The transition matrix of the shift register with shift polynomial s = X³ + X + 1 is a zero of f = X⁴ - s = X⁴ - X³ - X - 1.

Let us consider the coefficients in R = Z/2Z. Then this polynomial factors as (X + 1)²(X² + X + 1) and so is a divisor of X⁶ - 1. Therefore, the period is a divisor of 6.

In fact, the period is precisely 6. But there are exceptional initial states, leading to sequences with periods which are proper divisors of 6. For instance, the initial state (g₀, ..., g₃) = (1,0,1,1) gives the sequence

1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, ...

of period 3.

For the Fibonacci shift register the transition matrix is

A =

0 1 1 1

with minimal polynomial f(X) = X² - X - 1. The numbers in the sequence keep growing, at least when we take R = Z. Consequently, no finite period is to be expected. But if we perform arithmetic modulo a natural number m, a finite period is bound to show up. For, the number of distinct sequences of length 2 is finite: at most m², and hence the period is at most m².

If m = 2, then f as an element of (Z/2Z)[X] divides the polynomial X³ - 1, so the period is 1 or 3.

If m=3, then f divides X⁸ - 1, so the period is a divisor of 8.