20181221, 15:12  #1 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Modified Form of LL Test?
Iterated function on Wikipedia has two examples that are relevant:
\[f(x)=ax+b\to f^n(x)=a^nx+ \frac{a^n1}{a1}b\] and \[f(x)=ax^2+bx+\frac{b^22b8}{4a}\to f^n(x)=\frac{2\alpha^{2^n}+2\alpha^{2^n}b}{2a}\] where: \[\alpha=\frac{2ax+b\pm\sqrt{(2ax+b)^216}}{4}\] The first is the way Mersenne numbers iterate with a=2,b=1 . The second is how the the LucasLehmer test variants iterate. The normal LucasLehmer test sequence (prior to mod) is a=1,b=0. The reduced version using \(2x^21\) 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 b1 without changing the value of the first nth iterate gives the following: \[a^nx+\frac{a^n1}{a1}b+\frac{a^n1}{a1}\] Now we can eliminate all the b values and any direct multiplies by b from both nth iterates giving: \[f^n(x)=a^nx+\frac{a^n1}{a1}\] and \[f^n(x)=\frac{2\alpha^{2^n}+2\alpha^{2^n}}{2a}\] where: \[\alpha=\frac{2ax\pm\sqrt{(2ax)^216}}{4}\] Next thing to note is that a is equal in both cases (namely a=2). Plugging that fact, plus \(a=\frac{x1}{3}\) for our common x value into the first case, and the whole set of equations in the second reduces to : \[f^n(x)=\left ( \frac{x1}{3}\right )^nx+\left ( \frac{x1}{3}\right )^n1\] and \[f^n(x)=\frac{\alpha^{2^n}+\alpha^{2^n}}{2}\] where: \[\alpha=\frac{4x\pm\sqrt{(4x)^216}}{4}=x\pm\sqrt{x^21}\] 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 20181221 at 15:21 
20181221, 15:46  #2 
"Serge"
Mar 2008
Phi(3,3^1118781+1)/3
2^{2}×3×5×151 Posts 
Look no farther than Wikipedia  https://en.wikipedia.org/wiki/Lucas%...of_correctness
and you will find the same, only with \(\omega = \alpha\) 
20181221, 17:31  #3  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


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