Answer:

f(3k) = f(3k+1) - 1 = f(3k-1)

f(3k+1) = f(3k-1) + 1 = f(3k) + 1

f(3k+2) = f(3k) + 1

here you can see that at the values of 3k+1 (4,7,...) the value increments by one and same. The first pattern is due to initial conditions you set for <=1 values.

You can replace the function with 2+int((k-1)/3) for k>2.