mersenneforum.org Modified Form of LL Test?
 Register FAQ Search Today's Posts Mark Forums Read

 2018-12-21, 15:12 #1 science_man_88     "Forget I exist" Jul 2009 Dumbassville 100000110000002 Posts Modified Form of LL Test? Iterated function on Wikipedia has two examples that are relevant: $f(x)=ax+b\to f^n(x)=a^nx+ \frac{a^n-1}{a-1}b$ and $f(x)=ax^2+bx+\frac{b^2-2b-8}{4a}\to f^n(x)=\frac{2\alpha^{2^n}+2\alpha^{-2^n}-b}{2a}$ where: $\alpha=\frac{2ax+b\pm\sqrt{(2ax+b)^2-16}}{4}$ The first is the way Mersenne numbers iterate with a=2,b=1 . The second is how the the Lucas-Lehmer test variants iterate. The normal Lucas-Lehmer test sequence (prior to mod) is a=1,b=0. The reduced version using $$2x^2-1$$ is a=2,b=0. I chose to look at the latter as it shares a possible x value with the Mersenne numbers. Now getting the b values to equate was a priority for me ( because it simplifies the expressions if they both equal 0, as then b can be taken out of the variables). Allowing b to go to b-1 without changing the value of the first n-th iterate gives the following: $a^nx+\frac{a^n-1}{a-1}b+\frac{a^n-1}{a-1}$ Now we can eliminate all the b values and any direct multiplies by b from both n-th iterates giving: $f^n(x)=a^nx+\frac{a^n-1}{a-1}$ and $f^n(x)=\frac{2\alpha^{2^n}+2\alpha^{-2^n}}{2a}$ where: $\alpha=\frac{2ax\pm\sqrt{(2ax)^2-16}}{4}$ Next thing to note is that a is equal in both cases (namely a=2). Plugging that fact, plus $$a=\frac{x-1}{3}$$ for our common x value into the first case, and the whole set of equations in the second reduces to : $f^n(x)=\left ( \frac{x-1}{3}\right )^nx+\left ( \frac{x-1}{3}\right )^n-1$ and $f^n(x)=\frac{\alpha^{2^n}+\alpha^{-2^n}}{2}$ where: $\alpha=\frac{4x\pm\sqrt{(4x)^2-16}}{4}=x\pm\sqrt{x^2-1}$ My main question is: What's the simplified version of their modular remainder ( second mod first)? that can allow us to solve for a relation n+3 must have to get 0 out and therefore all odd Mersenne prime exponents. (Idea for this thread, was from: ewmayer and CRGreathouse) Last fiddled with by science_man_88 on 2018-12-21 at 15:21
 2018-12-21, 15:46 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·4,813 Posts Look no farther than Wikipedia - https://en.wikipedia.org/wiki/Lucas%...of_correctness and you will find the same, only with $$\omega = \alpha$$
2018-12-21, 17:31   #3
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by Batalov Look no farther than Wikipedia - https://en.wikipedia.org/wiki/Lucas%...of_correctness and you will find the same, only with $$\omega = \alpha$$
I'm looking for n that cause it to be 0. yes I know they will be such that n+3 is prime but I'm looking for a property that isn't so generic.

 Similar Threads Thread Thread Starter Forum Replies Last Post devarajkandadai Number Theory Discussion Group 14 2017-11-12 20:04 devarajkandadai Number Theory Discussion Group 0 2017-06-24 12:11 devarajkandadai Number Theory Discussion Group 2 2017-06-23 04:39 a1call Information & Answers 10 2017-02-05 22:18 Xyzzy Soap Box 27 2015-02-17 21:42

All times are UTC. The time now is 12:36.

Tue Dec 7 12:36:57 UTC 2021 up 137 days, 7:05, 0 users, load averages: 2.16, 2.03, 1.66

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.