%I
%S 0,1,0,0,1,0,1,1,0,0,0,1,1,0,1,1,1,0,0,0,0,1,0,1,0,1,1,1,0,1,1,1,1,0,
%T 0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0,0,0,1,0,0,1,1,1,0,1,0,0,1,1,0,0,1,
%U 1,1,0,1,0,1,1,1,1,0,1,1,1,1,1,0,0,1
%N Lexicographic blockfractal zeroone word with initial block 01.
%C To the initial block, 01, append the lexicographically ordered missing 2letter words (00,10,11) to get 01001011. To that, append the missing 3letter words to get 01001011000110111. To that, append the missing 4letter words to get 010010110001101110000101011101111, etc. In the limiting word, every finite binary word occurs infinitely many times; thus, the word (or sequence) is blockfractal, as defined at A280511.
%H Clark Kimberling, <a href="/A282244/b282244.txt">Table of n, a(n) for n = 1..10000</a>
%t str = "01"; t = Table[str = str <> StringJoin[Map[#[[1]] &,
%t Select[Map[{#, Length[StringPosition[str, #, 1]] > 0} &,
%t Table[StringJoin[Map[ToString, IntegerDigits[n, 2, k]]], {n,
%t 0, 2^k  1}]], ! #[[2]] &]]], {k, 7}]
%t ToExpression[Characters[Last[t]]] (* _Peter J. C. Moses, Mar 11 2017 *)
%Y Cf. A280511.
%K nonn,easy
%O 1
%A _Clark Kimberling_, Mar 16 2017
