# Average length of partitions

I'm wondering if there's a known asymptotic for an average number of terms for partitions of a given number? (I mean, given all the partitions of a given number, how many terms do they have on average?)

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.

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 $$Y_1=O\left(\frac{\sqrt{6n}}{\pi}\log \frac{\sqrt{6n}}{\pi}\right)=O(\sqrt{n}\log n)$$

• Could you please explain how this translates into asymptotics for $n\propto\ldots$? Should I solve the inequality for $Y_1$, replacing $v$ with $n/2$?.. – mavzolej Apr 7 at 12:55
• Thank you so much!! – mavzolej Apr 7 at 20:13

Here is a graph of the average number of partitions as a function of the source number: