It is well known that the divisor summatory function can be calculated in $O(x^{1/2})$ via $$D(x)=\sum_{n\le x} d(n) = 2 \sum_{k=1}^{\lfloor \sqrt{x}\rfloor} \lfloor\frac{x}{k}\rfloor - \lfloor \sqrt{x}\rfloor^2$$
Is there an efficient algorithm (complexity less than $O(x)$) to calculate $$T(x)=\sum_{n\le x} \tau(n)$$ where $\tau(n)$ is the Ramanujan tau function?
'Intuition': although there's no interesting way to interpret $\tau(n)$ as the lattice points under the hyperbola or something similar, it is multiplicative and the coefficients of a generating function with multiple special properties, so there might be a special property that allows 'faster' calculation.