Weighted Sum of Divisor Function For the first N Natural Numbers

Question

Good day to you.

The Function :

$ F(N) =\sum_{i=1}^{N} \sum_{d|i} 1 $,

i.e the function that summarizes the divisors of the first $N$ natural numbers can also be expressed as :

$ F(N)=\sum_{i=1}^{N} $ Floor($\frac{N}{i} $).

Now, the function :

$ G(N)=\sum_{i=1}^{N} i \cdot \sum_{d|i} 1 $. This is a weighted version of the above function.

Is it possible to expand / simplify this weighted version in a similar manner, or using some other way ? I tried to do so, but have been unable to make any progress.

Any Help would be useful. Thank You in Advance !

Answer 1

There is this: \begin{align*} \sum_{n\le x} n \tau(n) &= \sum_{n\le x} n \sum_{d \mid n} 1 \\ &= \sum_{d\le x} \sum_{n\le x:d \mid n} n \\ &= \sum_{d \le x} \sum_{k: dk\le x} dk \\ &= \sum_{d \le x} d \sum_{k \le x/d} k \\ &= \sum_{d \le x} \frac{d}{2} \left \lfloor \frac{x}{d}\right \rfloor \left \lfloor \frac{x}{d} + 1 \right \rfloor \\ &= \frac{1}{2} \sum_{d \le x} {d} \left \lfloor \frac{x}{d}\right \rfloor \left \lfloor \frac{x}{d} + 1 \right \rfloor. \\ \end{align*}