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

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    Do 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

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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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    Huh? There are a lot of geometric series with no term equal to 1, e.g., $2+6+18+54+\cdots$, or $1/2+3/2+9/2+27/2+\cdots$, and the ones like these that are just numbers don't have any coefficients, even if you raise them to the $k$th power. You are far too subtle for me. But I have upvoted your comment - OP has to do better.2012-02-08
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    @Gerry: You're right, it wasn't helpful like I felt while posting it. I've taken a different tact instead.2012-02-09
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    @Gerry: By ‘no 1 term’ anon meant ‘no constant term’. (I see that the answer has now been expanded to make this clear.)2012-02-09
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    This version’s very clear (and much more helpful).2012-02-09
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    Why do the sums start at 1, and not at zero? I guess $a_3$ has to be positive (for $k=3$), but why $a_1$ and $a_2$?2012-02-09
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    @Gerry: Thanks, I was thinking about the Ferrers diagrams incorrectly.2012-02-09