A PuzzleCritic Original:

*A sequence contains every positive integer exactly once, and no other terms. Must there exist, somewhere in the sequence:*

*(i) an odd number immediately followed by an even number;*

*(ii) a multiple of two immediately followed by a multiple of three?*

(i) Yes. Find the number 1 in the sequence. Only finitely many even numbers appear before it, so some even number will appear after it. Just keep moving through the sequence from the number 1 until you first hit an even number. This term is even, and the one immediately before it will be odd.

(ii) No. Consider the usual ordering of the positive integers; now imagine the sequence split into blocks of six terms: 1-6, 7-12, 13-18 and so on. Then, in each block, swap the second and third terms. The new sequence looks something like this:

1, 3, 2, 4, 5, 6, 7, 9, 8, 10, 11, 12, …

Now, every multiple of three immediately follows an odd number, and we have our counterexample.

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Richard RJanuary 15, 2018 / 12:10 amI’m not entirely convinced by the answer to i). It asserts that “only finitely many even numbers” appear before the first odd number in the sequence, so there must be a subsequent even number.

How about this, though? My sequence is (2,4,6….5,3,1) counting forward from the beginning in even numbers and back from the end in odd numbers, with each even number at the beginning of the sequence being matched by an odd number at the end. It is an infinite sequence of all the integers, but there will never be an even number following an odd number.

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