Coordinate-wise average of all integral partitions of $n$

Question

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.

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

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 — Fedor Petrov
– Fedor Petrov, Commented Sep 8, 2024 at 11:08
@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? — Jineon Baek
– Jineon Baek, Commented Sep 8, 2024 at 11:34
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} $. — Fedor Petrov
– Fedor Petrov, Commented Sep 8, 2024 at 14:37

Stephen DeSalvo · Accepted Answer · 2024-12-18 08:56:31Z

This is a well-studied problem with multiple types of answers.

$n=50$ is quite small for observing the asymptotic limit shape. Uniform sampling of integer partitions is quite easy and scalable, and will more closely match the limit shape below for modest values of $n$ and sample sizes. If you really want to dig deeper on the large deviations see the paper by Dembo-Vershik-Zeitouni, Large Deviations for Integer Partitions (1998).

The comprehensive answer to your question lies in 2 principle papers:

Bert Fristedt, The Structure of Random Partitions of Large Integer (1993).
Boris Pittel, On a Likely Shape of the Random Ferrers Diagram (1997).

Fristedt worked with Prohorov Distance whereas Pittel used Total Variation Distance. They are both worth reading thoroughly.

One can look to earlier results, e.g., Erdos-Lehner, The Distribution of the Number of Summands in the Partitions of a Positive Integer (1941), for the largest part size asymptotics (i.e., $b_1$ in your notation).

The parameterization in terms of decreasing part sizes is unnatural for most limit shape results. Instead, we typically use counts, as in $C_i(n)$ = # parts of size~$i$, for $i=1, \ldots, n$, and consider the joint distribution $(C_1(n), \ldots, C_n(n))$, where $\sum_{i=1}^n i\, C_i(n) = n$.

There is another comprehensive paper which covers the more general area of partitioning and contains many useful heuristics and references: Arratia-Tavare, Independent Process Approximations for Random Combinatorial Structures (1994).

Finally, for partitions under various restrictions and some nontrivial bijections which allow for transforming limit shapes geometrically, see DeSalvo-Pak, Limit Shapes via Bijections (2018).

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