Timeline for A sequence potentially consisting of only integers
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 8 at 23:12 | answer | added | Dave Benson | timeline score: 3 | |
| Jul 8 at 8:41 | answer | added | Antoine de Saint Germain | timeline score: 7 | |
| Aug 2, 2019 at 8:42 | comment | added | John Blythe Dobson | In a similar vein to the observation of @BrianHopkins, for $n=0$ through $n=35$ (which is as far as I can carry the calculations in Mathematica on my system) the residues of $b_n \bmod{20925}$ are 1, 1, 1, 1, 1, 16, 136, 9316, 136, 9316, 1, 11626, 1, 1, 1, 9316, 136, 16, 136, 9316, 1, 1, 1, 11626, 1, 9316, 136, 9316, 136, 16, 1, 1, 1, 1, 1, 16. If this apparent pattern persists, then $b_n \equiv b_{n-30} \equiv b_{34-n} \bmod{20925}$. | |
| Mar 11, 2019 at 7:49 | comment | added | Kurisuto Asutora | Alright - I assumed you would know that, but I thought it may be reasonable to mention it just in case. | |
| Mar 11, 2019 at 2:40 | comment | added | John Machacek | but perhaps some other techniques applicable to Somos sequences can be used, but I am less familiar with such techniques. | |
| Mar 11, 2019 at 2:31 | comment | added | John Machacek | @KurisutoAsutora I do know of Somos sequences. This sequence does not have Laurentness like the Somos sequence. So, it seems to behave differently yet still have integrality. | |
| Mar 5, 2019 at 20:03 | comment | added | Kurisuto Asutora | I guess you know what a Somos sequences is - they are defined by somewhat similar recurrences, and also surprisingly consists of integers in some setups. See en.wikipedia.org/wiki/Somos_sequence. Maybe some methods from there can be used? | |
| Feb 27, 2019 at 21:46 | comment | added | John Machacek | @SamHopkins I agree with you. I have tried to look more at the accepted answer on Math.SE in hopes it would address this question too, but I have not been able to completely follow it either. | |
| Feb 27, 2019 at 16:24 | comment | added | Sam Hopkins | I'm not 100% convinced the $k=4$ case is really resolved. | |
| Feb 26, 2019 at 15:54 | comment | added | John Machacek | Kevin, thanks for the computations. It's always nice to have additional verification. Brian, thanks for the interesting observation. @RenéGy for k=5 one can computer mod 32 and find 1, 1, 1, 1, 1, 16, 8, 4, 26, 15. For k>5 it will be an initial segment of 1s, followed by some 0s, followed by 16, 8, 4, 26, 15. Note the 0s don't really change things since we multiply by 0+1=1 in the recurrence. | |
| Feb 25, 2019 at 19:37 | comment | added | René Gy | I can't find a proof that for $k\ge5$, we have $(a^{(k)}_{2k-1} + 1) \equiv 2^4 \pmod{2^5}$. Would you give more hints? thanks! | |
| Feb 24, 2019 at 19:03 | comment | added | Brian Hopkins | These numbers get big fast! I noticed that the last digits seem to be always be 1 or 6. Moreover, after some initial "noise," the repeating pattern is 6, 6, 6, 6, 1, 1, 1 (starting at $n=11$). Looking modulo 30, from the same point the repeating pattern is 16, 16, 16, 16, 1, 1, 1. | |
| Feb 24, 2019 at 4:37 | comment | added | Kevin O'Bryant | The obvious Mathematica code (and an hour on my laptop) confirmed integrality up to $n=41$. | |
| Feb 23, 2019 at 21:18 | history | asked | John Machacek | CC BY-SA 4.0 |