Timeline for When does a Catalan number equal a Fibonacci number?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 30, 2014 at 12:21 | vote | accept | Joseph O'Rourke | ||
| Nov 30, 2014 at 1:59 | comment | added | Yaakov Baruch | Well, Jeremy Rouse's neat answer below does vindicate the question! | |
| Nov 30, 2014 at 1:52 | comment | added | Joseph O'Rourke | @YaakovBaruch: Essentially, No. It just so happened that I was employing both sequences within the same hour (there is some similarity to their recursive definitions), and then I wondered if they shared elements beyond $5$. Then it did not seem a trivial question to answer... | |
| Nov 30, 2014 at 1:51 | answer | added | Jeremy Rouse | timeline score: 38 | |
| Nov 30, 2014 at 1:29 | comment | added | Yaakov Baruch | Any reason why the overlap of these two sequences, among many others, would be either useful to know or easy/interesting to compute/prove? | |
| Nov 30, 2014 at 1:26 | comment | added | Anthony Quas | @Will: I don't think that a coincidence of Catalan and Fibonacci numbers would give a better rational approximation to $\log 4/\log\phi$ than occurs `in nature'. You know you can find $m$ and $n$ such that $|m\log 4-n\log\phi|<C/m$ by a pigeonhole argument. Since the $m$th Catalan number is about $4^m/(m+1)$, a coincidence would give $|m\log 4-n\log\phi|\lesssim\log m$. | |
| Nov 30, 2014 at 0:37 | comment | added | Will Sawin | the $n$th Catalan number is about $4^n$, and the $n$th Fibonacci number is about $\phi^n$. So a coincidence between Catalan and Fibonacci numbers would give a very good rational approximation to $\log 4/\log \phi$. | |
| Nov 30, 2014 at 0:32 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 2 characters in body
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| Nov 30, 2014 at 0:26 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |