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Jul 27, 2017 at 19:39 comment added Nate Eldredge @YellowPig: Try a sequence where you take a number $n_k$ and repeat it $\sqrt{n_k}$ times, then take the number $n_{k+1} = n_k+\sqrt{n_k}$ and repeat it $\sqrt{n_{k+1}}$ times, etc. If my back-of-the-envelope calculation is right, you should get $c=1$, but the first $N$ terms contain about $N^{2/3}$ distinct values.
Jul 25, 2017 at 17:35 history edited Nate Eldredge CC BY-SA 3.0
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Jul 25, 2017 at 17:28 history answered Nate Eldredge CC BY-SA 3.0