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Given $a_{n+1}=a_{n}+\frac{1}{a_{n}}$, and $a_0=5,$ how can I find $\lfloor a_{1000} \rfloor$? I've tried coming up with reasonable bounds, but to no avail (ex. ones like $\sqrt{x^2+1} < x + \frac{1}{x}$ are terrible).

I have looked at the only similar post I can find on this site (Solving $ a(n+1) = a(n) + \frac{1}{a(n)}$ with $a(1) = 1 $), but it doesn't seem to help.

Thanks in advance!

Freeman Cheng
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  • Thanks a bunch! For future notice how did you find this post? I spent quite a long time searching. – Freeman Cheng Aug 23 '19 at 19:19
  • Probably more than just one – rtybase Aug 23 '19 at 19:24
  • The square root formula from that referenced post seems very close to what is needed. A simple program using double precision numbers in java yields these approximate values: f(100) = 14.9701; sqrt(2 * 100) = 14.1421; f(1000) = 45.0023; sqrt(2 * 1000) = 44.7213; f(10000) = 141.5144; f(100000) = 447.2443; sqrt(2 * 100000) = 447.2135. – Thomas Bitonti Aug 23 '19 at 21:08
  • @FreemanCheng My search query: "x_n + frac 1 x_n" ... I figured a hit was more likely with $x_n$ than with $a_n$. – grand_chat Aug 23 '19 at 21:55

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