Timeline for Floors of powers of reals, how much do the first few determine the next?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2014 at 15:14 | comment | added | Marco Golla | Ok, I can't resist adding these two: $(1,2,2)$ is forcing (notice that this is the shortest possible forcing sequence); $(1,2,3,4,7,11)$ forces the next two elements! In fact, this possibility follows from my answer (to which I will add a comment now). | |
| Jan 9, 2014 at 23:38 | comment | added | Marco Golla | Take the sequence generated by that $r$, and consider its first 8 elements. According to octave, this sequence is forcing. I think octave is accurate enough for this computation (the 9-th element has 18 digits). | |
| Jan 9, 2014 at 22:24 | comment | added | Aaron Meyerowitz | I'm not sure how your example with $109.19...$ works out. I added an explanation of how to chance on forced terms to the end of the question. | |
| Jan 9, 2014 at 22:23 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Jan 9, 2014 at 15:00 | comment | added | Marco Golla | In fact, also $r \sim 109.19432559052055$ gives an even larger example. | |
| Jan 9, 2014 at 14:49 | comment | added | Marco Golla | A small correction to my comment above: the sequence I gave does not force the next term, but $(1,3,5,10,17,32)$ does. Still in the spirit of answering the question "How much [does] an initial segment of a feasible sequence limit the next term?", octave finds many forcing sequences even for $r>2$: $(2,6,17,47)$ and $(3,9,28,89)$ among them. This is despite of very unfavourable heuristics. | |
| Jan 8, 2014 at 9:44 | answer | added | Aaron Meyerowitz | timeline score: 3 | |
| Jan 8, 2014 at 8:52 | history | edited | Aaron Meyerowitz | CC BY-SA 3.0 |
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| Jan 7, 2014 at 23:48 | vote | accept | Aaron Meyerowitz | ||
| Jan 7, 2014 at 1:33 | comment | added | Richard Stanley | I believe that there was a Putnam problem that answered this question, but I don't have a good way to search through all previous exams. | |
| Jan 6, 2014 at 20:44 | answer | added | Sidney Raffer | timeline score: 6 | |
| Jan 6, 2014 at 20:10 | answer | added | Qiaochu Yuan | timeline score: 5 | |
| Jan 6, 2014 at 16:46 | answer | added | Marco Golla | timeline score: 4 | |
| Jan 6, 2014 at 15:01 | comment | added | Marco Golla | If the sequence $(\lfloor r^n\rfloor)_{n\ge 1}$ starts with $1,3,5,10,18,32$, then $32^\frac16 \le r < 18^\frac15$, and we have that $57 < 32^\frac76 < 18^\frac75 < 58$, so $57$ is the next element, regardless of $r$. | |
| Jan 6, 2014 at 12:03 | answer | added | joro | timeline score: 15 | |
| Jan 6, 2014 at 10:45 | history | asked | Aaron Meyerowitz | CC BY-SA 3.0 |