\input a4new \magnification=1200 \overfullrule=0pt \def\boxit#1#2{\setbox1=\hbox{\kern#1{#2}\kern#1}% \dimen1=\ht1 \advance\dimen1 by #1 \dimen2=\dp1 \advance\dimen2 by #1 \setbox1=\hbox{\vrule height\dimen1 depth\dimen2\box1\vrule}% \setbox1=\vbox{\hrule\box1\hrule}% \advance\dimen1 by .4pt \ht1=\dimen1 \advance\dimen2 by .4pt \dp1=\dimen2 \box1\relax} \def\tab#1{\boxit{1pt}{$\matrix{#1}$}} \def\bitab#1{\boxit{1pt}{\boxit{1pt}{$\matrix{#1}$}}} \def\bo#1{\boxit{1pt}{$#1$}} \def\bol#1{\boxit{4pt}{$#1$}} \def\bop#1{\boxit{3pt}{$#1$}} \def\he{\hskip 1mm} \def\th{\noalign{\smallskip}\noalign{\hrule}\noalign{\smallskip}} \def\tvi{\vrule height 1pt depth 0pt width 0pt} \font\petcap=cmcsc10 \font\tpetcap=cmcsc8 \font\petit=cmti7 \font\twelvebf=cmbx12 \centerline{\twelvebf POWERS OF STAIRCASE SCHUR FUNCTIONS} \centerline{\twelvebf AND SYMMETRIC ANALOGUES OF BESSEL POLYNOMIALS} \vskip 1cm \centerline{Bernard LECLERC \footnote{$\dagger$}{ \sevenrm Institut Gaspard Monge, Universit\'e de Marne-la-Vall\'ee, 2 rue de la Butte-Verte 93160 Noisy-le-Grand Cedex France}} \vskip 1cm {\sevenbf Abstract. }{\sevenrm We present several identities involving staircase Schur functions. These identities are then interpreted in terms of a sequence of orthogonal polynomials in one variable ${\scriptstyle x}$, with coefficients in the ring of symmetric functions. By an appropriate specialization these polynomials reduce to Bessel polynomials. This leads to a new determinantal expression for Bessel polynomials and suggests that their combinatorics might be linked to Young tableaux or shifted Young tableaux.}\bigskip \bf1 Introduction \rm\medskip Schur functions $S_I$ and skew Schur functions $S_{I/J}$ are indexed by partitions $I$ or skew partitions $I/J$, which are visualized graphically by a diagram of boxes. Several families of Schur functions associated with special diagrams are known to satisfy particular identities. Thus, Lascoux and Pragacz provided a determinantal expression of Schur functions and skew Schur functions in terms of ribbon functions, which generalizes the classical decomposition into hook functions given by Giambelli \bf[LP]\rm. \smallskip In this paper we are interested in Schur functions whose diagram is a staircase. % like %\smallskip %$$\matrix{ %\bol{}& \bol{}& & & & & & \cr %\bol{}&\bol{}&\bol{}& \bol{}& & & &\cr %\bol{}&\bol{}&\bol{}&\bol{}&\bol{}&\bol{}& &\cr %\bol{}&\bol{}&\bol{}&\bol{}&\bol{}&\bol{}&\bol{}&\bol{}\cr %}\ \ .$$ %\medskip Our starting point is the following formula for the square of a staircase Schur function $$S_{12\ldots n}^2=S_{12\ldots n+1/1}S_{12\ldots n-1} - S_{12\ldots n+1}S_{12\ldots n-1/1}\ ,$$ which first arose in the context of the Korteweg-de-Vries equation {\bf[AM]}. It shows that the polynomials $\theta_d$ constructed by Adler and Moser to represent the rational solutions of the KdV equation are, up to normalization, nothing but staircase Schur functions \footnote{*}{\sevenrm We thank D. Svrtan for pointing out this fact to us and kindly providing references on this subject.}. In a recent article Goulden has provided a combinatorial proof of this identity \bf[Go]\rm. We shall present more general formulae, expressing any power of a staircase Schur function. \smallskip An interpretation of these formulae in terms of orthogonal polynomials is then given. Indeed, it is shown that the family of staircase Schur functions $S_{12\ldots n}({\bb{E}}+x),\ n\geq0$, is a family of orthogonal polynomials in the variable $x$, which may be seen as a symmetric analogue of the family of Bessel polynomials. This provides a new determinantal expression for Bessel polynomials. It also shows that the powerful machinery of symmetric functions might be of some help in the study of this sequence of polynomials. Thus, using the fact that staircase $S$-functions are special cases of Schur $Q$-functions, Thibon and the author derived from a recent multiplication formula due to Worley the computation of the linearization coefficients \bf[LT]\rm which had been conjectured by Favreau \bf[Fa]\rm. \smallskip Finally, we show that our formulae are also linked to an old problem first studied by Borchardt \bf[Bo] \rm and Laguerre \bf[La]\rm, and further investigated by numerous authors including P\'olya \bf[Po] \rm and Foulkes \bf[Fo]\rm. The problem is to express any symmetric polynomial of $n$ indeterminates as a rational function of the power sums of odd degree. Set ${\bb{X}}=\{x_1,\ldots ,x_n\}$. The expression given by Polya and Foulkes for the $k^{th}$ elementary symmetric polynomial $\Lambda_k({\bb{X}})$ is the coefficient of $(-x)^{n-k}$ in the ratio $S_{12\ldots n}({\bb{X}}-x)/S_{12\ldots n-1}({\bb{X}})$. We obtain a similar expression for the complete symmetric polynomial $S_k({\bb{X}})$. \bigskip \bf2 Schur functions and minors identities \rm\medskip In this section we shall define our notations and recall some basic facts about Schur functions and minors identities. General references are \bf[LS]\rm, \bf[Mc] \rm and \bf[Le]\rm. \smallskip Let ${\bb{E}}$ denote a set of indeterminates or \it alphabet\rm. The \it complete symmetric functions \rm$S_i({\bb{E}})$ and the \it elementary symmetric functions \rm$\Lambda_i({\bb{E}})$ are defined by means of the generating series $$\sum_iS_i({\bb{E}})t^i = \prod_{e\in{\bb{E}}}(1-te)^{-1}\ ,\eqno (1)$$ $$\sum_i\Lambda_i({\bb{E}})t^i = \prod_{e\in{\bb{E}}}(1+te)\ .\eqno (2)$$ When there is no danger of confusion we shall omit the mention of the alphabet. For $I=(i_1,\ldots ,i_n)\in{\bb{N}}^n$, the \it Schur function \rm$S_I$ is defined by Jacobi--Trudi identity $S_I={\rm det}[S_{i_l+l-k}]_{1\leq k,l\leq n}\ .$ In other words, denoting by $\bf S\rm$ the infinite Toeplitz matrix $[S_{j-i}]_{i,j\geq0}$ where $S_k=0$ for $k<0$, $S_I$ is the minor of $\bf S\rm$ taken on the lines $0,\ 1,\ldots ,\ n-1$ and the columns $i_1,\ i_2+1,\ldots ,\ i_n+n-1$. More generally, given $J=(j_1,\ldots,\ j_n)\in{\bb{N}}^n$ one defines the \it skew Schur function \rm$S_{I/J}$ as the minor of $\bf S\rm$ taken on the lines $j_1,\ j_2+1,\ldots ,\ j_n+n-1$ and the columns $i_1,\ i_2+1,\ldots ,\ i_n+n-1$ , \it i. e. \rm $S_{I/J}={\rm det}[S_{i_l+l-j_k-k}]_{1\leq k,l\leq n}\ .$ Note that this definition makes sense even if the sequences $I$ and $J$ are not arranged in increasing order. However, by permutation of columns and rows one can always assume that $I$ and $J$ are \it partitions \rm i.e. weakly \it increasing \rm sequences of nonnegative integers. A partition $I$ is represented by a \it diagram \rm of boxes having $i_1$ boxes on its first row, $i_2$ boxes on its second row, etc.. $|I|=i_1+\ldots +i_n$ is called the \it weight \rm of $I$. The \it skew diagram \rm associated with $S_{I/J}$ is the complement of the diagram $J$ in the diagram $I$. \smallskip Schur functions may also be expressed in terms of elementary symmetric functions $\Lambda_i$. Indeed, defining by analogy the functions $\Lambda_{I/J}$ by $\Lambda_{I/J}={\rm det}[\Lambda_{i_l+l-j_k-k}]_{1\leq k,l\leq n}\ ,$ one has the relation $$\Lambda_{I/J}=S_{I^{\sim}/J^{\sim}},\eqno (3)$$ where $I^{\sim}$ denotes the \it conjugate partition \rm of $I$ i.e. the partition whose diagram is obtained by interchanging the rows and columns of the diagram of $I$. \smallskip Suppose ${\bb{F}}$ is a second alphabet and denote by ${\bb{E}}+{\bb{F}}$ the union of ${\bb{E}}$ and ${\bb{F}}$. It follows from (1), (2) that $$S_n({\bb{E}}+{\bb{F}})=\sum_{0\leq i\leq n}S_i({\bb{E}})S_{n-i}({\bb{F}})\ ,\eqno (4)$$ $$\Lambda_n({\bb{E}}+{\bb{F}})=\sum_{0\leq i\leq n}\Lambda_i ({\bb{E}})\Lambda_{n-i}({\bb{F}})\ .\eqno (5)$$ We shall also define the difference ${\bb{E}}-{\bb{F}}$ by setting $$\sum_iS_i({\bb{E}}-{\bb{F}})t^i = {\prod_{f\in{\bb{F}}}(1-tf)\over \prod_{e\in{\bb{E}}}(1-te)}\ , \qquad \sum_i\Lambda_i({\bb{E}}-{\bb{F}})t^i = {\prod_{e\in{\bb{E}}}(1+te)\over \prod_{f\in{\bb{F}}}(1+tf)}\ ,$$ so that in particular $$\Lambda_i({\bb{E}}) = (-1)^i S_i(-{\bb{E}})\ ,\eqno (6)$$ which is another way of expressing the duality between complete and elementary symmetric functions. This yields a compact expression for the polynomial whose set of zeros is the alphabet ${\bb{A}}$ of cardinality $n$ $$\prod_{a\in{\bb{A}}} (x-a) = \sum_ix^{n-i}(-1)^i\Lambda_i({\bb{A}}) = \sum_iS_{n-i}(x)S_i(-{\bb{A}}) = S_n(x-{\bb{A}})=(-1)^n\Lambda_n({\bb{A}}-x)\ .$$ \smallskip Taking different alphabets ${\bb{E}}_1,\ldots ,\ {\bb{E}}_n$ in the columns of the determinant of Jacobi--Trudi formula, one gets a \it multi-Schur function \rm$S_I({\bb{E}}_1,\ldots ,\ {\bb{E}}_n)= {\rm det}[S_{i_l+l-k}({\bb{E}}_l)]_{1\leq k,l\leq n}\ .$ An example of multi-Schur function considered in this paper is the $n^{th}$ orthogonal polynomial associated with the linear functional $\mu$ of moments $\mu(x^i)=S_i({\bb{E}})$. (Note that, since the $S_i$ are algebraically independent, one may always define a formal alphabet ${\bb{E}}$ by assigning to each $S_i$ a given value $S_i({\bb{E}})$ in a commutative ring.) Indeed this polynomial $p_n(x)$ has, up to a constant factor, the expression $$p_n(x)=S_{n^n0}({\bb{E}},{\bb{E}},\ldots ,{\bb{E}},x)= \left|\matrix{ S_n({\bb{E}})& S_{n+1}({\bb{E}})& \ldots & S_{2n-1}({\bb{E}})& x^n\cr S_{n-1}({\bb{E}})& S_n({\bb{E}})& \ldots & S_{2n-2}({\bb{E}})& x^{n-1}\cr \vdots&\vdots& & \vdots&\vdots\cr S_0({\bb{E}})&S_1({\bb{E}})&\ldots&S_{n-1}({\bb{E}}) & 1\cr }\right|\ ,\eqno (7)$$ (see \bf[Sz] \rm p.27). By subtraction of lines in the determinant, this polynomial may also be written as an ordinary Schur function on the alphabet ${\bb{E}}-x$. Indeed, one has $p_n(x)=S_{n^n}({\bb{E}}-x)\ .$ This is a consequence of the following important lemma. \smallskip {\petcap Lemma 2.1} {\it Let $m,\ n$ be two integers $m\leq n$, ${\bb{F}}$ an alphabet of cardinality $m$, and $I=(i_1,\ldots ,\ i_n)$ a partition. Then} $$S_I({\bb{E}}_1,\ldots ,{\bb{E}}_n)= \left|\matrix{ S_{i_1}({\bb{E}}_1-{\bb{F}})&\ldots &S_{i_n+n-1}({\bb{E}}_n-{\bb{F}})\cr \vdots& &\vdots\cr S_{i_1-n+m+1}({\bb{E}}_1-{\bb{F}})&\ldots &S_{i_n+m}({\bb{E}}_n-{\bb{F}})\cr S_{i_1-n+m}({\bb{E}}_1)&\ldots &S_{i_n+m-1}({\bb{E}}_n)\cr \vdots & &\vdots\cr S_{i_1-n+1}({\bb{E}}_1)&\ldots &S_{i_n}({\bb{E}}_n)\cr }\right|\ .$$ \smallskip \it Proof. \rm By (4), we have $S_k({\bb{E}}_j-{\bb{F}})= S_k({\bb{E}}_j)+S_{k-1}({\bb{E}}_j)S_1(-{\bb{F}})+\ldots +S_k(-{\bb{F}})\ .$ But $S_k(-{\bb{F}})=(-1)^k\Lambda_k({\bb{F}})=0$ if $k>m$. Thus, the transformation consists in adding to each of the first $n-m$ rows a linear combination of the next $m$ rows.\quad $\qed$ \medskip The Jacobi--Trudi formula expresses Schur functions as minors of the matrix $\bf S\rm$ of complete symmetric functions $S_i$. This shows that many algebraic relations for Schur functions may be obtained by merely specializing the numerous identities satisfied by minors of a generic matrix. We have shown elsewhere [{\bf Le}] that a great number of these identities are easily derived from a single one by Turnbull. We shall briefly recall this identity for the convenience of the reader. We first need some appropriate notations. \smallskip Let $M$ be a $n \times \infty$ matrix. We shall be interested in identities satisfied by maximal minors of $M$. Let $a,b,\ldots,c$ be $n$ column vectors of $M$. The maximal minor of $M$ taken on these $n$ columns will be denoted by either a bracket or a one--line tableau $$[ab\ldots c] = \tab{a &b &\ldots &c\cr}\he .$$ \smallskip A product of $p$ minors of $M$ will be designated by a $p \times n$ tableau $$[ab\ldots c].[de\ldots f].[gh\ldots i] = \tab{ a &b &\ldots &c\cr\th d &e &\ldots &f\cr\th g &h &\ldots &i\cr }\he.$$ To denote certain alternating sums of products of minors, we shall use tableaux with boxes enclosing some particular vectors. They are defined in the following way. \smallskip Let $T$ be a $p \times n$ tableau and $A$ a subset of elements of $T$. Given a permutation $\sigma$ of the elements of $A$, denote by $\sigma(T)$ the tableau in which the elements of $A$ are permuted by $\sigma$. Now, boxing in $T$ the elements of $A$, we get a new tableau $\tau$ which will represent the alternating sum of all the tableaux $\sigma(T)$, taking into account the fact that a permutation of elements of the same row gives a trivial action. More precisely, $\tau=\sum_\sigma sign(\sigma).\sigma(T) \he ,$ where $\sigma$ runs through the cosets of the symmetric group ${\goth{S}}(A)$ modulo the subgroup of the permutations which leave unchanged the rows of $T$.\par Let $i_k,\ k=1,\ldots p$ denote the number of elements of $A$ lying in the $k^{th}$ row of $\tau$. The number of terms in this sum will be $(i_1+\ldots +i_p)!/i_1!\ldots i_p!$. In particular, if all the enclosed elements of $\tau$ lie in the same row, it reduces to the single tableau $T$. \smallskip To illustrate these notations, let us write a well--known minor identity (Pl\"ucker's relations) and then its transcription with tableaux. \smallskip Let $M$ be the matrix $$M=\pmatrix{ a_1 &b_1 &c_1 &d_1 &e_1 &f_1 &\ldots\cr a_2 &b_2 &c_2 &d_2 &e_2 &f_2 &\ldots\cr a_3 &b_3 &c_3 &d_3 &e_3 &f_3 &\ldots\cr }\he .$$ We have $$\displaylines{\qquad\left|\matrix{ a_1 &b_1 &c_1\cr a_2 &b_2 &c_2\cr a_3 &b_3 &c_3\cr }\right| . \left|\matrix{ d_1 &e_1 &f_1\cr d_2 &e_2 &f_2\cr d_3 &e_3 &f_3\cr }\right|\he -\he \left|\matrix{ e_1 &b_1 &c_1\cr e_2 &b_2 &c_2\cr e_3 &b_3 &c_3\cr }\right| . \left|\matrix{ d_1 &a_1 &f_1\cr d_2 &a_2 &f_2\cr d_3 &a_3 &f_3\cr }\right|\hfill\cr \hfill -\he \left|\matrix{ f_1 &b_1 &c_1\cr f_2 &b_2 &c_2\cr f_3 &b_3 &c_3\cr }\right| . \left|\matrix{ d_1 &e_1 &a_1\cr d_2 &e_2 &a_2\cr d_3 &e_3 &a_3\cr }\right| \he =\he\left|\matrix{ a_1 &e_1 &f_1\cr a_2 &e_2 &f_2\cr a_3 &e_3 &f_3\cr }\right| . \left|\matrix{ d_1 &b_1 &c_1\cr d_2 &b_2 &c_2\cr d_3 &b_3 &c_3\cr }\right|\he .\qquad\cr}$$ \smallskip Denoting the column vectors of $M$ by $a,b,c,d,e,f$, this identity written using tableaux takes the following form \smallskip $$\tab{ \bo{a} &b &c\cr\th d &\bo{e} &\bo{f}\cr }\he = \he \tab{ a &e &f\cr\th d &b &c\cr }\he .$$ \smallskip It is readily deduced from Turnbull's identity, that we shall now state. \smallskip Let $\tau$ be a tableau and denote by $A$ the set of its boxed elements. Let $B$ be the set of elements of a given row, for example the first one, which do not belong to $A$. Finally let $C$ denote the set of all other elements of $\tau$. One can build another tableau $\upsilon$ equal to $\tau$ by merely exchanging the roles of $A$ and $B$ as shown by the following diagram: \setbox3=\hbox{$\vcenter{\offinterlineskip\cleartabs \def\cc#1{\hfill#1\hfill} \def\tvi{\vrule height 12pt depth 5pt width 0pt} \def\tv{\tvi\vrule} \def\th{\vrule height 0.4pt width 2em} \+ \th&\th&\th&\th&\th&\th&\cr \+ \tv&\tv& & &\cc{B}& &\tv\cr \+ &\th&\th&\th&\th&\th&\cr \+ \tv&\cc{\bf A\rm} &\tv&&&&\tv\cr \+ &\th& & & & &\cr \+ \tv&\tv&&&\cc{C}&&\tv\cr \+ \th&\th&\th&\th&\th&\th&\cr }$} \setbox4=\hbox{ = } \setbox5=\hbox{$\vcenter{\offinterlineskip\cleartabs \def\cc#1{\hfill#1\hfill} \def\tvi{\vrule height 12pt depth 5pt width 0pt} \def\tv{\tvi\vrule} \def\th{\vrule height 0.4pt width 2em} \+ \th&\th&\th&\th&\th&\th&\cr \+ \tv& &\cc{A} & &\tv &\cc{\bf B\rm} &\tv\cr \+ \th&\th&\th&\th&\th&\th&\cr \+ \tv&\cc{\bf B\rm} &\tv&&&&\tv\cr \+ &\th& & & & &\cr \+ \tv&\tv&&&\cc{C}&&\tv\cr \+ \th&\th&\th&\th&\th&\th&\cr }$} $$\line{\hfill{$\tau =$}\box3\box4\box5 {$= \nu$}\hfill}$$ More precisely, we have \medskip\ {\petcap Theorem} 2.1. {\it (Turnbull's identity, \bf[Tu]\rm, p.209.)} \smallskip {\it Let $k$ be the number of elements of $A$.} \smallskip (i) {\it If $k>n, \he \tau = 0.$} \smallskip (ii) {\it If $k\leq n$, and $\upsilon$ is built by}\par \hskip 1cm \bf a\rm. {\it exchanging each element of $A$ which does not belong to the first row, with any element of $B$;}\par \hskip 1cm \bf b\rm. {\it boxing the elements of $B$;}\par \hskip 1cm \bf c\rm. {\it removing the boxes of the elements of $A$;}\par {\it then $\tau=\upsilon$.} \smallskip For instance, taking $n=6,\ k=4,\ A=\{ a, b, c, d \},\ B= \{ \alpha,\beta,\gamma,\delta,\varepsilon\},$ one has $$\tau=\tab{ \bo{ a}& \alpha& \beta& \gamma& \delta& \varepsilon\cr\th \bo{ b}& \bo{ c}& f& g&h& i\cr\th \bo{ d}& j& k& l& m& o\cr } = \tab{ { a}& { b}& { c}& { d}& \bo{\delta}& \bo{\varepsilon}\cr\th \bo{\alpha}& \bo{\beta}& f& g& h& i\cr\th \bo{\gamma}& j& k& l& m& o\cr }=\upsilon\ .$$ \smallskip We shall now deduce from this identity some formulae for the power of a staircase Schur function. \bigskip \bf3 Powers of a staircase Schur function and symmetric analogues of Bessel polynomials \rm\medskip Let $\rho(k,m)$ denote the staircase partition $(m,\ 2m,\ldots ,\ km)$. The $n^{th}$ power of $S_{\rho(k,m)}$ admits the following expression: \medskip {\petcap Theorem 3.1. } $$(S_{\rho(k,m)})^n=\sum_{I\subset\rho(n-1,m)}(-1)^{|\rho(n-1,m)|-|I|}\ S_{\rho(n+k-1,m)/I}\ \Delta_I\ ,$$ {\it where $\Delta_I$ is the minor taken on the lines $i_1+1,\ i_2+2,\ldots ,\ i_{n-1}+n-1$ of the following $(m+1)(n-1)\times (n-1)$ matrix:} $$M=\pmatrix{ S_{\rho(k-1,m)/1^m}& S_{\rho(k-1,m)/1^{2m+1}}&\ldots& S_{\rho(k-1,m)/1^{(n-1)m+n-2}} \cr S_{\rho(k-1,m)/1^{m-1}}& S_{\rho(k-1,m)/1^{2m}}&\ldots& S_{\rho(k-1,m)/1^{(n-1)m+n-3}} \cr \vdots&\vdots&&\vdots\cr S_{\rho(k-1,m)}& S_{\rho(k-1,m)/1^{m+1}}&\ldots& S_{\rho(k-1,m)/1^{(n-2)(m+1)}} \cr 0& S_{\rho(k-1,m)/1^{m}}&\ldots& S_{\rho(k-1,m)/1^{(n-2)m+n-3}} \cr 0& S_{\rho(k-1,m)/1^{m-1}}&\ldots& S_{\rho(k-1,m)/1^{(n-2)m+n-4}} \cr \vdots& \vdots& & \vdots\cr 0& S_{\rho(k-1,m)}&\ldots& S_{\rho(k-1,m)/1^{(n-3)(m+1)}} \cr 0& 0&\ldots& S_{\rho(k-1,m)/1^{(n-3)m+n-4}}\cr \vdots& \vdots& &\vdots\cr 0& 0&\ldots& S_{\rho(k-1,m)}\cr }\ .$$ Note that we allow the first parts of $I$ to be zero. Before proving the theorem, let us illustrate it by a few examples. \medskip {\petcap Example 3.2. } We take $n=3$, $m=1$ and $k=4$. The theorem states that $$(S_{1234})^3=\sum_{I\subset12}(-1)^{3-|I|}\ S_{123456/I}\ \Delta_I\ ,$$ where $I$ runs over the set $\{00,\ 01, \ 02,\ 11,\ 12\}$ and $\Delta_I$ is the minor taken on the lines $i_1+1$, $i_2+2$ of the $4\times 2$ matrix: $$M=\pmatrix{ S_{123/1}& S_{123/1^3}\cr S_{123}& S_{123/1^2}\cr 0& S_{123/1}\cr 0& S_{123}\cr }\ =\ \pmatrix{ S_{1234/1112}& S_{1234/1114}\cr S_{1234/1111}& S_{1234/1113}\cr S_{1234/1110}& S_{1234/1112}\cr S_{1234/111(-1)}& S_{1234/1111}\cr }\ .$$ The second expression of $M$ shows more clearly the regularity of its columns (we recall that for any partition $J$, the determinants $S_{J/1110}$ and $S_{J/111(-1)}$ are null having two identical rows). Expanding this sum and the determinants $\Delta_I$ yields the following relation: $$\displaylines{\quad (S_{1234})^3=S_{123456}\ S_{123}\ S_{123/111}-S_{123456}\ S_{123/1} \ S_{123/11}+S_{123456/1}\ S_{123/1}\ S_{123/1}\hfill\cr \hfill -S_{123456/11}\ S_{123}\ S_{123/1}- S_{123456/2}\ S_{123}\ S_{123/1}+S_{123456/12}\ S_{123}\ S_{123}\ .\quad\cr}$$ \medskip {\petcap Example 3.3. } We take $n=2$, so that theorem 3.1 reads $$(S_{\rho(k,m)})^2=\sum_{0\leq i\leq m}(-1)^{m-i}\ S_{\rho(k+1,m)/i}\ S_{\rho(k-1,m)/1^{m-i}}\ ,$$ which is the formula investigated by Goulden \bf[Go] \rm for the square of a staircase Schur function. \medskip {\it Proof of Theorem} 3.1. We shall apply Turnbull's identity 2.1. Set $N=k+(n-1)(m+1)$ and consider the $N\times \infty$ matrix $$M=\pmatrix{ S_0 &S_1 &S_2 &S_3 &\ldots &1 &0 &0 &\ldots &0\cr 0 &S_0 &S_1 &S_2 &\ldots &0 &1 &0 &\ldots &0\cr 0 &0 &S_0 &S_1 &\ldots &0 &0 &1 &\ldots &0\cr \vdots&\vdots&\vdots& \vdots&&\vdots&\vdots &\vdots &\ddots &\vdots\cr 0 &0 &0 &0 &\ldots &0 &0 &0 &\ldots &1\cr }\he \it.\rm$$ Let $0,\ 1,\ 2,\ 3,\ldots $ denote the column vectors of the left part of $M$, and $1^*,\ 2^*,\ 3^*,\ldots N^*,$ denote the last $N$ column vectors of $M$. For brevity we shall write $m_p=pm+p-1$. The Schur function $S_{\rho(k,m)}$ may be expressed as a maximal minor of $M$ in $n$ different ways $$\eqalign{ S_{\rho(k,m)} &=[m_1\ m_2\ \ldots \ m_k\ (k+1)^*\ (k+2)^*\ \ldots \ N^*]\cr &=[0\ 1\ \ldots \ (m_1-1)\ m_1\ m_2\ \ldots \ m_{k+1}\ (k+m_1+1)^*\ (k+m_1+2)^*\ \ldots \ N^*]\cr &\vdots\cr &= [0\ 1\ 2\ \ldots \ (m_{n-1}-1)\ m_{n-1}\ m_n\ \ldots \ m_{k+n-1}]\ ,\cr }$$ so that its $n^{th}$ power is equal to the product of these $n$ minors $$\displaylines{ (S_{\rho(k,m)})^n= \hfill\cr \hfill\tab{ m_1&\ldots &m_k &(k+1)^*&\ldots & & &\ldots &N^*\cr\th 0 &\ldots &(m_1-1)& m_1 &\ldots &m_{k+1}& (k+m_1+1)^*& \ldots & N^*\cr\th \vdots& &&&\vdots &&& &\vdots\cr\th 0 &\ldots& & & \ldots &(m_{n-1}-1)&m_{n-1}& \ldots &m_{k+n-1}\cr }\ .\cr }$$ Now, recalling that a tableau having two equal letters on the same row is equal to zero, we may box the letters $m_1,\ m_2,\ \ldots, m_k$ in the first row, the letter $m_{k+1}$ in the second row, $\ldots $, the letter $m_{k+n-1}$ in the $n^{th}$ row, and apply Turnbull's identity. This yields $$\displaylines{ (S_{\rho(k,m)})^n\hfill\cr \hfill =\ \tab{ \bo{m_1}&\ldots &\bo{m_k} &(k+1)^*&\ldots & & &\ldots &N^*\cr\th 0 &\ldots &(m_1-1)& m_1 &\ldots &\bo{m_{k+1}}& (k+m_1+1)^*& \ldots & N^*\cr\th \vdots& &&& \vdots&&& &\vdots\cr\th 0 &\ldots& & &\ldots &(m_{n-1}-1)&m_{n-1}& \ldots &\bo{m_{k+n-1}}\cr }\ \ \cr \ \cr \hfill =\ \tab{ m_1&\ldots &m_k &m_{k+1}&\ldots &m_{k+n-1} &\bo{(k+n)^*} &\ldots &\bo{N^*}\cr\th 0 &\ldots &(m_1-1)& m_1 &\ldots &\bo{(k+1)^*}& (k+m_1+1)^*& \ldots & N^*\cr\th \vdots& &&& \vdots&&& &\vdots\cr\th 0 &\ldots& & &\ldots &(m_{n-1}-1)&m_{n-1}& \ldots &\bo{(k+n-1)^*}\cr }\ .\cr }$$ This last tableau represents a sum of products of Schur functions of the following type $$\pm S_{\rho(n+k-1,m)/I}\ S_{\rho(k-1,m)/1^{j_1}}\ldots \ S_{\rho(k-1,m)/1^{j_{n-1}}} \ .$$ The coefficient of a given $S_{\rho(n+k-1,m)/I}$ in this sum is in fact a tableau having $n-1$ rows with only one letter boxed in each row, that is a determinant of order $n-1$. There is no difficulty in recognizing this determinant as the minor $\Delta_I$ defined above. $\qed$ \medskip The formula for the square of $S_{\rho(k,1)}$ may be presented as the computation of a determinant of Schur functions $$\left|\matrix{ S_{\rho(k-1,1)}& S_{\rho(k+1,1)}\cr S_{\rho(k-1,1)/1}& S_{\rho(k+1,1)/1}\cr }\right| =(S_{\rho(k,1)})^2\ .\eqno (8)$$ The same may be done for theorem 3.1, which has also a determinantal formulation. We shall write for brevity $\rho(k)$ in place of $\rho(k,1)$. \medskip {\petcap Theorem 3.4.} $$\left|\matrix{ S_{\rho(k-1)}& 0& \ldots & 0& \ldots & 0& S_{\rho(k+1)}\cr S_{\rho(k-1)/1}& 0& \ldots & 0& \ldots & S_{\rho(k+2)}& S_{\rho(k+1)/1}\cr S_{\rho(k-1)/1^2}& S_{\rho(k,1)}& \ldots & 0& \ldots & S_{\rho(k+2)/1}& S_{\rho(k+1)/1^2}\cr \vdots & \vdots & & \vdots & & \vdots &\vdots\cr S_{\rho(k-1)/1^{2n-3}}& S_{\rho(k-1)/1^{2n-5}}&\ldots & S_{\rho(k+n-1)/1^{n-1}}& \ldots & S_{\rho(k+2)/1^{2n-4}}& S_{\rho(k+1)/1^{2n-3}}\cr }\right|$$ $$=(S_{\rho(k)})^n\ S_{\rho(k+1)}\ S_{\rho(k+2)}\ldots S_{\rho(k+n-2)}\ .$$ \medskip {\petcap Example 3.5.} For $n=3$ and $k=4$ one has $$\left|\matrix{ S_{123}& 0& 0& S_{12345}\cr S_{123/1}& 0&S_{123456}& S_{12345/1}\cr S_{123/1^2}& S_{123}& S_{123456/1}& S_{12345/1^2}\cr S_{123/1^3}& S_{123/1}& S_{123456/1^2}& S_{12345/1^3}\cr }\right| =(S_{1234})^3\ S_{12345}\ .$$ \medskip {\petcap Example 3.6.} For $n=4$ and $k=6$ one has $$\left|\matrix{ S_{12345} &0 &0 & 0& 0& S_{1234567}\cr S_{12345/1} &0 &0 & 0&S_{12345678}& S_{1234567/1}\cr S_{12345/1^2}&S_{12345} &0 &S_{123456789}& S_{12345678/1}& S_{1234567/1^2}\cr S_{12345/1^3}&S_{12345/1} &0 &S_{123456789/1}&S_{12345678/1^2}& S_{1234567/1^3}\cr S_{12345/1^4}&S_{12345/1^2}&S_{12345}&S_{123456789/1^2}&S_{12345678/1^3}& S_{1234567/1^4}\cr S_{12345/1^5}&S_{12345/1^3}&S_{12345/1}&S_{123456789/1^3}&S_{12345678/1^4}& S_{1234567/1^5}\cr }\right|$$ $$=(S_{123456})^4\ S_{1234567}\ S_{12345678}\ .$$ \medskip {\it Proof.} The proof of this last example will illustrate enough the principle of our computation. We consider the $12\times 30$ matrix $$M=\pmatrix{ S_0 &S_1 &S_2 &S_3 &\ldots &S_{17}&1 &0 &0 &\ldots &0\cr 0 &S_0 &S_1 &S_2 &\ldots &S_{16}&0 &1 &0 &\ldots &0\cr 0 &0 &S_0 &S_1 &\ldots &S_{15}&0 &0 &1 &\ldots &0\cr \vdots&\vdots&\vdots& \vdots&&\vdots&\vdots &\vdots &\ddots &\vdots\cr 0 &0 &0 &0 &\ldots &S_6&0 &0 &0 &\ldots &1\cr }\he ,$$ whose column vectors are denoted $0,\ 1,\ 2,\ldots 17,\ 1^*,\ 2^*,\ldots\ 12^*$. The determinant $D$ of example 3.6 is represented by the following tableau $$D= \tab{ 0& 1& 3& 5& 7& 9& 11& \bo{7^*}& 9^*& 10^*& 11^*& 12^*\cr\th 0& 1& 2& 3& 5& 7& 9& 11 & 13& \bo{ 8^*}& 11^*& 12^*\cr\th 0& 1& 2& 3& 4& 5& 7& 9& 11& 13& 15& \bo{9^*} \cr\th 1& 3& 5& 7& 9& 11& 13& 15& 17& \bo{10^*}& 11^*& 12^*\cr\th 1& 3& 5& 7& 9& 11& 13& 15& 17& 0& \bo{11^*}& 12^*\cr\th 1& 3& 5& 7& 9& 11& 13& 15& 17& 0& 2& \bo{12^*}\cr }\ .$$ Now we transform this tableau by Turnbull's identity and note that all boxes may be removed because of the presence of identical letters $$D= \tab{ 0& 1& 3& 5& 7& 9& 11& \bo{13}& 9^*& 10^*& 11^*& 12^*\cr\th 0& 1& 2& 3& 5& 7& 9& 11 & 13& \bo{15}& 11^*& 12^*\cr\th 0& 1& 2& 3& 4& 5& 7& 9& 11& 13& 15& \bo{17} \cr\th 1& 3& 5& 7& 9& 11& 13& 15& 17& \bo{0}& 11^*& 12^*\cr\th 1& 3& 5& 7& 9& 11& 13& 15& 17& 0& \bo{2}& 12^*\cr\th 1& 3& 5& 7& 9& 11& 7^*& 8^*& 9^*& 10^*& 11^*& 12^*\cr }$$ $$=\tab{ 0& 1& 3& 5& 7& 9& 11& 13& 9^*& 10^*& 11^*& 12^*\cr\th 0& 1& 2& 3& 5& 7& 9& 11 & 13& 15& 11^*& 12^*\cr\th 0& 1& 2& 3& 4& 5& 7& 9& 11& 13& 15& 17 \cr\th 1& 3& 5& 7& 9& 11& 13& 15& 17& 0& 11^*& 12^*\cr\th 1& 3& 5& 7& 9& 11& 13& 15& 17& 0& 2& 12^*\cr\th 1& 3& 5& 7& 9& 11& 7^*& 8^*& 9^*& 10^*& 11^*& 12^*\cr }$$ $$=(S_{123456})^4\ S_{1234567}S_{12345678}\qquad \qed$$ \medskip %{\it Remark.} For convenience we limited ourself to staircases of type %$\rho(k,1)$ in the statement of theorem 3.4. However a similar formula holds for %general staircases $\rho(k,m),\ m\geq1$, and it may be easily induced from the %following example: %$$\left|\matrix{ %S_{24} &0 &0 & 0& 0& S_{1357}\cr %S_{24/1} &0 &0 & 0&S_{2468}& S_{1357/1}\cr %S_{24/1^2}&0 &0 &S_{13579}& S_{2468/1}& S_{1357/1^2}\cr %S_{24/1^3}&S_{24} &S_{2468.10} &S_{13579/1}&S_{2468/1^2}& S_{1357/1^3}\cr %S_{24/1^4}&S_{24/1}&S_{2468.10/1}&S_{13579/1^2}&S_{2468/1^3}&S_{1357/1^4}\cr %S_{24/1^5}&S_{24/1^2}&S_{2468.10/1^2}&S_{13579/1^3}&S_{2468/1^4}& %S_{1357/1^5}\cr %}\right|$$ %$$=(S_{246})^3\ S_{1357}\ S_{2468}\ S_{13579}\ .$$ %\medskip We shall end this section with an interpretation of theorem 3.4 in terms of orthogonal polynomials. First we shall give an immediate generalization of formula (8). \medskip {\petcap Proposition 3.7. }{\it For $1\leq p\leq k+1$ we have} $$\left|\matrix{ S_{\rho(k-1)}& S_{\rho(k+1)}\cr S_{\rho(k-1)/1^p}& S_{\rho(k+1)/1^p}\cr }\right| =S_{\rho(k)}\ S_{\rho(k)/1^{p-1}}\ ,$$ {\it where $\rho(k)$ is short for the partition $12\ldots k$.} \smallskip {\it Proof. } This is an instance of Pl\"ucker relations. We consider again the matrix $M$ of the proof of theorem 3.1, taking $N=k+2$. The determinant to be computed is then represented by the $2\times (k+2)$ tableau $$(-1)^p\ \tab{ 0 &1 &3 &5 &\ldots &2k-1 &\bo{N^*}\cr \th 1 &3 &5 &7 &\ldots &2k+1 &\bo{(N-p)^*}\cr }\ .$$ Applying Turnbull's identity, this tableau is transformed into $$(-1)^p\ \tab{ 0 &1 &3 &5 &\ldots &2k-1 &\bo{2k+1}\cr \th \bo{1} &\bo{3} &\bo{5} &\bo{7} &\ldots &N^* &(N-p)^*\cr }$$ $$=\ (-1)^p\ \tab{ 0 &1 &3 &5 &\ldots &2k-1 &2k+1\cr \th 1 &3 &5 &7 &\ldots &N^* &(N-p)^*\cr } =S_{\rho(k)}\ S_{\rho(k)/1^{p-1}}\ .\qquad \qed$$ \smallskip >From this proposition we deduce the following result: \medskip {\petcap Theorem 3.8. }{\it Let ${\bb{E}}$ be an alphabet and let us denote by $\pi_k$ the polynomial of degree $k$ in $x$ with coefficients in the ring of symmetric functions of ${\bb{E}}$: $\pi_k(x)=S_{\rho(k)}({\bb{E}}-x)$. Then $\pi_k$ satisfies the three term recurrence relationship $(k\geq1)$ $$S_{\rho(k-1)}({\bb{E}})\pi_{k+1}(x)+xS_{\rho(k)}({\bb{E}})\pi_{k}(x)- S_{\rho(k+1)}({\bb{E}})\pi_{k-1}(x)=0\ .$$ In other words $\big(\pi_k(x)\big)_{k\geq0}$ is a family of orthogonal polynomials. } \medskip {\it Proof. } The expansion of $\pi_k$ is obtained by considering it as a multi-Schur function and expanding this determinant along its last column. Indeed, by lemma 2.1, $$\pi_k(x)=S_{12\ldots k0}({\bb{E}},\ldots ,{\bb{E}},x)=\sum_{0\leq i\leq k}S_{12\ldots k/1^i}({\bb{E}})(-x)^i\ .$$ By linearity we deduce from proposition 3.7 that $$\left|\matrix{ S_{\rho(k-1)}& S_{\rho(k+1)}\cr \pi_{k-1}(x)& \pi_{k+1}(x)\cr }\right| =-xS_{\rho(k)}\ \pi_k(x)\ ,$$ which is the required relationship. \ \quad $\qed$ \smallskip Thus, theorem 3.8 states that the elimination of $x^0$ between $\pi_{k-1}(x)$ and $\pi_{k+1}(x)$ produces a polynomial proportional to $x\pi_k(x)$. As a consequence the elimination of $x^0,\ x^1,\ x^2$ between $\pi_{k-1}(x),\ \pi_{k+1}(x),\ x\pi_{k+2}(x),\ x^2\pi_{k-1}(x)$ will produce a polynomial proportional to $x^3\pi_k(x)$. This elimination is expressed by the following formula $$\left|\matrix{ S_{\rho(k-1)}& 0& 0& S_{\rho(k+1)}\cr S_{\rho(k-1)/1}& 0&S_{\rho(k+2)}& S_{\rho(k+1)/1}\cr S_{\rho(k-1)/1^2}& S_{\rho(k-1)}& S_{\rho(k+2)/1}& S_{\rho(k+1)/1^2}\cr \pi_{(k-1)}(x)&x^2\pi_{k-1}(x)& -x\pi_{k+2}(x)& \pi_{k+1}(x)\cr }\right| =-(S_{\rho(k)})^2\ S_{\rho(k+1)}\ x^3\pi_k(x).$$ Picking up the coefficient of $x^3$ on each side we find again the case $n=3$ of theorem 3.4. And this process may clearly be carried out for larger values of $n$. \medskip The computation of the moments of the sequence of orthogonal polynomials $\big(\pi_k(x)\big)_{k\geq0}$ will result from the following identity. We shall denote by $\rho(n)+m$ the partition $(1,\ 2,\ldots ,\ n-1,\ n+m)$. \medskip {\petcap Theorem 3.9. }{\it Let k, n be integers such that $1\leq k\leq n$. There holds: $$\left|\matrix{ S_{\rho(k)/1^{k-1}}& S_{\rho(k+1)/1^{k-1}}& \ldots & S_{\rho(n)/1^{k-1}}\cr S_{\rho(k)/1^k}& S_{\rho(k+1)/1^k}& \ldots & S_{\rho(n)/1^k}\cr \vdots&\vdots&&\vdots \cr S_{\rho(k)/1^{n-1}}& S_{\rho(k+1)/1^{n-1}}& \ldots & S_{\rho(n)/1^{n-1}}\cr }\right| =S_{\rho(k-1)}P_{\rho(k)+(n-k+1)}S_{\rho(k+1)}S_{\rho(k+2)}\ldots S_{\rho(n-1)}\ ,$$ where $P_{\rho(k)+(n-k+1)}=\sum_{0\leq2i\leq n-k+1}\Lambda_{\rho(k)+2i}S_{n-k+1-2i}$.} \medskip {\petcap Example 3.10. } For $n=6,\ k=3$ one has $$\left|\matrix{ S_{123/1^2}& S_{1234/1^2}& S_{12345/1^2} & S_{123456/1^2}\cr S_{123/1^3}& S_{1234/1^3}& S_{12345/1^3} & S_{123456/1^3}\cr S_{123/1^4}& S_{1234/1^4}& S_{12345/1^4} & S_{123456/1^4}\cr S_{123/1^5}& S_{1234/1^5}& S_{12345/1^5} & S_{123456/1^5}\cr }\right| =S_{12}P_{127}S_{1234}S_{12345}\ ,$$ where $P_{127}=\Lambda_{123}S_4+\Lambda_{125}S_2+\Lambda_{127}$. \smallskip \it Proof. \rm The proof is similar to the proof of theorem 3.4, the only difference being that one has to apply Turnbull's identity several times in order to obtain the required factorization. \medskip {\petcap Corrolary 3.11. }{\it The $n^{th}$ moment of the sequence of orthogonal polynomials $\big(\pi_k(x)\big)_{k\geq0}$ is the sum of all hook Schur functions of weight $n+1$:} $\mu(x^n)=\sum_{0\leq i\leq n}S_{1^i(n+1-i)}({\bb{E}})\ .$ \smallskip \it Proof. \rm Let $q_k(x)$ denote the monic polynomial $q_k(x)=S_{\rho(k)}({\bb{E}}-x)/(-1)^kS_{\rho(k-1)}({\bb{E}})$. Theorem 3.9 may be seen as the resolution of the linear system in the unknown $a_{nk}$: $$ x^n = \sum_k a_{nk}q_k(x)\ . \eqno (9)$$ The result is that $a_{nk}=P_{\rho(k+1)+n-k}({\bb{E}})/ S_{\rho(k+1)}\ .$ Denote by $Sym({\bb{E}})$ the ring of symmetric functions of the alphabet ${\bb{E}}$ and let $\mu$ be the functional on $Sym({\bb{E}})[x]$ associated with the sequence of orthogonal polynomials $\big(\pi_k(x)\big)_{k\geq0}$. In fact $\mu$ is defined up to a constant factor, so that we can add the condition $\mu(1)=S_1({\bb{E}})$. Then, recalling that by definition $\mu(q_k(x))=0,\ k\geq1$ and applying $\mu$ to the equality (9), one obtains that $\mu(x^n)=P_{n+1}({\bb{E}})=\sum_{0\leq i\leq n}S_{1^i(n+1-i)}({\bb{E}})\ ,$ the last equality resulting easily from Pieri rule for the multiplication of a Schur function by an elementary symmetric function. $\ \qed$ \medskip \it Remarks. (i) \rm It can be shown that the symmetric function $$\sum_{0\leq2i\leq n-k+1}\Lambda_{\rho(k)+2i}S_{n-k+1-2i}$$ is nothing but the Schur $P$-function $P_{\rho(k)+n-k+1}$, hence the notation used above. More generally it has been shown in [{\bf LLT}] that there exists for every Schur $P$-function a similar quadratic expansion in terms of ordinary Schur $S$-functions. \smallskip \it(ii) \rm It is easily checked that $S_{\rho(n)/1^m}=S_{\rho(n)/m}$. It follows that the polynomials $\phi_k(x)=S_{\rho(k)}({\bb{E}}+x)$ are equal to the polynomials $\pi_k(-x)$, and therefore also satisfy a three term recurrence relationship: $$S_{\rho(k-1)}({\bb{E}})\phi_{k+1}(x)-xS_{\rho(k)}({\bb{E}})\phi_{k}(x)- S_{\rho(k+1)}({\bb{E}})\phi_{k-1}(x)=0\ .$$ They are orthogonal for the moments $\nu(x^n)=(-1)^n\sum_{0\leq i\leq n}S_{1^i(n+1-i)}({\bb{E}})\ .$ \medskip We must now emphasize on the differences between the two Schur function expressions of orthogonal polynomials hitherto encountered. As recalled in section 2, the family of orthogonal polynomials associated with a given functional $\mu$ may \it always \rm be represented up to a constant factor by the sequence of rectangle Schur functions $S_{n^n}({\bb{E}}-x),\ n\geq0,$ where ${\bb{E}}$ is the alphabet formally defined by $S_k({\bb{E}})=\mu(x^k),\ k\geq0$. In contrast Schur functions of the type $S_{12\ldots n}({\bb{E}}-x),\ n\geq0$ or $S_{12\ldots n}({\bb{E}}+x),\ n\geq0$ represent only particular families of orthogonal polynomials. Indeed, the recurrence relationship of theorem 3.8 is not generic, for the coefficient of $\pi_n(x)$ in this relationship is of the type $\alpha_nx$ while the general form of this coefficient is $\alpha_nx+\beta_n$. The following example shows that these polynomials may be seen as symmetric analogues of Bessel polynomials. \medskip {\petcap Example 3.12.} \smallskip Let ${\cal{E}}$ be the alphabet formally defined by $S_k({\cal{E}})=1/ k!,\ k\geq0.$ It is a classical result that for any partition $I$ of weight $n$, there holds $S_I({\cal{E}})=f_I/n!$, where $f_I$ is the number of standard Young tableaux of shape $I$, \it i.e. \rm the dimension of the irreducible representation of ${\goth{S}}_n$ indexed by $I$. It follows that the moments associated to the sequence $\big(\phi_k(x)\big)_{k\geq0}$ are equal to $\nu(x^n)=(-2)^n/ (n+1)!\ ,$ that is, are equal to the moments of Bessel polynomials (see for example \bf[Ch])\rm. The corresponding polynomials (suitably normalized) $$y_n(x)=S_{12\ldots n}({\cal{E}}+x)/S_{12\ldots n}({\cal{E}}),\quad n\geq0,\eqno (10)$$ are therefore the \it Bessel polynomials\rm. The first ones are $$\eqalign{ y_0(x)=&1,\cr y_1(x)=&1+x,\cr y_2(x)=&1+3x+3x^2,\cr y_3(x)=&1+6x+15x^2+15x^3,\cr y_4(x)=&1+10x+45x^2+105x^3+105x^4.\cr }$$ Formula (10) provides the following determinantal expression $$y_n(x)=\prod_{1\leq i\leq n}(2(n-i)+1)^i \left|\matrix{ {1/1!}& {1/3!}& {1/5!}& \ldots & {1/(2n-1)!}& (-x)^n\cr {1/0!}& {1/2!}& {1/4!}& \ldots & {1/(2n-2)!}& (-x)^{n-1}\cr 0& {1/1!}& {1/3!}& \ldots & {1/(2n-3)!}& (-x)^{n-2}\cr \vdots&\vdots &\vdots &&\vdots&\vdots \cr 0 & & & & {1/n!}& -x\cr 0 & & & & {1/(n-1)!}& 1\cr }\right|\ .$$ There exists a fairly extensive literature devoted to these polynomials. For a detailed account up to 1978, the interested reader is referred to \bf[Gr]\rm. We shall also mention the recent work of Dulucq and Favreau who have presented a combinatorial model for these polynomials, based upon weighted involutions \bf[DF]\rm. \smallskip Note that our point of view provides at once a q-analogue for Bessel polynomials, by merely replacing ${\cal{E}}$ by the alphabet ${\cal{E}}_q=\{1,\ q,\ q^2,\ldots \}$. More precisely, setting $v=x/(1-q)$ we define $$y_n(x,q)=q^{n\choose 2}S_{12\ldots n}({\cal{E}}_q+v)/S_{12\ldots n}({\cal{E}}_q).$$ The $y_n(x,q)$ are orthogonal polynomials in $x$, whose coefficients are polynomials in $q$. The first ones are $$\eqalign{ y_0(x,q)=&1,\cr y_1(x,q)=&1+x,\cr y_2(x,q)=&q+(1+q+q^2)x+(1+q+q^2)x^2,\cr y_3(x,q)=&q^3+q(1+q^2)(1+q+q^2)x+(1+q+q^2)(1+q+q^2+q^3+q^4)x^2\cr &\hfill +(1+q+q^2)(1+q+q^2+q^3+q^4)x^3.\cr }$$ These $q$-analogues are equal to those obtained by Dulucq by means of the combinatorial model of Dulucq and Favreau {\bf [D]}. \bigskip \bf4 Expression of a symmetric polynomial in terms of the power sums of odd degree \rm\medskip Let ${\bb{X}}$ denote a finite set of variables ${\bb{X}}=\{x_1,\ldots ,x_n\}$. It is well known that any symmetric polynomial $F({\bb{X}})$ may be expressed as a polynomial function of the power sums $\psi_k({\bb{X}})=\sum_{x\in{\bb{X}}}x^k,\ k=1,2\ldots n$. It was shown in the last century that $F({\bb{X}})$ may also be expressed as a rational function of $\psi_k({\bb{X}}),\ k=1,3,5,\ldots 2n-1$ (\bf[Bo]\rm, \bf[La]\rm). P\'olya \bf[Po] \rm and Foulkes \bf[Fo]\rm, among others, have proposed explicit expressions for some particular symmetric polynomials $F({\bb{X}})$. For instance \medskip {\petcap Proposition 4.1.} {\it The $k^{th}$ elementary symmetric function is equal to $\Lambda_k({\bb{X}})=S_{\rho(n)/1^{n-k}}({\bb{X}})/ S_{\rho(n-1)}({\bb{X}})\ \it,\rm$ the right-hand side being a rational function of the power sums of odd degree.} \smallskip \it Proof. \rm For sake of completeness we sketch the proof. It is known that $S_{\rho(n-1)}({\bb{X}})=\prod_{i0$. $\ \qed$ \bigskip \centerline{\petcap References} \bigskip \bf[AM] \rm\hskip 5mm {\petcap M. Adler and J. Moser}, {\it On a class of polynomials connected with the Korteweg-de-Vries equation}, Commun. math. Phys. 61, (1978), 1--30. \medskip \bf[Bo] \rm\hskip 5mm {\petcap C.W. Borchardt}, {\it \"Uber eine Eigenschaft der Potenzsummen ungerader Ordnung}, Monastber. Akad. Berlin 1857, 301--311. \medskip \bf[Ch]\rm\hskip 5mm {\petcap T.S. 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