# 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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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." –  anon Feb 8 '12 at 20:09
Also, if this is homework, there's nothing wrong with that, but please add the homework tag. –  Gerry Myerson Feb 8 '12 at 23:57

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. –  Gerry Myerson Feb 8 '12 at 23:55
@Gerry: You're right, it wasn't helpful like I felt while posting it. I've taken a different tact instead. –  anon Feb 9 '12 at 0:33
@Gerry: By ‘no 1 term’ anon meant ‘no constant term’. (I see that the answer has now been expanded to make this clear.) –  Brian M. Scott Feb 9 '12 at 0:36
This version’s very clear (and much more helpful). –  Brian M. Scott Feb 9 '12 at 0:37
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 $a_2$? –  Gerry Myerson Feb 9 '12 at 2:45
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