Let $\alpha \vdash d$ be a partition of $d$, i.e. $\alpha = (\alpha_1 \geq \alpha_2 \geq …\geq \alpha_l)$, where $\sum_k \alpha_k = d$. Define a Laurent polynomial in $Q$ as follows:

$$ P_\alpha(Q) = \sum_{k=1}^\infty Q^{-\alpha_k + k-1}(1-Q) \\ \quad \quad \quad \quad \quad \quad = Q^{l} + \sum_{k=1}^{l} (Q^{-\alpha_k + k-1} - Q^{-\alpha_k + k}) $$

**Question 1:** Has anyone encountered such polynomials and know a nice formula for them and/or a reference to a place where they have appeared?

I can prove that $P_\alpha(Q) = P_{\alpha'}(Q^{-1})$ where $\alpha'$ is the conjugate partition.

The sum of these polynomials over all $\alpha \vdash d$ appears to be nice but I have not discovered a good closed formula. I would especially like a formula for the coefficients of the following generating series in $v$, which has arisen in a computation in Donaldson-Thomas theory. Define Laurent polynomials $C_d(Q)$ by the following formula:

$$ \sum_{d=0}^\infty C_d (Q)\,\, v^d = \frac{Q}{(1-Q)^2}\left( \sum_\alpha v^{|\alpha|} P_\alpha \right) \prod_{m=1}^\infty (1-v^m). $$

(Note that $C_0$ is not a polynomial, but for $d>0$, $C_d$ seems to be a polynomial). Computation for $d$ up to 40 suggests that

**$C_d(Q)$ is some sort of $Q$ deformation of $\sigma(d)$, the sum of divisors function.**

**Question 2:** Prove that $C_d(1)=\sigma(d)$. Is there a formula for $C_d(Q)$ which looks like a $Q$-deformation of the formula $\sigma(d) = \sum_{k|d}k$? Or is there some other natural explanation for this phenomenon?

Below is a list of $C_d$ for $d$ up to 9:

$$C_1 = 1$$ $$C_2 = Q+1+1/Q$$ $$C_3 = Q^2+Q+1/Q+1/Q^2$$ $$C_4 = Q^3+Q^2+Q+1+1/Q+1/Q^2+1/Q^3$$ $$C_5 = Q^4+Q^3+Q^2+1/Q^2+1/Q^3+1/Q^4$$ $$C_6 = Q^5+Q^4+Q^3+Q^2+Q+2+1/Q+1/Q^2+1/Q^3+1/Q^4+1/Q^5$$ $$C_7 = Q^6+Q^5+Q^4+Q^3+1/Q^3+1/Q^4+1/Q^5+1/Q^6$$ $$C_8 = Q^7+Q^6+Q^5+Q^4+Q^3+Q^2+Q+1+1/Q+1/Q^2+1/Q^3+1/Q^4+1/Q^5+1/Q^6+1/Q^7$$ $$C_9 = Q^8+Q^7+Q^6+Q^5+Q^4+Q+1+1/Q+1/Q^4+1/Q^5+1/Q^6+1/Q^7+1/Q^8$$

polynomialso my equation makes sense. The infinite sum is really a finite sum since for large $k$ the sum telescopes. I'll edit the question to make that more clear. $\endgroup$ – Jim Bryan May 5 '14 at 2:53