TITLE: Is there a Lucas-Lehmer equivalent test for primes of the form ${3^p-1 \over 2}$?
QUESTION [7 upvotes]: I'm reviewing the cyclotomic form $f_b(n)= {b^n-1 \over b-1}$ for various properties to extend an older treatize of mine on that form.    
With respect to primality there is the Lucas-Lehmer-test for primeness of $f_2(p)$ where of course $p$ itself must be a prime. 
I was now looking, whether I can say some things for primes of the form $f_3(p) = {3^p-1 \over 2}$ ,for instance $f_3(3)=13, f_3(7)=1093, f_3(13)=797161, ...$ (more terms see bottom). For this I was looking for a comparable test, similar to the scheme in the Lucas-Lehmer test. 
There is a short remark at Weisstein's mathworld involving the concept of Lucas-sequences for a generalized primality test (eq (2) to (4)), of which then the Lucas-Lehmer-test is only a special case, but I could not decode the formulae & recipes into some algorithm.  

So Q: Is there a primality test for numbers of the form $f_3(p) = {3^p-1 \over 2}$  similar to the scheme in the Lucas-Lehmer test?


More terms for $(3^{a(n)}-1)/2 \in \Bbb P$                
 a(n)=[3, 7, 13, 71, 103, 541, 1091, 1367, 1627, 4177,
 9011, 9551, 36913, 43063, 49681, 57917, 483611, 877843 ]

source: OEIS:A028491

REPLY [2 votes]: There is the basic idea behind the Lucas-Lehmer primality test. For $n= 2^p-1$ we can choose $d= 3$ (quadratic reciprocity) and $\alpha= 2+\sqrt{3}$

If $n$ is prime and $d$ is not a square $\bmod n$ then $$\mathbb{Z}/n\mathbb{Z}[\sqrt{d}] = \{ a+b \sqrt{d}, (a,b) \in \mathbb{Z}/n\mathbb{Z}\}$$ is a field with $n^2$ elements, and its multiplicative group is cyclic with $n^2-1$ elements. If we find an element $\alpha$ of multiplicative order $n+1$ (ie. $\alpha^{n+1} \equiv 1 \bmod n, \alpha^{(n+1)/p} \not\equiv 1 \bmod n $ for every prime divisors $p | n+1$) then $n$ is prime.
(since otherwise with $q$ the least prime divisor of $n$ then $\mathbb{Z}/q\mathbb{Z}[\sqrt{d}]^\times$ is a group with less than $q^2-1 \le n-1$ elements, so the order of $\alpha\bmod q$ can't be $n+1$)

Thus all we need is to know the prime divisors of $n+1$ and compute $\alpha^{(n+1)/p} \bmod n$ for many $\alpha$. The same idea works in $\mathbb{Z}/n\mathbb{Z}$ if we know the prime divisors of $n-1$, and in $\mathbb{Z}/n\mathbb{Z}[x]/(f(x))$ for some irreducible polynomial $f$ of degree $k$ if we know a large part of the factorization of $n^k-1$.
$$\boxed{\ \ \text{Thus it doesn't work for }\ \frac{3^a-1}{2} \quad(\text{but it does for }2\cdot 3^a-1)\ \ }$$