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 science_man_88 2018-12-21 15:12

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)

 Batalov 2018-12-21 15:46

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 [$]\omega = \alpha[/$]

 science_man_88 2018-12-21 17:31

[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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