Summation with Ceilinged Logarithmic Function According to Johann Blieberger's paper - "Discrete Loops and Worst Case Performance" (1994):
$$
\sum_{i = 1}^{n}\left \lceil \log_2{(i)} \right \rceil = n\left \lceil \log_2{(n)} \right \rceil - 2^{\left \lceil \log_2{(n)} \right \rceil} + 1
$$
Now, I was wondering if someone knows what the following may equal?
$$
\sum_{i = 1}^{n}i\left \lceil \log_2{(i)} \right \rceil = ?
$$

Use sum by parts:
$$
\sum_{1 \le k \le n} x_k \Delta y_k
  = x_{n + 1} y_{n + 1} - x_1 y_1
       - \sum_{1 \le k \le n} y_{k + 1} \Delta x_k
$$
In your case $\Delta y_k = \lceil \log_2 k \rceil$ and $x_k = k$. You'll get a sum similar to the original, you should be able to solve for that one.
