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 A362160 #96 Dec 07 2024 19:16:18 %S A362160 0,0,1,0,2,3,1,0,4,6,2,3,7,5,1,0,8,12,4,6,14,10,2,3,11,15,7,5,13,9,1, %T A362160 0,16,24,8,12,28,20,4,6,22,30,14,10,26,18,2,3,19,27,11,15,31,23,7,5, %U A362160 21,29,13,9,25,17,1,0,32,48,16,24,56,40,8,12,44,60,28 %N A362160 Irregular triangle read by rows: The n-th row contains 2^n integers corresponding to the words of the n-bit Gray code with the most significant bits changing fastest. %C A362160 The n-th row of the triangle is defined recursively as row(0) = 0 which corresponds to the empty word, and row(n) = row(n-1)0, row^r(n-1)1, for n > 0. Here row(n-1)0 is the sequence of words of the (n-1)-bit Gray code of this type suffixed with 0, and row^r(n-1)1 means the sequence of reflected words (i.e., words are taken in reverse order) of the (n-1)-bit Gray code of this type and then each word is suffixed with 1. %C A362160 Another way to obtain row(n) is by applying the transition sequence A001511(n), which indicates which bit to flip in the current word to get the next word - see the FORMULA section. %C A362160 If we reverse the internal order of bits in each word of row(n), we obtain the binary reflected n-bit Gray code (see A003188) and vice versa. %D A362160 W. Lipski Jr, Combinatorics for programmers, Mir, Moscow, 1988, (in Russian), p. 31, Algorithm 1.13. %D A362160 F. Ruskey, Combinatorial Generation. Working Version (1j-CSC 425/520), 2003, pp. 120-121. %H A362160 Valentin Bakoev, <a href="/A362160/b362160.txt">Rows n = 0..15, flattened</a> %H A362160 Valentin Bakoev, <a href="https://doi.org/10.53656/math2023-6-1-mir">Mirror (left-recursive) Gray Code</a>, Mathematics and Informatics, Vol. 66, Number 6, pp. 559-578, (2023). %F A362160 T(n,k) = 2*T(n-1,k) for 0 <= k < 2^(n-1), and %F A362160 T(n,2^(n-1)+k) = 2*T(n-1,2^(n-1)-k-1) + 1 = T(n,2^(n-1)-k-1) + 1 for 0 <= k < 2^(n-1). %F A362160 T(n,k+1) = T(n,k) XOR 2^(n-A001511(k)). %F A362160 A000120(T(n,n)) = A005811(n). - _Alois P. Heinz_, Jun 27 2023 %F A362160 T(n, k) = A088208(n, k) - 1. - _Andrey Zabolotskiy_, Dec 06 2024 %e A362160 Triangle begins: %e A362160 k = 0 1 2 3 4 5 6 7 ... %e A362160 n=0: 0, %e A362160 n=1: 0, 1, %e A362160 n=2: 0, 2, 3, 1, %e A362160 n=3: 0, 4, 6, 2, 3, 7, 5, 1, %e A362160 n=4: 0, 8, 12, 4, 6, 14, 10, 2, 3, 11, 15, 7, 5, 13, 9, 1, %e A362160 n=5: 0, 16, 24, 8, 12, 28, 20, 4, 6, 22, 30, 14, 10, 26, 18, 2, 3, 19, 27, 11, 15, 31, 23, 7, 5, 21, 29, 13, 9, 25, 17, 1, %e A362160 ... %e A362160 In row n=3, the corresponding binary words of length 3 are 000, 100, 110, 010, 011, 111, 101, and 001 - notice that the most significant bits change the fastest. %p A362160 with(ListTools): with(Bits): %p A362160 T:= (n, k)-> Join(Reverse(Split(Xor(k, iquo(k, 2)), bits=n))): %p A362160 seq(seq(T(n, k), k=0..2^n-1), n=0..6); # _Alois P. Heinz_, Jun 05 2023 %o A362160 (PARI) T(n,k) = fromdigits(Vecrev(binary(bitxor(k,k>>1)), n), 2); \\ _Kevin Ryde_, Apr 17 2023 %Y A362160 Cf. A003188 (Gray code), A006516 (row sums), A000004 (column k=0), A000079 (column k=1), A088208. %Y A362160 T(n,2^n-1) gives A057427. %Y A362160 Cf. also A000120, A005811, A363674. %K A362160 nonn,easy,tabf,base %O A362160 0,5 %A A362160 _Valentin Bakoev_, Apr 10 2023