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 $a\_n$.

—

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 $a\_n= 4^{-n} b\_n$.

—

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.

—

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 

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

—

How to make this rigorous? 

—

(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.)