Um not wrong

5 DEFINETLY DOES go 5,16,8,4,2,1,4,2,1,4,2,1,4,2,1,4,2,1.... The issue is that there are only 6, 4 or 3 numbers there that matter (depending on how you count)
You are either counting the 5,16,8 (so 3), the 5,16,8,4 (so 4, counting until you reach a 4) or 5,16,8,4,2,1 (so 6, counting until the repetition would start)

The sequence NEVER ENDS, but all (needs proof) eventually repeat 4,2,1,4,2,1 and so each sequence, even for the number 4 is infinitely long, we just don't care about after the repetition.

Perhaps you should re-read the article, and consider the original commentators work. The reason you are disagreeing seems as simple as how are you counting. You each are counting in a different way, yet seem to be saying the exact same thing.

The question isn't whether definitionally if 0 is or is not a natural number, you have to define whether it is and then do the math accordingly. I always asked my professors if they considered 0 in N or not for their classes. The guy you are saying is wrong, just defined his algorithm differently than you did. It isn't that you are right and he is wrong, or vice versa, it is that you both defined what you are looking for differently, but still validly.

Same holds for Fibonacci, does Fib start at 0 or at 1? If 0 is in N than the explicit formula for fibonacci is also different, yet they all are the same if you change the initial definitions to match.

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