integers which are sums of binomial coefficients: $\sum_i {n \choose k_i}$ Let $n$ be an integer. For $S$ a subset of $\{0,\dots,n\}$, define
$$m(S) = \sum_{k \in S} {n \choose k}.$$
Let $M_n$ be the set of integers of the form $m(S)$ for all sets $S \subset \{0,\dots,n\}$. Obviously $M_n \subset \{0,\dots,2^n\}$ and for $n=1,2,3$ this inclusion is an equality. For $n \geq 4$ however, this inclusion is not an equality, as for instance $n-1$ is not in $M_n$. 

What can be said about $M_n$? Is there a simple way to recognize if a number less than $2^n$ is in $M_n$? Can we compute or estimate the proportion $|M_n|/(2^n+1)$ of elements of $\{0,\dots,2^n\}$ that are in $M_n$?

I ask this question mainly out of curiosity, but elements of the form $m(S)$ appeared recently in paper I am working on. 
 A: As Gerry notes, it is in the OEIS if we also allow the empty sum of $0$. As I commented, the description does seem a bit off. 
The counts (out to $n=30$) do indeed seem to be close to the upper bound of  $3^{(n+1)/2}$ (for odd $n$) or $2\ 3^{(n-2)/2}$ (for $n$ even). The bound is obtained for $n=9$ and seldom goes much below half of that. The counts $M_n$ for $n=18,19,20$ are $21345, 22005, 64917$ so ratios of 1.0309 and then 2.95. In general prime $n$ results in a relatively small increase although $n=19$ is much more dramatic than $23$ or $29.$ In that case of $n=19$ we have only about $30\%$ of the upper bound.
The counts are always odd (including $1$ for $n=0$) with the exception $M_1=2$. This is because (as observed) the attainable values come in pairs $m,2^n-m$ except that the attainable value $2^{n-1}$ is unpaired. ($n=0,1$ are special)
The values mod $3$  are 
$2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 1$
I suppose the high number of multiples of $3$ can be explained by the observation that if we can attain a sum of $m$ without making use of $\binom{n}{1}$ or $\binom{n}{n-1}$ then we can also obtain $m+1$ and $m+2$. Often the values $m$ of the first type are spread out enough that distinct values are more than $2$ apart (for example when $n \gt 2$ is prime and all these sums are multiples of $n$). An exception is $n=10.$ There we have $250=\binom{10}{1}+\binom{10}{3}+\binom{10}{7}$ and $252=\binom{10}{5}$ among others.
