# Extracting constant terms: is there a direct way?

$$\DeclareMathOperator\CT{CT}$$ Let $$\CT_t(f(t))$$ denote the constant term of the Laurent polynomial of $$f(t)$$.

Define the two functions $$F(x_1,\dots,x_n)$$ and $$G(y)$$ by $$F:=\prod_{i=1}^nx_i^{-1}(1-x_i)^{-2}\prod_{1\leq i I like to ask:

QUESTION. Is the following true? It would be great if there is a direct way to compare these two. $$\CT_{x_1}\CT_{x_2}\cdots\CT_{x_n}\left(F(x_1,\dots,x_n)\right)=\CT_y\left(G(y)\right).$$

NOTE 1. The sequence on the right-hand side is available at the OEIS as A301741 with an explicit evaluation.

NOTE 2. Incidentally, we also have (a consequence of Han's formula proved here) $$\CT_{x_1}\CT_{x_2}\cdots\CT_{x_n}\left(F(x_1,\dots,x_n)\right)= \sum_{\lambda\vdash n}f^{\lambda}\prod_{u\in\lambda} \frac{(n+2)^{h_u}+n^{h_u}}{(n+2)^{h_u}-n^{h_u}};$$ where $$h_u$$ is the hook-length of cell $$u$$ (in the Young diagram of $$\lambda$$) and $$f^{\lambda}$$ is the number of Standard Young Tableau of shape $$\lambda$$ (given by the hook-length formula).

NOTE 3. A cute analogue: let $$f:=\prod_{i=1}^nx_i^{-1}(1-x_i)^{-1}\prod_{1\leq i and $$g:=n!\cdot y^{-n}e^{ny-y^2/2}$$. Then, $$\CT_{x_1}\CT_{x_2}\cdots\CT_{x_n}\left(f(x_1,\dots,x_n)\right) =\CT_y(g(y)).$$ Proof. Fedor's reasoning applies (it'd be nice to employ Richard's too) \begin{align*} {\rm CT}\, f&= [x_1\ldots x_n] \prod_i (1-x_i)^{-1}\prod_{i

For power series $$u(x_1,\ldots,x_n),v(x_1,\ldots,x_n)$$ call $$u,v$$ similar and write $$u\sim v$$ if all monomials $$\prod x_i^{c_i}$$ with $$c_i\in \{0,1\}$$ have equal coefficients in $$u,v$$. In other words, if $$u$$ is congruent to $$v$$ modulo the ideal generated by $$x_i^2$$'s. Note that this similarity respects addition and multiplucation, and that $$(1-x_i)^{-1}\sim \exp(x_i)$$ and $$(1-x_i-x_j)^{-1}\sim 1+x_i+x_j+2x_ix_j\sim\exp(x_i+x_j+x_ix_j)$$. Thus \begin{align*} {\rm CT}\, F&= [x_1\ldots x_n] \prod_i (1-x_i)^{-2}\prod_{i where $$S=x_1+\ldots+x_n$$ (since $$S^2/2\sim \sum_{i). Now if we expand $$\exp\left((n+1)S+S^2/2\right)$$ as a power series in $$S$$, we get $$[x_1\ldots x_n]S^n=n!$$ and $$[x_1\ldots x_n]S^k=0$$ for $$k\ne n$$, thus your identity.

• Wonderful argument, Fedor. Thanks. Dec 20, 2021 at 16:09

By the exponential formula, the constant term of $$G(y)$$ equals $$\sum_w (n+1)^{\mathrm{fix}(w)}$$, where $$w$$ ranges over all involutions in the symmetric group $$S_n$$ and $$\mathrm{fix}(w)$$ is the number of fixed points of $$w$$. In other words, this is the number of involutions in $$S_n$$ whose fixed points are colored by one of the colors $$1,2,\dots,n+1$$.

The constant term of $$F$$ is the coefficient of $$x_1\cdots x_n$$ in $$\hat{F} = x_1\cdots x_n F$$. Hence we can ignore all exponents greater than 1 in the expansion of $$\hat{F}$$, so we want the coefficient of $$x_1\cdots x_n$$ in $$\prod_{i=1}^n (1+x_i)^2 \prod_{1\leq i

Given an involution $$w\in S_n$$ with each fixed point colored by one of the colors $$1,2,\dots,n+1$$, we can obtain a monomial $$x_1\cdots x_n$$ as follows: if the fixed point $$a$$ is colored 1, choose $$x_a$$ from the first factor of $$(1+x_a)^2$$ in (*). If $$a$$ is colored 2, choose $$x_a$$ from the second factor of $$(1+x_a)^2$$. If $$a$$ is colored $$k>2$$ then consider the $$(k-2)$$nd factor of $$\prod_{1\leq i (where we order the $${n\choose 2}$$ factors in some way) containing a term $$x_a$$. Let $$x_b$$ be the other variable appearing in this factor. Say that this is the $$(m-2)$$nd factor containing $$x_b$$. If $$b$$ is not colored $$m$$, then choose the term $$x_a$$ from the factor $$1+x_a+x_b+x_ax_b+x_ax_b$$. If $$b$$ is colored $$m$$, then choose the first of the two terms equal to $$x_ax_b$$.

It remains to account for the 2-cycles. If $$(a,b)$$ is a 2-cycle, then choose the second term equal to $$x_ax_b$$ from the factor $$1+x_a+x_b+x_ax_b+x_ax_b$$. From all remaining factors we choose the term 1. This sets up a bijection between the colored involutions and the monomials $$x_1\cdots x_n$$ appearing in the expansion of $$\hat{F}$$.

• Wonderful interpretation, Richard. Thanks. Dec 20, 2021 at 16:09