sum of floor ("weighted" divisor summary function) is there a closed form for 
$$
\sum_{i=1}^N\left\lfloor\frac{N}{i}\right\rfloor i
$$
Or is there any faster way to calculate this for any given N value?
 A: Maybe you can simplify something from here $n = m\lfloor \frac{n}{m}\rfloor + ( n \mod m)$, i.e. you can factor $N$ and this may simplify the sum for $N$ not too large.
A: This answer does not give closed form formula (for reasons mentioned in comments), but it gives a way to calculate the sum more efficiently than just going through all $i\leq N$.
The trick is that values of $\left\lfloor\frac{N}{i}\right\rfloor$ do not change for long ranges of $i$, for example:
$$
\left\lfloor\frac{10}{6}\right\rfloor = \left\lfloor\frac{10}{7}\right\rfloor = \dots = \left\lfloor\frac{10}{10}\right\rfloor = 1.
$$
This can be used to skip over these and sum them all in one iteration. Since you have
$$
\left\lfloor\frac{N}{i}\right\rfloor = k \Leftrightarrow \frac{N}{k+1} < i \leq \frac{N}{k} 
$$ 
all you need to do is iterate from current $i$ to next different value which will be $\left\lfloor\frac{N}{k}\right\rfloor+1$. 
For example one can define
\begin{align*}
k_n&=\left\lfloor\frac{N}{i_{n-1}}\right\rfloor\\
i_n&=\left\lfloor\frac{N}{k_n}\right\rfloor+1\\
s_n&=s_{n-1}+\sum_{i=\left\lfloor\frac{N}{k_n+1}\right\rfloor+1}^{\left\lfloor\frac{N}{k_n}\right\rfloor} k_n = s_{n-1}+\frac{1}{2}k_n\Bigl(\left\lfloor\frac{N}{k_n+1}\right\rfloor+\left\lfloor\frac{N}{k_n}\right\rfloor+1\Bigr)\Bigl(\left\lfloor\frac{N}{k_n}\right\rfloor-\left\lfloor\frac{N}{k_n+1}\right\rfloor\Bigr)
\end{align*}
with $k_0=0,s_0=0,i_0=1$. Then just iterating while $i_n \leq N$, resulting sum is then $s_{n+1}$. 
Example for $N=10$:
\begin{align*}
k_0&=0&s_0&=0&i_0&=1\\
k_1&=10&s_1&=10&i_1&=2\\
k_2&=5&s_2&=20&i_2&=3\\
k_3&=3&s_3&=29&i_3&=4\\
k_4&=2&s_4&=47&i_4&=6\\
k_5&=1&s_5&=87&i_5&=11\\
\end{align*}
So result is $87$ and only $5$ iterations were needed. For $N=100$ it takes $19$ iterations, for $N=1000$ it takes $62$, etc...
