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Irregular triangle read by rows where row n contains the balanced binary words of length 2n interpreted as binary numbers.
6

%I #46 Jan 09 2024 10:28:26

%S 1,2,3,5,6,9,10,12,7,11,13,14,19,21,22,25,26,28,35,37,38,41,42,44,49,

%T 50,52,56,15,23,27,29,30,39,43,45,46,51,53,54,57,58,60,71,75,77,78,83,

%U 85,86,89,90,92,99,101,102,105,106,108,113,114,116,120,135

%N Irregular triangle read by rows where row n contains the balanced binary words of length 2n interpreted as binary numbers.

%C Within a row, strings are ordered lexicographically, which means the resulting values are ordered numerically.

%C This is from an idea of David Lovler, which he calls "zigzags". It is a rearrangement of A072601. A072603 lists all the numbers that are not in this sequence. A000984 gives the number of coin flip sequences of length 2,4,6, etc.

%C Not a permutation of the integers. E.g. 8 never occurs. When there are more 0's than 1's, adding 0's doesn't bring it to balance. - _Kevin Ryde_, Aug 31 2023

%H Paolo Xausa, <a href="/A362030/b362030.txt">Table of n, a(n) for n = 1..17576</a> (rows 1..8 of the triangle, flattened).

%e The first few terms written as binary words with leading 0's: 01, 10, 0011, 0101, 0110, 1001, 1010, 1100, 000111, 001011, 001101, 001110, ... (cf. A368804).

%e Triangle T(n,k) begins:

%e 1, 2;

%e 3, 5, 6, 9, 10, 12;

%e 7, 11, 13, 14, 19, 21, 22, 25, 26, 28, 35, 37, 38, ...;

%e 15, 23, 27, 29, 30, 39, 43, 45, 46, 51, 53, 54, 57, ...;

%e ...

%p T:= n-> sort(map(Bits[Join], combinat[permute]([0$n, 1$n])))[]:

%p seq(T(n), n=1..4); # _Alois P. Heinz_, Apr 13 2023

%t T[n_] := Sort[FromDigits[#, 2] & /@ Permutations[Join[ConstantArray[0, n], ConstantArray[1, n]]]]; Flatten[Table[T[n], {n, 1, 4}]][[1 ;; 64]] (* _Robert P. P. McKone_, Aug 29 2023 *)

%Y Columns k=1-2 give: A000225, A083329.

%Y Row sums give A131568.

%Y Main diagonal gives A036563(n+1).

%Y Cf. A000984 (row lengths), A072601, A072603, A368804 (binary).

%K nonn,tabf,easy

%O 1,2

%A _Louis Conover_, Apr 05 2023