obtaining all  vectors of given length and with with $+-1$ entries from a given one We can "travel" on all the vector space $V =GF(2)^n$ by doing the following
(a) choose a primitive polynomial $P(t)$ of degree $n$ over $GF(2)$.
(b) change vector $ X = (x_1, \ldots,x_{n-1}) \in V$ into vector $Y = (y_1, \ldots, y_{n-1}) \in V$.
(c) repeat until $V$ is exhausted (2^n times)
where
$y_1+y_2z+ \cdots + y_nz^{n-1} = z(x_1+x_2z+ \cdots + x_nz^{n-1})$
and $z$ is a zero of $P$, i.e., $P(z)=0.$
I want to do the same with integral vectors containing only 1 and -1
I.e.: "travel" on all possible vectors $(r_1, \ldots, r_{n-1})$
with $r_i^2=1$ 
How to do that ???
I do some trys without success...
reason of the question:  I have only limited time on  a computer 
((five days per job, two jobs allowed))
and I need to try some computations on all such vectors with moderately large $n$
the loop:
from r_1=-1 to 1 by 2 do;
from r_2=-1 to 1 by 2 do
$\cdots$
from r_{n-1}=-1 to 1 by 2 do;
do not "fit" in my allowed time.
following suggestion (thanks) let consider the following:
I need to examine each of the $2^n$ vectors.
To fit time allowed suffices to break the $2^n$ in smaller parts and apply to each of them the method I am asking for here !
I tried:
(a) $r_i \in \{−1,1\}$ go to $si=(r_i+1)/2$ in $\{0,1\}$
(b) apply idea with primitive polynomial, to the $s_i$'s
(So forced to take some reduction modulo $2$ in some coordinates)
(c) recover $R_j$ the new $r_j$, by $R_j=2s_j−1$
so that from vector
$(r_1,…,r_n)$ we get new vector $(R_1,…,R_n)$
and applying this $2^n$ times we should (hopefully) get all the $2^n$ vectors
but this does NOT work since I ended, e.g. to the cycle
$(−1,−1,…,−1)$ going to itself indefinitely
In other words: Can I write these $2^n$ vectors as a sequence
$v_1,…,v_{2^n}$ in such a manner
that I can with some simple algebraic computation,
(similar to the use of the primitive polynomial in case the vectors are in $GF(2)^n$))
get the vector
$v_k$ from the vector $v_{k−1}$
beginning with any fixed vector
$v_1$
???
 A: There are like $2^n$ of these vectors.  If you "need to try some computations on all such vectors" then, unless you can factor something out of your computation, you will need an amount of time roughly proportional to $(2^n) \times $ (time it takes to do each computation).
A: see new edited version of the question
A: Simply represent each vector as a binary integer:
$v_0 = "0 0 0 ... 0 0 0" = b_{n-1} b_{n-2}... b_2 b_1 b_0 = 0$
so the binary digit sequence representing $i$ is a decimal number $d$
$$d = \sum_{j=0}^{j=n-1} b_j \cdot 2^{j}$$
Given such a binary representation, you can transform the binary digits into the coefficients by mapping $0 \to -1$ and $1 \to 1$
$b_j = 0 \to r_j = (-1)$ and
$b_j = 1 \to r_j = (+1)$
Then, given a $v_j$ as a binary digit, generate $v_{j+1}$ by adding $1$ to the binary representation.
There is no shortcut around having to test each of the $2^n$ possibilities, so this does not make your overall calculations faster.  It just makes it easier to generate the binary digits.
A simpler way is to iterate your index as an integer (call it $z$) from $0$ to $n-1$ and generate the binary representation of each $z$.  There are many ways to do that quickly which you can find with simple searches on the internet or by asking at stackexchange.com
