
Corollary 12

The matrix 𝒢=[g(n,k)]n,k0\mathcal{G}=[g(n,k)]_{n,k\geqslant 0} is the Riordan array defined by (t(z),t(z)1)(t(z),t(z)-1), where

t(z)=1+z(1z)2(13z+z3)(1+z+z3)2z(1z+z2).t(z)=\frac{1+z(1-z)^{2}-\sqrt{(1-3z+z^{3})(1+z+z^{3})}}{2z(1-z+z^{2})}.

Proof. It follows from the relation t(z)=1zr+1t(z)=\frac{1}{zr}+1. \Box

It seems difficult to obtain a close form for the coefficient g(n,k)g(n,k) using the Lagrange Inversion Formula. Indeed, t(z)t(z) can be expressed in terms of C(u)C(u), where u=14(z6+2z42z3+3z2+2z)u=\frac{1}{4}(-z^{6}+2z^{4}-2z^{3}+3z^{2}+2z) is a polynomial of degree 6, which complicates the calculations.

4.2 A bijective approach

Corollary 11 proves that the set of UUUU-less Motzkin paths with air pockets of length nn (ending on the xx-axis) is equinumerous to the set of Motzkin paths of length nn avoiding UHUUHU. Below, we provide a bijection between these two sets.

Any UUUU-less Motzkin path with air pockets is either of the form (ii) ϵ\epsilon, (iiii) HαH\alpha, (iiiiii) UDαUD\alpha, (iviv) UHDαUHD\alpha, (vv) UHαHDβUH\alpha HD\beta, or (vivi) UHkγDiβUH^{k}\gamma D_{i}\beta, where α,β\alpha,\beta are UUUU-less MAP (possibly empty), and γDi1\gamma D_{i-1} is a UUUU-less MAP and k1k\geqslant 1. According to all these cases, we define the map χ\chi.

Definition 3

We recursively define the map χ\chi from UUUU-less MAP of length nn to UHUUHU-less Motzkin paths of length nn. Let PP be a UUUU-less MAP, we set:

χ(P)={ϵif P=ϵ,(i)Hχ(α)if P=Hα,(ii)UDχ(α)if P=UDα,(iii)UHDχ(α)if P=UHDα,(iv)UHHχ(α)Dχ(β)if P=UHαHDβ,(v)Uχ(γDi1)Hk1Dχ(β)if P=UHkγDiβ,(vi)\chi(P)=\left\{\begin{array}[]{llr}\epsilon&\text{if }P=\epsilon,&(i)\\ H\chi(\alpha)&\text{if }P=H\alpha,&(ii)\\ UD\chi(\alpha)&\text{if }P=UD\alpha,&(iii)\\ UHD\chi(\alpha)&\text{if }P=UHD\alpha,&(iv)\\ UHH\chi(\alpha)D\chi(\beta)&\text{if }P=UH\alpha HD\beta,&(v)\\ U\chi(\gamma D_{i-1})H^{k-1}D\chi(\beta)&\text{if }P=UH^{k}\gamma D_{i}\beta,&(vi)\\ \end{array}\right.

Due to the recursive definition, the image of UUUU-less MAP of length nn by χ\chi is a UHUUHU-less Motzkin path of length nn. The recursive definition naturally induces that χ\chi is a bijection. For instance, if P=UHUHD2UDUHUHHUD3HP=UHUHD_{2}UDUHUHHUD_{3}H then we obtain χ(P)=χ(UH1UHD2)χ(UD)χ(UHUHHUD3)χ(H)=UUHDDUDUUUDHDDH\chi(P)=\chi(UH^{1}UHD_{2})\chi(UD)\chi(UHUHHUD_{3})\chi(H)=UUHDDUDUUUDHDDH (see Figure 6 for an illustration of this example).

Refer to caption
Figure 6: χ(P)=UUHDDUDUUUDHDDH\chi(P)=UUHDDUDUUUDHDDH.

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