# An upper bound for integer partitions with unique summands

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

• 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. – Ross Millikan Dec 29 '18 at 5:00
• Lots of information here Don't see the result you're looking for, though. – saulspatz Dec 29 '18 at 5:57