Timeline for Iterations of $2^{n-1}+5$: the strong law of small numbers, or something bigger?
Current License: CC BY-SA 4.0
33 events
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| S Oct 26, 2020 at 17:31 | history | bounty ended | R.P. | ||
| S Oct 26, 2020 at 17:31 | history | notice removed | R.P. | ||
| Oct 21, 2020 at 17:17 | history | edited | Michael Hardy | CC BY-SA 4.0 |
edited title
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| S Oct 19, 2020 at 15:35 | history | bounty started | R.P. | ||
| S Oct 19, 2020 at 15:35 | history | notice added | R.P. | Draw attention | |
| Oct 25, 2018 at 20:20 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
+update #4
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| Jan 25, 2018 at 3:15 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
added update #3
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| Dec 30, 2017 at 17:39 | answer | added | Dan Brumleve | timeline score: 5 | |
| Nov 5, 2016 at 21:31 | comment | added | Terry Tao | A small comment that may speed up your search for prime factors of $n_5$ (though only by a factor of 2): for any such factor $p$, $-5$ must be a quadratic residue, so by reciprocity we must have $p=1,3,7,9 \hbox{ mod } 20$. | |
| Nov 4, 2016 at 23:04 | comment | added | Michael | Are there analogous rules for different $n_1$ that work for small $k$? I tried $n_1=1$ and $n_1=2$ with the identical rule and the property $n_2|n_3$ was broken immediately. | |
| Nov 4, 2016 at 20:31 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Nov 4, 2016 at 20:28 | comment | added | Max Alekseyev | @PietroMajer: I see now what you mean. It is easier to explain via $$10^{\frac{n_{k}-n_{k-1}}{n_{k-1}}}\equiv (-2^{n_{k-1}})^{\frac{n_{k}-n_{k-1}}{n_{k-1}}}\equiv 2^{n_{k}-n_{k-1}}\pmod{n_k}.$$ So, $10^{\frac{n_{k}-n_{k-1}}{n_{k-1}}} \equiv 1\pmod{n_k}$ iff $2^{n_{k}-n_{k-1}}\equiv 1\pmod{n_k}$. Added this to the question. | |
| Nov 4, 2016 at 20:13 | comment | added | Pietro Majer | (so if $a$ divides $b:=2^{a-1}+5$ the above congruence applies with $b=ac$). | |
| Nov 4, 2016 at 19:42 | comment | added | Pietro Majer | Actually it is necessary and sufficient. Just an elementary congruence: $2^{ac-1}+5=-5(10^{c-1}-1)\; \text {mod}\; 2^{a-1}+5$ for all positive integer $a$ and odd positive integer $c$. | |
| Nov 4, 2016 at 16:32 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Nov 4, 2016 at 16:20 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Oct 12, 2016 at 12:18 | comment | added | Pietro Majer | A further observation (trying to reduce $n_{k+1}$ modulo $n_k$). Assume $a$ divides $b:=2^{a-1}+5$. Then $b$ divides $2^{b-1}+5$ if and only if $b$ divides $ 10^{\frac{b-a}{a}}-1$. For instance, to check that $n_3$ divides $n_4\sim 2\cdot10^{78}$ it is sufficient to check that it divides $10^{28}-1$. | |
| Oct 11, 2016 at 20:24 | comment | added | Pace Nielsen | @MaxAlekseyev This reminds me of the sequence of repeated Mersenne primes: 2^2-1=3, 2^3-1=7, 2^7-1=127, 2^127-1=big prime P, 2^P-1= who knows! The guess there is similar to your "Outcome #1", this is probably just the law of small(ish) numbers. | |
| Oct 11, 2016 at 19:10 | comment | added | Max Alekseyev | @SebastienPalcoux: I doubt, but this statement may be hard to disprove either. | |
| Oct 11, 2016 at 18:03 | comment | added | Sebastien Palcoux | Assuming $n_k \mid n_{k+1}$, can $\frac{n_{k+1}}{n_k}$ be always prime? | |
| Oct 11, 2016 at 17:38 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
typo
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| Oct 11, 2016 at 17:29 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Oct 11, 2016 at 17:23 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
added remarks on computing n_k modulo m
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| Oct 9, 2016 at 14:20 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
extended search of primes to 10^8
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| Oct 9, 2016 at 13:09 | comment | added | Max Alekseyev | Just for the record: dxdy.ru/topic111861.html artofproblemsolving.com/community/q1h1316016p7069861 | |
| Oct 9, 2016 at 13:08 | comment | added | Jack Tiger Lam | @MaxAlekseyev I came across it on AOPS and backtraced it to dydx. | |
| Oct 9, 2016 at 13:03 | comment | added | Max Alekseyev | They copied it from forums dxdy.ru and artofproblemsolving.com where I posted it earlier. | |
| Oct 9, 2016 at 12:59 | comment | added | Julian Rosen | This question is also on math.SE, posted by someone else. | |
| Oct 9, 2016 at 10:36 | review | Suggested edits | |||
| Oct 9, 2016 at 11:25 | |||||
| Oct 9, 2016 at 9:22 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
clarified concern in outcome #1
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| Oct 9, 2016 at 4:23 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Oct 9, 2016 at 4:02 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Oct 9, 2016 at 3:54 | history | asked | Max Alekseyev | CC BY-SA 3.0 |