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Let $p_\neq (n)$ be the number of all partitions of $n$ such that all summands are distinct (for example $p_\neq (6)=4$).

How do we show that $p_\neq (n) \leq e^{2\sqrt n}$?

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    $\begingroup$ My first thought is that the largest summand needs to be greater than $2\sqrt n$ because the sum of all the numbers up to that is too small. So we have $p_{\neq}(n)=\sum_{k=2\sqrt n+1}^np_{\neq}(n-k)$. I don't know if that works but the coincidence of the exponent is promising. $\endgroup$ – Ross Millikan Dec 29 '18 at 5:00
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    $\begingroup$ Lots of information here Don't see the result you're looking for, though. $\endgroup$ – saulspatz Dec 29 '18 at 5:57

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