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Dec 22, 2021 at 15:51 comment added Fedor Petrov Hello! Of course, fixed.
Dec 22, 2021 at 15:51 history edited Fedor Petrov CC BY-SA 4.0
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Dec 22, 2021 at 13:59 comment added Notamathematician Hello Fedor! Thank you for answer! Of course you mean $a(n)=a(\lfloor k(1+\sqrt{2}\rfloor)=k$ instead of $a(n)=a(\lfloor k(1+\sqrt{2}\rfloor)=2$?
Dec 22, 2021 at 10:38 vote accept Notamathematician
Dec 22, 2021 at 10:18 history answered Fedor Petrov CC BY-SA 4.0