Timeline for Simply generated sequences with mysterious differences
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 27, 2018 at 2:43 | vote | accept | Clark Kimberling | ||
| Jan 24, 2018 at 21:32 | comment | added | მამუკა ჯიბლაძე | @MartinRubey Since $a_n-a_{n-1}=a_0(b_{n-1}-b_{n-2})+a_1(b_{n-2}-b_{n-3})+a_2(b_{n-3}-b_{n-4})+q$, arguing as in Pietro Majer's answer, if almost all $a_n−a_{n-1}$ are $>1$ then almost all $b_n−b_{n-1}$ are either $1$ or $2$, so that there are only finitely many possibilities for $a_n−a_{n−1}$. It is however not entirely clear what happens if (and how) infinitely many $a_n−a_{n-1}$ are $1$... | |
| Jan 24, 2018 at 8:54 | comment | added | Martin Rubey | Is it (intuitively) clear that $|D|$ is also finite for longer "convolutions", eg. $a_n=\sum_{k=0}^2 a_k b_{n-3+k} + qn + r$? | |
| Jan 23, 2018 at 13:36 | comment | added | Martin Rubey | @მამუკაჯიბლაძე: Yes, I agree. | |
| Jan 23, 2018 at 13:26 | comment | added | მამუკა ჯიბლაძე | @MartinRubey In the referred question it was formulated differently, with $a_0=1$ and $a_n=a_1b_{n-1}\color{red}{-}a_0b_{n-2}+2n$, then it started with $a_*=(1,2,9,12,...)$ and $b_*=(3,4,5,6,...)$. Not really essential, I believe the main thing is to explain why the set of possible differences is finite. | |
| Jan 23, 2018 at 6:55 | comment | added | Martin Rubey | I think in example 1 we have $|D|=8$, I cannot see $(1,1)$ occurring. | |
| Jan 22, 2018 at 17:12 | answer | added | Pietro Majer | timeline score: 4 | |
| Jan 22, 2018 at 15:55 | comment | added | მამუკა ჯიბლაძე | I agree with @Pietro in that it might be more convenient to keep them all positive and set $a_n=e_1a_1b_{n-1}+e_0a_0b_{n-2}+qn+r$ with $e_0,e_1=\pm1$ (or maybe even more general) | |
| Jan 22, 2018 at 15:33 | comment | added | Clark Kimberling | Pietro - As a quick fix, I added absolute values. | |
| Jan 22, 2018 at 15:32 | history | edited | Clark Kimberling | CC BY-SA 3.0 |
added 4 characters in body
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| Jan 22, 2018 at 15:19 | comment | added | Pietro Majer | In Example 1, it is not clear $a_0=-1$, since it is required that $(a_n)$ and $(b_n)$ make a partition of the positive integers. After all, is there a reason why the coefficients in front of $b_{n-1}$ and $b_{n-2}$ must be exactly $a_1$ and $a_0$ and not some other integers $p$ and $s$ like $q$ and $r$? | |
| Jan 22, 2018 at 14:30 | history | asked | Clark Kimberling | CC BY-SA 3.0 |