2 Added phi example. Clarified other direction.

Here is an approach which may give some better estimates for particular values of $\alpha$:

$$\sum_{i= 1}^N i((i\alpha)) = \sum_{i=1}^N \sum_{j=i}^N ((j \alpha)) = \sum_{i=1}^N \sum_{k=0}^{N-i}((N\alpha -k\alpha))$$.

So, if you can estimate

$$\sup_{\substack{x\in (0,1) \\ M \le N}} \bigg|\sum_{k=0}^{M} ((x-k\alpha))\bigg|,$$

then you can crudely multiply by $N$ to get an estimate for $|\sum i((i\alpha))|$.

Specifically, my guess is that for quadratic irrationals $\alpha$, there is an upper bound for

$$\bigg|\sum_{k=0}^M ((x-k\alpha))\bigg|$$

which is $O (\sqrt{\log M})$ \log M)$, which would give you a bound of$O(N \sqrt{\log N})$log N)$, and more generally that there is a bound in terms of the coefficients of the simple continued fraction for $\alpha$, so that if those are bounded, then you still get $O(N \sqrt {\log N})$log N)$. Again For the particular value$\phi = (\sqrt5 + 1)/2$, that's a rough guess based on some heuristics$\sum_{i=0}^{M} ((i \phi))$has logarithmic growth$c + (5\sqrt5 - 11)/4 \log_\phi M$(achieved at indices in the sequences A064831 (+) and A059840 (-)), which suggests that$\sup \sum_{i=0}^M ((x-i\phi))$also has logarithmic growth, which would give an$N \log N$bound for the sum. In the opposite direction, for all$\alpha \not\in \frac 12\mathbb Z$, $$\limsup \bigg(\log_N \bigg|\sum_{i=0}^N i((i\alpha))\bigg|\bigg) \ge 1$$ since there are terms proportional to$N$. The sum can be greater than$N^{2-\epsilon}$infinitely often by choosing$\alpha$so that it is extremely well approximated by infinitely many rational numbers, the sum can be greater than$N^{2-\epsilon}$infinitely often. When$\alpha$is very closely approximated by rational$p/q$, then for$N$a small multiple of$q$(where "small" is relative to how well$p/q$approximates$\alpha$), about$1/q$of the terms can be moved past integers with a small perturbation of$\alpha$to$\alpha'$, which causes a jump of about$N^2/q$in the sum. So, either the sum for$\alpha$or$\alpha'$is large. We can choose a sequence$p_n/q_n$which converges to an$\alpha$which produces large sums infinitely often, so that for these$\alpha,$$$\limsup \bigg(\log_N \bigg|\sum_{i=0}^N i((i\alpha))\bigg|\bigg) = 2$$. 1 Here is an approach which may give some better estimates for particular values of$\alpha$: $$\sum_{i= 1}^N i((i\alpha)) = \sum_{i=1}^N \sum_{j=i}^N ((j \alpha)) = \sum_{i=1}^N \sum_{k=0}^{N-i}((N\alpha -k\alpha))$$. So, if you can estimate $$\sup_{\substack{x\in (0,1) \\ M \le N}} \bigg|\sum_{k=0}^{M} ((x-k\alpha))\bigg|,$$ then you can crudely multiply by$N$to get an estimate for$|\sum i((i\alpha))|$. Specifically, my guess is that for quadratic irrationals$\alpha$, there is an upper bound for $$\bigg|\sum_{k=0}^M ((x-k\alpha))\bigg|$$ which is$O (\sqrt{\log M})$would give you a bound of$O(N \sqrt{\log N})$, and more generally that there is a bound in terms of the coefficients of the simple continued fraction for$\alpha$, so that if those are bounded, then you still get$O(N \sqrt {\log N})$. Again, that's a rough guess based on some heuristics. In the opposite direction, by choosing$\alpha$so that it is extremely well approximated by infinitely many rational numbers, the sum can be greater than$N^{2-\epsilon}$infinitely often. When$\alpha$is very closely approximated by rational$p/q$, then for$N$a small multiple of$q$(where "small" is relative to how well$p/q$approximates$\alpha$), about$1/q$of the terms can be moved past integers with a small perturbation of$\alpha$to$\alpha'$, which causes a jump of about$N^2/q$in the sum. So, either the sum for$\alpha$or$\alpha'$is large. We can choose a sequence$p_n/q_n$which converges to an$\alpha\$ which produces large sums infinitely often.