Multiplier theorem for $1$-perfect sequences

 

Belevitch [19xx]. Let $s$ be a $1$-perfect sequence of period $v=2\ell$ satisfying $$s_{i+\ell} = -s_{i},\quad\forall i$$ excepted $s_0=s_{\ell}=1$. Let us set $s_0=0$ and $s_\ell=0$. We obtain a sequence of type II of weight$v-2$ that satisfies $$s_\tau s_{-\tau} (-1)^{\tau} = (-1)^{\ell\over 4}$$

 

Assume $0 < \tau <{\ell\over 2}$

$$\begin{split}s\times s(\tau) &=0\\&=\sum_{i=0}^{v-1} s_i\......tau}\\&=2 \sum_{i=0}^{\ell-1} s_i\,s_{i+\tau}\\\end{split}$$

In the sum

$$\sum_{i=0}^{\ell-1} s_i\,s_{i+\tau}$$

we have 2 terms equals to $0$ and $\ell-2$ terms equal to $\pm 1$.The product of the non-zero terms is equal to

$$(-1)^{{\ell-2}\over 2}$$

Let us replace $s_{i+\tau}$ by $-s_{i+\tau-\ell}$ when $i+\tau\geq \ell$. For $1\leq i\leq \ell-1$, $s_i$ appears twice except $s_\tau$ and$s_{\ell-\tau}$ which appears only once. Because of the signs the products of all the terms is equal to

$$-(-1)^{\tau}s_{\ell-\tau}s_{\tau}$$

Finally,

$$s_\tau s_{-\tau} (-1)^{\tau} = (-1)^{\ell\over 4}$$

 

Transformation of Legendre sequences
 

 If $v\equiv 3 \mod 4$ then we can tranform the Legendre sequence $s$ in a binary sequence $r$ by

$$r(x) = \delta_0(x) + s(x)$$

the out-phase correlations are unchanged since

$$r\times r(\tau) = \delta_0(\tau) + s(\tau) + s(-\tau) +s\times s(\tau)$$