Are linear shift register sequences corresponding to reciprocal polynomials equivalent? I am looking into sequences generated by LFSRs (linear shift register sequences). I was wondering if sequences corresponding to reciprocal connection polynomials (that is, corresponding to shift registers with the taps reversed) are equivalent. 
When I say "equivalent" I mean that they are shift-equivalent (one is equal to some shift of the other) and produced by maybe different states.
All the baby-examples that I did by hand were confirming that this is true, however I don't know if that is the case with bigger examples.. I also could not prove something theoretically.
 A: Expanding on @JyrkiLahtonen's comment, suppose that an LFSR with feedback polynomial 
$\Lambda(z) = 1 + \lambda_1z + \cdots + \lambda_Lz^L$ of degree $L$
(meaning that $\lambda_L\neq 0$) generates a sequence $S_0, S_1, \ldots, S_{2L-1}, \ldots$. This sequence satisfies the linear recurrence: 
$$S_{i} + \lambda_{1}S_{i-1} + \lambda_2S_{i-2} 
+ \cdots + \lambda_LS_{i-L} = 0, ~ i = L, L+1, \ldots
$$
with $S_L = -\left(\lambda_Ls_0 + \lambda_{L-1}s_1 + \cdots + \lambda_1s_{L-1}\right)$ being the first element that is computed from
the initial contents $(S_0, S_1, \ldots, S_{L-1})$ of the LFSR
and fed back into the right end of the LFSR as the contents shift
left by one place. The symbol $S_0$ on the left falls off the end
of the register and is the output of the shift register.
Note that the output will in succession have value $S_0, S_1, S_2, \ldots$. In a finite field,
the sequence generated by an LFSR is periodic with the period
depending on the irreducible factors of $\Lambda(z)$ as well as
the initial loading of the LFSR.
The reverse polynomial of $\Lambda(z)$ is
$$\tilde{\Lambda}(z) = z^L\Lambda(z^{-1}) = \lambda_L + \lambda_{L-1}z
+ \cdots + \lambda_1z^{L-1} + z^L$$ which has the same coefficients 
as $\Lambda(z)$ but running in
reverse order.  Now suppose that $K \gg L$ is some fixed positive
integer. Then, we have that 
$$\begin{align}
S_{K} + \lambda_{1}S_{K-1} + \lambda_2S_{K-2} 
+ \cdots + \lambda_{L-1}S_{K-L+1} + \lambda_LS_{K-L} &= 0\\
(\lambda_L^{-1})S_K + (\lambda_L^{-1}\lambda_{1})S_{K-1} 
+ (\lambda_L^{-1}\lambda_2)S_{K-2} 
+ \cdots + (\lambda_L^{-1}\lambda_{L-1})S_{K-L+1} + S_{K-L} &= 0\\
\end{align}$$
and so $S_{K-L} = -\left((\lambda_L^{-1})S_K + (\lambda_L^{-1}\lambda_{1})S_{K-1} 
+ (\lambda_L^{-1}\lambda_2)S_{K-2} 
+ \cdots + (\lambda_L^{-1}\lambda_{L-1})S_{K-L+1}\right).$
Working our way out of this thicket of subscripts, let us
consider an LFSR whose feedback polynomial is the scalar
multiple
$$\lambda_L^{-1}\tilde{\Lambda}(z)= 1 + (\lambda_L^{-1}\lambda_{L-1})z 
+ \cdots + (\lambda_L^{-1}\lambda_1)z^{L-1} + (\lambda_L)^{-1}z^L$$
of $\tilde{\Lambda}(z)$, and whose initial loading is 
$(S_K, S_{K-1}, \ldots, S_{K-L+1})$. The term $S_{K-L}$ is computed
and fed back into the right end of the LFSR register, etc. Thus
the output of this LFSR whose feedback polynomial is a scalar multiple of
the reverse polynomial of $\Lambda(z)$ will, in succession, be $$S_K, S_{K-1}, S_{K-2}, \ldots, S_1, S_0, \ldots$$
that is, a sequence that is shift-equivalent to the reverse
of the sequence $S_0, S_1, \ldots, $ generated by the LFSR
with feedback polynomial $\Lambda(z)$.
