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(I started working on this problem after trying to get any "interesting" pattern out of the number that Gowers randomly wrote while answering:What is realistic mathematics?.)

The number was 123871205412470874297947938271423698765734564756028492656.

Take any number, for instance:

123871205412470874297947938271423698765734564756028492656

3484756955 (in the preceding number there are three 0's, four 1's,..., and finally five 9's)

0001231111 (in the preceding number there are no 0's, no 1's,..., and finally one 9)

3511000000

6201010000

.... clearly the list won't end as there will always be some no 0's or 1's or 2's, etc.

A variant of this sequence is discussed here(Conway's look-and-say sequence).

This a fairly simple obsevation, so is there a literature about such sequences from which I can learn more?

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The 5th number above should be 6201010000. Then 6210001000. And then it's constant. –  Gerry Myerson Aug 8 '10 at 13:24
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e.g. 1->0100000000->9100000000->8100000001->....->6201001000->6210001000 So, 1 culminates in 6210001000. Maybe all numbers end at this number, any heuristics? –  Unknown Aug 8 '10 at 13:49
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You might try to think about a more informative title ... –  Tom Goodwillie Aug 8 '10 at 14:04
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What convention do you choose when the initial number has more than 9 copies of a digit? –  S. Carnahan Aug 8 '10 at 14:45
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Sorry, giving a downvote. I think projects like these on MO need to be extremely well thought out, and contain significant mathematical content and motivation. –  Cam McLeman Aug 8 '10 at 14:49
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1 Answer

up vote 4 down vote accepted

You have to be a bit more precise - for instance what happens if there are more than nine of a particular digit ? Regardless, a cursory literature search comes up with an article by Sauerberg and Shu which studies the Conway sequence as well as ones similar to yours, which are called factor-free counting sequences. The final section shows they are eventually periodic and gives a list of all possible cycles.

In particular, when counting the digits 0,1,...,9 as in the question, one ends up either at the fixed point 6210001000 or at the 2-cycle 6300000100-->7101001000.

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