$\DeclareMathOperator{\leg}{\operatorname{leg}}\DeclareMathOperator{\arm}{\operatorname{arm}}$The following situation arose from the study of some localization computations on Hilbert schemes of points.

Let $\phi$, $\psi$ and $\gamma$ be power series (with coefficients in $\mathbb Q$). Assume $\gamma$ has valuation 1 (i.e. it has no constant term but has an $X^1$ term). Fix $k$ a non-negative integer. For each $n$, let $F_n$ be the $X^0$ coefficient of $$ \sum_{\Lambda \vdash n} \sum_{\lambda_i \vdash \Lambda_i} (-1)^{\ell(\Lambda)} \frac{(\ell(\Lambda)-1)!}{\operatorname{Aut}\Lambda} \left(\sum_i \sum_{\square \in \lambda_i} \phi(X(\square_1t_1 + \square_2t_2))\right)^k \prod_i \prod_{\square \in \lambda_i} \frac{\psi(X(\square_1t_1 + \square_2t_2))}{\gamma(XT_1^\square) \gamma(X T_2^\square)} $$ Here $\Lambda$ ranges over all partitions of $n$, and $\ell(\Lambda)$ is the number of parts of $\Lambda$. The second sum is over tuples of partitions of the parts $\Lambda_i$ of $\Lambda$. The sum $\square \in \lambda_i$ is over the boxes in the Young diagram of $\lambda$ with coordinates $(\square_1,\square_2)$. and $$ T_1^\square = (\leg(\square) + 1) t_1 - \arm(\square)t_2 ,\;\;\;\; T_2^\square= -\leg(\square)t_1 + (\arm(\square) +1)t_2 $$ (coming from tangent weights of a torus action on the Hilbert scheme).

It is clear that $F_n$ is a symmetric rational function of $t_1$ and $t_2$ of degree 0.

Conjecture (After being simplified,) $F_n$ has denominator $t_1t_2$. In other words, it is of the form $$ \frac{at_1^2 + bt_1t_2 + at_2^2}{t_1t_2} $$

Bonus points if you can say something about the generating function of the $F_n$.

Here is a link to a SageMath worksheet with some examples. I haven't found a counterexample yet. It is false if $\gamma$ has valuation 2.

https://github.com/uberparagon/sum-over-parts-conjecture/blob/master/check.ipynb

Here's the easiest non-trivial special case, which I still can't do: Let $k=1$, $\phi(X) = X^{2m}$, $\psi =1$, $\gamma =X$. Then it is not hard to check that $F_n=0$ unless $n=m$. Show that $F_m$ satisfies the conjecture.

The question is a bit complicated, but the answer is simple enough that I hope there is some way to manage it.

$\DeclareMathOperator{\leg}{\operatorname{leg}}\DeclareMathOperator{\arm}{\operatorname{arm}}$The following situation arose from the study of some localization computations on Hilbert schemes of points.

Let $\phi$, $\psi$ and $\gamma$ be power series (with coefficients in $\mathbb Q$). Assume $\gamma$ has valuation 1 (i.e. it has no constant term but has an $X^1$ term). Fix $k$ a non-negative integer. For each $n$, let $F_n$ be the $X^0$ coefficient of $$ \sum_{\Lambda \vdash n} \sum_{\lambda_i \vdash \Lambda_i} (-1)^{\ell(\Lambda)} \frac{(\ell(\Lambda)-1)!}{\operatorname{Aut}\Lambda} \left(\sum_i \sum_{\square \in \lambda_i} \phi(X(\square_1t_1 + \square_2t_2))\right)^k \prod_i \prod_{\square \in \lambda_i} \frac{\psi(X(\square_1t_1 + \square_2t_2))}{\gamma(XT_1^\square) \gamma(X T_2^\square)} $$ Here $\Lambda$ ranges over all partitions of $n$, and $\ell(\Lambda)$ is the number of parts of $\Lambda$. The second sum is over tuples of partitions of the parts $\Lambda_i$ of $\Lambda$. The sum $\square \in \lambda_i$ is over the boxes in the Young diagram of $\lambda$ with coordinates $(\square_1,\square_2)$. and $$ T_1^\square = (\leg(\square) + 1) t_1 - \arm(\square)t_2 ,\;\;\;\; T_2^\square= -\leg(\square)t_1 + (\arm(\square) +1)t_2 $$ (coming from tangent weights of a torus action on the Hilbert scheme).

It is clear that $F_n$ is a symmetric rational function of $t_1$ and $t_2$ of degree 0.

Conjecture (After being simplified,) $F_n$ has denominator $t_1t_2$. In other words, it is of the form $$ \frac{at_1^2 + bt_1t_2 + at_2^2}{t_1t_2} $$

Bonus points if you can say something about the generating function of the $F_n$.

Here is a link to a SageMath worksheet with some examples. I haven't found a counterexample yet. It is false if $\gamma$ has valuation 2.

https://github.com/uberparagon/sum-over-parts-conjecture/blob/master/check.ipynb

Here's the easiest non-trivial special case, which I still can't do: Let $k=1$, $\phi(X) = X^{2m}$, $\psi =1$, $\gamma =X$. Then it is not hard to check that $F_n=0$ unless $n=m$. Show that $F_m$ satisfies the conjecture.

The question is a bit complicated, but the answer is simple enough that I hope there is some way to manage it.

Edit: I noticed that if you make the substitution like $q=\exp(t_1)$ and $t = \exp(t_2)$ and pick $\gamma = 1 - \exp(x)$, then the denominator looks like the denominator in for the $qt$-Catalan numbers. There is a non-trivial denominator-cancelling theorem there too, so I wonder if it is related.

The following situation arose from the study of some localization computations on Hilbert schemes of points.

Let $\phi$, $\psi$ and $\gamma$ be power series (with coefficients in $\mathbb Q$). Assume $\gamma$ has valuation 1 (i.e. it has no constant term but has an $X^1$ term). Fix $k$ a non-negative integer. For each $n$, let $F_n$ be the $X^0$ coefficient of $$ \sum_{\Lambda \vdash n} \sum_{\lambda_i \vdash \Lambda_i} (-1)^{\ell(\Lambda)} \frac{(\ell(\Lambda)-1)!}{\operatorname{Aut}\Lambda} \left(\sum_i \sum_{\square \in \lambda_i} \phi(X(\square_1t_1 + \square_2t_2))\right)^k \prod_i \prod_{\square \in \lambda_i} \frac{\psi(X(\square_1t_1 + \square_2t_2))}{\gamma(XT_1^\square) \gamma(X T_2^\square)} $$ Here $\Lambda$ ranges over all partitions of $n$, and $\ell(\Lambda)$ is the number of parts of $\Lambda$. The second sum is over tuples of partitions of the parts $\Lambda_i$ of $\Lambda$. The sum $\square \in \lambda_i$ is over the boxes in the Young diagram of $\lambda$ with coordinates $(\square_1,\square_2)$. and $$ T_1^\square = (leg(\square) + 1) t_1 - arm(\square)t_2 ,\;\;\;\; T_2^\square= -leg(\square)t_1 + (arm(\square) +1)t_2 $$ (coming from tangent weights of a torus action on the Hilbert Scheme).

It is clear that $F_n$ is a symmetric rational function of $t_1$ and $t_2$ of degree 0.

Conjecture (After being simplified,) $F_n$ has denominator $t_1t_2$. In other words, it is of the form $$ \frac{at_1^2 + bt_1t_2 + at_2^2}{t_1t_2} $$

Bonus points if you can say something about the generating function of the $F_n$.

Here is a link to a SageMath worksheet with some examples. I haven't found a counterexample yet. It is false if $\gamma$ has valuation 2.

https://github.com/uberparagon/sum-over-parts-conjecture/blob/master/check.ipynb

Here's the easiest non-trivial special case, which I still can't do: Let $k=1$, $\phi(X) = X^{2m}$, $\psi =1$, $\gamma =X$. Then it is not hard to check that $F_n=0$ unless $n=m$. Show that $F_m$ satisfies the Conjecture.

The question is a bit complicated, but the answer is simple enough that I hope there is some way to manage it.

$\DeclareMathOperator{\leg}{\operatorname{leg}}\DeclareMathOperator{\arm}{\operatorname{arm}}$The following situation arose from the study of some localization computations on Hilbert schemes of points.

Let $\phi$, $\psi$ and $\gamma$ be power series (with coefficients in $\mathbb Q$). Assume $\gamma$ has valuation 1 (i.e. it has no constant term but has an $X^1$ term). Fix $k$ a non-negative integer. For each $n$, let $F_n$ be the $X^0$ coefficient of $$ \sum_{\Lambda \vdash n} \sum_{\lambda_i \vdash \Lambda_i} (-1)^{\ell(\Lambda)} \frac{(\ell(\Lambda)-1)!}{\operatorname{Aut}\Lambda} \left(\sum_i \sum_{\square \in \lambda_i} \phi(X(\square_1t_1 + \square_2t_2))\right)^k \prod_i \prod_{\square \in \lambda_i} \frac{\psi(X(\square_1t_1 + \square_2t_2))}{\gamma(XT_1^\square) \gamma(X T_2^\square)} $$ Here $\Lambda$ ranges over all partitions of $n$, and $\ell(\Lambda)$ is the number of parts of $\Lambda$. The second sum is over tuples of partitions of the parts $\Lambda_i$ of $\Lambda$. The sum $\square \in \lambda_i$ is over the boxes in the Young diagram of $\lambda$ with coordinates $(\square_1,\square_2)$. and $$ T_1^\square = (\leg(\square) + 1) t_1 - \arm(\square)t_2 ,\;\;\;\; T_2^\square= -\leg(\square)t_1 + (\arm(\square) +1)t_2 $$ (coming from tangent weights of a torus action on the Hilbert scheme).

It is clear that $F_n$ is a symmetric rational function of $t_1$ and $t_2$ of degree 0.

Conjecture (After being simplified,) $F_n$ has denominator $t_1t_2$. In other words, it is of the form $$ \frac{at_1^2 + bt_1t_2 + at_2^2}{t_1t_2} $$

Bonus points if you can say something about the generating function of the $F_n$.

Here is a link to a SageMath worksheet with some examples. I haven't found a counterexample yet. It is false if $\gamma$ has valuation 2.

https://github.com/uberparagon/sum-over-parts-conjecture/blob/master/check.ipynb

Here's the easiest non-trivial special case, which I still can't do: Let $k=1$, $\phi(X) = X^{2m}$, $\psi =1$, $\gamma =X$. Then it is not hard to check that $F_n=0$ unless $n=m$. Show that $F_m$ satisfies the conjecture.

The question is a bit complicated, but the answer is simple enough that I hope there is some way to manage it.