Timeline for The p-adic valuation of a linear recurrence
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Aug 20, 2017 at 14:52 | history | edited | Joe Silverman | CC BY-SA 3.0 |
Updated reference
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| Sep 16, 2016 at 19:39 | comment | added | Joe Silverman | @LucGuyot Yes, sorry, in both cases "infinitely many $n$" should be "infinitely many $p$", and ditto "almost all $n$" should be "almost all $p$". | |
| Sep 16, 2016 at 19:02 | comment | added | Luc Guyot | @Joe Silverman: I am confused by your last paragraph because the (in)equalities have no dependency on $n$. Is it about an infinite number of primes $p$? | |
| Nov 1, 2015 at 22:27 | comment | added | Vesselin Dimitrov | Are you considering an arbitrary binary linear recurrence? This applies to $a_n = q^n - 1$, and if I understand correctly, a similar result holds for elliptic divisibility sequences. But take $p = 3$ and $a_n = 5^n - 2$. The rank of apparition is $n = 1$, with corresponding valuation $1$, but the $3$-adic valuation of $a_n$ is zero for all even $n$. Moreover, for $n = 5$ we have $3^2 \mid 5^5 - 2 = a_5$, and then for $n = 11$ we have $3^3 \mid 5^{11} - 2 = a_{11}$. So, in this example, the $3$-adic valuation is rather subtle; indeed it depends on the rational approximations to a $3$-adic log. | |
| Nov 1, 2015 at 20:15 | history | answered | Joe Silverman | CC BY-SA 3.0 |