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A110305
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Factors of alternators which produce least alternating multiples.
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3
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1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 4, 1, 2, 1, 2, 1, 2, 0, 1, 19, 1, 3, 1, 2, 1, 2, 1, 1, 11, 1, 5, 1, 2, 1, 2, 1, 2, 0, 1, 5, 1, 14, 1, 2, 1, 2, 1, 1, 11, 1, 4, 1, 3, 1, 8, 1, 4, 0, 1, 7, 1, 4, 1, 13, 1, 4, 1, 1, 11, 1, 4, 1, 6, 1, 5, 1, 6, 0, 1, 5, 1, 3, 1, 3, 1, 7, 1, 1, 12, 1, 18, 1, 23, 1, 34, 1, 111, 0
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,11
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COMMENTS
| An alternating integer is a positive integer for which, in base-10, the parity of its digits alternates. E.g. 121 is alternating because its consecutive digits are odd-even-odd, 1 being odd and 2 even. Of course, 1234567890 is also alternating. An alternator is a positive integer which has a multiple which is alternating.
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REFERENCES
| 45th International Mathematical Olympiad (45th IMO), Problem #6 and Solution, Mathematics Magazine, 2005, Vol. 78, No. 3, pp. 247, 250-251.
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LINKS
| Walter Nissen, Home Page (listed in lieu of email address)
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EXAMPLE
| a(13) is 4 because the least multiple of 13 which is alternating is 52, which is 13 * 4. Of course 13, 26 and 39 are not alternating. 52 is alternating because 5 is odd and 2 is even.
For n congruent to 0 mod 20, a(n) is shown as zero to indicate that n is not an alternator.
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CROSSREFS
| Cf. A030141, A110303, A110304.
Sequence in context: A010198 A204846 A127991 * A010199 A113492 A010200
Adjacent sequences: A110302 A110303 A110304 * A110306 A110307 A110308
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KEYWORD
| base,easy,nonn
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AUTHOR
| Walter Nissen Jul 18 2005
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