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An integer partition $\lambda$ of $n$ can be represented as a tuple $\lambda = (a_1, \cdots, a_n)$ in $\mathbb{Z}^n$ of $n$ nonnegative numbers $a_1 \geq a_2 \geq \cdots \geq a_n \geq 0$ such that $\sum_{i=1}^n a_i = n$. Note that I use exactly $n$ numbers and allow zeroes in the entries of $\lambda$.

Now take all partitions $\lambda$ of $n$, consider them as points in $\mathbb{Z}^n$ and take their arithmetic mean $\lambda_\text{avg}$ in $\mathbb{R}^n$. How should $\lambda_\text{avg} = (b_1, \cdots, b_n)$ look like?

I guessed that it may look like the limiting shape of a random partition as calculated by Vershik (as explained in here). The heuristic calculation tells that, after normalizing the shape by $\sqrt{n}$ so that we plot the points $(i / \sqrt{n}, b_i / \sqrt{n})$ for $1 \leq i \leq n$, then the points should be near the curve $C$ described by $e^{-\pi x / \sqrt{6}}+e^{-\pi y / \sqrt{6}} = 1$. Below, you can compare $\lambda_\text{avg}$ (normalized) and the curve $C$. While they seem to loosely match, they don't seem to match completely.

Average partition calculated in Mathematica

Are there any results of this kind, or are there any possible suggestions to literature for which I can investigate what is happening further?

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    $\begingroup$ For $i$ comparable to $\sqrt{n} $ the average of $a_i$ is given by Vershik's curve. Simply because almost all $a_i$`s are given by Vershik, and others are uniform $O(\sqrt{n}) $, thus they do not contribute to the average $\endgroup$ Commented Sep 8 at 11:08
  • $\begingroup$ @FedorPetrov I think I don't really know how much a general partition diverges from Vershik's curve. I should confess that I don't understand Vershik's paper well yet. Could you elaborate a bit more if possible? $\endgroup$ Commented Sep 8 at 11:34
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    $\begingroup$ It suffices that for fixed constants $0<c_1<c_2$ and fixed $\varepsilon>0$ the proportion of bad partitions goes to 0. Here bad means that there exists $k\in (c_1\sqrt{n},c_2\sqrt{n})$ such that $a_k$ differs from Vershik's curve prediction at least by $\varepsilon \sqrt{n} $. $\endgroup$ Commented Sep 8 at 14:37

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