This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A356120 #30 Jan 08 2023 16:29:42
%S A356120 0,0,1,0,2,3,1,0,4,6,7,5,2,3,1,0,8,12,14,15,13,10,11,9,4,6,7,5,2,3,1,
%T A356120 0,16,24,28,30,31,29,26,27,25,20,22,23,21,18,19,17,8,12,14,15,13,10,
%U A356120 11,9,4,6,7,5,2,3,1,0,32,48,56,60,62,63,61,58,59,57,52,54,55,53,50,51,49,40,44,46,47,45,42
%N A356120 Irregular triangle read by rows: the n-th row lists the values 0..2^n-1 representing all subsets of a set of n elements. When its elements are linearly ordered, the subsets are lexicographically ordered.
%C A356120 The sequence in the n-th row of the triangle is denoted by row(n). It contains the values in the range 0..2^n-1 corresponding to the binary vectors of length n. The integers in row (n), considered as characteristic vectors of the set B_n = {b_n, b_(n-1), ..., b_2, b_1}, whose elements are linearly ordered: b_n < b_(n-1) < ... < b_2 < b_1, define all subsets of B_n in lexicographic order. To obtain row(n) we reason inductively as follows. Obviously, row(0) = 0 = a(0) and corresponds to the empty set {}. Assume that the sequence row(n-1) = i_0, i_1, ..., i_(2^(n-1)-1) is obtained. It defines a lexicographic order of the subsets of B_(n-1) = {b_(n-1) , ..., b_2, b_1} - note that its elements linearly ordered. Then row (n) = i_0, 2^(n-1) + i_0, 2^(n-1) + i_1, ..., 2^(n-1) + i_(2^(n-1)-1), i_1, ..., i_(2^(n-1)-1) defines all subsets of B_n in lexicographic order. In other words, row(n) = 0, (row(n-1) + 2^(n-1)), (row(n-1) without the first term 0), i.e., row(n-1) is taken twice: first, all terms of row(n-1) are incremented by 2^(n-1) and second, the resulting sequence is inserted after the first term of row(n-1), which is always 0 and corresponds to {}.
%D A356120 Donald E. Knuth, The Art of Computer Programming, Volume 4A, Section 7.2.1.3 Exercise 19 (Binomial tree traversed in post-order).
%H A356120 Valentin Bakoev, <a href="/A356120/b356120.txt">Table of n, a(n) for n = 0..1023</a>
%H A356120 Joerg Arndt, <a href="https://www.jjj.de/fxt/#fxtbook">Matters Computational (The Fxtbook)</a>, section 1.26 "Binary words in lexicographic order for subsets", pp. 70-74.
%H A356120 Joerg Arndt, <a href="https://arxiv.org/abs/1405.6503">Subset-lex: did we miss an order?</a>, arXiv:1405.6503 [math.CO], 2014-2015.
%H A356120 Valentin Bakoev, <a href="https://doi.org/10.1109/ICAI55857.2022.9959993">An Algorithm for Generating All Subsets in Lexicographic Order</a>, ICAI 2022, pp. 271-275.
%F A356120 row(n) is defined as:
%F A356120 row(0) = 0;
%F A356120 row(n) = 0, (2^(n-1)+row(n-1)), (row(n-1)\{0}),
%F A356120 where (2^(n-1) + row(n-1)) means a subsequence obtained by adding 2^(n-1) to every term of row(n-1), and (row(n-1)\{0}) means a subsequence row(n-1) without its first term 0, for n = 0, 1, 2, ...).
%F A356120 Then a(n) is defined as:
%F A356120 a(2^m - 1) = 0, for m = 0, 1, 2, ..., and
%F A356120 a(2^m + k) = 2^(m-1) + a(2^(m-1) + k - 1), for k = 0, 1, ..., 2^(m-1) - 1, and
%F A356120 a(2^m + 2^(m-1) + k) = a(2^(m-1)+k), for k = 0, 1, ..., 2^(m-1) - 2.
%e A356120 For n = 1, 2, 3, the sets B_n, their subsets (the column under B_n), binary characteristic words (column bin.) and corresponding integers (column dec.) are:
%e A356120 B_1 = {c} bin. dec. | B_2 = {b, c} bin. dec. | B_3 = {a, b, c} bin. dec.
%e A356120 {} 0 0 | {} 00 0 | {} 000 0
%e A356120 {c} 1 1 | {b} 10 2 | {a} 100 4
%e A356120 | {b, c} 11 3 | {a, b} 110 6
%e A356120 | {c} 01 1 | {a, b, c} 111 7
%e A356120 | {a, c} 101 5
%e A356120 | {b} 010 2
%e A356120 | {b, c} 011 3
%e A356120 | {c} 001 1
%e A356120 As seen, when B = {a, b, c}, its subsets {}, {a}, {a, b}, {a, b, c}, {a, c}, {b}, {b, c}, {c} are in lexicographic order, the corresponding binary words of length 3 are 000, 100, 110, 111, 101, 010, 011, 001, and so row(3) = 0, 4, 6, 7, 5, 2, 3, 1.
%e A356120 Triangle T(n,k) begins:
%e A356120 k=0 1 2 3 4 5 6 7 ...
%e A356120 n=0: 0;
%e A356120 n=1: 0, 1;
%e A356120 n=2: 0, 2, 3, 1;
%e A356120 n=3: 0, 4, 6, 7, 5, 2, 3, 1;
%e A356120 n=4: 0, 8, 12, 14, 15, 13, 10, 11, 9, 4, 6, 7, 5, 2, 3, 1;
%e A356120 n=5: 0, 16, 24, 28, 30, 31, 29, 26, 27, 25, 20, 22, 23, 21, 18, 19, 17, 8, 12, 14, 15, 13, 10, 11, 9, 4, 6, 7, 5, 2, 3, 1,
%e A356120 ...
%t A356120 (* computing row(n) *)
%t A356120 n = 5;
%t A356120 Array[row, 2^n];
%t A356120 row[0] = 0; row[1] = 1;
%t A356120 len = 2;
%t A356120 For[i = 2, i <= n, i++,
%t A356120 For[j = 1, j < len, j++, row[j + len] = row[j]];
%t A356120 For[j = len, j > 0, j--, row[j] = row[j - 1] + len];
%t A356120 len = len*2;
%t A356120 ];
%o A356120 (C++) // To generate a(0), a(1), ..., a(2^m-1) of A356120
%o A356120 void a356120 (int m) {
%o A356120 a[0] = 0;
%o A356120 uint clen = 2; // length of the current row (= 2^i)
%o A356120 uint plen = 1; // length of the previous row (= 2^(i-1))
%o A356120 for (int i = 1; i <= m; i++) {
%o A356120 a[clen-1] = 0;
%o A356120 for (int k = 0; k < plen; k++)
%o A356120 a[clen + k] = plen + a[plen + k - 1];
%o A356120 for (int k = 0; k < plen-1; k++)
%o A356120 a[clen + plen + k] = a[plen + k];
%o A356120 plen = clen;
%o A356120 clen *= 2;
%o A356120 }
%o A356120 }
%Y A356120 Cf. A006516 (row sums), A000225 (main diagonal, the n-th term of row(n)).
%Y A356120 row(n) without leading 0, when read in reverse order, gives the first 2^n-1 terms of A108918.
%Y A356120 Starting with the n-th term of row(n), columns 0..4 are: A000004, A131577 without 0, A007283, A135092, A110286.
%K A356120 nonn,tabf
%O A356120 0,5
%A A356120 _Valentin Bakoev_, Jul 27 2022