# Conjectured primality test for numbers of the form $N=4 \cdot 3^n-1$

This is a repost of this question.

Can you provide proof or counterexample for the claim given below?

Inspired by Lucas-Lehmer primality test I have formulated the following claim:

Let $$P_m(x)=2^{-m}\cdot((x-\sqrt{x^2-4})^m+(x+\sqrt{x^2-4})^m)$$ . Let $$N= 4 \cdot 3^{n}-1$$ where $$n\ge3$$ . Let $$S_i=S_{i-1}^3-3 S_{i-1}$$ with $$S_0=P_9(6)$$ . Then $$N$$ is prime if and only if $$S_{n-2} \equiv 0 \pmod{N}$$ .

You can run this test here .

Numbers $$n$$ such that $$4 \cdot 3^n-1$$ is prime can be found here .

I was searching for counterexample using the following PARI/GP code:

CE431(n1,n2)=
{
for(n=n1,n2,
N=4*3^n-1;
S=2*polchebyshev(9,1,3);
ctr=1;
while(ctr<=n-2,
S=Mod(2*polchebyshev(3,1,S/2),N);
ctr+=1);
if(S==0 && !ispseudoprime(N),print("n="n)))
}


P.S.

Partial answer can be found here.

• Is this a partial case of mathoverflow.net/q/308886 ? – Max Alekseyev Jul 11 at 20:20
• @MaxAlekseyev No, it isn't. In post you mentioned we treat numbers of the form $N =k \cdot b^n-1$ with base $b$ being an even number not divisible by $3$ . – Peđa Terzić Jul 12 at 3:50