Maximum Period $[$Law$,2007]$

For $m$ a power of $2,$ say $m=2^b,$ and $c0,$ the longest possible period is $P=m=2^b,$ which is achieved whenever $c$ is relatively prime to $m($that is$,$ the greatest common factors of $c$ and $m$ is $1)$ and $a=1+4k,$ where $k$ is an integer.

For $m$ a power of $2,$ say $m=2^b,$ and $c=0,$ the longest possible period is $P=\frac{m}{4}=2^{b-2},$ which is achieved if the seed $(x_0)$ is odd and if the multiplier $a$ is given by $a=3+8k$ or $a=5+8k,$ where $k$ is an integer $=0,1,2$dots

For $m$ a prime number and $c=0,$ the longest possible period is $P=m-1,$ which is achieved whenever the multiplier a has the property that the smallest integer $k$ such that $a^k-1$ is divisible by $m$ is $k=m-1.$

${?}^2=$

$x_i+1=(a{x}_i+c{)}^{\text{m}\text{a}\text{d}\text{m}}$

$x_i+1=1$

Jsing the multiplicative congruential method, find the period of the generator for $a=13,m=2^6=64,$ and $0=1,2,3,4.$