Laurent polynomials associated to partitions and a $Q$-deformation of $\sigma(d)$ 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:
$$C_1 = 1$$
$$C_2 = Q+1+1/Q$$
$$C_3 = Q^2+Q+1/Q+1/Q^2$$
$$C_4 =  Q^3+Q^2+Q+1+1/Q+1/Q^2+1/Q^3$$
$$C_5 =  Q^4+Q^3+Q^2+1/Q^2+1/Q^3+1/Q^4$$
$$C_6 =  Q^5+Q^4+Q^3+Q^2+Q+2+1/Q+1/Q^2+1/Q^3+1/Q^4+1/Q^5$$
$$C_7 =  Q^6+Q^5+Q^4+Q^3+1/Q^3+1/Q^4+1/Q^5+1/Q^6$$
$$C_8 =  Q^7+Q^6+Q^5+Q^4+Q^3+Q^2+Q+1+1/Q+1/Q^2+1/Q^3+1/Q^4+1/Q^5+1/Q^6+1/Q^7$$
$$C_9 = Q^8+Q^7+Q^6+Q^5+Q^4+Q+1+1/Q+1/Q^4+1/Q^5+1/Q^6+1/Q^7+1/Q^8$$
 A: 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.
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
$$\sum_{n\geq 1}\frac{q^n}{(1-q^n)^2}=\sum_{n\geq 1} \sigma(n)q^n.$$
