I think VSJWe will prove that $\lim_{n\to\infty} a_i^*/n = 1/2^{i+1}$.
Also let $s_*=\sum a_i^*$
To do this, note that if $a^*$ is right aboutoptimal, adding $1$ to $a_i$ and $a_j$ and subtracting one from $a_{i+j}$ must reduce the limits but wrong aboutnumber of unordered partitions, so
$$ \frac{ (s^*+1) (a_{i+j} ^*)} { (a_i^*+1)(a_j^*+1) } \leq 1$$
or if $j=i$,
$$ \frac{ (s^*+1) (a_{2i}^* )} { (a_i^*+1)(a_i^*+2) } \leq 1$$
Similarly, subtracting one from $a_i$ and $a_j$ and adding one to $a_{i+j}$ must also reduce the number of orderedunordered partitions, so
$$ \frac{ a_i^* a_j^*} { s^* (a_{i+j}^*+1)} \leq 1$$
or if $j=i$,
$$ \frac{ a_i^* (a_i^*-1)} { s^* (a_{2i}^*+1)} \leq 1$$
Using these inequalities, one could probably give explicit bounds for the $a_i^*$ and use that to determine the convergence rate, but that sounds messy, so we can prove convergence using soft methods instead.
If $\lim_{n\to\infty} a_1^*/s_*=1/2$, we can conclude that $\lim_{n\to\infty} a_i^*/s_*=1/2^i$ $n_*/s_*= \sum_i i a_i^*/s_i^*$, so if we can interchange the sum and the limit we get $n_*/s_*=2$ and the desired result.
Conversely, if $\lim_{n\to\infty} a_1^*/s_*$ does not exist or exists and isn't $1/2$, a subsequence exists with limit $x\neq 1/2$, so by these inequalities the limit of $a_i^*/s_*= x^i$ Since $1 = \sum_i a_i^*/s_i^*$, if we can interchange the limits we get $1=x/(1-x)$ and $x=1/2$ and a contradiction.
So the key step is to interchange the sum and the limit. We will do this using the Dominated Convergence theorem and the estimate $a_i^*/s_* = O(1/i^3)$, which makes both sums converge.
We obtain the estimate as follows: Setting $j=1$ in the first inequality, we see that $a_{i+1}^*< a_i^*+1$, so $a_{i+1}^* \leq a_i^*$, so the function is nonincreasining Thus $a_i^* \leq s^*/i$. So
$$a_{2i}^* \leq \frac{(s^*/i+1) (s^*/i+2)}{(s^*+1)}$$
and
$$a_{3i}^* \leq \frac{ (s^*/i+1) (s^*/i+2)(s^*/i+3)}{ (s^*+1) (s^*+2) }$$
Also $a_{3i}^*=0$ unless $s^* \geq 3i$. So $a_{3i}^*/s^*= O(1/i^3)+O(1/i^2s^*)+O(1/is^{*2})+O(1/s^{*3})=O(1/i^3)$.
Then the same holds for every $a_i$ and we have the desired result.
To work on this problem, it is helpful to have a method to generate a random ordered partition. This is very easy - flip $n$ coins, and divide them into streaks. So if the flips are HTTHHHTHTTTTH, the corresponding ordered partition is $1+2+3+1+4+1$, and the corresponding unordered partition is $1+1+1+2+3+4$. Since each ordered partition is equally likely to occur (it comes from exactly two sequences of coin flips), this question is equivalent to the question: Which unordered partition is most likely to occur, and what is its probability of occurrence?
The law of large numbers suggestsCentral Limit Theorem shows that the most likely unordered partition is exactly what VSJ said - the maximum likelihood $a_i$ should be close to the expected value of $a_i$, which is $n/2^{i+1}$.
But since the probability of any individual unordered partition occurring, even the most likely one, is clearlygetting a given $o(1)$, the number of occurrence of any unordered partition$a_1,a_2,\dots, a_k$ is $o(2^n)$, and thus can't be $c2^n$.
More precisely$O(1/n^{k/2})$, the Central limit Theorem shows thatso the probability of getting a given $a_1,a_2,\dots, a_k$ isunordered partition must be $O(1/n^{k/2})$ for all $k$, so the number of ordered partitions corresponding to a given full unordered partition must be $o(2^n/n^{k/2})$$O(2^n/n^{k/2})$ for all $k$.