Parity of hooklengths in a partition diagram Main Question
Let $\lambda\vdash n$ be a partition, with hooklengths $\{h_1,\dots,h_n\}$ in its partition diagram. Is there a formula for determining
$$\#\{h_i\text{ even}\}-\#\{h_i\text{ odd}\}?$$
For example, the partition $\lambda=(4,2)$ has hooklengths:
$$\begin{matrix}5 & 4 & 2 & 1 \\ 2 & 1\end{matrix},$$
and so its hooklength parity is zero (three even and three odd).
Background
In the above setup, define the $q$-hooklength formula
$$f^\lambda(q)=\frac{[n]_q!}{\prod_{x\in\lambda}[h(x)]_q}\in\mathbb{Z}[q]$$
where $[k]_q=1+q+\cdots+q^{k-1}$. Suppose $m=\lfloor n/2\rfloor$ and $k$ is the number of even hooklengths. One can show that
$$f^\lambda(-1)=\begin{cases}0 & m>k \\ a_\lambda> 0 & m=k \\ \infty & m<k\end{cases}$$
Moreover, by a result of Stembridge, this value is related to the number of standard tableaux fixed by evacuation. Furthermore, since $f^\lambda(q)\in\mathbb{Z}[q]$, this seems to suggest that we always have $m\geq k$, with equality if and only if there exist standard tableaux of shape $\lambda$ which are fixed under the evacuation map. I am wondering if there is an elementary way to calculate this parity.
 A: If you look at the hooklengths in the $i^{\text{th}}$ row of the diagram, you will see a descending sequence that skips some numbers. Specifically, starting from the left, we will:

*

*See the numbers $1, \dots, \lambda_{i+1} - \lambda_i$, but then skip $\lambda_{i+1} - \lambda_i + 1$.

*See $\lambda_{i+2} - \lambda_{i+1}$ more numbers, but then skip $\lambda_{i+2} - \lambda_i + 2$.

*And so on, skipping $\lambda_{i+j} - \lambda_i + j$ for each $j$.

When we skip a number $k$, we subtract $(-1)^k$ from the hooklength parity you're measuring, compared to if we didn't. So we accumulate a total of
$$
   -\sum_{i<j} (-1)^{\lambda_j - \lambda_i + j - i}
$$
from the skipped numbers. (We could also write this as $(-1)^{\lambda_i + \lambda_j + i + j}$: adding or subtracting doesn't affect parity.)
That's not the whole story. If we imagine a hypothetical situation where we didn't skip any numbers, the hooklength parity is not $0$: it is the negative of the number of odd values in the first column. The hooklength in the $i^{\text{th}}$ row (out of $k$) and the first column is equal to $\lambda_i + k - i$. We want a contribution of $0$ if this number is even, and $-1$ if this number is odd, which is given by $\frac12((-1)^{\lambda_i + k - i} - 1)$.
Overall, the hooklength parity of a partition $(\lambda_1, \lambda_2, \dots, \lambda_k)$ will be
$$
    -\sum_{i=1}^k \sum_{j=i+1}^k (-1)^{\lambda_i + \lambda_j + i + j} + \frac12 \sum_{i=1}^k (-1)^{\lambda_i + i + k} - \frac k2.
$$
But there's more simplifying to be done, because the terms in the first sum can be factored as $(-1)^{\lambda_i + i} (-1)^{\lambda_j + j}$, so they are half of the off-diagonal terms in $\left(\sum_{i=1}^k (-1)^{\lambda_i + i}\right)^2$. The diagonal terms in this sum are all $1$, so we can rewrite the first sum as $\frac12\left(\sum_{i=1}^k (-1)^{\lambda_i + i}\right)^2 - \frac k2$, and the whole expression as
$$
  \frac12 \left(\sum_{i=1}^k (-1)^{\lambda_i + i + k}\right) -  \frac12 \left(\sum_{i=1}^k (-1)^{\lambda_i + i}\right)^2.
$$
For example, for the partition $(4,2)$ we get
$$
   \frac12 \left((-1)^{4+1+2} + (-1)^{2+2+2}\right) - \frac12\left((-1)^{4+1} + (-1)^{2+2}\right)^2 = 0.
$$
