A new Wagstaff PRP test ? - Page 3

T.Rex · 2023-01-10, 21:01

Quote:

Originally Posted by kijinSeija

Let $W_p = (2^p+1)/3$ where $p$ is a prime number $>3$

Let the sequence $S_i = S_{i-1}^2-2$ with $S_0 = 1/4 = (2^{(p-2)}+1)/3$

$W_p$ is prime if $S_{p-2} \equiv (W_p-1)/2+2 \equiv 2 \cdot S_0 + 1$

2*S0+1 = 1/2+1 = 3/2 and (3/2)^2-2 = 1/4 .

Nothing new. Just start with 3/2.

T.Rex · 2023-01-10, 21:10

Quote:

Originally Posted by kijinSeija

Let $Np=2^p+1$ and $Wp=(2^p+1)/3$ for Wagstaff numbers with $p$ a prime number $> 3$ .

Then $W_p$ is prime if $S_{p-2} \equiv 10 \cdot S_0 - 1 (mod N_p)$ and $p \equiv 2 (mod 3)$ or $S_{p-2} \equiv 2 \cdot S_0 + 1 (mod N_p)$ and $p \equiv 1 (mod 3)$

Can't get what is S0.

Anyway, use modulo Np is a good idea.

About LLT good for N numbers such that N+1 can be easily factored, yes it is written in books. But that's wrong.
The LLT (with a different seed than 4 : 5) can be used for proving that a Fermat number is prime. There are at least 3 proofs for this, including mine.
See: PrimalityTest2FermatNumbers.pdf

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