2 slight corrections, for clarity: ... 2) ... smallest even integer in that sequence greater than 8 ... 3) ... by definition less than or equal to 8 ... will

2 slight corrections, for clarity: ... 2) ... smallest even integer in that sequence greater than 8 ... 3) ... by definition less than or equal to 8 ... will still be less than or equal to 8, and all inputs less than or equal to 8 terminate ... Everything hinges on odds being as or less likely than evens which I believe I show initially "for 2n-1 (clearly odd), f(2n-1) = 3(2n-1)+1 = 2(3n-1), which is clearly even"; at least one even must follow every odd and, additionally, since there are only evens and odds, every odd must have been immediately preceded by an even. It's definitely an innocuous observation, but it justifies a walk all the way down to 8 by exhausting the (3) possible routes through the evens.