A110305 Factors of alternators which produce least alternating multiples.
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
Offset: 1
Examples
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.
Links
- 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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