%I #27 Aug 22 2015 19:48:53
%S 1,2,12,21,121,122,211,212,1211,1221,1222,2111,2112,2122,12111,12112,
%T 12211,12212,12221,12222,21111,21112,21121,21122,21221,21222,121111,
%U 121112,121122,122111,122112,122121,122211,122212,122221,122222,211111,211112
%N Triangle read by rows in which row n lists the binary words of length n over the alphabet {1,2} with no initial repeats.
%C As usual in the OEIS, binary alphabets are encoded with {1,2} over the alphabet {0,1} the entries contain nonzero "numbers" beginning with 0.
%H Alois P. Heinz, <a href="/A211029/b211029.txt">Rows n = 1..14, flattened</a>
%H <a href="/index/Cu#curling_numbers">Index entries for sequences related to curling numbers</a>
%e The fourth row of triangle of binary sequences is
%e 0100, 0110, 0111, 1000, 1001, 1011 (see section example of A122536) therefore the fourth row of this triangle is
%e 1211, 1221, 1222, 2111, 2112, 2122.
%e The first six rows of triangle are:
%e 1, 2;
%e 12, 21;
%e 121, 122, 211, 212;
%e 1211, 1221, 1222, 2111, 2112, 2122;
%e 12111, 12112, 12211, 12212, 12221, 12222, 21111, 21112, 21121, 21122, 21221, 21222;
%e 121111, 121112, 121122, 122111, 122112, 122121, 122211, 122212, 122221, 122222, 211111, 211112, 211121, 211122, 211212, 211221, 211222, 212211, 212221, 212222;
%p s:= proc(n) s(n):= `if`(n=0, [[]], map(x->
%p [[x[], 1], [x[], 2]][], s(n-1))) end:
%p T:= proc(n) map(x-> parse(cat(x[])), select(proc(l) local i;
%p for i to iquo(nops(l), 2) do if l[1..i]=l[i+1..2*i]
%p then return false fi od; true end, s(n)))[] end:
%p seq(T(n), n=1..7); # _Alois P. Heinz_, Dec 02 2012
%Y Row n has length A122536(n).
%Y Cf. A093371, A211027, A213969, A216955.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Nov 29 2012
%E More terms and name improved by _R. J. Mathar_, Nov 30 2012