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%I #14 Jan 14 2017 11:41:31
%S 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,
%T 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,
%U 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
%N Factors of alternators which produce least alternating multiples.
%C 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.
%C For n congruent to 0 mod 20, a(n) is shown as zero to indicate that n is not an alternator.
%H 45th International Mathematical Olympiad (45th IMO), <a href="http://www.jstor.org/stable/30044168">Problem #6 and Solution</a>, Mathematics Magazine, 2005, Vol. 78, No. 3, pp. 247, 250-251.
%e 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.
%Y Cf. A030141, A110303, A110304.
%K base,easy,nonn
%O 1,11
%A _Walter Nissen_, Jul 18 2005
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