A351939 Irregular triangle read by rows: the n-th row contains the values 0..2^n-1 sorted first by Hamming weight and then by position in reflected Gray code.

0, 0, 1, 0, 1, 2, 3, 0, 1, 2, 4, 3, 6, 5, 7, 0, 1, 2, 4, 8, 3, 6, 5, 12, 10, 9, 7, 13, 14, 11, 15, 0, 1, 2, 4, 8, 16, 3, 6, 5, 12, 10, 9, 24, 20, 18, 17, 7, 13, 14, 11, 25, 26, 28, 21, 22, 19, 15, 27, 30, 29, 23, 31, 0, 1, 2, 4, 8, 16, 32, 3, 6, 5, 12, 10, 9, 24, 20, 18, 17, 48, 40, 36, 34, 33, 7, 13, 14, 11, 25

Offset

0,6

Comments

Values in the range 0..2^n-1 correspond to length n binary vectors. The Hamming weight is the number of 1 bits. Reflected Gray code is described in A003188 and A014550.

Rows can be subdivided into subsequences of vectors having the same Hamming weight. Within each subsequence, adjacent values will differ by 2 bits. For example, the subsequence of length 5 vectors with Hamming weight 2 is 00011, 00110, 00101, 01100, 01010, 01001, 11000, 10100, 10010, 10001 (in decimal 3, 6, 5, 12, 10, 9, 24, 20, 18, 17).

A revolving door algorithm can be used to enumerate values of the same weight in the required order. See Knuth ("Gray codes for combinations", p. 362) for additional information.

References

D. Knuth, The art of computer programming, Volume 4A: Combinatorial Algorithms, Part 1, Addison-Wesley, 2011.

F. Ruskey, Combinatorial Generation. Working Version (1j-CSC 425/520), 2003.

Links

Valentin Bakoev, Table of n, a(n) for n = 0..65535

Valentin Bakoev, Some problems and algorithms related to the weight order relation on the n-dimensional Boolean cube, Discrete Mathematics, Algorithms and Applications, Vol. 13 No 3, 2150021 (2021); arXiv preprint, arXiv:1811.04421 [cs.DM], 2018-2020.

Valentin Bakoev, Ordering the Boolean Cube Vectors by Their Weights and with Minimal Change, Int'l Conf. Algebraic Informatics (CAI 2022) Lecture Notes Comp. Sci. (LNCS) Vol. 13706, 43-54.

Sage Reference Manual, Gray codes

Formula

The n-th row is the concatenation of the subsequences g(n, 0), ..., g(n, n), where the subsequences are defined as follows:

g(n, 0) = (0),

g(n, n) = (2^n - 1),

g(n, k) = g(n-1, k) concatenate (g^r(n-1, k-1) + 2^(n-1)) for 0 < k < n.

In the above, g^r(n-1, k-1) + 2^(n-1) means the 2^(n-1) is added to each member of the subsequence g(n-1, k-1) in reversed order.

Example

Triangle T(n,k) begins:
n=0: 0;
n=1: 0, 1;
n=2: 0, 1, 2, 3;
n=3: 0, 1, 2, 4, 3, 6, 5, 7;
n=4: 0, 1, 2, 4, 8, 3, 6, 5, 12, 10, 9, 7, 13, 14, 11, 15;
n=5: 0, 1, 2, 4, 8, 16, 3, 6, 5, 12, 10, 9, 24, 20, 18, 17, 7, 13, 14, 11, 25, 26, 28, 21, 22, 19, 15, 27, 30, 29, 23, 31;
...
For row n = 3, the binary words of length 3 in reflected Gray code order are 000, 001, 011, 010, 110, 111, 101, 100. Arranging these by Hamming weight but otherwise preserving the order gives 000, 001, 010, 100, 011, 110, 101, 111. As decimal numbers these are 0, 1, 2, 4, 3, 6, 5, 7, which is row 3.

Maple

b:= proc(n) b(n):= `if`(n<2, n, Bits[Xor](n, b(iquo(n, 2)))) end:
h:= proc(n) h(n):= add(i, i=Bits[Split](n)) end:
T:= n-> sort([$0..2^n-1], (x, y)-> h(x)<h(y) or h(x)=h(y) and b(x)<b(y))[]:
seq(T(n), n=0..6);  # Alois P. Heinz, Mar 01 2022

Mathematica

b[n_] := If[n < 2, n, BitXor[n, b[Quotient[n, 2]]]];
h[n_] := DigitCount[n, 2, 1];
T[n_] := SortBy[{h, b}][Range[0, 2^n - 1]];
Table[T[n], {n, 0, 6}] // Flatten (* Jean-François Alcover, Oct 21 2022, after Alois P. Heinz *)

Prog

(PARI) row(n)=vecsort(vector(2^n, i, i--; bitxor(i, i>>1)), (x, y) -> cmp(hammingweight(x), hammingweight(y)))
{ for(n=0, 5, print(row(n))) } \\ Andrew Howroyd, Feb 28 2022

Crossrefs

Row sums are A006516.

Right border gives A000225.

Left border gives A000004.

T(n,n) gives A131577.

Cf. A000120 (Hamming weight), A003188, A014550 (reflected Gray code), A294648 (weight-lexicographic order).

Sequence in context: A356784 A359941 A360302 * A170899 A221321 A367959

Adjacent sequences: A351936 A351937 A351938 * A351940 A351941 A351942

Keyword

nonn,tabf

Author

Valentin Bakoev, Feb 26 2022

Status

approved