My question is a follow up on Binary sequence count of unique patterns
I understand how to count the number of periodic sequences that can be generated using a specific number of bits (as solved in the link above).
However, I would like to know if there exist some way to generate all the possible unique periodic sequences (without solving for all possible permutations). For example:
given 3 digits, the unique periodic sequences will be:
000
100 (equivalent to: 010,001)
110 (equivalent to: 011,101)
111
Thus, there are four unique periodic sequences that are possible. I have generated those using all possible permutations and reducing them to "canonical" forms. However, given 12 digits, this becomes a heavy task.
Thus, I would like to know if there exist some general representation of all possible unique periodic sequences for a given number of bits (or for a given period) such that by a change of parameter I can get them all.
So far, I have found the following result:
The defining function of periodic sequence of N bits is defined as:
$ f(k) = \sum_{j=1}^{N} a_j(e^{\frac{2\pi}{N}i})^{jk} $
where $i = \sqrt{-1}$ and $k \in [1,2,3,4,...]$.
In order to find the appropriate $a_j$ one may solve the following:
$ \mathbf{E} \mathbf{a} = \text{any combination of 1 and 0 in vector form}$
Here:
$ E_{jk} = (e^{\frac{2\pi}{N}i})^{jk}$ ($j$ and $k$ are integers in $[1,N]$) and $\mathbf{a}$ gathers $a_j$ into vector form.
The interesting thing is that: for any unique periodic binary sequence there will be a distnct $|\mathbf{a}|$ (i.e a vector with absolute values of the components of $\mathbf{a}$ and not its norm) relating to it.
It can be used on permutations but I cannot go on and define a single parameter or equation such that it will result in unique periodic sequences. (and I believe such one exists)