Help with Proving : Period estimation for for concatenated sequences Assuming I have two 8-bit random number sequences $s[n]$ and $d[n]$ which each have a period of $X$ and $Y$ respectively. Therefore:
$$s[n+X] = s[n]\\
d[n+Y] = d[n]$$
If they were concatenated bitwise to form a 16-bit random number sequence $e[n] = s[n]||d[n]$, they would have a period of $Z$:
$$e[n+Z] = e[n]$$
If my assumption is correct,
If $X=Y$, then $Z=X=Y$.
Else
$$Z = \text{LCM}(X,Y)$$
So my question is, is this assumption correct? It seems correct logically, but I wonder if there is any proof of theorem out there which I can use to support this claim. Any help would be much appreciated.
 A: Yes. I assume both $s$ and $d$ have the same length, otherwise it's not clear how to zip them.
If $X$ is a period, then any integer multiple $kX$ is also a period: $s[n+kX] = s[n+(k-1)X] = ... = s[n]$.
Since $\text{LCM}(X,Y)$ is by definition both a multiple of $X$ and $Y$, it is also a period of both $s$ and $d$. If two strings share a period $Z$, then their bitwise concatenation also has this period: $e[n+Z] = (s[n+Z], d[n+Z]) = (s[n], d[n]) = e[n]$.

If $X=Y$, then $Z=X=Y$. Else...

You don't need this special case; the formula $Z=\text{LCM}(X,Y)$ also holds for $X=Y$, since $\text{LCM}(X,X)=X$.

Without more information about strings $s$ and $d$, it can be shown that $Z=\text{LCM}(X,Y)$ is the best that can be said. Consider the strings of length $X\cdot Y$
$s[N] = \begin{cases} 1 &\mbox{if } X \mid N \\ 0 &\mbox{otherwise} \end{cases}$
$d[N] = \begin{cases} 1 &\mbox{if } Y \mid N \\ 0 &\mbox{otherwise} \end{cases}$
Their periods are $X$ and $Y$.
Let $e$ be their bitwise concatenation. For $Z$ to be a period of $e$, $e[Z]$ must be equal to $e[0]$. For this to happen, $s[Z]$ and $d[Z]$ must be 1, which happens exactly when $X \mid Z$ and $Y \mid Z$; therefore, $Z$ must be the common multiple, so LCM is optimal.
