cp's OEIS Frontend

A Thue-Morse generator using the "Ordering of Iterates" algorithm.

n-th row = (r(1),r(2),...,r(m)), where m=2^(n-1), satisfies r(r(k))=k for k=1,2,...,m and has exactly 2^[ n/2 ] solutions of r(k)=k. (The function r(k) reverses bits. Or rather, r(k)=revbits(k-1)+1.)

In a knockout competition with m players, arranging the competition brackets (see links) in r(k) order, where k is the rank of the player, ensures that highest ranked players cannot meet until the later stages of the competition. None of the top 2^p ranked players can meet earlier than the p-th from last round of the competition. At the same time the top ranked players in each match meet the highest ranked player possible consistent with this rule. The sequence for the top ranked players meeting the lowest ranked player possible is A131271. - Colin Hall, Jul 31 2011, Feb 29 2012

Row n contains one of A003407(2^(n-1)) non-averaging permutations of [2^(n-1)], i.e., a permutation of [2^(n-1)] without 3-term arithmetic progressions. - Alois P. Heinz, Dec 05 2017

In a knockout competition with 2^n players, arranging the competition brackets (see Wikipedia) in T(n,k) order, where T(n,k) is the rank of the k-th player, ensures that highest ranked players cannot meet until the later stages of the competition. None of the top 2^p ranked players can meet earlier than the p-th from last round of the competition. At the same time the top ranked players in each match meet the lowest ranked player possible consistent with this rule. The sequence for the top ranked players meeting the highest ranked player possible is A049773. - Colin Hall, Feb 28 2012

Ranks in natural order of 2^n increasing real numbers appearing in limit cycles of interval iterations, or Median Spiral Order. [See the works by Demongeot & Waku]

From Andrey Zabolotskiy, Dec 06 2024 (Start):

For n>0, row n-1 is the permutation relating row n of Kepler's tree of fractions with row n of the left half of Stern-Brocot tree. Specifically, if K_n(k) [resp. SB_n(k)] is the k-th fraction in the n-th row of A294442 [resp. A057432], where 1/2 is in row 1 and k=1..2^(n-1), then K_n(k) = SB_n(T(n-1, k)).

The inverse permutation is row n of A088208.

When 1 is subtracted from each term, rows 3-5 become A240908, A240909, A240910. (End)

Let s(0)=1; s(n+1)=s(n),ri(n), where ri(n) is s(n) reversed and incremented. Each s(n) is an initial part of this sequence.

For each m, a(1 to 2^m) is a permutation of A063787(1 to 2^m). For k=1 to 2^m, a(2^m+1-A088372(m,k)) = A063787(k).

Partial sums give A164910: (1, 3, 6, 8, 11, 15, 20, ...).

a(0) = 1, then using the dragon curve sequence A014577: (1, 1, 0, 1, 1, ...) as a code: (1 = add to current term, 0 = subtract from current term, to get the next term), see example.

Rows of A088696 tend to this sequence.

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.

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.

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.