Average length of partitions

This was apparently answered in 1941 by Erdos and Lehner see here; however, I only have access to a 1995 paper by Fristedt, which cites the result as (2.2):

$$\lim_{n\to\infty} P_n \left(\frac{\pi}{\sqrt{6n}} Y_1 - \log \frac{\sqrt{6n}}{\pi}\le v\right)=e^{-e^{-v}}$$

Here, $P_n$ denotes probability measure, with all partitions of $n$ equiprobable. $Y_1$ denotes the size of the largest part of a partition. By considering conjugation, the average size of the largest part of a partition equals the average number of parts of a partition.

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More details, by request. This formula gives more than just the average of $Y_1$, it gives a lot of information about the probability distribution of $Y_1$. For example, taking $v=0$, we get $Y_1 \le \frac{\sqrt{6n}}{\pi} \log \frac{\sqrt{6n}}{\pi}$ with limiting probability $e^{-1}\approx 0.37$. Taking instead $v=2$, we get $Y_1\le 2\frac{\sqrt{6n}}{\pi}+ \frac{\sqrt{6n}}{\pi}\log \frac{\sqrt{6n}}{\pi}$ with probability $e^{-e^{-2}}\approx 0.87$. Hence, subtracting, we get $$Y_1\in \left[\frac{\sqrt{6n}}{\pi}\log \frac{\sqrt{6n}}{\pi},2\frac{\sqrt{6n}}{\pi}+ \frac{\sqrt{6n}}{\pi}\log \frac{\sqrt{6n}}{\pi}\right]$$ with probability $e^{-e^{-2}}-e^{-e^{-0}}\approx 0.51$.

If you just want the answer to the original question, it is $$Y_1=O\left(\frac{\sqrt{6n}}{\pi}\log \frac{\sqrt{6n}}{\pi}\right)=O(\sqrt{n}\log n)$$