Question

What sequence of pseudorandom numbers is generated using the linear congruential generator \(xn + 1 = \left( {3xn + 2} \right) mod 13\)with seed \(x0 = 1\)?

Short answer

It is easily observed that this linear congruential generator is inherently pseudo-random as it only produces the numbers\(5,4\), and \(1\).

Step-by-step solution

Step 1

\({x_{n + 1}} = \left( {3{x_n} + 2} \right)mod(13),{x_0} = 1\)

The given recursive function and the initial value.

Step 2

\({x_0} = 1\)

\(\begin{array}{*{20}{c}}{n = 0,{x_1} = (5)mod(13) = 5}\\{n = 1,{x_2} = (17)mod(13) = 4}\\{n = 2,{x_3} = (14)mod(13) = 1}\\{n = 3,{x_4} = (5)mod(13) = 5}\end{array}\)

\(\begin{array}{*{20}{c}}{n = 4,{x_5} = (17)mod(13) = 4}\\{n = 5,{x_6} = (14)mod(13) = 1}\end{array}\)

It is easily observed that this linear congruential generator is inherently pseudo-random as it only produces the numbers\(5,4\) , and \(1\).