Denote by $\sigma_1(n)$ the sum of the divisors of $n$. For example, when $n=9$ we get $\sigma_1(9) = 1+3+9=13$. Define $D(x) = \sum_{n=1}^x\sigma_1(n)$. Varius methods are known for computing $D(x)$, most rely upon the crucial observation that summing the divisors themselves rather than looking at each number and its divisors specifically, we get a faster computation. That is: $$D(x) = \sum_{d=1}^x d \cdot \bigg \lfloor \frac xd \bigg \rfloor$$ Simply put, for each divisor, we multiply the divisor by the amount of times it appears. This form addresses an $\mathcal{O}(n)$ computation. But noticing the floor function is constant for large ranges, it admits a faster, $\mathcal{O}(n^{1/2})$ computation: $$D(x) = \sum_{d=1}^{ \big\lfloor \frac {x}{\sqrt x+1} \big \rfloor} d \cdot \bigg \lfloor \frac xd \bigg \rfloor + \sum_{d=1}^{\sqrt x} d \cdot \Bigg (T \bigg(\bigg \lfloor \frac nd \bigg \rfloor \bigg ) - T \bigg(\bigg \lfloor \frac n{d+1} \bigg \rfloor \bigg ) \Bigg )$$ where $T(n)= \frac {n\cdot(n+1)}2$ is the $n$'th triangular number. Further simplifaction can result in an algorithm of similiar time complexity, but still noticably faster: $$D(x) = \sum_{k=1}^{\lfloor \sqrt{x} \rfloor} \left( k \cdot \left \lfloor {\frac{x}{k}} \right \rfloor + T \left( \left \lfloor{\frac{x}{k}} \right \rfloor \right) \right) - T(\lfloor \sqrt{x} \rfloor) \cdot \lfloor \sqrt{x} \rfloor$$
Using the last formula, I can compute the correct result for $D(10^{18})$ in about 3.5 minutes (using Java). However, in https://oeis.org/search?q=87%2C8299&language=english&go=Search , (in the links section) terms up to $D(10^{36})$ are known. Extrapolating how long my algorithm would take, it seems to me that even when using a faster computer and a faster programming language, this would still be unfeasible using the last formula, which is $\mathcal{O}(n^{1/2})$. I would assume an algorithm with a better time complexity was used (my guess would be $\mathcal{O}(n^{1/3})$).
Am I wrong or is there really an even faster algorithm based on a more sophisticated mathematical observation behind it? If there is, I would like to know about it.