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1:54 PM
Hi all.
marry Xmas
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Q: $a_n = a_{n-1}^3 - a_{n-2} $ conjectured inf and sup

mickMy mentor tommy1729 told me : Let $b_1 = 1.$ Let $ b_2 = x.$ Also $ \frac{1}{2} < x < 2 $ Define for $ n>2 $: $$ b_n = \frac{\exp(\ln(b_{n-1})^3)}{b_{n-2}} $$ Then $$ \sup_{n>2} b_n = x , \inf_{n>2} b_n = \frac{1}{x} $$ Is this true ? How to show that ? We can rewrite the sequence $b_n$ ...

 

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