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Suppose we have a natural number $n \ge 0$.

Given natural numbers $\alpha_1,\ldots,\alpha_k$ such that

  • $k\le n$
  • $\sum_i \alpha_i = n$

what is the maximum value that $\Pi_i \alpha_i$ can take?

I'm quite sure that there is a theorem telling me the result, but I cannot find it. For sure an upper bound is $n^k$ but I'm searching for a real upper bound. I'm pretty sure that upper the bound should be $n^2$, but I don't know I could prove it.

  • 1
    The upper bound is much bigger than $n^2$. Suppose $n$ is even. Then $n = 2+2+\cdots+2$, and the product is $2^{n/2}$. If $n$ is odd you can take $n = 2+2+\cdots+3$ and get $3\cdot2^{(n-1)/2}$.2012-03-27
  • 0
    If you partition $n$ into about $\sqrt n$ parts of size about $\sqrt n$, then the product is $n^{{1\over2}{\sqrt n}}$, which is pretty big, even bigger than $2^{n/2}$ for certain $n$.2012-03-27
  • 0
    [OEIS A000792](https://oeis.org/A000792)2017-09-12

2 Answers 2