Timeline for A possibly surprising appearance of $\sqrt{2}.$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2018 at 0:12 | comment | added | Gjergji Zaimi | Yes, oops, I meant $d_n$ of course. | |
| Jan 17, 2018 at 0:12 | history | edited | Gjergji Zaimi | CC BY-SA 3.0 |
edited body
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| Jan 17, 2018 at 0:11 | comment | added | Pietro Majer | btw the substitution map from the sequence $a_{n+1}-a_n$ to itself I got was $2\to4$, $3\to25$, $4\to235$, $5\to2335$, with prefix $3333{\bf5}$ (say with country code $17$, which we may forget) | |
| Jan 16, 2018 at 23:56 | comment | added | Pietro Majer | Is there a typo? Since $b_n$ takes values in $\{1,2\}$, then $d_n$ take values in $\{-1,0,1\}$, but the substitution you have, that should reproduce $d_n$, in fact gives strings of $1$'s and $2$'s only. | |
| Jan 16, 2018 at 19:55 | history | edited | Gjergji Zaimi | CC BY-SA 3.0 |
added some details
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| Jan 16, 2018 at 14:22 | vote | accept | Clark Kimberling | ||
| Jan 27, 2018 at 2:43 | |||||
| Jan 16, 2018 at 8:20 | comment | added | Pietro Majer | As a consequence, $a_n$ can't satisfy any linear (or polynomial) $r$ terms recurrence, otherwise $c_n$ would also satisfy some finite terms recurrence; but then $c_n$, being also finite-valued, would be (eventually) periodic, and the limit $a_n/n$ would be a rational number! | |
| Jan 16, 2018 at 2:03 | history | answered | Gjergji Zaimi | CC BY-SA 3.0 |