Find a closed form for the generating function for the number of partitions of n into 3 parts

Question

Find a closed form for the generating function for the number of partitions of n into 3 parts.

Answer 1

Note that any $k$-part partition's conjugate (obtained by flipping the Ferrers diagram) is a partition with parts of sizes less than or equal to $k$. Of course, this means we may write the conjugates as

$$a_11+a_22+\cdots+a_kk=n,$$

with $a_j$'s nonnegative integers counting the number of $j$'s in the conjugate partition of $n$. According to the diagram, there is at least one $k$ in the conjugate partition, so $a_j\ge1$. From this we may deduce that, since $k$-part partitions are in bijection with all-parts-of-size-$\le k$ partitions,

$$\sum_{n=1}^\infty p(k,n)x^n=\left(\sum_{a_1=0 }^\infty x^{a_1}\right)\left(\sum_{a_2=0}^\infty x^{a_2 2}\right)\cdots\left(\sum_{a_k=1}^\infty x^{a_k k}\right)=$$

$$=\frac{1}{1-x}\frac{1}{1-x^2}\cdots\frac{1}{1-x^{k-1}}\frac{x^k}{1-x^k}=\frac{x^k}{(1-x)(1-x^2)\cdots(1-x^k)}.$$

Of course here we are looking at $k=3$.