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%I #13 Dec 02 2015 00:32:07
%S 1,4,3,4,13,12,28,10,9,13,13,12,13,112,20
%N First occurrence of the smallest value subsequence of length n in the Kolakoski sequence (A000002).
%C The sequence of minimal sums begins 1, 2, 4, 5, 6, 8, 9, 11, 13, 14, 15, 17, 18, 19, 21, ...
%e a(3) = 3 because the Kolakoski sequence starting at position 3 is 2, 1, 1, which sums to 4, which is the smallest possible sum of 3 consecutive terms.
%e a(8) = 10 because the Kolakoski sequence starting at position 10 is 1, 2, 2, 1, 1, 2, 1, 1, which sums to 11, which is the smallest possible sum of 8 consecutive values in the Kolakoski sequence. Note that we cannot find a sequence of length eight with a sum of 10 because it would have to be of the form 1, 1, 2, 1, 1, 2, 1, 1, which would mean that 2, 1, 2, 1, 2 would have to appear earlier in the sequence, which would mean that 1, 1, 1 would have to appear even earlier in the sequence, which is impossible.
%Y Cf. A000002, A074297, A074298.
%K nonn,more
%O 1,2
%A _Jon Perry_, Sep 21 2002
%E a(8)-a(15) from and edited by _Nathaniel Johnston_, May 02 2011
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