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I am interested in knowing whether an exact formula (analogous to the Hardy-Ramanujan-Rademacher formula for $p(n)$) for the number of partitions of a positive integer into k parts is known. I tried searching on the internet but could not find such a formula. Can someone confirm whether such a formula is known and provide a link to its proof if possible.


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You want $k$ equal parts? What about $[k\mid n]$ (using the Iverson bracket). – Marc van Leeuwen Nov 19 '12 at 9:26
No I do not form not equal parts. I am looking for the analogous formula for number of compositions in k parts which is $\binom{n-1}{k-1}$. – Shahab Nov 19 '12 at 9:31
You have a cryptic way of expressing yourself. Saying you "do not form not equal parts" means you do form equal parts only. I dont't think that is what you mean, but it is what you say. And the question says equal parts too. – Marc van Leeuwen Nov 19 '12 at 9:37
I'm sorry. I do not want equal parts. That was a mistake in typing. – Shahab Nov 19 '12 at 9:45
up vote 1 down vote accepted

In Chapter 6 of Andrews and Eriksson, Integer Partitions, formulas are given for $k=1,2,3,4,5$. Full proofs are given for $k=1,2,3,4$, while $k=5$ is left as an exercise.

[EDIT: By the way, this is a very nice little book. "The aim of this introductory textbook is to provide an accessible and wide-ranging introduction to partitions, without requiring anything more of the reader than some familiarity with polynomials and infnite series."]

Actually, the function studied in that text is a bit different from what you want, but it's easy to get back and forth between the two functions. The number of partitions into (exactly) $k$ parts equals the number of partitions with greatest part $k$. The text studies the number of partitions with greatest part at most $k$. But partitions of $n$ with greatest part $k$ correspond to partitions of $n-k$ with greatest part at most $k$.

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Thank you. Is a formula for a general $k$ known on the lines of the analogous formula for compositions $\binom{n-1}{k-1}$? – Shahab Nov 22 '12 at 2:51
Ask me again, after you have looked at Andrews and Eriksson. – Gerry Myerson Nov 22 '12 at 5:32
I have looked at the reference. The basic idea is using the corresponding generating function. However I am still in doubt whether a general formula, howsoever complicated but in closed form, is known or not? – Shahab Nov 22 '12 at 11:27
Well, it seems unlikely. They have to use different methods for different $k$, and the formulas look very different, and they don't say anything about there being a general formula. I deduce from that that there isn't a general formula, but I recognize that's not a proof. – Gerry Myerson Nov 22 '12 at 11:51

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