Thread: Modified Form of LL Test? View Single Post
 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