A088372 -id:A088372 - 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

A088748 a(n) = 1 + Sum_{k=0..n-1} 2 * A014577(k) - 1.

Original entry on oeis.org

1, 2, 3, 2, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 5, 6, 7, 6, 5, 6, 5, 4, 3, 4, 5, 4, 5, 6, 5, 4, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 5, 6, 5, 4, 5, 6, 7, 6, 5, 6, 5, 4, 5, 6, 7, 6, 7, 8, 7, 6, 5, 6, 7, 6, 5, 6, 5, 4, 3, 4, 5, 4, 5, 6

Offset: 0

Gary W. Adamson, Oct 14 2003

nonn

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 first 8 terms of the sequence = (1, 2, 3, 2, 3, 4, 3, 2), where the first four terms = (1, 2, 3, 2). Reverse, add 1, getting (3, 4, 3, 2), then append.
The sequence begins with "1", then using the dragon curve coding, we get:
1...2...3...2...3...4... = A088748
....1...1...0...1...1... = A014577, the dragon curve.

Michael De Vlieger, Table of n, a(n) for n = 0..16383
J.-P. Allouche, G.-N. Han, and J. Shallit, On some conjectures of P. Barry, arXiv:2006.08909 [math.NT], 2020.
J.-P. Allouche and J. Shallit, On three conjectures of P. Barry, arxiv preprint arXiv:2006.04708 [math.NT], June 8 2020.
Paul Barry, Some observations on the Rueppel sequence and associated Hankel determinants, arXiv:2005.04066 [math.CO], 2020.
Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
Paul Barry, Conjectures and results on some generalized Rueppel sequences, arXiv:2107.00442 [math.CO], 2021.

Cf. A014577, A063787, A088208, A088372, A088696, A164910, A005811.

Array[1 + Sum[2 (1 - (((Mod[#1, 2^(#2 + 2)]/2^#2)) - 1)/2) - 1 &[k, IntegerExponent[k, 2]], {k, # - 1}] &, 102] (* Michael De Vlieger, Aug 26 2020 *)

a(n) = 1 + A005811(n). [Joerg Arndt, Dec 11 2012]

Edited by Don Reble, Nov 15 2005
Additional comments from Gary W. Adamson, Aug 30 2009
Edited by N. J. A. Sloane, Sep 06 2009

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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