Here is what I observed :

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

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

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 $p \equiv 1 (mod 3)$

for example with $p = 17$ we get this on PARI/GP :

```
Mod(10923, 131073)
Mod(35497, 131073)
Mod(32258, 131073)
Mod(121088, 131073)
Mod(84743, 131073)
Mod(17450, 131073)
Mod(19919, 131073)
Mod(8588, 131073)
Mod(90716, 131073)
Mod(105422, 131073)
Mod(118412, 131073)
Mod(129713, 131073)
Mod(14576, 131073)
Mod(121514, 131073)
Mod(16598, 131073)
Mod(109229, 131073)
```

$17 \equiv 2 (mod 3)$ and we get $10 \cdot (2^{17-2}+1)/3 - 1$ at the last residue and $(2^{17}+1)/3 = 43691$ and this is indeed prime.

You can run the test here.

And if it's true for all prime, is there a way to proving this ?