A generalization of Chebyshev polynomials and non rooted posets

TL;DR

This paper generalizes Chebyshev polynomials and applies the results to determine the M"obius function of a specific poset, providing an affirmative answer to a conjecture by Bj"orner, Sagan, and Vatter.

Contribution

It introduces a new generalization of Chebyshev polynomials and uses it to analyze the M"obius function of a particular poset structure.

Findings

Derived a formula for the M"obius function of the generalized subword order

Confirmed the conjecture by Bj"orner, Sagan, and Vatter for the specific poset

Established a connection between Chebyshev polynomial generalizations and poset combinatorics

Abstract

In this paper we give a generalization of Chebyshev polynomials and using this we describe the M\"obius function of the generalized subword order from a poset {a1,...as,c |ai<c}, which contains an affirmative answer for the conjecture by Bj\"orner, Sagan, Vatter.[5,10]

Equations

$$T_{n}^{s}(X)=\Sigma_{m\leq n,\ n-m:even}(-1)^{(n-m)/2}\Bigl{(}(\begin{array}[]{c}(n+m)/2\\ (n-m)/2\end{array})\cdot s^{m}-(\begin{array}[]{c}(n+m)/2-1\\ (n-m)/2\end{array})\cdot s^{m-1}\Bigr{)}\cdot X^{m}$$

$$=\Sigma_{m\leq n,\ n-m:even}(-1)^{(n-m)/2}\Bigl{(}(\begin{array}[]{c}(n+m)/2\\ (n-m)/2\end{array})\cdot s^{m}-(\begin{array}[]{c}(n+m)/2-1\\ (n-m)/2\end{array})\cdot s^{m-1}\Bigr{)}\cdot X^{m}$$

$$+\Sigma_{m\leq n+2,\ n+2-m:even}(-1)^{(n+2-m)/2}\Bigl{(}(\begin{array}[]{c}(n+2+m)/2\\ (n+2-m)/2\end{array})\cdot s^{m}-(\begin{array}[]{c}(n+2+m)/2-1\\ (n+2-m)/2\end{array})\cdot s^{m-1}\Bigr{)}\cdot X^{m}$$

$$=\Bigl{(}(\begin{array}[]{c}n+2\\ 0\end{array})\cdot s^{n+2}-(\begin{array}[]{c}n+2\\ 0\end{array})\cdot s^{n+1}\Bigr{)}\cdot X^{n+2}$$

$$+\Sigma_{m\leq n,\ n-m:even}(-1)^{(n-m)/2+1}\Bigl{(}(\begin{array}[]{c}(n+m)/2\\ (n-m)/2+1\end{array})\cdot s^{m}-(\begin{array}[]{c}(n+m)/2-1\\ (n-m)/2+1\end{array})\cdot s^{m-1}\Bigr{)}\cdot X^{m}$$

$$=s^{n+1}(s-1)X^{n+2}$$

$$+\Sigma_{m\leq n-1,\ n-1-m:even}(-1)^{(n-m-1)/2+1}\Bigl{(}(\begin{array}[]{c}(n+m+1)/2\\ (n-m-1)/2+1\end{array})\cdot s^{m+1}-(\begin{array}[]{c}(n+m+1)/2-1\\ (n-m-1)/2+1\end{array})\cdot s^{m}\Bigr{)}\cdot X^{m+1}$$

$$=sX\biggl{(}s^{n}(s-1)X^{n+1}$$

$$+\Sigma_{m\leq n-1,\ n-1-m:even}(-1)^{(n+1-m)/2}\Bigl{(}(\begin{array}[]{c}(n+1+m)/2\\ (n+1-m)/2\end{array})\cdot s^{m}-(\begin{array}[]{c}(n+1+m)/2-1\\ (n+1-m)/2\end{array})\cdot s^{m-1}\Bigr{)}\cdot X^{m}\Biggr{)}$$

$$MobPat\{p_{1}\cdots p_{m},c^{n}\}=\left\{\begin{array}[]{rl}-\Sigma_{i=0}^{n-p}(-1)^{n-p-i}M((m,p,),(n,p+i))&ifm<n\\ (-1)^{n-m}&ifm=n\end{array}\right.$$

$$M((m,p),(n,q))=\Sigma_{k=0}^{n-m}M((m-i,p-i),(n-i-k,q-i))\cdot s^{k}\cdot(\begin{array}[]{c}i+k-1\\ i-1\end{array}).$$

$$M((m,p),(n,q))=\Sigma_{k=0}^{n-m}M((m-i+1,p-i+1),(n-i-k+1,q-i+1))\cdot s^{k}(\begin{array}[]{c}i+k-2\\ i-2\end{array})$$

$$=\Sigma_{k=0}^{n-m}(\Sigma_{l=0}^{n-m}M((m-i,p-i),(n-i-k-l,q-i))\cdot 2^{l})\cdot s^{k}(\begin{array}[]{c}i+k-2\\ i-2\end{array})$$

$$=\Sigma_{k,l=0}^{n-m}M((m-i,p-i),(n-i-(k+l),q-i))\cdot s^{k+l}(\begin{array}[]{c}i+k-2\\ i-2\end{array})$$

$$=\Sigma_{k+l=0}^{n-i}M((m-i,p-i),(n-i-(k+l),q-i))\cdot s^{k+l}(\begin{array}[]{c}i+k-2\\ i-2\end{array})$$

$$=\Sigma_{\alpha=0}^{n-i}M((m-i,p-i),(n-i-\alpha,q-i))(\Sigma_{j=0}^{\alpha}(\begin{array}[]{c}i+j-2\\ i-2\end{array}))\cdot s^{\alpha}.$$

$$\Sigma_{x=0}^{\alpha}(\begin{array}[]{c}x+i\\ i\end{array})=(\begin{array}[]{c}i+\alpha+1\\ i+1\end{array}),$$

$$=\Sigma_{\alpha=0}^{n-i}M((m-i,p-i),(n-i-\alpha,q-i))(\Sigma_{j=0}^{\alpha}(\begin{array}[]{c}i+j-2\\ i-2\end{array}))\cdot s^{\alpha}$$

$$=\Sigma_{\alpha=0}^{n-i}M((m-i,p-i),(n-i-\alpha,q-i))\cdot s^{\alpha}(\begin{array}[]{c}n+\alpha-1\\ \alpha-1\end{array}).$$

$$M((m,p),(n,q))=\Sigma_{k=0}^{n-m}M((m-p,0),(n-p-k,q-p))\cdot s^{k}\cdot(\begin{array}[]{c}p+k-1\\ p-1\end{array}).$$

$$=-\Sigma_{i=0}^{n-p}(-1)^{n-p-i}\Sigma_{k=0}^{n-m}M((m-p,0),(n-p-k,i))\cdot s^{k}(\begin{array}[]{c}p+k-1\\ p-1\end{array})$$

$$=(-1)^{n-p-1}\Sigma_{i=0}^{n-p}\Sigma_{k=0}^{n-m}M((m-p,0),(n-p-k,i))(-1)^{i}\cdot s^{k}(\begin{array}[]{c}p+k-1\\ p-1\end{array})$$

$$=(-1)^{n-p-1}\Sigma_{k=0}^{n-m}\{\displaystyle\underbrace{\Sigma_{i=0}^{n-p-k}M((m-p,0),(n-p-k,i))(-1)^{i}}_{=0}\}\cdot s^{k}(\begin{array}[]{c}p+k-1\\ p-1\end{array})$$

$$T(k,n)=-\Sigma_{i=k}^{n}(\begin{array}[]{c}n\\ i\end{array})\cdot s^{n-i}\cdot(-1)^{n-i}.$$

$$M((k,k),(n,i))=(\begin{array}[]{c}n\\ i\end{array})\cdot s^{n-i}.$$

$$=-\Sigma_{i=k}^{l}(\begin{array}[]{c}l\\ i\end{array})\cdot s^{l-i}\cdot(-1)^{l-i}+\Sigma_{i=k-1}^{l-1}(\begin{array}[]{c}l-1\\ i\end{array})\cdot s^{l-i-1}\cdot(-1)^{l-i-1}$$

$$=-\Sigma_{i=k}^{l-1}(\begin{array}[]{c}l\\ i\end{array})\cdot s^{l-i}\cdot(-1)^{l-i}-1+\Sigma_{i=k-1}^{l-2}(\begin{array}[]{c}l-1\\ i\end{array})\cdot s^{l-i-1}\cdot(-1)^{l-i-1}+1$$

$$=-\Sigma_{i=k}^{l-1}\{(\begin{array}[]{c}l\\ i\end{array})-(\begin{array}[]{c}l-1\\ i-1\end{array})\}\cdot s^{l-i}\cdot(-1)^{l-i}$$

$$=-s\cdot\Sigma_{i=k}^{l-1}(\begin{array}[]{c}l-1\\ i\end{array})\cdot s^{l-i-1}\cdot(-1)^{l-i-1}=-sT(k,l-1).$$

$$=\Sigma_{k=1}^{n}\bigl{(}-\Sigma_{i=0}^{k}s^{i}(\begin{array}[]{c}k\\ i\end{array})(-1)^{i}\bigr{)}\mu(c^{k},c^{n})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$

Full text

A generalization of Chebyshev polynomials and non rooted posets

Masaya Tomie

(tomie@math.tsukuba.ac.jp)

In this paper we give a generalization of Chebyshev polynomials and using this we describe the Mobius function of the generalized subword order derived from a poset $\{a_{1},\cdots a_{s},c\ |\ \ a_{i}<c\ for\ i=1,\cdots s\}$, which contains an affirmative answer for the conjecture by Björner, Sagan and Vatter.(cf,[5] [10])

1 INTRODUCTION

Björner was the first to determine the Möbius functions of factor orders and subword orders. To determine the Möbius functions, he used involutions, shellability, and generating functions. [2][3][4]

Björner and Stanley found an interesting relation among the subword order derived from a two point set $\{a,\ b\}$ , symmetric groups and composition orders. [6]

Factor orders, subword orders, and generalized subword orders were studied in the context of Möbius functions derived from word orders.

In [10] Sagan and Vatter gave a description of the Möbius function of the generalized subword order derived from positive integers in two ways, namely the sign reversing involution and the discrete Morse theory. More generally they gave a combinatorial description of the Möbius functions derived from rooted forests. And in [5][10] they gave a very interesting conjecture which connects with the relation between a non-rooted forest $P_{2}$ as in Notation.1 and Chebyshev polynomials.

Conjecture 1** **([5][10])

We put $P:=\{a,\ b,\ c,\ |\ a<c,\ b<c\ \}$, and consider the poset $P^{*}$ consisting of finite words of $P$ with its generalized subword order. Let $\mu$ be a Möbius function of $P^{*}$. Suppose $0\leq i\leq j$.

Then $\mu(a^{i},c^{j})$ is the coefficient of $X^{j-i}$ in $T_{i+j}(X)$.

Now we call $\{T_{n}(X)\ |\ n\in\mathbb{N}\}$ Chebyshev polynomials of first kind.

A series of Chebyshev polynomials $\{T(X)\ |\ n\in\mathbb{N}\}$ is a system of orthogonal polynomials and induces a special case of hypergeometric functions as a generalization of a binomial series. And this polynomial series is an example of the best approximation polynomials.

Not only in analysis, but in combinatorics, Chebyshev polynomials appear in permutation pattern avoidances [7] and Chebyshev posets, Chebyshev transformations defined by Hetyei which are related to $cd$-indeices, $f$-vectors and $h$-vectors respectively.

In this paper we give a natural generalization of Chebyshev polynomials in the following way.

Definition 1** **(generalized Chebyshev polynomials)

We define the polynomial $T^{s}_{k}(X)\ for\ s,k\in\mathbb{N}$ as follows:

(1)   $T^{s}_{0}(X)=1,\ T^{s}_{1}(X)=(s-1)X$,

(2)   $T^{s}_{k+2}(X)+T^{s}_{k}(X)=sX\cdot T^{s}_{k+1}(X)$.

Now the $T^{2}_{n}(X)$ are Chebyshev polynomials of first kind. And notice $deg(T^{s}_{k}(X))=k$.

Then, using generalized Chebyshev polynomials, we generalize the conjecture as follows.

Theorem 1

Let $P_{s}$ be a poset as Notation.1 and $\mu$ be the Möbius function of $P_{s}^{*}$. Then for $0\leq m\leq n$ , $\mu(a_{1}^{m},c^{n})$ is the coefficient of $X^{n-m}$ in $T^{s}_{m+n}(x)$.

2 PRELIMINERIES

In this section, we give some basic definitions and notations used in this paper. For the basic definitions of posets and Möbius functions, see [12] and for the definitions of subword orders and generalized subword orders, see [2] [3] [4] [5] [10]. First we recall a path in a poset $P$.[12]

Definition 2** **([12])

Let $P$ be a poset. An arranged sequence $(\theta_{1},\cdots,\theta_{r})$ with $\theta_{i}\in P$ and $\theta_{1}<\cdots<\theta_{r},$ is called a path of length $r-1$ . We denote it by $(\theta_{1}\rightarrow\cdots\rightarrow\theta_{r})_{p}$

Proposition 1** **([12])

Let $P$ be a locally finite poset, $\mu$ be a Möbius function of $P$ and $C_{k}$ be the number of paths of $P$ of length $k$ respectively. Then we have

$\mu(u,v)=\Sigma_{k}(-1)^{k}C_{k}$, for all $u\leq v\in P$.

Definition 3** **([10])

Let $P^{*}$ be the poset with the subword order derived from a poset $P$. We take $p_{1},\cdots p_{k}$ and $q_{1}\cdots q_{l}$ from $P^{*}$. If $p_{1}\cdots p_{k}\leq q_{1}\cdots q_{l}$ as a subword order, we call $S(j_{1},\cdots,j_{k})$ ($j_{1}<\cdots<j_{k}$), an embedding of $p_{1}\cdots p_{k}$ into $q_{1}\cdots q_{l}$ if $p_{i}\leq q_{j_{i}}$ for $1\leq i\leq k$

And an embedding $S(j_{1},\cdots,j_{k})$ is called the right most embedding of $p_{1}\cdots p_{k}$ into $q_{1}\cdots q_{l}$, if for any embedding $S^{\prime}(j^{\prime}_{1},\cdots,j^{\prime}_{k})$ , we have $j^{\prime}_{i}\leq j_{i}$ for all $1\leq i\leq k$.

Notation 1

In this paper we fix a poset $P_{s}$ for $s\in\mathbb{N}$ as follows.

$P_{s}:=\{a_{1},\cdots a_{s},c\ |\ \ a_{i}<c,\ for\ i=1,\cdots s\}\}$

Definition 4

We define as follows.

Let $P_{s}^{*}$ be a poset with the generalized subword order derived from a poset $P_{s}$ as in Notation 1 and let $X$ be a set of the paths of $P$.

Put $Mob(X):=\Sigma_{k\geq 1}C_{k}$, where $C_{k}$ is the number of paths in $X$ whose length is $k$. Also we define

$\{a^{k}c^{l}\}:=\{p_{1}\cdots p_{k+l}\ |\ \ \sharp\{p_{1},\cdots,p_{k+l}\}\cap\{a_{1},\cdots a_{s}\}=k,\ \sharp\{p_{1},\cdots,p_{k+l}\}\cap\{c\}=l\}$ for $k,l\in\mathbb{N}\cup\{0\}$,

$<p_{1}\cdots p_{k},\{a^{l}c^{m}\}>:=\{q_{1}\cdots q_{l+m}\in\{a^{l}c^{m}\}\ |\ \ p_{1}\cdots p_{k}\leq q_{1}\cdots q_{l+m}\}$ for $k,l\in\mathbb{N}\cup\{0\}$, and

$Pat\{p_{1}\cdots p_{k},q_{1}\cdots q_{l}\}:=\{(p_{1}\cdots p_{k}\rightarrow\theta_{1}\rightarrow,\cdots\rightarrow\theta_{r}\rightarrow q_{1}\cdots q_{l})_{p}\ |\ p_{1}\cdots p_{k}<\theta_{1}<\cdots<\theta_{r}<q_{1}\cdots q_{l},\ |\theta_{i}|=l\}$ respectively. Here $|\theta|$ is the number of letters of $\theta$.

Definition 5** **(generalized Chebyshev polynomial)

We define the polynomial $T^{s}_{k}(X)\ s,k\in\mathbb{N}$ as follows:

(1)   $T^{s}_{0}(X)=1,\ T^{s}_{1}(X)=(s-1)X$,

(2)   $T^{s}_{k+2}(X)+T^{s}_{k}(X)=sX\cdot T^{s}_{k+1}(X)$.

Now the $\{T^{2}_{n}(X)\ |\ n\in\mathbb{N}\}$ are Chebyshev polynomials of first kind. And notice $deg(T^{s}_{k}(X))=k$.

Here we give a simple expression of the generalized Chebyshev polynomials.

Proposition 2

For $s,n\in\mathbb{N}$, we have

[TABLE]

PROOF

We show the above formula by induction. It is easy to see $T_{0}^{s}(X)=1,\ T_{1}^{s}(X)=(s-1)X$. Now we have

$T_{n}^{s}+T_{n+2}^{s}$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

$=sX\cdot T_{n+1}^{s}$.

Hence we obtain the derived result.                   $\Box$

3 MAIN RESULTS

In this section, we give a proof of Theorem 1

Lemma 1

Let $P$ be a finite poset and we take an element $x\in P$. We put as follows:

$P_{\leq x}:=\{y\ |\ y\leq x\}$ , $\widehat{P_{\leq x}}:=\{(\theta_{1}\rightarrow\cdots\rightarrow\theta_{r-1}\rightarrow x)_{p}\ |\ \theta_{i}\in P\}$ , $P_{\geq x}:=\{y\ |\ y\geq x\}$ , $\widehat{P_{\geq x}}:=\{x\rightarrow\theta_{1}\rightarrow\cdots\rightarrow\theta_{r-1})_{p}\ |\ \theta_{i}\in P\}$ and $\mathfrak{P}_{x}:=\{(\cdots\rightarrow\tau_{r}\rightarrow x\rightarrow\sigma_{1}\rightarrow\cdots)\ |\ \tau_{i}\leq x,\ \sigma_{i}\geq x\ \}$. Now a path $(x)\in\widehat{P_{\leq x}},\ \widehat{P_{\geq x}},\ \mathfrak{P}_{x}$. Then we have $Mob\mathfrak{P}_{x}=Mob\widehat{P_{\leq x}}\ Mob\widehat{P_{\geq x}}$.

PROOF

Notice that a path which passes through $x$ splits into the two paths, one starts from $x$ and the other one ends $x$. From that we obtain the derived result.      $\Box$

Lemma 2

For $m,n,p,q\in\mathbb{N}\cup\{0\}$ such that $0\leq m\leq n,\ 0\leq p\leq m,\ 0\leq q\leq n$, we take $p_{1}\cdots p_{m},\tilde{p_{1}}\cdots\tilde{p_{m}}\in\{a^{m-p}c^{p}\}$. Then we have

$\sharp<p_{1}\cdots p_{m},\{a^{n-q}c^{q}\}>=\sharp<\tilde{p_{1}}\cdots\tilde{p_{m}},\{a^{n-q}c^{q}\}>$.

PROOF

Claim 1   We have $\sharp<p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\cdots p_{m},\{a^{n-q}c^{q}\}>=\sharp<p_{1}\cdots\displaystyle\overbrace{{a_{y}}}^{i-th}\cdots p_{m},\{a^{n-q}c^{q}\}>$.

(Proof of claim1)

We take $\forall q_{1}\cdots q_{n}\in<p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\cdots p_{m},\{a^{n-q}c^{q}\}>$. And we consider the right most embedding into $q_{1}\cdots q_{n}$. Notice that the right most embedding is unique. Here we put $S(j_{1},j_{2},\cdots,j_{m})$ as the right most embedding $p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\cdots p_{m}$ into $q_{1}\cdots q_{n}$ .

Now we define the map $\Phi$ as follows.

$\Phi$               $<p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\cdots p_{m},\{a^{n-q}c^{q}\}>\longrightarrow<p_{1}\cdots\displaystyle\overbrace{a_{y}}^{i-th}\cdots p_{m},\{a^{n-q}c^{q}\}>$

$\Phi(q_{1}q_{2},\cdots\displaystyle\overbrace{a_{x}}^{j_{i}-th}a_{x_{1}}\cdots a_{x_{t}}\displaystyle\overbrace{p_{i+1}}^{j_{i+1}-th}\cdots q_{n})=q_{1}q_{2}\cdots\displaystyle\overbrace{a_{y}}^{j_{i}-th}a_{y_{1}}\cdots a_{y_{t}}\displaystyle\overbrace{p_{i+1}}^{j_{i+1}-th}\cdots q_{n}$,

Here we put $a_{y_{k}}=a_{x_{k}+y-x}$, and $a_{k+s}=a_{k}$.

It is easy to see the right most embedding of $p_{1}\cdots\displaystyle\overbrace{a_{y}}^{i-th}\cdots p_{m}$ into $\Phi(q_{1}\cdots q_{n})$ is $S(j_{1},j_{2},\cdots,j_{m})$. And by the construction of $\Phi$, we can easily define the inverse map of $\Phi$. Hence we prove this claim.

Claim 2   We have $\sharp<p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\displaystyle\overbrace{c}^{(i+1)-th}\cdots p_{m},\{a^{n-q}c^{q}\}>=\sharp<p_{1}\cdots\displaystyle\overbrace{c}^{i-th}\displaystyle\overbrace{a_{x}}^{(i+1)-th}\cdots p_{m},\{a^{n-q}c^{q}\}>$.

(Proof of claim2)

We take $\forall q_{1}\cdots q_{n}\in<p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\displaystyle\overbrace{c}^{(i+1)-th}\cdots p_{m},\{a^{n-q}c^{q}\}>$ and put $S(j_{1},j_{2},\cdots,j_{m})$ as the right most embedding $p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\displaystyle\overbrace{c}^{(i+1)-th}\cdots p_{m}$ into $q_{1}\cdots q_{n}$.

Now we define the map $\Phi$ as follows.

$\Phi$               $<p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\displaystyle\overbrace{c}^{(i+1)-th}\cdots p_{m},\{a^{n-q}c^{q}\}>\longrightarrow<p_{1}\cdots\displaystyle\overbrace{c}^{i-th}\displaystyle\overbrace{a_{x}}^{(i+1)-th}\cdots p_{m},\{a^{n-q}c^{q}\}>$,

$\Phi(q_{1}\cdots\displaystyle\underbrace{\displaystyle\overbrace{q_{j_{i}}}^{=a_{x}}\cdots}_{A}\displaystyle\underbrace{\displaystyle\overbrace{q_{j_{i+1}}}^{=c}\cdots}_{B}q_{j_{i+2}}\cdots q_{n})=q_{1}\cdots\displaystyle\underbrace{\displaystyle\overbrace{q_{j_{i+1}}}^{=c}\cdots}_{B}\displaystyle\underbrace{\displaystyle\overbrace{q_{j_{i}}}^{=a_{x}}\cdots}_{A}q_{j_{i+2}}\cdots q_{n}$.

Here the right most embedding of $p_{1}\cdots\displaystyle\overbrace{c}^{i-th}\displaystyle\overbrace{a_{x}}^{(i+1)-th}\cdots p_{m}$ into $q_{1}\cdots\displaystyle\underbrace{\displaystyle\overbrace{q_{j_{i+1}}}^{=c}\cdots}_{B}\displaystyle\underbrace{\displaystyle\overbrace{q_{j_{i}}}^{=a}\cdots}_{A}q_{j_{i+2}}\cdots q_{n}$ is $S(j_{1}\cdots j_{i},j_{i+2}+j_{i}-j_{i+1},j_{i+2}\cdots j_{m})$. By the construction, all of the elements of

$<p_{1}\cdots\displaystyle\overbrace{a_{x}}^{i-th}\displaystyle\overbrace{c}^{(i+1)-th}\cdots p_{m},\{a^{n-q}c^{q}\}>$ whose right most embedding are $S(j_{1},j_{2},\cdots,j_{m})$, have one to one correspondence to the elements of $<p_{1}\cdots\displaystyle\overbrace{c}^{i-th}\displaystyle\overbrace{a_{x}}^{(i+1)-th}\cdots p_{m},\{a^{n-q}c^{q}\}>$ whose right most embedding are $S(j_{1}\cdots j_{i},j_{i+2}+j_{i}-j_{i+1},j_{i+2}\cdots j_{m})$. Hence the $\Phi$ is bijeciton. Therefore we have this claim2. By these claims we obtain the derived result.        $\Box$

From Lemma 2 , if $p_{1}\cdots p_{m}\in\{a^{m-p}c^{p}\}$, then

$\sharp<p_{1}\cdots p_{m},\{a^{n-q}c^{q}\}>=\sharp<\displaystyle\underbrace{a_{1}\cdots a_{1}}_{(m-p)times}\displaystyle\underbrace{c\cdots c}_{p-times},\{a^{n-q}c^{q}\}>$. So we denote the number as $M((m,p),(n,q))$ for all $0\leq m\leq n,\ 0\leq p\leq m,\ 0\leq q\leq n$.

Lemma 3

Let $k,l\in\mathbb{N}\cup\{0\},0\leq k\leq l$ and $p_{1}\cdots p_{l}\in\{a^{l-k}c^{k}\}$, then we have $[p_{1}\cdots p_{l},c^{l}]\simeq B_{l-k}$. Now $B_{l-k}$ is a Boolean algebra of rank $l-k$.

Lemma 4

For $m,n,p,k\in\mathbb{N}\cup\{0\}$ such that $0\leq m\leq n,\ 0\leq p\leq m$, we take $p_{1}\cdots p_{m}\in\{a^{m-p}c^{p}\}$, then the number of paths in $Pat\{p_{1}\cdots p_{m},c^{n}\}$ whose length are $k$ equals to the number of paths in $Pat\{\displaystyle\underbrace{a_{1}\cdots a_{1}}_{(m-p)times}\displaystyle\underbrace{c\cdots c}_{p-times},c^{n}\}$.

PROOF

Notice that if we take $q_{1}\cdots q_{n}\in\{a^{n-q}c^{q}\}$, then the number of length $l$ paths from $q_{1}\cdots q_{n}$ to $c^{n}$ equals to the number of length $l$ paths from $\displaystyle\underbrace{a_{1}\cdots a_{1}}_{(n-q)times}\displaystyle\underbrace{c\cdots c}_{q-times}$ to $c^{n}$.

Hence we have

$\sharp\{p_{1}\cdots p_{m}\rightarrow\theta_{1}\rightarrow\cdots\rightarrow\theta_{k}=c^{n}\ |\ |\theta_{i}|=n\ for\ i=1,\cdots k\}$

$=\Sigma_{p\leq r\leq n}M((m,p),(n,r))\sharp\{\displaystyle\underbrace{a_{1}\cdots a_{1}}_{(n-r)times}\displaystyle\underbrace{c\cdots c}_{r-times}\rightarrow\tau_{1}\rightarrow\cdots\rightarrow\tau_{k-1}=c^{n}\}$.         (By Lemma 3)

Hence we obtain the derived result.

$\Box$

Lemma 5

For $m,n,p\in\mathbb{N}\cup\{0\}$, such that $0\leq m\leq n,\ 0\leq p\leq m$, we take $p_{1},\cdots p_{m}\in\{a^{m-p}c^{p}\}$. Then we have

[TABLE]

PROOF

From Lemma 4 we have $MobPat\{p_{1}\cdots p_{m},c^{n}\}=MobPat\{\displaystyle\underbrace{a_{1}\cdots a_{1}}_{(m-p)times}\displaystyle\underbrace{c\cdots c}_{p-times},c^{n}\}$.

Then we have

$MobPat\{\displaystyle\underbrace{a_{1}\cdots a_{1}}_{(m-p)times}\displaystyle\underbrace{c\cdots c}_{p-times},c^{n}\}$

$=(-1)\Sigma_{i=0}^{n-p}M((m,p),(n,p+i))\mu(\displaystyle\underbrace{a_{1}\cdots a_{1}}_{(n-p-i)times}\displaystyle\underbrace{c\cdots c}_{(p+i)times},c^{n})$           (By Proposition 1)

$=(-1)\Sigma_{i=0}^{n-p}M((m,p),(n,p+i))(-1)^{n-p-i}$.

Hence we obtain the derived result.           $\Box$

Lemma 6

For $m,n,p,q\in\mathbb{N}\cup\{0\}$ such that $1\leq m\leq n,\ 1\leq p\leq m,\ 1\leq q\leq n,\ 1\leq p\leq q$, we have

$M((m,p),(n,q))=\Sigma_{i=0}^{n-m}M((m-1,p-1),(n-1-i,q-1))\cdot s^{i}$.

PROOF

We have

the left hand side

$=\sharp\{x_{1}\cdots x_{n}\in\{a^{n-q}c^{q}\}\ |\ a_{1}^{m-p}c^{p}\leq x_{1}\cdots x_{n}\}$         (By Lemma 2)

$=\sharp\{x_{1}\cdots x_{n}\in\{a^{n-q}c^{q}\}\ |\ a_{1}^{m-p}c^{p}\leq x_{1}\cdots x_{n},x_{n}=c\}$

$+\sharp\{x_{1}\cdots x_{n}\in\{a^{n-q}c^{q}\}\ |\ a_{1}^{m-p}c^{p}\leq x_{1}\cdots x_{n},x_{n-1}=c,x_{n}\neq c\}$

$\cdots$

$+\sharp\{x_{1}\cdots x_{n}\in\{a^{n-q}c^{q}\}\ |\ a_{1}^{m-p}c^{p}\leq x_{1}\cdots x_{n},x_{n-i}=c,x_{n-i+k}\neq c(1\leq k\leq i)\}$ $\vdots$

$+\sharp\{x_{1}\cdots x_{n}\in\{a^{n-q}c^{q}\}\ |\ a_{1}^{m-p}c^{p}\leq x_{1}\cdots x_{n},x_{m}=c,x_{m+k}\neq c(1\leq k\leq n-m)\}$

$=\sharp\{\displaystyle\underbrace{\cdots}_{A}\ |\ a_{1}^{m-p}c^{p-1}\leq A\ A\in\{a^{n-q}c^{q-1}\}\}$

$+\sharp\{\displaystyle\underbrace{\cdots}_{A}\displaystyle\underbrace{c}_{n-1}\ |\ a_{1}^{m-p}c^{p-1}\leq A\ A\in\{a^{n-q-1}c^{q-1}\}\}\cdot s^{1}$

$\vdots$

$+\sharp\{\displaystyle\underbrace{\cdots}_{A}\displaystyle\underbrace{c}_{n-i}\ |\ a_{1}^{m-p}c^{p-1}\leq A\ A\in\{a^{n-q-i}c^{q-1}\}\}\cdot s^{i}$

$\vdots$

$+\sharp\{\displaystyle\underbrace{\cdots}_{A}\displaystyle\underbrace{c}_{n-i}\ |\ a_{1}^{m-p}c^{p-1}\leq A\ A\in\{a^{m-q}c^{q-1}\}\}\cdot s^{n-m}$

(In case of $n-q-i<0$ , we recognize $\sharp\{\displaystyle\underbrace{\cdots}_{A}\displaystyle\underbrace{c}_{n-i}\ |\ a^{m-p}c^{p-1}\leq A\ A\in a^{n-q-i}c^{q-1}\}$ as [math].)

$=M((m-1,p-1),(n-1,q-1))\cdot s^{0}+M((m-1,p-1),(n-2,q-1))\cdot s^{1}+\cdots$

$+M((m-1,p-1),(n-1-i,q-1))\cdot s^{i}\cdots M((m-1,p-1),(m-1,q-1))\cdot s^{n-m}$

Hence we obtain the derived result.           $\Box$

Lemma 7

For $m,n,i\in\mathbb{N}\cup\{0\}$ such that $i\leq m\leq n,\ i\leq p\leq q,\ i\leq p\leq m,\ i\leq q\leq n$, we have

[TABLE]

PROOF

In case of $i=1$, it is shown by Lemma 6.

We show the above formula by induction. We suppose this lemma holds for $i-1$. Now we see

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Now we remark the following formula.

[TABLE]

hence we have

[TABLE]

[TABLE]

$\Box$

Lemma 8

For $m,n,p,q\in\mathbb{N}$, $1\leq m\leq n,\ 1\leq p\leq q,1\leq p\leq m,\ 1\leq q\leq n,$ we have

[TABLE]

Lemma 9

For $1\leq\alpha\leq\beta$, we have

$\Sigma_{i=0}^{\beta}M((\alpha,0),(\beta,i))\cdot(-1)^{i}=0$.

PROOF

We give a combinatorial proof. We put $\widehat{M}_{i}:=<\displaystyle\underbrace{a_{1}\cdots a_{1}}_{\alpha-times},\{a^{\beta-i}c^{i}\}>$, $\widehat{M}:=\biguplus_{0\leq i\leq\beta}\widehat{M}_{i}$, $\widehat{M}_{ev}:=\biguplus_{0\leq i\leq\beta\ i;\ even}\widehat{M}_{i}$ and $\widehat{M}_{odd}:=\biguplus_{0\leq i\leq\beta\ i;\ odd}\widehat{M}_{i}$.

Then we have $\Sigma_{i=0}^{\beta}M((\alpha,0),(\beta,i))\cdot(-1)^{i}=\sharp\widehat{M}_{ev}-\sharp\widehat{M}_{odd}$.

We consider the map $\Psi$ as follows.

$\Psi$               $\widehat{M}\longrightarrow\widehat{M}$

$\Psi(\displaystyle\underbrace{\cdots}_{A}a_{1}a_{x_{1}}\cdots a_{x_{t}})=\displaystyle\underbrace{\cdots}_{A}ca_{x_{1}}\cdots a_{x_{t}})$

$\Psi(\displaystyle\underbrace{\cdots}_{A}ca_{x_{1}}\cdots a_{x_{t}})=\displaystyle\underbrace{\cdots}_{A}a_{1}a_{x_{1}}\cdots a_{x_{t}}$

$\Psi(\displaystyle\underbrace{\cdots}_{A}a_{1})=\displaystyle\underbrace{\cdots}_{A}c$

$\Psi(\displaystyle\underbrace{\cdots}_{A}c)=\displaystyle\underbrace{\cdots}_{A}a_{1}$

Here $\Psi$ changes $a_{1}$ into $c$ and $c$ into $a_{1}$ which appears right most position of each elements. Since $\alpha$ not being [math], each element of $\widehat{M}$ contains $a_{1}$ or $c$. From that the map $\Psi$ is well-defined. Therefore obviously $\Psi^{-1}=\Psi$ and $\Psi(\widehat{M}_{ev})=(\widehat{M}_{odd})\ \Psi(\widehat{M}_{odd})=\widehat{M}_{ev}$. Hence $\Psi$ is a bijection and $\sharp\widehat{M}_{ev}=\sharp\widehat{M}_{odd}$. Hence we obtain the derived result.           $\Box$

Lemma 10

For $m,n,p\in\mathbb{N}\cup\{0\}$, $0\leq m<n,\ 0\leq p<m$, we have

$p_{1}\cdots p_{m}\in\{a^{m-p}c^{p}\}\Longrightarrow MobPat\{p_{1}\cdots p_{m},c^{n}\}=0$.

PROOF

We have

$MobPat\{p_{1}\cdots p_{m},c^{n}\}=-\Sigma_{i=0}^{n-p}M((m,p),(n,p+i))\cdot(-1)^{n-p-i}$

[TABLE]

[TABLE]

[TABLE]

$=0$. Hence we obtain the derived result.            $\Box$

Lemma 11

For $m,n\in\mathbb{N},\ 1\leq m\leq n$, we put

$P:=\{a_{1}^{m}\rightarrow\tau_{1}\rightarrow\cdots\rightarrow c^{m}\rightarrow\theta^{k_{1}}_{1}\cdots\rightarrow c^{k_{1}}\rightarrow\theta^{k_{2}}_{1}\cdots\rightarrow c^{k_{2}}\rightarrow\cdots\rightarrow c^{k_{r}}\ |\ m<k_{1}<\cdots<k_{r}=n,\ |\tau_{i}|=m,\ |\theta^{k_{i}}_{j}|=k_{i}\ \}$, and $p_{1}\cdots p_{m}\in\{a^{m}\ \}$.

Then we have $\mu(p_{1}\cdots p_{m},c^{n})=\mu(a_{1}^{m},c^{n})=Mob(P)=(-1)^{m}\mu(c^{m},c^{n})$.

PROOF

We have

$\mu(p_{1}\cdots p_{m},c^{n})$ $=Mob(\{(p_{1}\cdots p_{m}\rightarrow\theta_{1}\rightarrow\cdots\theta_{r}\rightarrow c^{n})\ |\ p_{1}\cdots p_{m}<\theta_{1}<\cdots\theta_{r}<c^{n}\})$.

Now we put

$X^{l_{1}}_{l_{2}}$

$:=\{(p_{1}\cdots p_{m}\rightarrow\cdots\theta_{r}\rightarrow\tau_{1}\rightarrow\cdots\tau_{s}\rightarrow c^{l_{2}}\rightarrow\sigma^{k_{1}}_{1}\rightarrow\cdots\rightarrow c^{k_{1}}\rightarrow\sigma^{k_{2}}_{1}\rightarrow\cdots\rightarrow c^{k_{2}}\cdots\rightarrow c^{n})\ |\ |\theta_{r}|=l_{1},\ \theta_{r}\neq c^{l_{1}},\ |\tau_{1}|,\cdots|\tau_{s}|=l_{2},\ |\sigma^{k_{i}}_{j}|=k_{i}\}$.

Then we have

$\{(p_{1}\cdots p_{m}\rightarrow\theta_{1}\rightarrow\cdots\theta_{r}\rightarrow c^{n})\ |\ p_{1}\cdots p_{m}<\theta_{1}<\cdots\theta_{r}<c^{n}\}$

$=\biguplus_{m\leq l_{1}<l_{2}\leq n}X^{l_{1}}_{l_{2}}$

$\biguplus\{(p_{1}\cdots p_{m}\rightarrow\cdots\rightarrow c^{m}\rightarrow\sigma^{k_{1}}_{1}\rightarrow\cdots\rightarrow c^{k_{1}}\rightarrow\sigma^{k_{2}}_{1}\rightarrow\cdots\rightarrow c^{k_{2}}\cdots\rightarrow c^{n})\ |\ |\sigma^{k_{i}}_{j}|=k_{i}\}$.

Now for $m\leq l_{1}<l_{2}\leq n$,

$Mob(X^{l_{1}}_{l_{2}})=$

$\Sigma_{q_{1}\cdots q_{l_{1}}\neq c^{l_{1}}}Mob(\{(p_{1}\cdots p_{m}\rightarrow\theta_{1}\rightarrow\cdots\rightarrow\theta_{r}\rightarrow q_{1}\cdots q_{l_{1}})\ |\ p_{1}\cdots p_{m}<\theta_{1}<\cdots<q_{1}\cdots q_{l_{1}}\})\cdot$

$Mob(\{(q_{1}\cdots q_{l_{1}}\rightarrow\tau_{1}\rightarrow\cdots\rightarrow\tau_{s}\rightarrow c^{l_{2}})\ |\ |\tau_{i}|=l_{2}\})\cdot$

$Mob(\{(c^{l_{2}}\rightarrow\sigma^{k_{1}}_{1}\rightarrow\cdots\rightarrow c^{k_{1}}\rightarrow\sigma^{k_{2}}_{1}\rightarrow\cdots\rightarrow c^{k_{2}}\rightarrow\cdots\rightarrow c^{n})\ |\ |\sigma^{k_{i}}_{j}|=k_{i}\})$.

By Lemma 10 , we have

$Mob(\{(q_{1}\cdots q_{l_{1}}\rightarrow\tau_{1}\rightarrow\cdots\rightarrow\tau_{s}\rightarrow c^{l_{2}})\ |\ |\tau_{i}|=l_{2}\})=0$.

Hence we have $Mob(X^{l_{1}}_{l_{2}})=0$

Therefore we have $\mu(p_{1},\cdots p_{m},c^{n})=$

$Mob(\{(p_{1},\cdots p_{m}\rightarrow\cdots\rightarrow c^{m}\rightarrow\sigma^{k_{1}}_{1}\rightarrow\cdots\rightarrow c^{k_{1}}\rightarrow\sigma^{k_{2}}_{1}\rightarrow\cdots\rightarrow c^{k_{2}}\cdots\rightarrow c^{n})\ |\ |\sigma^{k_{i}}_{j}|=k_{i}\})$

$=Mob(\{(p_{1}\cdots p_{m}\rightarrow\theta_{1}\cdots\theta_{r}\rightarrow c^{m})\ |\ |\theta_{i}|=m\})\cdot$

$Mob(\{(c^{m}\rightarrow\sigma^{k_{1}}_{1}\rightarrow\cdots\rightarrow c^{k_{1}}\rightarrow\sigma^{k_{2}}_{1}\rightarrow\cdots\rightarrow c^{k_{2}}\cdots\rightarrow c^{n})\ |\ |\sigma^{k_{i}}_{j}|=k_{i}\})$

$=(-1)^{m}\cdot\mu(c^{m},c^{n})$.

Hence we obtain the derived result.           $\Box$

Lemma 12

We have

$\mu(\phi,c)=s-1,\ \ \mu(a_{i},c)=-1,\ \mu(c,c^{2})=2s-1,\ \mu(a_{i},c^{2})=-2s+1$.

Now we put $T(k,n):=MobPat\{c^{k}\rightarrow\theta_{1}\rightarrow\cdots\rightarrow\theta_{r}=c^{n}\ |\ |\theta_{i}|=n\}$, $T(n,k)=0$ for $0\leq k<n$, and $T(n,n):=-1$ for $0\leq n$ .

Lemma 13

For $0\leq k\leq n$, we have

[TABLE]

PROOF

In case when $k=n$, it is trivial.

In case when $0\leq k<n$, we have

$T(k,n)=-\Sigma_{i=k}^{n}M((k,k),(n,i))\cdot(-1)^{n-i}$.

Now we have

[TABLE]

Hence we obtain the derived result.

Lemma 14

For $0\leq k\leq l$, we have $T(k,l)-T(k-1,l-1)=-sT(k,l-1)$.

PROOF

In case when $k=l$, it is trivial.

It is enough to show the case of $k<l$. Then we have

$T(k,l)-T(k-1,l-1)$

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Hence we obtain the derived result.           $\Box$

Lemma 15

Suppose $1\leq m<n$, then we have

$\mu(c^{m},c^{n})=\Sigma_{k=m}^{n-1}\mu(c^{m},c^{k})T(k,n)$.

PROOF

We heve $\mu(c^{m},c^{n})$

$=Mob(\{(c^{m}\rightarrow\sigma^{m_{1}}_{1}\rightarrow\cdots\rightarrow c^{m_{1}}\rightarrow\sigma^{m_{2}}_{1}\rightarrow\cdots\rightarrow c^{m_{2}}\rightarrow\cdots c^{m_{r-1}}\rightarrow\sigma^{m_{r-1}}_{1}\cdots\rightarrow c^{m_{r}}=c^{n})\ |\ |\sigma^{m_{i}}_{j}|=m_{i}\})$.

In case of $2\leq r$, we put $k=m_{r-1}$ and in case of $r=1$, we put $k=m$. By Lemma 1, we obtain the derived result.

$\Box$

Lemma 16

For $1\leq m<n$, we have

$\mu(c^{m},c^{n})-\mu(c^{m-1},c^{n-1})=s\mu(c^{m},c^{n-1})$.

PROOF

In case of $n=2$, we see $\mu(c,c^{2})-\mu(\phi,c)=(2s-1)-(s-1)=s,\ \mu(c,c)=1$ . Hence this lemma holds for $n=2$. holds.

We prove by induction on $n$. Suppose that the relation holds for $n-1$. Then we have

$\mu(c^{m},c^{n})-\mu(c^{m-1},c^{n-1})$

$=\Sigma_{k=m}^{n-1}\mu(c^{m},c^{k})T(k,n)-\Sigma_{k=m-1}^{n-2}\mu(c^{m-1},c^{k})T(k,n-1)$

$=\Sigma_{k=m}^{n-1}\mu(c^{m},c^{k})T(k,n)-\mu(c^{m},c^{k})T(k-1,n-1)+\mu(c^{m},c^{k})T(k-1,n-1)$

$-\mu(c^{m-1},c^{k-1})T(k-1,n-1)$

$=\Sigma_{k=m}^{n-1}\mu(c^{m},c^{k})(-s)T(k,n-1)+s\mu(c^{m},c^{k-1})T(k-1,n-1)$

$=\Sigma_{k=m}^{n-1}\mu(c^{m},c^{k})(-s)T(k,n-1)+s\Sigma_{k=m+1}^{n-1}\mu(c^{m},c^{k-1})T(k-1,n-1)$

$=(-s)\mu(c^{m},c^{n-1})T(n-1,n-1)$

$=s\mu(c^{m},c^{n-1})$.

Lemma 17

For $1\leq m\leq n$, we have

$\mu(a_{1}^{m},c^{n})+\mu(a_{1}^{m-1},c^{n-1})=s\mu(a_{1}^{m},c^{n-1})$.

PROOF

In case of $m<n$, we have by Lemma 16. In case of $m=n$, from $a_{1}^{m}\nleqq c^{n-1}$, $\mu(a_{1}^{m},c^{n})=(-1)^{m}$ and $\mu(a_{1}^{m-1},c^{n-1})=(-1)^{m-1}$, therefore the right hand side $=0$. Hence we obtan the derived result.

Lemma 18

For $1\leq m\leq n$, $\mu(a_{1}^{m},c^{n})$ is coefficient of $X^{n-m}$ in $T^{s}_{m+n}(X)$.

PROOF

If $m+n=2,i.e\ m=n=1$, we have $T^{s}_{2}(X)=s(s-1)X^{2}-1,\ \mu(a_{1},c)=-1$. Hence this lemma holds.

If $3\leq m+n$, by the relation $T^{s}_{k+2}(X)+T^{s}_{k}(X)=sX\cdot T^{s}_{k+1}(X)$ and Lemma 17 we obtain the derived result.

Lemma 19

For $n\in\mathbb{N}$ we heve $\mu(\phi,c^{n})=s^{n-1}(s-1)$.

PROOF

If $n=1$, our claim follows from Lemma 12. We show by induction. We suppose that $\mu(\phi,c^{k})=s^{k-1}(s-1)$ when $k\leq n-1$

Now we have

$\mu(\phi,c^{n})$

$=\Sigma_{k=1}^{n}MobPat(\phi,c^{k})\mu(c^{k},c^{n})$              (by Lemma10)

=\Sigma_{k=1}^{n}\bigl{(}-\Sigma_{i=0}^{k}s^{i}M((0,0),(k,k-i))(-1)^{i}\bigr{)}\mu(c^{k},c^{n})

[TABLE]

$=\Sigma_{k=1}^{n}-(1-s)^{k}\mu(c^{k},c^{n})$.

And we have

$\mu(\phi,c^{n})-s\mu(\phi,c^{n-1})$

$=\Sigma_{k=1}^{n}-(1-s)^{k}\mu(c^{k},c^{n})-\Sigma_{k=1}^{n-1}-(1-s)^{k}\mu(c^{k},c^{n-1})$

=\Sigma_{k=1}^{n-1}-(1-s)^{k}\bigl{(}\mu(c^{k},c^{n})-s\mu(c^{k},c^{n-1})\bigr{)}-(1-s)^{n}

$=\Sigma_{k=1}^{n-1}-(1-s)^{k}\mu(c^{k-1},c^{n-1})-(1-s)^{n}$            (by Lemma15)

=-(1-s)\bigl{(}\Sigma_{k=1}^{n-1}(1-s)^{k-1}\mu(c^{k-1},c^{n-1})\bigr{)}-(1-s)^{n}

=-(1-s)\bigl{(}\Sigma_{k=0}^{n-2}(1-s)^{k}\mu(c^{k},c^{n-1})\bigr{)}-(1-s)^{n}

=-(1-s)\bigl{(}-\mu(\phi,c^{n-1})-\mu(\phi,c^{n-1})-(1-s)^{n-1})\bigr{)}-(1-s)^{n}

$=0$.

So we have $\mu(\phi,c^{n})=s\mu(\phi,c^{n-1})$. Hence we obtain the derived result. $\Box$

Lemma 20

For $n\in\mathbb{N}$, $\mu(\phi,c^{n})$ is the coefficient of $X^{n}$ in $T^{s}_{n}(X)$.

Therefore we have the following theorem.

Theorem 2

For $0\leq m\leq n$ , $\mu(a_{1}^{m},c^{n})$ is the coefficient of $X^{n-m}$ in $T^{s}_{m+n}(x)$.

Corollary 1

Conjecture 1 is true.

ACKNOWLEDGEMENT

The author wishes to thank Professor Jun Morita for his valuable advice. And he is also grateful to Professor Daisuke Sagaki, Sho Matsumoto for their helpful comments.

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