A059893 -id:A059893 - cp's OEIS frontend

A352515 Number of weak nonexcedances (parts on or below the diagonal) of the n-th composition in standard order.

0, 1, 0, 2, 0, 1, 2, 3, 0, 1, 1, 2, 1, 3, 3, 4, 0, 1, 1, 2, 0, 2, 2, 3, 1, 2, 3, 4, 3, 4, 4, 5, 0, 1, 1, 2, 0, 2, 2, 3, 0, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 3, 4, 4, 5, 2, 4, 4, 5, 4, 5, 5, 6, 0, 1, 1, 2, 0, 2, 2, 3, 0, 1, 2, 3, 2, 3, 3, 4, 0, 1, 1, 2, 2, 3, 3

Offset: 0

Gus Wiseman, Mar 23 2022

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See also A000120, A059893, A070939, A114994, A225620.

			The 89th composition in standard order is (2,1,3,1), with weak nonexcedances {2,3,4}, so a(89) = 3.

MathOverflow, Why 'excedances' of permutations? [closed].

Positions of first appearances are A000225.
The strong version is A352514, counted by A352521 (first column A219282).
The strong opposite version is A352516, counted by A352524 (first A008930).
The opposite version is A352517, counted by A352525 (first column A177510).
Triangle A352522 counts these comps (first col A238874), partitions A115994.
A008292 is the triangle of Eulerian numbers (version without zeros).
A011782 counts compositions.
A173018 counts permutations by number of excedances, weak A123125.
A238349 counts comps by fixed points, first col A238351, rank stat A352512.
A352488 is the weak nonexcedance set of A122111.
A352523 counts comps by unfixed pts, first col A010054, rank stat A352513.
Cf. A088218, A098825, A114088, A131577, A238352.

stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
paw[y_]:=Length[Select[Range[Length[y]],#>=y[[#]]&]];
Table[paw[stc[n]],{n,0,30}]

A352516 Number of excedances (parts above the diagonal) of the n-th composition in standard order.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1

Offset: 0

Gus Wiseman, Mar 23 2022

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See also A000120, A059893, A070939, A114994, A225620.

			The 5392th composition in standard order is (2,2,4,5), with excedances {1,3,4}, so a(5392) = 3.

MathOverflow, Why 'excedances' of permutations? [closed].

Positions of first appearances are A104462.
The opposite version is A352514, counted by A352521 (first column A219282).
The weak opposite version is A352515, counted by A352522 (first A238874).
The weak version is A352517, counted by A352525 (first column A177510).
The triangle A352524 counts these compositions (first column A008930).
A008292 is the triangle of Eulerian numbers (version without zeros).
A011782 counts compositions.
A173018 counts permutations by number of excedances, weak A123125.
A238349 counts comps by fixed points, first col A238351, rank stat A352512.
A352487 is the excedance set of A122111.
A352523 counts comps by unfixed points, first A010054, rank stat A352513.
Cf. A088218, A098825, A114088, A115994, A131577, A238352, A352513, A352523.

stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
pd[y_]:=Length[Select[Range[Length[y]],#

A352517 Number of weak excedances (parts on or above the diagonal) of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2

Offset: 0

Gus Wiseman, Mar 23 2022

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See also A000120, A059893, A070939, A114994, A225620.

			The 169th composition in standard order is (2,2,3,1), with weak excedances {1,2,3}, so a(169) = 3.

MathOverflow, Why 'excedances' of permutations? [closed].

Positive positions of first appearances are A164894.
The version for partitions is A257990.
The strong opposite version is A352514, counted by A352521 (first A219282).
The opposite version is A352515, counted by A352522 (first column A238874).
The strong version is A352516, counted by A352524 (first column A008930).
The triangle A352525 counts these compositions (first column A177510).
A008292 is the triangle of Eulerian numbers (version without zeros).
A011782 counts compositions.
A173018 counts permutations by number of excedances, weak A123125.
A238349 counts comps by fixed points, first col A238351, rank stat A352512.
A352489 is the weak excedance set of A122111.
A352523 counts comps by unfixed points, first A010054, rank stat A352513.
Cf. A088218, A098825, A114088, A115994, A131577, A238352, A352513, A352523.

stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
pdw[y_]:=Length[Select[Range[Length[y]],#<=y[[#]]&]];
Table[pdw[stc[n]],{n,0,30}]

A162911 Numerators of drib tree fractions, where drib is the bit-reversal permutation tree of the Bird tree.

Original entry on oeis.org

1, 1, 2, 2, 3, 1, 3, 3, 5, 1, 4, 3, 4, 2, 5, 5, 8, 2, 7, 4, 5, 3, 7, 4, 7, 1, 5, 5, 7, 3, 8, 8, 13, 3, 11, 7, 9, 5, 12, 5, 9, 1, 6, 7, 10, 4, 11, 7, 11, 3, 10, 5, 6, 4, 9, 7, 12, 2, 9, 8, 11, 5, 13, 13, 21, 5, 18, 11, 14, 8, 19, 9, 16, 2, 11, 12, 17, 7, 19, 9, 14, 4, 13, 6, 7, 5, 11, 10, 17, 3, 13

Offset: 1

Ralf Hinze (ralf.hinze(AT)comlab.ox.ac.uk), Aug 05 2009

The drib tree is an infinite binary tree labeled with rational numbers. It is generated by the following iterative process: start with the rational 1; for the left subtree increment and then reciprocalize the current rational; for the right subtree interchange the order of the two steps: the rational is first reciprocalized and then incremented. Like the Stern-Brocot and the Bird tree, the drib tree enumerates all the positive rationals (A162911(n)/A162912(n)).
From Yosu Yurramendi, Jul 11 2014: (Start)
If the terms (n>0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
   1,
   1, 2,
   2, 3,1, 3,
   3, 5,1, 4, 3, 4,2, 5,
   5, 8,2, 7, 4, 5,3, 7,4, 7,1, 5, 5, 7,3, 8,
   ...
then the sum of the m-th row is 3^m (m = 0,1,2,), each column k is a Fibonacci-type sequence.
If the rows are written in a right-aligned fashion:
                                                                           1
                                                                        1, 2
                                                                   2, 3,1, 3
                                                        3, 5,1, 4, 3, 4,2, 5
                                   5, 8,2, 7,4, 5,3, 7, 4, 7,1, 5, 5, 7,3, 8
                                                                         ...
then each column k also is a Fibonacci-type sequence.
If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are the reverses of blocks of A162912 (a(2^m+k) = A162912(2^(m+1)-1-k), m = 0,1,2,..., k = 0..2^m-1).
(End)
From Yosu Yurramendi, Jan 12 2017: (Start)
a(2^(m+2m'  )     + A020988(m'))   = A000045(m+1), m>=0, m'>=0
a(2^(m+2m'+1)     + A020989(m'))   = A000045(m+3), m>=0, m'>=0
a(2^(m+2m'  ) - 1 - A002450(m'))   = A000045(m+1), m>=0, m'>=0
a(2^(m+2m'+1) - 1 - A072197(m'-1)) = A000045(m+3), m>=0, m'>0
a(2^(m+1) -1) = A000045(m+2), m>=0. (End)

			The first four levels of the drib tree:
  [1/1],
  [1/2, 2/1],
  [2/3, 3/1, 1/3, 3/2],
  [3/5, 5/2, 1/4, 4/3, 3/4, 4/1, 2/5, 5/3].

Ralf Hinze, Functional pearls: the bird tree, J. Funct. Programming 19 (2009), no. 5, 491-508.
Index entries for fraction trees

This sequence is the composition of A162909 and A059893: a(n) = A162909(A059893(n)). This sequence is a permutation of A002487(n+1).

import Ratio; drib :: [Rational]; drib = 1 : map (recip . succ) drib \/ map (succ . recip) drib; (a : as) \/ bs = a : (bs \/ as); a162911 = map numerator drib; a162912 = map denominator drib

a(n) = my(x = 0, y = 1); forstep(i = logint(n, 2), 0, -1, [x, y] = if(bittest(n, i), [y, x + y], [x + y, x])); y \\ Mikhail Kurkov, Oct 12 2023

blocklevel <- 6 # arbitrary
a <- 1
for(m in 0:blocklevel) for(k in 0:(2^m-1)){
  a[2^(m+1)+2*k  ] <- a[2^(m+1)-1-k]
  a[2^(m+1)+2*k+1] <- a[2^(m+1)-1-k] + a[2^m+k]
}
a
# Yosu Yurramendi, Jul 11 2014

a(n) where a(1) = 1; a(2n) = b(n); a(2n+1) = a(n) + b(n); and b(1) = 1; b(2n) = a(n) + b(n); b(2n+1) = a(n).
a(2^(m+1)+2*k) = a(2^(m+1)-k-1), a(2^(m+1)+2*k+1) = a(2^(m+1)-k-1) + a(2^m+k), a(1) = 1, m>=0, k=0..2^m-1. - Yosu Yurramendi, Jul 11 2014
a(2^(m+1) + 2*k) = A162912(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^(m+1) + 2*k + 1) = a(2^m + k) + A162912(2^m + k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Mar 30 2016
a(n*2^m + A176965(m)) = A268087(n), n > 0, m > 0. - Yosu Yurramendi, Feb 20 2017
a(n) = A002487(A258996(n)), n > 0. - Yosu Yurramendi, Jun 23 2021

A162912 Denominators of drib tree fractions, where drib is the bit-reversal permutation tree of the Bird tree.

Original entry on oeis.org

1, 2, 1, 3, 1, 3, 2, 5, 2, 4, 3, 4, 1, 5, 3, 8, 3, 7, 5, 5, 1, 7, 4, 7, 3, 5, 4, 7, 2, 8, 5, 13, 5, 11, 8, 9, 2, 12, 7, 9, 4, 6, 5, 10, 3, 11, 7, 11, 4, 10, 7, 6, 1, 9, 5, 12, 5, 9, 7, 11, 3, 13, 8, 21, 8, 18, 13, 14, 3, 19, 11, 16, 7, 11, 9, 17, 5, 19, 12, 14, 5, 13, 9, 7, 1, 11, 6, 17, 7, 13, 10, 15

Offset: 1

Ralf Hinze (ralf.hinze(AT)comlab.ox.ac.uk), Aug 05 2009

The drib tree is an infinite binary tree labeled with rational numbers. It is generated by the following iterative process: start with the rational 1; for the left subtree increment and then take the reciprocal of the current rational; for the right subtree interchange the order of the two steps: take the reciprocal and then increment. Like the Stern-Brocot and the Bird tree, the drib tree enumerates the positive rationals: A162911(n)/A162912(n).
From Yosu Yurramendi, Jul 11 2014: (Start)
If the terms (n>0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
   1,
   2, 1,
   3, 1, 3,2,
   5, 2, 4,3,4,1, 5,3,
   8, 3, 7,5,5,1, 7,4,7,3,5,4, 7,2, 8,5,
   13,5,11,8,9,2,12,7,9,4,6,5,10,3,11,7,11,4,10,7,6,1,9,5,12,5,9,7,11,3,13,8,
then the sum of the m-th row is 3^m (m = 0,1,2,), and each column k is a Fibonacci sequence (a(2^(m+2)+k) = a(2^(m+1)+k) + a(2^m+k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
If the rows are written in a right-aligned fashion:
                                                                          1,
                                                                        2,1,
                                                                   3,1, 3,2,
                                                          5,2,4,3, 4,1, 5,3,
                                        8,3, 7,5,5,1,7,4, 7,3,5,4, 7,2, 8,5,
  13,5,11,8,9,2,12,7,9,4,6,5,10,3,11,7,11,4,10,7,6,1,9,5,12,5,9,7,11,3,13,8,
then each column k also is a Fibonacci sequence.
If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are the reverses of blocks of A162911 ( a(2^m+k) = A162911(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1). (End)

			The first four levels of the drib tree:
  [1/1],
  [1/2, 2/1],
  [2/3, 3/1, 1/3, 3/2],
  [3/5, 5/2, 1/4, 4/3, 3/4, 4/1, 2/5, 5/3].

R. Hinze, Functional pearls: the bird tree, J. Funct. Programming 19 (2009), no. 5, 491-508.
Index entries for fraction trees

This sequence is the composition of A162910 and A059893: a(n) = A162910(A059893(n)). This sequence is a permutation of A002487(n+2).

Cf. A096773.

import Ratio; drib :: [Rational]; drib = 1 : map (recip . succ) drib \/ map (succ . recip) drib; (a : as) \/ bs = a : (bs \/ as); a162911 = map numerator drib; a162912 = map denominator drib

blocklevel <- 6 # arbitrary
a <- 1
for(m in 0:blocklevel) for(k in 0:(2^m-1)){
  a[2^(m+1)+2*k]   <- a[2^(m+1)-1-k] + a[2^m+k]
  a[2^(m+1)+2*k+1] <- a[2^(m+1)-1-k]
}
a
# Yosu Yurramendi, Jul 11 2014

b(n) where a(1) = 1; a(2n) = b(n); a(2n+1) = a(n) + b(n); and b(1) = 1; b(2n) = a(n) + b(n); b(2n+1) = a(n).
a(2^(m+1)+2*k) = a(2^m+k) + a(2^(m+1)-1-k) , a(2^(m+1)+2*k+1) = a(2^(m+1)-1-k) , a(1) = 1 , m=0,1,2,3,... , k=0,1,...,2^m-1. - Yosu Yurramendi, Jul 11 2014
a(2^(m+1) + 2*k + 1) = A162911(2^m + k), m >= 0, 0 <= k < 2^m.
a(2^(m+1) + 2*k) = A162911(2^m + k) + a(2^m + k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Mar 30 2016
a(n*2^(m+1) + A096773(m)) = A268087(n), n > 0, m >= 0.  - Yosu Yurramendi, Feb 20 2017
a(n) = A002487(1+A258996(n)), n > 0. - Yosu Yurramendi, Jun 23 2021

Edited by Charles R Greathouse IV, May 13 2010

A030109 Write n in binary, reverse bits, subtract 1, divide by 2.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 3, 0, 4, 2, 6, 1, 5, 3, 7, 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15, 0, 16, 8, 24, 4, 20, 12, 28, 2, 18, 10, 26, 6, 22, 14, 30, 1, 17, 9, 25, 5, 21, 13, 29, 3, 19, 11, 27, 7, 23, 15, 31, 0, 32, 16, 48, 8, 40, 24, 56, 4, 36, 20, 52, 12, 44, 28, 60, 2, 34, 18, 50

Offset: 1

David W. Wilson

The sequence divides naturally into blocks of length 2^k, k = 0, 1, 2, ... On block k, let n go from 0 to 2^k-1, write n in binary using k bits and reverse the bits. - N. J. A. Sloane, Jun 11 2002
For example: the 3-bit strings are 000, 001, 010, 011, 100, 101, 110 and 111. When they are bit-reversed, we get 000, 100, 010, 110, 001, 101, 011, 111. Or, in decimal representation 0,4,2,6,1,5,3,7.
In other words: Given any n>1, the set of numbers A030109[i] for indexes i ranging from 2^n to 2^(n+1)-1 is a permutation of the set of consecutive integers {0,1,2,...,2^n-1}. Example: for n=2, we have the permutation of {0,2,1,3} of {0,1,2,3} This is important in the standard FFT algorithms requiring a starting (or ending) bit-reversal permutation of indices. - Stanislav Sykora, Mar 15 2012

			As an irregular triangle, first few rows are:
  0;
  0,1;
  0,2,1,3;
  0,4,2,6,1,5,3,7;
  0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15;
  ...

Alois P. Heinz, Table of n, a(n) for n = 1..8191
Matteo Ceregini, Florian Kurpicz, and Rossano Venturini, Faster Wavelet Tree Queries, IEEE Data Compression Conference (DCC 2024). See also arXiv:2302.09239 [cs.DS], 2023. See p. 3.
Léo Ducas and Thomas Prest, Fast Fourier Orthogonalization, IACR, Report 2015/1014, 2015-2016.

Cf. A030101. A049773 is another version.

a030109 = flip div 2 . subtract 1 . a030101
-- Reinhard Zumkeller, Mar 14 2015

% To get the irregular triangle
function T = ITriang(rows)
  T = cell(rows, 1);
  T{1} = [0];
  for r = 1:rows - 1;
    T{r + 1} = [2*T{r} (2*T{r} + 1)];
  end
end
% Miguel Vargas, May 04 2024

a:= proc(n) option remember; local r; `if`(n<3, 0,
      `if`(irem(n, 2, 'r')=0, a(r), a(r) +2^ilog2(r)))
    end:
seq(a(n), n=1..127);  # Alois P. Heinz, Oct 08 2012

Table[(FromDigits[Reverse[IntegerDigits[n,2]],2]-1)/2,{n,90}] (* Harvey P. Dale, Oct 26 2013 *)

a(n) = (fromdigits(Vecrev(binary(n)), 2) - 1)/2; \\ Michel Marcus, Oct 01 2019

maxrow <- 10 # by choice
a <- 0
for(m in 0:maxrow) for(k in 0:(2^m-1)) {
  a[2^(m+1)+    k] <- 2*a[2^m+k]
  a[2^(m+1)+2^m+k] <-   a[2^(m+1)+k]+1
}
a
# Yosu Yurramendi, Jan 24 2015

a(n) = A059893(n) - A053644(n). If 2*2^k<= n<3*2^k then a(n) = 2*a(n-2^k); if 3*2^k<= n<4*2^k then a(n) = 1+ a(n-2^k) starting with a(1) = 0. - Henry Bottomley, Sep 13 2001
a(2n) = a(n), a(2n+1) = a(n) + 2^[log_2(n)]. - Ralf Stephan, Aug 22 2003
a(2^m*(2*A072758(n)+1)) = n for m and n >= 0. - Yosu Yurramendi, Jan 24 2015

More terms from Patrick De Geest, Jun 15 1998

A080541 In binary representation: keep the first digit and left-rotate the others.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 8, 10, 12, 14, 9, 11, 13, 15, 16, 18, 20, 22, 24, 26, 28, 30, 17, 19, 21, 23, 25, 27, 29, 31, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 64, 66, 68, 70, 72, 74, 76, 78, 80

Offset: 1

Reinhard Zumkeller, Feb 20 2003

Permutation of natural numbers: let r(n,0)=n, r(n,k)=a(r(n,k-1)) for k>0, then r(n,floor(log_2(n))) = n and for n>1: r(n,floor(log_2(n))-1) = A080542(n).

Discarding their most significant bit, binary representations of numbers present in each cycle of this permutation form a distinct equivalence class of binary necklaces, thus there are A000031(n) separate cycles in each range [2^n .. (2^(n+1))-1] (for n >= 0) of this permutation. A256999 gives the largest number present in n's cycle. - Antti Karttunen, May 16 2015

			a(20)=a('10100')='11000'=24; a(24)=a('11000')='10001'=17.

Inverse: A080542.
Cf. A000031, A003986, A053644, A053645, A000523, A007088, A079944, A080413, A080543, A054429, A256999, A257250.
The set of permutations {A059893, A080541, A080542} generates an infinite dihedral group.

f:= proc(n) local d;
   d:= ilog2(n);
   if n >= 3/2*2^d then 2*n+1-2^(d+1) else 2*n - 2^d fi
end proc:
map(f, [$1..100]); # Robert Israel, May 19 2015

A080541[n_] := FromDigits[Join[{First[#]}, RotateLeft[Rest[#]]], 2] & [IntegerDigits[n, 2]];
Array[A080541, 100] (* Paolo Xausa, May 13 2025 *)

def A080541(n): return ((n&(m:=1< 1 else n  # Chai Wah Wu, Jan 22 2023

maxlevel <- 6 # by choice
a <- 1:3
for(m in 1:maxlevel) for(k in 0:(2^(m-1)-1)){
a[2^(m+1)       + 2*k    ] = 2*a[2^m           + k]
a[2^(m+1)       + 2*k + 1] = 2*a[2^m + 2^(m-1) + k]
a[2^(m+1) + 2^m + 2*k    ] = 2*a[2^m           + k] + 1
a[2^(m+1) + 2^m + 2*k + 1] = 2*a[2^m + 2^(m-1) + k] + 1
}
a
# Yosu Yurramendi, Oct 12 2020

(define (A080541 n) (if (< n 2) n (A003986bi (A053644 n) (+ (* 2 (A053645 n)) (A079944off2 n))))) ;; A003986bi gives the bitwise OR of its two arguments. See A003986.
;; Where A079944off2 gives the second most significant bit of n. (Cf. A079944):
(define (A079944off2 n) (A000035 (floor->exact (/ n (A072376 n)))))
;; Antti Karttunen, May 16 2015

From Antti Karttunen, May 16 2015: (Start)
a(1) = 1; for n > 1, a(n) = A053644(n) bitwise_OR (2*A053645(n) + second_most_significant_bit_of(n)). [Here bitwise_OR is a 2-argument function given by array A003986 and second_most_significant_bit_of gives the second most significant bit (0 or 1) of n larger than 1. See A079944.]
Other identities. For all n >= 1:
a(n) = A059893(A080542(A059893(n))).
a(n) = A054429(a(A054429(n))).
(End)
A080542(a(n)) = a(A080542(n)) = n. [A080542 is the inverse permutation.]
From Robert Israel, May 19 2015: (Start)
Let d = floor(log[2](n)).  If n >= 3*2^(d-1) then a(n) = 2*n + 1 - 2^(d+1), otherwise a(n) = 2*n - 2^d.
G.f.: 2*x/(x-1)^2 + Sum_{n>=1} x^(2^n)+(2^n-1)*x^(3*2^(n-1)))/(x-1). (End)

A080542 In binary representation: keep the first digit and rotate right the others.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 7, 8, 12, 9, 13, 10, 14, 11, 15, 16, 24, 17, 25, 18, 26, 19, 27, 20, 28, 21, 29, 22, 30, 23, 31, 32, 48, 33, 49, 34, 50, 35, 51, 36, 52, 37, 53, 38, 54, 39, 55, 40, 56, 41, 57, 42, 58, 43, 59, 44, 60, 45, 61, 46, 62, 47, 63, 64, 96, 65, 97, 66, 98, 67, 99, 68

Offset: 1

Reinhard Zumkeller, Feb 20 2003

Permutation of natural numbers with inverse = A080541: A080541(a(n)) = a(A080541(n)) = n;
let r(n,0)=n, r(n,k)=a(r(n,k-1)) for k>0, then r(n,floor(log_2(n))) = n and for n>1: r(n,floor(log_2(n))-1) = A080541(n).
Discarding their most significant bit, binary representations of numbers present in each cycle of this permutation form a distinct equivalence class of binary necklaces, thus there are A000031(n) separate cycles in each range [2^n .. (2^(n+1))-1] (for n >= 0) of this permutation.  A256999 gives the largest number present in n's cycle. - Antti Karttunen, May 16 2015

			a(20) = a('10100') = '10010' = 18.
a(25) = a('11001') = '11100' = 28.

Inverse: A080541.
Cf. A000031, A000035, A000523, A004526, A007088, A053644, A053645, A054429, A072376, A080414, A080544, A256999, A257250.
The set of permutations {A059893, A080541, A080542} generates an infinite dihedral group.

kfd[n_]:=Module[{a,b},{a,b}=TakeDrop[IntegerDigits[n,2],1];FromDigits[ Join[a,RotateRight[b]],2]]; Array[kfd,80] (* The program uses the TakeDrop function from Mathematica version 10 *) (* Harvey P. Dale, Feb 12 2016 *)

def A080542(n): return (1+(n&1))*(1<>1) if n > 1 else n # Chai Wah Wu, Jan 22 2023

nmax <- 31 # by choice
a <- 1:3
for(n in 1:nmax) for(k in 0:3)
a[4*n + k] = 2*a[2*n + (k == 1 | k == 3)] + (k == 2 | k == 3)
a
# Yosu Yurramendi, Sep 05 2020

(define (A080542 n) (if (< n 2) n (+ (A053644 n) (+ (* (A000035 n) (A072376 n)) (A004526 (A053645 n))))))  ;; Antti Karttunen, May 16 2015

a(n) = 2^log2(n) + floor((n-2^log2(n))/2) + (n mod 2)*2^(log2(n)-1), where log2(n) is the integer part of base-2 logarithm.
From Antti Karttunen, May 16 2015: (Start)
a(1) = 1; for n > 1, a(n) = A053644(n) + (A000035(n)*A072376(n)) + A004526(A053645(n)). [Essentially the same formula but represented with A-numbers.]
Other identities. For all n >= 1:
a(n) = A059893(A080541(A059893(n))).
a(n) = A054429(a(A054429(n))).
(End)

A086592 Denominators in left-hand half of Kepler's tree of fractions.

Original entry on oeis.org

2, 3, 3, 4, 4, 5, 5, 5, 5, 7, 7, 7, 7, 8, 8, 6, 6, 9, 9, 10, 10, 11, 11, 9, 9, 12, 12, 11, 11, 13, 13, 7, 7, 11, 11, 13, 13, 14, 14, 13, 13, 17, 17, 15, 15, 18, 18, 11, 11, 16, 16, 17, 17, 19, 19, 14, 14, 19, 19, 18, 18, 21, 21, 8, 8, 13, 13, 16, 16, 17, 17, 17, 17, 22, 22, 19, 19, 23

Offset: 1

Antti Karttunen, Aug 28 2003

Form a tree of fractions by beginning with 1/1 and then giving every node i/j two descendants labeled i/(i+j) and j/(i+j).
Level n of the left-hand half of the tree consists of 2^(n-1) nodes: 1/2; 1/3, 2/3; 1/4, 3/4, 2/5, 3/5; 1/5, 4/5, 3/7, 4/7, 2/7, 5/7, 3/8, 5/8; ... .
The right-hand half is identical to the left-hand half. - Michel Dekking, Oct 05 2017
n>1 occurs in this sequence phi(n) = A000010(n) times, as it occurs in A007306 (Franklin T. Adams-Watters' comment), that is the sequence obtained by adding numerator and denominator in the Calkin-Wilf enumeration system of positive rationals. A020650(n)/A020651(n) is also an enumeration system of all positive rationals (Yu-Ting system), and in each level m >= 0 (ranks between 2^m and 2^(m+1)-1) rationals are the same in both systems. Thus a(n) has the same terms in each level as A007306. The same property occurs in all numerator+denominator sequences of enumeration systems of positive rationals, as, for example, A007306 (A007305+A047679), A071585 (A229742+A071766), and A268087 (A162909+A162910). - Yosu Yurramendi, Apr 06 2016

Johannes Kepler, Mysterium cosmographicum, Tuebingen, 1596, 1621, Caput XII.
Johannes Kepler, Harmonice Mundi, Linz, 1619, Liber III, Caput II.
Johannes Kepler, The Harmony of the World [1619], trans. E. J. Aiton, A. M. Duncan and J. V. Field, American Philosophical Society, Philadelphia, 1997, p. 163.

Johannes Kepler, Harmonices mundi libri V ... (A Latin original scanned in Internet Archive. The fraction-tree is illustrated on the page 27 of the third book (Liber III), which is on the page 117 of the PDF-document.)
Johannes Kepler, Excerpt from the Chapter II of the Book III of the Harmony of the World: On the seven harmonic divisions of the string (Illustrates the A020651/A086592-tree).
OEIS Wiki, Historical sequences
Pelegrí Viader, Jaume Paradís and Lluís Bibiloni, A New Light on Minkowski's ?(x) Function, J. Number Theory, 73 (2) (1998), 212-227. See p. 215.
Index entries for fraction trees
Index entries for sequences related to music

Bisection of A020650.
See A093873/A093875 for the full tree.
A020651 gives the numerators. Bisection: A086593. Cf. A002487, A004169.

(* b = A020650 *) b[1] = 1; b[2] = 2; b[3] = 1; b[n_] := b[n] = Switch[ Mod[n, 4], 0, b[n/2 + 1] + b[n/2], 1, b[(n - 1)/2 + 1], 2, b[(n - 2)/2 + 1] + b[(n - 2)/2], 3, b[(n - 3)/2]]; a[n_] := b[2n]; Array[a, 100] (* Jean-François Alcover, Jan 22 2016 *)

maxlevel <- 15
d <- c(1,2)
for(m in 0:maxlevel)
for(k in 1:2^m) {
   d[2^(m+1)    +k] <- d[k] + d[2^m+k]
   d[2^(m+1)+2^m+k] <- d[2^(m+1)+k]
}
b <- vector()
for(m in 0:maxlevel) for(k in 0:(2^m-1)) b[2^m+k] <- d[2^(m+1)+k]
a <- vector()
for(n in 1:2^maxlevel) {a[2*n-1] <- b[n]; a[2*n] <- b[n+1]}
a[1:128]
# Yosu Yurramendi, May 16 2018

a(n) = A020650(n) + A020651(n) = A020650(2n).
a(n) = A071585(A059893(n)), a(A059893(n)) = A071585(n), n > 0. - Yosu Yurramendi, May 30 2017
a(2*n-1) = A086593(n); a(2*n) = A086593(n+1), n > 0. - Yosu Yurramendi, May 16 2018
a(n) = A007306(A231551(n)), n > 0. - Yosu Yurramendi, Aug 07 2021

Entry revised by N. J. A. Sloane, May 24 2004

A245325 Numerators of an enumeration system of the reduced nonnegative rational numbers.

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 3, 3, 3, 2, 1, 5, 4, 5, 4, 5, 4, 5, 4, 3, 3, 2, 1, 8, 7, 7, 5, 8, 7, 7, 5, 8, 7, 7, 5, 8, 7, 7, 5, 5, 4, 5, 4, 3, 3, 2, 1, 13, 11, 12, 9, 11, 10, 9, 6, 13, 11, 12, 9, 11, 10, 9, 6, 13, 11, 12, 9, 11, 10, 9, 6, 13, 11, 12, 9, 11, 10, 9, 6, 8, 7, 7, 5, 8, 7, 7, 5, 5, 4, 5, 4, 3, 3, 2, 1, 21, 18, 19, 14, 19

Offset: 1

Yosu Yurramendi, Jul 18 2014

a(n)/A245326(n) enumerates all the reduced nonnegative rational numbers exactly once.
If the terms (n>0) are written as an array (in a left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
  1,
  1,2,
  2,1,3,3,
  3,3,2,1,5,4,5,4,
  5,4,5,4,3,3,2,1,8,7,7,5,8,7,7,5,
  8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,
then the sum of the m-th row is 3^m (m = 0,1,2,), and each column k is a Fibonacci sequence.
If the rows are written in a right-aligned fashion:
                                                                        1,
                                                                      1,2,
                                                                 2, 1,3,3,
                                                      3, 3, 2,1, 5, 4,5,4,
                                 5, 4, 5,4, 3, 3,2,1, 8, 7, 7,5, 8, 7,7,5,
8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,
then each column is an arithmetic sequence. The differences of the arithmetic sequences give the sequence A071585 (a(2^(m+1)-1-k) - a(2^m-1-k) = A071585(k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are permutations of terms of blocks from A002487 (Stern's diatomic series or the Stern-Brocot sequence), and, more precisely, the reverses of blocks of A229742 (a(2^m+k) = A229742(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1). Moreover, each block is the bit-reversed permutation of the corresponding block of A245327.

Index entries for fraction trees

Cf. A245326, A002487, A071585, A229742, A273494.

blocklevel <- 6 # arbitrary
a <- 1
for(m in 0:blocklevel) for(k in 0:(2^(m-1)-1)){
  a[2^(m+1)+k]             <- a[2^m+2^(m-1)+k]
  a[2^(m+1)+2^(m-1)+k]     <- a[2^m+k]
  a[2^(m+1)+2^m+k]         <- a[2^(m+1)+k] +  a[2^m+k]
  a[2^(m+1)+2^m+2^(m-1)+k] <- a[2^(m+1)+2^m+k]
}
a

a(n) = A002487(A059893(A180200(n))) = A002487(1+A059893(A154435(n))). - Yosu Yurramendi, Sep 20 2021

cp's OEIS Frontend

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A162912 Denominators of drib tree fractions, where drib is the bit-reversal permutation tree of the Bird tree.

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A030109 Write n in binary, reverse bits, subtract 1, divide by 2.

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A080541 In binary representation: keep the first digit and left-rotate the others.

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