4 ms·
I'm still thinking about your modular perspective. I believe that is an accurate restatement of one of my cases, but it is not all of them and would not be suff
by vy8vWJlco 14y ago
I'm still thinking about your modular perspective. I believe that is an accurate restatement of one of my cases, but it is not all of them and would not be sufficient alone. Whereas all the cases together, however, are exhaustive. To summarize, my above comments/claims boil down to:
1) any even number input taken will produce a sub-sequence ending in an even output, by exhaustion (see above), that is guaranteed to be smaller after a maximum number of steps
2) this new even can be taken again, as often as necessary, to reach the smallest integer in that sequence greater than 8
3) the next even number (at most 1 odd away) is by definition below 8 and, while it may not have a next even by my arguement of necessity, it will still be below 8, and all inputs below 8 terminate
4) there is nothing about this reasoning that limits it to a particular sequence/input (all initially-odd inputs adhere after 1 iteration), so it is true of every sequence
Am I saying anything untrue or ambiguous?
Thanks for taking the time!
- vy8vWJlco 14y ago2 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.