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I am interested in approaches which can be used to prove ( disprove ) conjecture given below .

Let's define number $N$ as : $N=3\cdot 2^n+1$ such that $n \equiv 2,6,8,10 \pmod{12}$ or $n \equiv 1 \pmod 4$ .

Let's define starting seed $S$ as :

$S = \begin{cases} 50542, & \text{if } n\equiv 8 \pmod {12} \\ 32672, & \text{if } n\equiv 1 \pmod 4 \\ 1692, & \text{if } n\equiv 2 \pmod {12} \\ 21868, & \text{if }n\equiv 6,10 \pmod {12} \end{cases}$

Let's define sequence $S_i$ as :

$S_i=S^2_{i-1}-2~$ with $~S_0=S$

Conjecture :

$N ; (n > 2)~$ is a prime iff $S_{n-2} \equiv 0 \pmod N$

I have checked this claim for all exponents under $100000$ from this list .

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    The best approach $f$or disproving such a conjecture would $b$e looking for a counterexample by testing composite numbers as well. The best approach for understanding the problem would be to familiarize yourself with the standard primality tests based on the factorizations of n+1 and n-1. But perhaps this is not an option as this would keep you from posting this question over and over again.2012-05-05

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