Timeline for Periods of Continued Fractions
Current License: CC BY-SA 2.5
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 6, 2010 at 16:49 | comment | added | Joseph Malkevitch | Ideas in this paper might be of use: math.princeton.edu/mathlab/jr02fall/Periodicity/periodmain.htm | |
| Jul 23, 2010 at 6:46 | history | edited | Gerry Myerson | CC BY-SA 2.5 |
corrected spelling
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| Jul 23, 2010 at 6:43 | comment | added | Gerry Myerson | @Franz, "$a$ or $b$" is equivalent to "$b$ or $a$", so the way I interpret it, the right side of the equation is certainly invariant under switching $p$ and $q$. | |
| Jul 23, 2010 at 0:00 | answer | added | Gerry Myerson | timeline score: 3 | |
| Jul 22, 2010 at 8:06 | comment | added | Franz Lemmermeyer | Let me add the remark that the left hand side of the equation in your question is invariant under switching p and q, but the right hand side is not. Thus you can't really expect it to hold. | |
| Jul 22, 2010 at 0:21 | comment | added | Will Jagy | Read the comments after Franz's answer, he revised that part of it. I will run some experiments. But bear in mind that sometimes your $l(pq)$ will be roughly the size of $l(p) \cdot l(q)$ which is consistent with some very simple bounds of Lagrange. But I bet that there are infinitely many instances of $$ p \cdot q = n^2 + 1 $$ which has about the shortest continued fraction period. That is, consider $$ \sqrt{10}, \; \sqrt{26}, \; \sqrt{65}, \; \sqrt{82}, \; \sqrt{122}, \; \sqrt{145}, \; \ldots \sqrt{901}, \; \ldots $$ where $901 = 17 \cdot 53$ was the first not divisible by 2 or 5. | |
| Jul 22, 2010 at 0:06 | comment | added | Gerry Myerson | It should be easy enough to check this for a few primes. Did you? | |
| Jul 21, 2010 at 23:53 | history | edited | Wadim Zudilin | CC BY-SA 2.5 |
improved; deleted 1 characters in body
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| Jul 21, 2010 at 23:25 | history | asked | Jerald Jetson | CC BY-SA 2.5 |