Optimizing method for counting length of elements between blocks
One gap can be calculated as tetrahedral number $P(x) = \sum \limits _{i = 1} ^{i=x} f(i) = \frac{x * (x + 1) * (x + 2)}{6}$ where $f(i) = \frac{i * (i + 1)}{2}$
After calculating all gaps to calculate chopped off sum you need to subtract $P(x) - P(x-1)$ from the sum.