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I have encountered a necessity to work with a series of the following form.

There are $N$ variables $x_1,\ldots x_N$. It is convenient to introduce monomial symmetric polynomials $m_{\lambda}$. They are defined for any power $r$ and a partition $\lambda$ of $r$ into $N$ parts, some of which can be 0, as sums of monomials with powers being elements of $\lambda$. For example, for $r=3$ and $N=3$ we have $m_{(3)}=x_1^3+x_2^3+x_3^3$, $m_{(2,1)}=x_1^2x_2+x_1^2x_3+x_2^2x_1+x_2^2x_3+x_3^2x_1+x_3^2x_2$, $m_{(1,1,1)}=x_1x_2x_3$. The power series is

$S(x_1,\ldots x_N)=\sum\limits_{r=0}^{\infty}\sum\limits_{\lambda_r}\frac{m_{\lambda_r}(x_1,\ldots x_N)}{L_{r,\lambda}}$.

Here the summation over $\lambda_r$ goes over all possible partitions of $r$ into $N$ parts with possible 0's, and $L_{r,\lambda}$ is a number of terms in $m_{\lambda_r}(x_1,\ldots x_N)$. In the example above, for $N=3$ and $r=3$ we will have, for $L_{3,(3)}=3$, $L_{3,(2,1)}=6$, $L_{3,(1,1,1)}=1$.

An alternative form of this series is

$S(x_1,\ldots x_N)=\frac{1}{N!}\sum\limits_{r=0}^{\infty}\sum\limits_{\lambda_r}N_{r,\lambda}m_{\lambda_r}(x_1,\ldots x_N)$

Here $N_{r,\lambda}$ is an order of the stabilizer of the parts of $\lambda$ under the action of the symmetric group $S_N$ acting on the variables $x_1,\ldots x_N$. Again, in the example above, for $N=3$ and $r=3$, $N_{3,(3)}=2$, $N_{3,(2,1)}=1$, $N_{3,(1,1,1)}=6$.

How can one compute a sum of such a power series? Or, maybe, a sum of some similar series with a factor of $1/L_{r,\lambda}$?

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1 Answer 1

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For a polynomial or a formal power series $F(x_1,\ldots,x_N)$ in $x_1,\ldots,x_N$, let $$ {\rm Sym} [F(x_1,\ldots,x_N)]=\frac{1}{N!}\sum_{\sigma\in \mathfrak{S}_N} F(x_{\sigma(1)},\ldots,x_{\sigma(N)}) $$ denote the (normalized) symmetrization, as an average over permutations in the symmetric group $\mathfrak{S}_N$.

For an integer partition $\lambda=(1^{a_1}2^{a_2}\cdots)$, with $a_1$ parts equal to 1, $a_2$ parts equal to 2, etc., the monomial symetric function $m_{\lambda}$ satisfies $$ {\rm Sym}[x_1^{\lambda_1}\cdots x_N^{\lambda_N}]=\frac{a_1!a_2!\cdots}{N!}\ m_{\lambda}(x_1,\ldots,x_N) $$ because the overcounting factor or size of the stabilizer of the monomial being symmetrized is $a_1! a_2!\cdots$. By the Orbit-Stabilizer Theorem, the number of terms in $m_{\lambda}$ is $$ L_{r,\lambda}=\frac{N!}{a_1!a_2!\cdots}\ . $$ As a result, the wanted series is $$ S(x_1,\ldots,x_N)={\rm Sym}[U(x_1,\ldots,x_N)] $$ with $$ U(x_1,\ldots,x_N)=\sum_{r\ge 0}\sum_{\lambda \vdash r} x_1^{\lambda_1}x_2^{\lambda_2}\cdots x_N^{\lambda_N} $$ Changing variables to $u_1=\lambda_1-\lambda_2$, $u_2=\lambda_2-\lambda_3$,..., $u_{N-1}=\lambda_{N-1}-\lambda_N$, $u_N=\lambda_N$, we get $$ U(x_1,\ldots,x_N)=\sum_{u_1,\ldots,u_N\ge 0}x_1^{u_1+\cdots+u_N} x_2^{u_2+\cdots+u_N}\cdots x_{N-1}^{u_{N-1}+u_{N}} x_N^{u_N} $$ This immediately gives $$ S(x_1,\ldots,x_N)={\rm Sym}\left[\frac{1}{(1-x_1)(1-x_1 x_2)\cdots(1-x_1\cdots x_N)}\right] $$


Edit:

Figuring out the result as a single fraction may perhaps be interesting. There are related formulas where, in the numerator of $U$, one has a product over descents of products of the $x$'s up to that descent. This would be the trivial or antichain poset case of formula (6.3) in the article "Linear extension sums as valuations of cones" by Boussicault, Féray, Lascoux, and Reiner.

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