# 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.

-

2
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

1
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

## 1 Answer

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 $k=3$.

-

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
show 1 more comment
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.953376829624176, "perplexity_flag": "head"}