$\sum_{k=1}^n m(k)$, where $m(k)$ is defined by $2^{m(k)} || k$. I'm looking at the sum:
$$f(n) = \sum_{k=1}^n m(k),$$
where $m(k)$ is defined by $2^{m(k)} || k$, i.e. $2^{m(k)}$ is the largest power of $2$ that divides $k$. For example, we have $f(8) = 0+1+0+2+0+1+0+3 = 7$. Here's a table of $n$, $m(n)$, and $f(n)$ for a few small $n$:
n     m(n)  f(n)
1     0     0
2     1     1
3     0     1
4     2     3
5     0     3
6     1     4
7     0     4
8     3     7

The values $m(n)$ in the table are the last $m(k)$-value for the given $n$, i.e. the value of $m(k)$ when $k=n$. I haven't spent much time trying to figure this out, mostly because it's a funny little problem and I thought I'd share it. It seems to me that we have $f(2^j) = 2^j-1 \forall j$ and $f(n)\sim n$.
 A: Let $m(k)$ be the exponent of $2$ occurring in $k$, i.e. $2^{m(k)}\|k$.
One quickly deduces $$\tag1f(2n+1)=f(2n)=n+f(n)$$ as only the even numbers contribute and $2k$ contributes one more to $2n$ and $2n+1$ than $k$ does to $n$, i.e. $m(2k)=m(k)+1$.
Since $f(1)=m(1)=0$, we find $f(3)=f(2)=1$, $f(5)=f(4)=3$, and so on.
The sequence goes
$$ 0, 1, 1, 3, 3, 4, 4, 7, 7, 8, 8, 10, 10, 11, 11, 15, 15, 16, 16, 18, 18, 19, 19, 22, 22, 23, 23, 25, 25, 26, 26,\ldots$$
And according to OEIS, this equals $m(n!)$ and also $n$ minus the number of $1$s in the binary expansion of $n$. One easily verifies that these two discriptions indeed follow the recursion formula $(1)$.
A: There is a name for your "m" function, it is called the 2-adic valuation, and is usually denoted by $\nu_2$. 
It is easy to show that 
$$f(2^j)=2^{j}-1$$
Indeed, for $1 \leq k \leq j-1$ there are exactly $2^{j-k-1}$ numbers between $1$ and $2^j$ with $\nu_2(n)=k$, namely
$$\{ 2^k\cdot 1, 2^k \cdot 3, 2^k \cdot 5,..., 2^k\cdot (2^{j-k}-1) \}$$
When $k=n$ there is exactly $1$ number with $\nu_2(n)=k$.
Then, by rearranging the terms of your sum into groups of equal terms you get
$$f(2^j)=j+\sum_{k=1}^{j-1} k 2^{j-k-1}$$
It is easy to show that this sum is exactly $2^j-1$.
Edit With the new fix, you need to add a 1 to every single term, thus the sum increases by exactly $2^j$.
A: As a bit of interesting trivia concerning $f(n)$ we see that the Mellin-Perron summation formula applies, giving
$$ f(n) = \frac{1}{2} v_2(n) + 
\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} L(s) n^s \frac{ds}{s}
\quad\text{where}\quad
L(s) = \sum_{n\ge 1} \frac{v_2(n)}{n^s}.$$
But we have 
$$ L(s) = \sum_{m\ge 1} \frac{v_2(2m)}{(2m)^s} =
\frac{1}{2^s} \sum_{m\ge 1} \frac{1+v_2(m)}{m^s} =
\frac{1}{2^s} (\zeta(s) + L(s)),$$
so that
$$ L(s) = \frac{\zeta(s)}{2^s-1},$$
giving
$$  f(n) = \frac{1}{2} v_2(n) + 
\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \frac{\zeta(s)}{2^s-1} n^s \frac{ds}{s}$$
We can apply the Cauchy Residue Theorem to this integral,
treating first the poles at $s=1$ and $s=0$, which yields
$$f(n) \sim \frac{1}{2} v_2(n) 
+ n - \frac{1}{2}\log_2 n - \frac{1}{4} - \frac{1}{2}\log_2 \pi.$$
The above approximation is excellent, giving for example $f(200) \sim 196.6023238$ when the correct value is $197$, or $f(411) \sim 405.5827546$ when the correct value is $405.$
Finally, $f(1000)$ is $994$ when the approximation gives $995.4413598.$
The complete asymptotic expansion is obtained by including all the poles on the imaginary axis at $\frac{2\pi i k}{\log 2}$, with $k$ a nonzero integer. These poles are all simple. We have
$$\operatorname{Res}\left(\frac{1}{2^s-1}; s = \frac{2\pi i k}{\log 2}\right) =
\frac{1}{\log 2}.$$
Now introducing $$\rho_k = \frac{2\pi i k}{\log 2}$$ we finally obtain
$$f(n) \sim \frac{1}{2} v_2(n) 
+ n - \frac{1}{2}\log_2 n - \frac{1}{4} - \frac{1}{2}\log_2 \pi +
\frac{1}{\log 2} 
\sum_{k\in \mathbb{Z}\setminus 0} \frac{\zeta(\rho_k)}{\rho_k} e^{2\pi i k\log_2 n},$$
where the sum is a Fourier series in $\log_2 n$ that describes the fluctuations.
Taking the terms up to $|k|=7$, we obtain $f(200) \sim  197.0498970.$ With $34$ terms, we obtain $f(411)\sim 405.3130370$ and $f(1000) \sim 993.9869087.$
A: One simplification of the sum is $f(n)=\sum\limits_{k=1}^\infty \left\lfloor\frac{n}{2^k}\right\rfloor$
(obviously, only terms for $k\leq \log_2(n)$ are going to be non-zero). This can be evaluated very easily on the computer, since each consecutive term corresponds to dividing the previous one by 2 and ignoring any remainder.
