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Using the Durfee square, prove that $$ \sum_{j=0}^n\left[\begin{array}{l} n \\ j \end{array}\right] \frac{t^j q^{j^2}}{(1-t q) \cdots\left(1-t q^j\right)}=\prod_{i=1}^n \frac{1}{1-t q^i} . $$

My thoughts:

Structure of the Partition

Each partition $\pi$ with a Durfee square of side $j$ decomposes as:

(a) the Durfee square;

(b) a partition $\pi_1$ that fits into a $j \times (n-j)$ box; and

(c) a partition $\pi_2$ that fits into a $j \times j$ box to the right of $\pi_1$.

Sum of the Parts of $\pi$

If $|\pi|$ represents the sum of the parts of $\pi$, then: $$ |\pi|=j^2 + |\pi_1| + |\pi_2| $$

Generating Function Interpretation

  • Durfee Square Contribution: The Durfee square itself contributes $q^{j^2}$, accounting for each square within the Durfee square.
  • Generating $\pi_1$:
    • $\pi_1$ consists of parts above the Durfee square. This can have at most $j$ parts, each of size at most $n-j$. The generating function for these partitions is represented by $\left[\begin{array}{c} n \\ j \end{array}\right]$.
  • Generating $\pi_2$:
    • $\pi_2$ consists of parts to the right of the Durfee square, involving parts that decrease in size from $j$ to $1$. The generating function for these parts, accommodating the gradation and inclusion of the parameter $t$ for counting purposes, is $\frac{t^j}{(1-tq)(1-tq^2) \cdots (1-tq^j)}$.

Combining Contributions

Putting it all together, the generating series for such partitions $\pi$ becomes: $$ q^{j^2}\left[\begin{array}{c} n \\ j \end{array}\right] \frac{t^j}{(1-tq)(1-tq^2) \cdots (1-tq^j)} $$

Summing Over All Durfee Squares

To cover all possible configurations of partitions up to a maximum part size of $n$: $$ \sum_{j=0}^n q^{j^2}\left[\begin{array}{c} n \\ j \end{array}\right] \frac{t^j}{(1-tq)(1-tq^2) \cdots (1-tq^j)} = \prod_{i=1}^n \frac{1}{1-t q^i}, $$ where the right-hand side directly generates all partitions with parts not exceeding $n$, each part contributing a factor of $t$ and $q^i$ to the partition's weight. This accounts for all possible partitions (with and without structural constraints like Durfee squares) fitting into a Ferrers diagram up to size $n$.

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  • $\begingroup$ So... What's the question? It seems that you already have a supposed proof? Do you ask to review it? $\endgroup$ Commented May 7 at 21:19
  • $\begingroup$ Also, what exactly is $\left[ \begin{matrix}n \\ j\end{matrix}\right]$ here? It's usually used to denote Stirling numbers of the first kind, but unlikely to match from the context. Did you mean $\binom{n}{j}_q$? $\endgroup$ Commented May 7 at 21:47

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