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A211968 Triangle of binary numbers with some initial repeats.

%I A211968 #35 Feb 12 2025 10:43:27
%S A211968 11,110,111,1010,1100,1101,1110,1111,10100,10101,11000,11001,11010,
%T A211968 11011,11100,11101,11110,11111,100100,101000,101001,101010,101011,
%U A211968 101101,110000,110001,110010,110011,110100,110101,110110,110111,111000,111001,111010,111011
%N A211968 Triangle of binary numbers with some initial repeats.
%C A211968 Triangle read by rows in which row n lists the binary numbers with n digits and with some initial repeats, n >= 2.
%C A211968 Also triangle read by rows in which row n lists the binary words of length n with some initial repeats and with initial digit 1, n >= 2.
%H A211968 Alois P. Heinz, <a href="/A211968/b211968.txt">Rows n = 2..14, flattened</a>
%H A211968 <a href="/index/Cu#curling_numbers">Index entries for sequences related to curling numbers</a>
%e A211968 Triangle begins, starting at row 2:
%e A211968   11;
%e A211968   110, 111;
%e A211968   1010, 1100, 1101, 1110, 1111;
%e A211968   10100, 10101, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111;
%p A211968 s:= proc(n) s(n):= `if`(n=1, [[1]], map(x->
%p A211968       [[x[], 0], [x[], 1]][], s(n-1))) end:
%p A211968 T:= proc(n) map(x-> parse(cat(x[])), select(proc(l) local i;
%p A211968       for i to iquo(nops(l), 2) do if l[1..i]=l[i+1..2*i]
%p A211968       then return true fi od; false end, s(n)))[] end:
%p A211968 seq(T(n), n=2..7);  # _Alois P. Heinz_, Dec 04 2012
%t A211968 T[n_] := FromDigits /@ Select[Range[2^(n-1), 2^n-1] // IntegerDigits[#, 2]&, FindTransientRepeat[Reverse[#], 2][[2]] != {}&];
%t A211968 Table[T[n], {n, 2, 7}] // Flatten (* _Jean-François Alcover_, Feb 12 2025 *)
%Y A211968 Complement in base 2 of A211027.
%Y A211968 Rows lengths give: A093370.
%Y A211968 Cf. A093370, A093371, A121880, A122536, A211967, A211969, A211973, A216955.
%K A211968 nonn,tabf,base
%O A211968 2,1
%A A211968 _Omar E. Pol_, Dec 03 2012