Formula for $\sum_{i=1}^n \lfloor i\sqrt{2} \rfloor $

Asked Sep 09 '20 at 17:38

Active Sep 09 '20 at 17:38

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I'm trying to find a faster way to compute this:

$$\sum_{i=1}^n \lfloor i\sqrt{2} \rfloor $$

I doubt that there is a compact form, but there must be some way to calculate this faster than just iterating over i. If it wasn't for the floor, it would be easy, but I don't know and can't find anything on how to solve this.

asked Sep 09 '20 at 17:38

Finn

@player3236 Yes, thank you! How, did I not come across that one, I've been searching for hours.. – Finn Sep 09 '20 at 17:57
what kind of approximation are looking for ? for large $n$ ? – G Cab Sep 09 '20 at 18:08

Formula for $\sum_{i=1}^n \lfloor i\sqrt{2} \rfloor $

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