
Sequence embedding?
Introduction:
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.
Conditions:
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}$$ with $\displaystyle \alpha$ a primitive element of $\displaystyle \mathbb{F}_{p}$ and $\displaystyle $n=1,3,5,\cdots ,p1$$. To be entirely correct $\displaystyle $n$$ belongs to a prime residue group, so $\displaystyle \left ( n,p1 \right )=1$ but since we are restricting ourselves to Fermat primes then we can just consider $\displaystyle $n$$ to be in the set of all the odd integers less than $\displaystyle $p$$.
A numerical example using Mathematica:
For a sequence of length 16 with primitive element 3,
Sort[Table[Zech[n,3,17],{n,1,171,2]]
{4, 6, 7, 10, 11, 12, 13, 15}
For a sequence of length 256 with primitive element 3, and taking values (mod 16)
Mod[Sort[Table[Zech[n,3,257],{n,1,2571,2]],16]
{4,6,7,9,13,...,5,10,11,12,13,15}
where we can see that the lengthier sequence is wrapped by the smaller seq.
I'm interested in the values of $\displaystyle $n$$ 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
$\displaystyle 3^{n} \in \left \{ 3^{2i}1:i=0,1,\cdots \frac{171}{2}1 \right \}\Rightarrow 3^{n\left ( \textup{mod}\: 16 \right )} \in \left \{ 3^{2j}1:j=0,1,\cdots \frac{2571}{2}1 \right \}$
but one of the problems is that of ordering. Any advise please?