- **science_man_88**
(*https://www.mersenneforum.org/forumdisplay.php?f=140*)

- - **Modified Form of LL Test?**
(*https://www.mersenneforum.org/showthread.php?t=23920*)

Modified Form of LL Test?[URL="https://en.m.wikipedia.org/wiki/Iterated_function"]Iterated function[/URL] 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) |

Look no farther than Wikipedia - [URL]https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test#Proof_of_correctness[/URL]
and you will find the same, only with [$]\omega = \alpha[/$] |

[QUOTE=Batalov;503541]Look no farther than Wikipedia - [URL]https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test#Proof_of_correctness[/URL]
and you will find the same, only with [$]\omega = \alpha[/$][/QUOTE] 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. |

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