# Is there a closed form of the sum $\sum _{n=1}^x\lfloor n \sqrt{2}\rfloor$

I need to find the n-th partial sum of: $$\sum _{n=1}^x\lfloor n \sqrt{2}\rfloor$$

I tried to expand as the following:

$$=\frac{\sqrt{2} x \left(x+1\right)}{2}-\frac{x}{2}+\frac{1}{\pi }\sum _{n=1}^x\sum _{k=0}^{\infty}\frac{\sin \left(2 \pi \sqrt{2} k\ n\right)}{k}$$

$$=\frac{\sqrt{2} x \left(x+1\right)}{2}-\frac{x}{2}+\frac{1}{\pi }\sum _{n=1}^x \arctan (\tan( \frac{\pi - 2 \pi \sqrt{2} n}{2}))$$

For $n ∈ \{0,\frac{1}{\sqrt{2}}\}$

$$=\frac{\sqrt{2} x \left(x+1\right)}{2}-\frac{x}{2}+\frac{1}{\pi }\sum _{n=1}^x \frac{\pi - 2 \pi \sqrt{2} n}{2}$$

But I still have the last term that oscillates around some value and I cant figure out how to find the result without doing the sum up to x. if there is no closed form, is it possible to find it for special values of x? Any hint is appreciated.

• I think your best chance is construct a monotonic recurrence relation for $\sqrt 2$ with the property to be related in some easy way to $n$, i.e. that any approximation to $\sqrt 2$ using the recurrence relation build rational or natural numbers that coincide with what you want. – Masacroso Nov 24 '16 at 23:04
• @Masacroso Thanks for the hint, so if maximum value of x is 10 an approximation of 11 decimal digits should be ok? – OpenI Nov 25 '16 at 8:40
• @Openl it is not a hint, just an idea (I dont know if it can work). If you have some function that approximate $\sqrt 2$ by rational numbers then, probably, it can be used to try to solve your problem. – Masacroso Nov 25 '16 at 9:39
• Related question: How to find $\sum_{i=1}^n\left\lfloor i*\sqrt{2}\right\rfloor$ (But no answers have been posted so far.) – Martin Sleziak Dec 10 '16 at 8:44