A057432 -id:A057432 - cp's OEIS frontend

A088208 Table read by rows where T(0,0)=1; n-th row has 2^n terms T(n,j),j=0 to 2^n-1. For j==0 mod 2, T(n+1,2j)=T(n,j) and T(n+1,2j+1)=T(n,j)+2^n. For j==1 mod 2, T(n+1,2j+1)=T(n,j) and T(n+1,2j)=T(n,j)+2^n.

1, 1, 2, 1, 3, 4, 2, 1, 5, 7, 3, 4, 8, 6, 2, 1, 9, 13, 5, 7, 15, 11, 3, 4, 12, 16, 8, 6, 14, 10, 2, 1, 17, 25, 9, 13, 29, 21, 5, 7, 23, 31, 15, 11, 27, 19, 3, 4, 20, 28, 12, 16, 32, 24, 8, 6, 22, 30, 14, 10, 26, 18, 2, 1, 33, 49, 17, 25, 57, 41, 9, 13, 45, 61, 29, 21, 53, 37, 5, 7, 39, 55, 23

Offset: 1

Gary W. Adamson, Sep 23 2003

Schroeder, p. 281 states "The ordering with which the iterates x_n fall into the 2^m different chaos bands [order as to magnitude] is also the same as the ordering of the iterates in a stable orbit of period length P = 2^m. For example, for both the period-4 orbit and the four chaos bands, the iterates, starting with the largest iterate x_1, are ordered as follows: x_1 > x_3 > x_4 > x_2."
From Andrey Zabolotskiy, Dec 06 2024: (Start)
For n>0, row n-1 is the permutation relating row n of the left half of Stern-Brocot tree with row n of Kepler's tree of fractions. 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 SB_n(k) = K_n(T(n-1, k)).
The inverse permutation is row n of A131271.
Equals A362160+1. (End)

			1
1 2
1 3 4 2
1 5 7 3 4 8 6 2
1 9 13 5 7 15 11 3 4 12 16 8 6 14 10 2

Manfred R. Schroeder, "Fractals, Chaos, Power Laws", W.H. Freeman, 1991, p. 282.

Reinhard Zumkeller, Rows n = 1..13 of triangle, flattened

Cf. A088372.

Cf. A049773, A362160, A131271, A294442, A057432.

a088208 n k = a088208_tabf !! (n-1) !! (k-1)
a088208_row n = a088208_tabf !! (n-1)
a088208_tabf = iterate f [1] where
   f vs = (map (subtract 1) ws) ++ reverse ws where ws = map (* 2) vs
-- Reinhard Zumkeller, Mar 14 2015

nmax = 6;
T[, 0] = 1; T[n, j_] /; j == 2^n = n;
Do[Which[
  EvenQ[j], T[n+1, 2j] = T[n, j]; T[n+1, 2j+1] = T[n, j] + 2^n,
  OddQ[j], T[n+1, 2j+1] = T[n, j]; T[n+1, 2j] = T[n, j] + 2^n],
{n, 0, nmax}, {j, 0, 2^n-1}];
Table[T[n, j], {n, 0, nmax}, {j, 0, 2^n-1}] // Flatten (* Jean-François Alcover, Aug 03 2018 *)

Edited by Ray Chandler and N. J. A. Sloane, Oct 08 2003

A131271 Triangular array T(n,k), n>=0, k=1..2^n, read by rows in bracketed pairs such that highest ranked element is bracketed with lowest ranked.

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 3, 1, 8, 4, 5, 2, 7, 3, 6, 1, 16, 8, 9, 4, 13, 5, 12, 2, 15, 7, 10, 3, 14, 6, 11, 1, 32, 16, 17, 8, 25, 9, 24, 4, 29, 13, 20, 5, 28, 12, 21, 2, 31, 15, 18, 7, 26, 10, 23, 3, 30, 14, 19, 6, 27, 11, 22, 1, 64, 32, 33, 16, 49, 17, 48, 8, 57

Offset: 0

J. Demongeot (Jacques.Demongeot(AT)imag.fr), Jun 24 2007

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)

			Triangle begins:
1;
1,  2;
1,  4, 2, 3;
1,  8, 4, 5, 2,  7, 3,  6;
1, 16, 8, 9, 4, 13, 5, 12, 2, 15, 7, 10, 3, 14, 6, 11;
...

Alois P. Heinz, Table of n, a(n) for n = 0..8190
John M. Campbell, A generalization of Deterministic Finite Automata related to discharging, arXiv:2506.14072 [cs.FL], 2025. See p. 7.
Jacques Demongeot and Jules Waku, Counter-Examples about Lower- and Upper-Bounded Population Growth, Math. Pop. Studies 12 (2005), 199-210.
Jacques Demongeot and Jules Waku, Application of interval iterations to the entrainment problem in respiratory physiology, Phil. Trans. R. Soc. A, 367 (2009), 4717-4739.
Wikipedia, Bracket (tournament)

Cf. A049773, A088208, A294442, A057432.
Cf. A005578 (last elements in rows), A155944 (T(n,2^(n-1)) for n>0).
Cf. A006519, A240908, A240909, A240910.

T:= proc(n,k) option remember;
      `if`({n, k} = {1}, 1,
      `if`(irem(k, 2)=1, T(n-1, (k+1)/2), 2^(n-1)+1 -T(n-1, k/2)))
    end:
seq(seq(T(n, k), k=1..2^(n-1)), n=1..7); # Alois P. Heinz, Feb 28 2012, with offset 1

T[0, 1] = 1;
T[n_, k_] := T[n, k] = If[Mod[k, 2] == 1, T[n, (k + 1)/2], 2^n + 1 - T[n, k/2]];
Table[T[n, k], {n, 0, 6}, {k, 2^n}] // Flatten (* Jean-François Alcover, May 31 2018, after Alois P. Heinz *)

T(0,1) = 1, T(n,2k-1) = T(n-1,k), T(n,2k) = 2^n+1 - T(n-1,k).

T(n,1) = 1; for 1 < k <= 2^n, T(n,k) = 1 + (2^n)/m - T(n,k-m), where m = A006519(k-1). - Mathew Englander, Jun 20 2021

Edited (with new name from Colin Hall) by Andrey Zabolotskiy, Dec 06 2024

A057431 Obtained by reading first the numerator then the denominator of fractions in full Stern-Brocot tree (A007305/A047679).

Original entry on oeis.org

0, 1, 1, 0, 1, 1, 1, 2, 2, 1, 1, 3, 2, 3, 3, 2, 3, 1, 1, 4, 2, 5, 3, 5, 3, 4, 4, 3, 5, 3, 5, 2, 4, 1, 1, 5, 2, 7, 3, 8, 3, 7, 4, 7, 5, 8, 5, 7, 4, 5, 5, 4, 7, 5, 8, 5, 7, 4, 7, 3, 8, 3, 7, 2, 5, 1, 1, 6, 2, 9, 3, 11, 3, 10, 4, 11, 5, 13, 5, 12, 4, 9, 5, 9, 7, 12, 8, 13, 7, 11, 7, 10, 8, 11, 7, 9, 5, 6, 6, 5

Offset: 0

N. J. A. Sloane, Sep 08 2000

When presented in this way, every row (e.g. row 3, 1 3 2 3 3 2 3 1) is a palindrome. - Joshua Zucker, May 11 2006

Alois P. Heinz, Table of n, a(n) for n = 0..10000
N. J. A. Sloane, Stern-Brocot or Farey Tree
Index entries for sequences related to Stern's sequences

Cf. A007305, A047679, A007306, A002487, A057432.

F:= proc(n) option remember; local t;
t:= L -> [[L[1], [L[1][1]+L[2][1], L[1][2]+L[2][2]], L[2]],
           [L[2], [L[2][1]+L[3][1], L[2][2]+L[3][2]], L[3]]][];
      if n=0 then [[[ ], [0, 1], [ ]], [[ ], [1, 0], [ ]]]
    elif n=1 then [[[0, 1], [1, 1], [1, 0]]]
             else map(t, F(n-1))
      fi
    end:
aa:= n-> map(x-> x[], [seq(map(x-> x[2], F(j))[], j=0..n)])[]:
aa(7);   # aa(n) gives the first 2^(n+1)+2 terms
# Alois P. Heinz, Jan 13 2011

sbt[n_] := Module[{R, L, Y, w, u},
   R = {{1, 0}, {1, 1}};
   L = {{1, 1}, {0, 1}};
   Y = {{1, 0}, {0, 1}};
   w[b_] := Fold[#1.If[#2 == 0, L, R]&, Y, b];
   u[a_] := {a[[2, 1]] + a[[2, 2]], a[[1, 1]] + a[[1, 2]]};
   Map[u, Map[w, Tuples[{0, 1}, n]]]];
Join[{0, 1, 1, 0}, Table[sbt[n], {n, 0, 5}]] // Flatten (* Jean-François Alcover, Sep 06 2022, after Peter Luschny in A007305 *)

More terms from Joshua Zucker, May 11 2006

cp's OEIS Frontend

A088208 Table read by rows where T(0,0)=1; n-th row has 2^n terms T(n,j),j=0 to 2^n-1. For j==0 mod 2, T(n+1,2j)=T(n,j) and T(n+1,2j+1)=T(n,j)+2^n. For j==1 mod 2, T(n+1,2j+1)=T(n,j) and T(n+1,2j)=T(n,j)+2^n.

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A131271 Triangular array T(n,k), n>=0, k=1..2^n, read by rows in bracketed pairs such that highest ranked element is bracketed with lowest ranked.

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A057431 Obtained by reading first the numerator then the denominator of fractions in full Stern-Brocot tree (A007305/A047679).

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