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 $p_\neq (n) \leq e^{2\sqrt n}$?

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ Lots of information here Don't see the result you're looking for, though. $\endgroup$ – saulspatz Dec 29 '18 at 5:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.