TryThisProblem

Today I looked at the problem titled, "TryThisProblem" on the Problemsolving Wiki.

Here is what I think. To prove that the statement is true for every positive natural number, we must just prove that it is possible to get the function, through its layers, down to 2 or some power of 2, because then it will ultimately get down to 1 by the definition of f(n).
I believe this is possible to show by testing the single-digit naturals. Larger natural numbers will eventually boil down to these (although I am unable to prove this), so by testing these it can show that the larger numbers follow the rule of the statement.

f(1) = 3(1) + 1 = 4 / 2 = 2 / 2 = 1. so F(3, 1) = 1
f(2) = 2 / 2 = 1. so F(1, 2) = 1
f(3) = 3(3) + 1 = 10 / 2 = 5 * 3 + 1 = 16 / 2 = 8 / 2 = 4 / 2 = 2 / 2 = 1. so F(7, 3) = 1
f(4) = 4 / 2 = 2 / 2 = 1. so F(2, 4) = 1
f(5) = 3(5) + 1 = 16 / 2 = 8 / 2 = 4 / 2 = 2 / 2 = 1. so F(5, 5) = 1.
f(6) = 6 / 2 = 3 * 3 + 1 = 10 / 2 = 5 * 3 + 1 = 16 / 2 = 8 / 2 = 4 / 2 = 2 / 2 = 1. so F(8 , 6) = 1
f(7) = 3(7) + 1 = 22 / 2 = 11 * 3 + 1 = 34 / 2 = 17 * 3 + 1 = 52 / 2 = 26 / 2 = 13 * 3 + 1 = 40 / 2 = 20 / 2 = 10 / 2 = 5 * 3 + 1 = 16 / 2 = 8 / 2 = 4 / 2 = 2 / 2 = 1. so F(16, 7) = 1
f(8) = 8 / 2 = 4 / 2 = 2 / 2 = 1. so F(3, 8) = 1
f(9) = 3(9) + 1 = 28 / 2 = 14 / 2 = 7 * 3 + 1 = 22 / 2 = 11 * 3 + 1 = 34 / 2 = 17 * 3 + 1 = 52 / 2 = 26 / 2 = 13 * 3 + 1 = 40 / 2 = 20 / 2 = 10 / 2 = 5 * 3 + 1 = 16 / 2 = 8 / 2 = 4 / 2 = 2 / 2 = 1. so F(19, 9) = 1

Although all that would be needed to disprove my hypothesis is a counterexample.