Find a closed form for the generating function for the number of partitions of n into 3 parts.
Find a closed form for the generating function for the number of partitions of n into 3 parts
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2Do they have to be distinct parts? Also, don't just transcribe the question like that - then it makes it look like you're commmanding us as if you're our teacher or at least don't have any intention of talking with us. Say, "I have this problem, `problem`, and here are my thoughts." – 2012-02-08
1 Answers
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 $k=3$.
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0@Gerry: Thanks, I was thinking about the Ferrers diagrams incorrectly. – 2012-02-09