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Fix a prime $p$, and $k$ a natural number. The question is then:

How many partitions of $p^k$ are there into powers of $p$?

So, for instance, if $p = 2$ and $k = 2$, there are 4, namely (4), (2, 2), (2, 1, 1) and (1, 1, 1, 1).

This seems like it could be a hard problem in number theory, but I don't really know much number theory (I'm a topologist), and so don't have a feel for such things.

Marc van Leeuwen
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Peter Nelson
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  • http://oeis.org/A002577 is p=2 – Jack Schmidt Jul 12 '13 at 02:05
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    http://oeis.org/A018818 is also cool (the equivalent question for all positive integers, not just prime powers) – Jack Schmidt Jul 12 '13 at 02:08
  • http://www.jstor.org/stable/2325075 has some bounds on the number – Jack Schmidt Jul 12 '13 at 02:14
  • http://www.ams.org/mathscinet-getitem?mr=469864 (online at http://www.dli.gov.in/rawdataupload/upload/insa/INSA_2/20005a84_1276.pdf) is related too, take m=n to be a prime power. if you are looking at conjugacy classes in p-groups, this could provide a very nice result when the centralizer sizes or conjugacy class sizes are bounded. – Jack Schmidt Jul 12 '13 at 02:22
  • (That's probably all I've got to contribute. Number theory isn't my thing, but this is a nice question.) – Jack Schmidt Jul 12 '13 at 02:23
  • Your example is confusing, since $2\cdot2=2+2$. Since you want an additive partition, I might suggest giving an example with $p=3$. – 2'5 9'2 Jul 12 '13 at 04:23
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    Let $f(n)$ count the number of partitions of $n$ into parts consisting of powers of $p$. Observe that in each such partition, the number of parts equal to $1$ is itself congruent to $n$ mod $p$. The number of such partitions with precisely $m$ ones (given $n\equiv m$ mod $p$) is $f\left(\frac{n-m}{p}\right)$. Therefore $$f(n)=f\left(\frac{n-r}{p}\right)+f\left(\frac{n-r}{p}-1\right)+f\left(\frac{n-r}{p}-2\right)+\cdots$$ is a recursive formula satisfied by $f$, where $0\le r<p$ is the residue of $n$ mod $p$. – anon Jul 12 '13 at 04:25
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    Furthermore a generating function for $f(n)$ is $$\sum_{n=1}^\infty f(n)x^n=\prod_{n=0}^\infty\left(1-x^{p^n}\right)^{-1}.$$ For a group theory interpretation, it is also the case that $f(n)$ counts the number of ${\bf Z}(p^\infty)$-sets of order $n$ (see group action), where ${\bf Z}(p^\infty)$ stands for the Prüfer $p$-group. @alex.jordan I don't see what $2\cdot2=2+2$ has to do with the question. – anon Jul 12 '13 at 04:26
  • @anon When I saw the question, at first I thought OP meant multiplicative partitions like $2^2$, $2^1\cdot2^1$. Not additive partitions like $2^2$, $2^1+2^1$, ... I'm just saying that an example with $3$ in place of $2$ would help prevent anyone else from that misinterpretation. – 2'5 9'2 Jul 12 '13 at 04:30
  • I think the examples (4), (2,2), (2,1,1), (1,1,1,1) collectively being highlighted as partitions of $4$ is perfectly sufficient to dispel that potential confusion. | Oops, I was confusing subgroups with quotients. I mean ${\bf Z}_p$-sets, where ${\bf Z}_p$ denotes the $p$-adic integers. Probably not relevant anyway. – anon Jul 12 '13 at 04:31

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My question 428904, "how many ways to make change, asymptotics" asked the same question and I found the answer in a 1948 paper in Indagationes Math X "On Mahler's Partition Problem". NG De Bruijn cited Karl Mahler's 1940 paper from Journ. London Math Soc. "On a special functional equation". His formula, for base $p=r$ and number $N=rh$ was $$ \log C(rh)=\frac{1}{2\log r}\left(\log\frac{h}{\log h}\right)^2 +\left(\frac{1}{2}+\frac{1}{\log r}+\frac{\log\log r}{\log r}\right)\log h\\-\left(1+\frac{\log\log r}{\log r}\right)\log\log h+O(1)$$ De Bruijn went on to calculate the O(1) which was a periodic but not differentiable function.
$C(rh)$ obeys the functional equation $C(rh)=C(rh-r)+C(h)$.

Empy2
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