What is limiting behavior for the median of $2^X+2^Y$, where $X$ and $Y$ are iid binomial variables? Suppose $X_n$ and $Y_n$ are iid variables, each distributed binomially as $B(n,1/2)$.
Let $M_n$ be the median of $2^{X_n}+2^{Y_n}$. Empirically
$$\sqrt{2}\, \le \left(\frac{M_n}{2}\right)^{1/n}\!\! \le \frac32$$
What can we say about $\lim_{n\to\infty}\left(\frac{M_n}{2}\right)^{1/n}$, whose graph looks like the below?



*

*Does the limit exist?

*Is the limit properly between $\sqrt{2}$ and $3/2$?

*Does the limit have a nice formula?


The sequence is in the OEIS as A288347, which has an interpretation, methods of calculation, and some calculated values.
The first inequality above says that $\text{median}(2^{X_n})+\text{median}(2^{Y_n})\le \text{median}(2^{X_n}+2^{Y_n})$, which looks easy enough to prove.
The second inequality above says that $\text{median}(2^{X_n}+2^{Y_n}) \le \text{mean}(2^{X_n}+2^{Y_n})$, which is a discretized special case of another question, and looks hard to prove.
 A: The limit is $\sqrt2$. Indeed, let $(h_n)$ be such that $h_n/\sqrt n\to\infty$ but $h_n/n\to0$ (as $n\to\infty$). 
Then, by the central limit theorem, with probability $\to1$ we have $n/2-h_n\le X_n\le n/2+h_n$ and $n/2-h_n\le Y_n\le n/2+h_n$ and hence 
\begin{equation}
 2^{n/2-h_n}\le 2^{X_n}+2^{Y_n}\le2^{n/2+h_n+1}.
\end{equation}
So, eventually (for all large enough $n$)
\begin{equation}
 2^{n/2-h_n}\le M_n\le2^{n/2+h_n+1}.
\end{equation}
So, 
\begin{equation*}
 \Big(\frac{M_n}2\Big)^{1/n}\to2^{1/2},
\end{equation*}
as claimed. 
A: I've accepted Iosif's answer, and am rewriting it here in a way that is more intuitive for me:
Let $a$ satisfy
$$P[X<a]<\dfrac{1}{\sqrt{2}}<P[X\le a]$$
Using the normal approximation to the binomial, and the $(1/\sqrt{2})^{th}$ quantile of the normal,
$a \sim \frac12 (n + .545\sqrt{n})$.
Then in two variables:
\begin{align}
P[2^X+2^Y\le 2^a]
& = P[X<a]P[Y<a] \\
& < 1/2 \\
& < P[X\le a]P[Y\le a]\\
& = P[2^X+2^Y\le 2^{a+1}]
\end{align}
And for the median:
\begin{align}
2^a  &< \ \ \ \ M_n \ \ \ \ \ \ \, <\, 2^{a+1}\\
2^{(a-1)/n} &<  \left(\frac{M_n}{2}\right)^{\!1/n\!} <\, 2^{a/n} \sim \sqrt{2}\,(1.459^{1/\sqrt{n}})\\
\end{align}
So the desired limit is indeed $\sqrt{2}$, but the graph above would not go below $1.42$ until $n\sim 8558$.
