For $n$ an integer, let $a_n$ be the number of ways in which one may partition the set $\{1, \ldots, 2n \}$ in two parts with:

- the same number of elements: $n$
- and the same sum: $2n(2n+1)/4$.

Find an equivalent of $a_n$.

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Put in probabilistic terms: let $(B_i)_{1\le i \le 2n}$ be independent Bernoulli($1/2$) random variables. Find an equivalent of the probability:

$b_n = \mathbf P\left[\sum_{i=1}^{2n} B_i =n , \sum_{i=1}^{2n} i B_i = 2n(2n+1)/4\right]$

The two quantities are related by $a_n= 4^{-n} b_n$.

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Edit: $n$ has to be even for the above quantities to be non-zero; otherwise the half-sum $2n(2n+1)/4$ is not an integer.

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The (numerically verified) conjecture is that, for some constant $c$, and for **even** $n$:

$$a_n \sim c \cdot \frac{4^n}{n^2}$$

Using bravely local central limit theorem (which amounts to assume the two random variables $(\sum_{i=1}^{2n} B_i, \sum_{i=1}^{2n}i B_i)$ form a gaussian vector), computing the mean vector and covariance matrix of that vector - the two random variables are indeed correlated in the limit - and taking into account aperiodicity (edit: no need!), one might even guess that

$c= \frac{\sqrt{3}}{\pi}$

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How to make this rigorous?

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(The question was raised in RMS as R772, where the conjecture is also stated; I feel there is some interest to share this with a broader audience.)

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