Suppose \(f:\N \to \N\) satisfies the recurrence relation
Note that with the initial condition \(f(0) = 1\text{,}\) the values of the function are: \(f(1) = 4\text{,}\) \(f(2) = 2\text{,}\) \(f(3) = 1\text{,}\) \(f(4) = 4\text{,}\) and so on, the images cycling through those three numbers. Thus \(f\) is NOT injective (and also certainly not surjective). Might it be under other initial conditions? 3
If \(f\) satisfies the initial condition \(f(0) = 5\text{,}\) is \(f\) injective? Explain why or give a specific example of two elements from the domain with the same image.
If \(f\) satisfies the initial condition \(f(0) = 3\text{,}\) is \(f\) injective? Explain why or give a specific example of two elements from the domain with the same image.
If \(f\) satisfies the initial condition \(f(0) = 27\text{,}\) then it turns out that \(f(105) = 10\) and no two numbers less than 105 have the same image. Could \(f\) be injective? Explain.
Prove that no matter what initial condition you choose, the function cannot be surjective.