|
%I #34 Feb 27 2021 11:18:19
%S 1,10,100,101,1000,1001,1011,10000,10001,10010,10011,10110,10111,
%T 100000,100001,100010,100011,100101,100110,100111,101100,101110,
%U 101111,1000000,1000001,1000010,1000011,1000100,1000101,1000110,1000111,1001010,1001011,1001100,1001101,1001110,1001111,1011000,1011001,1011100,1011101,1011110,1011111
%N Triangle of binary numbers >= 1 with no initial repeats.
%C Triangle read by rows in which row n lists the binary numbers with n digits and with no initial repeats.
%C Also triangle read by rows in which row n lists the binary words of length n with no initial repeats and with initial digit 1. See also A211029.
%H Alois P. Heinz, <a href="/A211027/b211027.txt">Rows n = 1..15, flattened</a>
%H <a href="/index/Cu#curling_numbers">Index entries for sequences related to curling numbers</a>
%e Triangle begins:
%e 1;
%e 10;
%e 100, 101;
%e 1000, 1001, 1011;
%e 10000, 10001, 10010, 10011, 10110, 10111;
%e 100000, 100001, 100010, 100011, 100101, 100110, 100111, 101100, 101110, 101111;
%e 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1011000, 1011001, 1011100, 1011101, 1011110, 1011111;
%p s:= proc(n) s(n):= `if`(n=1, [[1]], map(x->
%p [[x[], 0], [x[], 1]][], 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
%t T[n_] := If[n == 1, {1}, FromDigits /@ Select[Range[2^(n-1), 2^n-2] // IntegerDigits[#, 2]&, FindTransientRepeat[Reverse[#], 2][[2]] == {}&]];
%t Array[T, 7] // Flatten (* _Jean-François Alcover_, Feb 27 2021 *)
%Y Column 1 is A011557. Row n has length A093371(n).
%Y Cf. A122536, A211029, A211968, A211969, A216955.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Nov 30 2012
|
