%I #11 Jan 31 2023 08:50:25
%S 1,1,2,3,1,2,3,2,3,3,4,4,1,2,3,2,3,3,4,4,2,3,3,4,4,3,4,4,4,1,1,1,2,3,
%T 2,3,3,4,4,2,3,3,4,4,3,4,4,4,1,1,2,3,3,4,4,3,4,4,4,1,1,3,4,4,4,1,1,4,
%U 1,1,1,2,3,2,3,1,2,3,2,3,3,4,4,2,3,3,4,4,3,4,4,4,1,1,2,3,3,4,4,3,4,4,4,1,1
%N From the 4-symbol substitution 1->{2, 3}, 2->{3, 4}, 3->{4}, 4->{1}.
%t s[1] = {2, 3}; s[2] = {3, 4}; s[3] = {4}; s[4] = {1};
%t t[a_] := Join[a, Flatten[s /@ a]];
%t p[0] = {1}; p[1] = t[{1}]; p[n_] := t[p[n - 1]]
%t aa = Flatten[Table[p[n], {n, 0, 4}]]
%K nonn
%O 0,3
%A _Roger L. Bagula_, Apr 25 2005