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Well, that's all there is to it. To repeat myself, what's the greatest value you can find for a number $k$ for $P(n, k)$, where $n$ is fixed?

Please answer in algebraic, well, whatever level (I hope you understand). I'm still not out of high school! Anyway, thanks, Mr. Answerer!

Oh yes, there's an approximation for $P(n,k)$ - something like $\frac{1}{2\pi n} \left( \frac{e^2 n}{k^2} \right)^k$ or such - I'm not sure...

Edit:

The numbers are tabulated at OEIS A002569. Some references are given. At OEIS A046155 the value of $k$ maximizing $P(n,k)$ is given for all $n≤85$.

Um, that's right, Mr. Gerry. There seems to be a pattern in it, anyway...

Edit 2

Would this PDF be of any help?

Edit 3: Yep Gerry, that's right. I've done my research on the topic earlier (and didn't bother to mention it), and found out that for $P(100, k)$, 7 was the greatest value, even superceding $k=100(1/2)=50$. By the way, that formula is here (from http://mathoverflow.net/questions/72418/what-are-the-best-known-bounds-on-the-number-of-partitions-of-n-into-exactly-k);

$$P(n, k)\sim\frac{1}{2\pi n} \left( \frac{e^2 n}{k^2} \right)^k$$

And the one I gave earlier was a simplification of this equation. Also, I know about that triangular-looking grid for $P(n,k)$. But then, what is derived from it?

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You can partition $n$ into up to $n$ positive parts, but no more. –  Henry Mar 8 '12 at 18:05

1 Answer 1

The numbers are tabulated at http://oeis.org/A002569. Some references are given. At http://oeis.org/A046155 the value of $k$ maximizing $P(n,k)$ is given for all $n\le85$.

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