I'm an Electronics Engineer with a love for math so due to my background pardon me if my description is inadequate (I tried hard to make it formally correct).

Context and motivation:
I have been working with Sidel'nikov sequences and through exploratory analysis I found that some sequences are contained in lengthier ones.

Consider only binary sequences and $\displaystyle \mathbb{F}_{p}$ for $\displaystyle $p$$ a Fermat prime.

Although not the original definition, the positions of "1" in a Sidel'nikov sequence can be determined using the Zech log,

$\displaystyle $Z_{\alpha }^{\left ( p \right )}\left ( n \right ):\textup{ord}_{p}\left ( \alpha \right )=1+\alpha ^{n}$$ or alternatively $\displaystyle $\alpha ^{Z\left ( n \right )}\equiv _{p}1+\alpha ^{n}$$ with $\displaystyle \alpha$ a primitive element of $\displaystyle \mathbb{F}_{p}$ and $\displaystyle $n=1,3,5,\cdots ,p-1$$. To be entirely correct $\displaystyle $n$$ belongs to a prime residue group, so $\displaystyle \left ( n,p-1 \right )=1$ but since we are restricting ourselves to Fermat primes then we can just consider $\displaystyle $n$$ to be in the set of all the odd integers less than $\displaystyle $p$$.

A numerical example using Mathematica:
For a sequence of length 16 with primitive element 3,

{4, 6, 7, 10, 11, 12, 13, 15}

For a sequence of length 256 with primitive element 3, and taking values (mod 16)


where we can see that the lengthier sequence is wrapped by the smaller seq.

I'm interested in the values of $\displaystyle $n$$ that satisfy both congruences. So the question is: what values of $\displaystyle $n$$ satisfy simultaneously
$\displaystyle \alpha^{Z_\alpha ^{(257)} (\textup{mod} \: 16)}$ and $\displaystyle \alpha^{Z_\alpha ^{(17)}$?

I've tried to attack the problem using instead
$\displaystyle 3^{n} \in \left \{ 3^{2i}-1:i=0,1,\cdots \frac{17-1}{2}-1 \right \}\Rightarrow 3^{n\left ( \textup{mod}\: 16 \right )} \in \left \{ 3^{2j}-1:j=0,1,\cdots \frac{257-1}{2}-1 \right \}$
but one of the problems is that of ordering. Any advise please?