# Alternating sum of hook lengths: Part I

Given $$\lambda$$ an integer partition of $$n$$, let $$h_{ij}(\lambda)$$ denote the hook length of cell $$(i,j)$$ in the Young diagram of $$\lambda$$.

Is there a closed formula or a generating function for the following sequence? $$f_n=\sum_{\lambda\vdash n}\sum_{(i,j)\in\lambda}(-1)^{i+j}\cdot h_{ij}(\lambda)$$

• Why mention $\lambda'$? Apr 2 at 21:08
• You need $\lambda'$ to compute hook length. Apr 2 at 21:11
• Depends what you call "compute". I guess you meant writing an algebraic formula for $h_{ij}$ in terms of $\lambda_i$ and $\lambda'_j$. I was thinking of a visual definition of the hook length from the diagram. Apr 2 at 21:14
• I see. I will minimize the confusion by editing those lines in my post which I just noted that Matthieu did already. Apr 2 at 21:33

The hook length $$h_{ij}(\lambda)$$ counts the number of boxes directly below or directly to the right of box $$(i,j)$$. (I picture the Young diagram of $$\lambda$$ as having the corner $$(0,0)$$ located in the upper left.) With this in mind, the alternating sum $$\sum_{(i,j)\in \lambda}(-1)^{i+j}h_{ij}(\lambda)$$ can be expanded as a signed sum over all the boxes in $$\lambda$$. We see that a box $$(i,j)$$ contributes $$+1$$ if both $$i,j$$ are even, $$-1$$ if both $$i,j$$ are odd, and $$0$$ otherwise. Using this description we see that $$\sum_{(i,j)\in \lambda}(-1)^{i+j}h_{ij}(\lambda)=\left\lceil \frac{\lambda_1}{2}\right\rceil-\left\lfloor\frac{\lambda_2}{2}\right\rfloor+\left\lceil \frac{\lambda_3}{2}\right\rceil-\left\lfloor\frac{\lambda_4}{2}\right\rfloor+\cdots$$ where $$\lambda=(\lambda_1, \lambda_2, \dots)$$.
Now we can use a wonderful formula due to Boulet (Theorem 1 in "A Four-parameter Partition Identity") which tells us (after the appropriate specialization) $$\sum_{\lambda}x^{\sum_{(i,j)\in \lambda}(-1)^{i+j}h_{ij}(\lambda)}q^{|\lambda|}=\prod_{k\geq 1}\frac{(1+xq^{4k-1})(1+xq^{4k-3})}{(1-q^{4k})(1-xq^{4k-2})^2}$$ I will call this two variable generating function $$H(x,q)$$. The generating function $$F(q)=\sum_{n\geq 0}f_nq^n$$ is given by $$F(q)=\left[\frac{\partial}{\partial x} H(x,q)\right]_{x=1}$$ therefore we start by computing $$\frac{\partial}{\partial x} H(x,q)=H(x,q)\sum_{k\geq 1}\left(\frac{q^{4k-1}}{1+xq^{4k-1}}+\frac{2q^{4k-2}}{1-xq^{4k-2}}+\frac{q^{4k-3}}{1+xq^{4k-3}}\right)$$ and finally setting $$x=1$$ gives $$F(q)=\left(\prod_{k\geq 1}\frac{1}{1-q^k}\right)\cdot\left(\sum_{n\geq 1}A_nq^n\right)$$ where $$\sum_{n\geq 1}A_{n}q^n=\sum_k\left(\frac{q^{4k-1}}{1+q^{4k-1}}+\frac{2q^{4k-2}}{1-q^{4k-2}}+\frac{q^{4k-3}}{1+q^{4k-3}}\right)$$. If we expand this last sum, the terms simplify a little bit to give $$\sum_{n\geq 1}A_n q^n=\sum_{k\geq 1}\frac{q^{2k-1}}{1-q^{2k-1}}$$.
It seems that $$\sum_nA_nq^n=\sum_k\frac{q^{2k-1}}{1-q^{2k-1}}=\sum_n\frac{d(n)}{\nu_2(2n)}q^n,$$ where $$d(n)$$ is the sum of (positive) divisors of $$n$$ and $$\nu_2(m)$$ is the $$2$$-adic valuation of $$m$$.
Therefore, we obtain the convolution sum $$f_n=\sum_{k=1}^n\frac{d(k)\cdot p(n-k)}{\nu_2(2k)}$$ where $$p(m)$$ is the number of integer partition of $$m$$.