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:


$$C\_1 = 1$$
$$C\_2 = Q+1+1/Q$$
$$C\_3 = Q^2+Q+1/Q+1/Q^2$$
$$C\_4 = Q^3+Q^2+Q+1+1/Q+1/Q^2+1/Q^3$$
$$C\_5 = Q^4+Q^3+Q^2+1/Q^2+1/Q^3+1/Q^4$$
$$C\_6 = Q^5+Q^4+Q^3+Q^2+Q+2+1/Q+1/Q^2+1/Q^3+1/Q^4+1/Q^5$$
$$C\_7 = Q^6+Q^5+Q^4+Q^3+1/Q^3+1/Q^4+1/Q^5+1/Q^6$$
$$C\_8 = Q^7+Q^6+Q^5+Q^4+Q^3+Q^2+Q+1+1/Q+1/Q^2+1/Q^3+1/Q^4+1/Q^5+1/Q^6+1/Q^7$$
$$C\_9 = Q^8+Q^7+Q^6+Q^5+Q^4+Q+1+1/Q+1/Q^4+1/Q^5+1/Q^6+1/Q^7+1/Q^8$$



Laurent polynomials associated to partitions and a $Q$-deformation of $\sigma(d)$
Closely related infinite series appear in relation to the infinite wedge; this then gives a proof of your second question.


**Warnings**: I skimp out on the details of a slightly laborious Calculus I exercise at the end, but the answer was already getting long enough.


Also, it looks like I probably made a sign error, presumably related to $Q^{1/2}-Q^{-1/2}$ transforming under $Q\to 1/Q$ and how your invariants behave under conjugation.


A related infinite series
=========================


In the fermionic viewpoint for partitions, the natural thing to consider is not your Laurent polynomials, but the infinite series


$$f\_\alpha(Q)=\sum\_{k=1}^\infty Q^{\alpha\_k-k+1/2}.$$


In the bijection between partitions and charge zero elements of Dirac's electron sea, the exponents here are exactly the energies of the electrons.


Your function simply differs by a factor: $P\_\alpha(Q^{-1})=(Q^{1/2}-Q^{-1/2})f\_\alpha(Q)$.


The $f\_\alpha$ occur naturally as eigenvalues of operators on the infinite wedge; in the notation of Okounkov and Pandharipande's trilogoy on the Gromov-Witten theory of curves, for instance, we have:
$$\mathcal{E}\_0(\ln(Q))v\_\lambda=f\_\lambda(Q)v\_\lambda$$


But for your second question, the source you want is Bloch and Okounkov [The character of the infinite wedge](http://arxiv.org/abs/alg-geom/9712009).


A result of Bloch and Okounkov
==============================


The pertinent bit to you is the easiest example of their Theorem 0.5, which they prove as Theorem 6.5 as a warm-up to the main event.


Some notation: for a function on polynomials, they define
$$\langle f\rangle\_q=\sum\_\lambda f(\lambda)q^{|\lambda|}/\sum\_{\lambda} q^{|\lambda|}=\left(\sum\_\lambda f(\lambda)q^\lambda\right)\prod (1-q^n),$$the expected value of $f$ over partitions if a partition $\lambda$ has probability $q^{|\lambda|}$.


Then, Theorem 6.5 of Bloch-Okounkov gives:
$$\left\langle \sum q^{\alpha\_k-k+1/2}\right\rangle\_q=\frac{1}{\Theta(t)}$$
where
$$\Theta(t)=\frac{(t^{1/2}-t^{-1/2})(qt)\_\infty (q/t)\_\infty}{(q)\_\infty^2}$$
with the standard Pochhammer symbol $(a)\_\infty=\prod\_{n\geq 0} (1-aq^n)$.


Product formula for $\sum C\_d(Q)v^d$
=====================================


Bloch and Okounkov's result immediately gives an infinite product formula for your $\sum C\_d(Q)v^d$. I am going to change your notation to be consistent with Bloch-Okounkov and Pochhammer symbols -- $Q$ becomes $t$, and $v$ becomes $q$.


The right hand side of your formula for $\sum C\_d(t) q^d$ becomes


$$\frac{t}{(1-t)^2} \left\langle (t^{1/2}-t^{-1/2})\sum t^{\alpha\_k-k+1/2}\right\rangle\_q $$


We can pull the $(t^{1/2}-t^{-1/2})$ outside the expectation, and simplify the resulting factors involving $t$ to $1/(t^{1/2}-t^{-1/2})$. Then we can substitute Bloch-Okounkov's result, to get


$$\sum C\_d(t)q^d=\frac{(q)\_\infty^2}{(t^{1/2}-t^{-1/2})^2(qt)\_\infty (q/t)\_\infty}$$


A $t$-deformation of $\sigma(d)$
================================


We now show that this implies $C\_d(1)=\sigma(d)$.


We would like to take the limit as $t\to 1$ of our product formula, but this limit does not exist because the constant in $q$ term is $$\frac{1}{(1-t)(1-t^{-1})}$$ (I think this is right, and there was a sign error in a previous formula). We subtract this term out by hand to remove the pole, and then take the limit:


$$\lim\_{t\to 1} \frac{(q)\_\infty^2-(qt)\_\infty(q/t)\_\infty}{(1-t)(1-1/t)(qt)\_\infty(q/t)\_\infty}$$


Note that the limit approaches $0/0$, and using l'Hôpital's rule once and simplifying, one sees that this limit is indeed


$$\sum\_{n\geq 1}\frac{q^n}{(1-q^n)^2}=\sum\_{n\geq 1} \sigma(n)q^n.$$


