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%I #13 Oct 19 2017 03:14:28
%S 11,31,199
%N a(n) = length k of longest binary sequence x(1) ... x(k) such that for no n <= i < j <= k/2 is x(i) ... x(2i) a subsequence of x(j) ... x(2j).
%C Doesn't the binary sequence 000010011001110011101010101010101010101100110 demonstrate that a(2)>=45 ? - _R. J. Mathar_, Jul 29 2007 Answer: No - see the following comment.
%C The sequence of length 45 above does not satisfy the requirements of the definition: Subsequences are not required to be consecutive. Therefore it cannot show a(2)>=45. In the sequence we find for i=2, j=3: x(i..2i) is 000; x(j..2j) is 001001; and 000 is a subsequence of 001001. - _Don Reble_, May 13 2008
%D a(1) - a(3) computed by R. Dougherty, who finds that a(4) >= 187205.
%H H. M. Friedman, <a href="http://dx.doi.org/10.1006/jcta.2000.3154">Long finite sequences</a>, J. Comb. Theory, A 95 (2001), 102-144.
%e a(1) = 11 from 01110000000.
%Y See A093383-A093386 for illustrations of a(2) and a(3). Cf. A014221, A094091.
%K nonn,bref,nice,more
%O 1,1
%A _N. J. A. Sloane_, Apr 29 2004
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