A231551 -id:A231551 - cp's OEIS frontend

A020651 Denominators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = f(n)+1, f(2n+1) = 1/(f(n)+1)).

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

Offset: 1

David W. Wilson

If we insert an initial 1, this is the sequence of numerators in left-hand half of Kepler's tree of fractions. 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). See A086592 for denominators. See A294442 for the Kepler tree itself.
Level n of the tree consists of 2^n 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; ... Fibonacci numbers occur at the right edge this tree, i.e., a(A000225(n)) = A000045(n+1). The fractions are given in their reduced form, thus gcd(A020650(n), A020651(n)) = 1 and gcd(A020651(n), A086592(n)) = 1 for all n. - Antti Karttunen, May 26 2004
A generalization which includes the "rabbit tree" (A226080) and "all rationals tree" (A226130) follows. Suppose that a,b,c,d,e,f,g,h are complex numbers. Let S be the set of numbers defined by these rules: (1) 1 is in S; (2) if x is in S and cx+d is not 0, then U(x) = (ax+b)/(cx+d) is in S; (3) if x is in S and gx+h is not 0, then D(x) = (ex+f)/(gx+h) is in S. If an infinite path in the resulting tree has convergent nodes, then there is some node after which the path is "updown zigzag" ((UoD)o(UoD)o ...) or "downup zigzag" (DoU)o(DoU)o ...). If ag+ch is not 0, then the updown zigzag limit is invariant of x and equals [ae + cf - bg - dh + sqrt(X)]/(2(ag + ch)), where X = (ae + cf - bg - dh)^2 + 4(be + df + ag + ch). If ce + dg is not 0, then the downup zigzag limit is invariant of x and equals [ae + bg - cf - dh + sqrt(Y)]/(2(ce + dg)), where Y = (ae + bg - cf - dh)^2 + 4(af + bh)(ce + dg)) = X. Thus, for the tree A020651, the updown zigzag limit is -1 + sqrt(2) and the downup zigzag limit, sqrt(2). - Clark Kimberling, Nov 10 2013
From Yosu Yurramendi, Jul 13 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,
   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,
   1,6,5,6,4,9,5,9,3,10,7,10,4,11,7,11,2,9,7,9,5,12,7,12,3,11,8,11,5,13,8,13,
then the sum of the m-th row is 3^m (m = 0,1,2,), and each column is an arithmetic sequence. The differences of the arithmetic sequences, except the first on the left, give the sequence A093873 (Numerators in Kepler's tree of harmonic fractions) (a(2^(m+1)-1-k) - a(2^m-1-k) = A093873(k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
If the rows are written in a right-aligned fashion:
                                                                          1,
                                                                       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,
  1,6,5,6,4,9,5,9,3,10,7,10,4,11,7,11,2,9,7,9,5,12,7,12,3,11,8,11,5,13,8,13,
then each column k is a Fibonacci sequence. (End)
For m >= 0, a(2^m) = 1 and a(3*2^m) = 2. For n >= 0, a(A070875(n)) = 3 (for m >= 0, a(5*2^m) = 3 and a(7*2^m) = 3). - Yosu Yurramendi, Jun 02 2016

			1, 2, 1/2, 3, 1/3, 3/2, 2/3, 4, 1/4, 4/3, ...

T. D. Noe, Table of n, a(n) for n = 1..10000
D. N. Andreev, On a Wonderful Numbering of Positive Rational Numbers, Matematicheskoe Prosveshchenie, Series 3, volume 1, 1997, pages 126-134 (in Russian). a(n) = denominator of r(n).
Godofredo Iommi and Mario Ponce, Odometers in non-compact spaces, arXiv:2404.03768 [math.DS], 2024. See p. 19.
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).
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.
Shen Yu-Ting, A Natural Enumeration of Non-Negative Rational Numbers -- An Informal Discussion, American Mathematical Monthly, volume 87, number 1, January 1980, pages 25-29. a(n) = denominator of gamma_n.
Index entries for fraction trees
Index entries for sequences related to music

See A294442 and A093873/A093875 for two different versions of the Kepler tree.

Cf. A020650, A086592.

import Data.List (transpose); import Data.Ratio (denominator)
a020651_list = map denominator ks where
   ks = 1 : concat (transpose [map (+ 1) ks, map (recip . (+ 1)) ks])
-- Reinhard Zumkeller, Feb 22 2014

A020651 := n -> `if`((n < 2),n,`if`(type(n,even), A020651(n/2), A020650(n-1)));

f[1] = 1; f[n_?EvenQ] := f[n] = f[n/2]+1; f[n_?OddQ] := f[n] = 1/(f[(n-1)/2]+1); a[n_] := Denominator[f[n]]; Table[a[n], {n, 1, 94}] (* Jean-François Alcover, Nov 22 2011 *)

N <- 25 # arbitrary
a <- c(1,1,2)
for(n in 1:N){
  a[4*n]   <- a[2*n]
  a[4*n+1] <- a[2*n] + a[2*n+1]
  a[4*n+2] <-          a[2*n+1]
  a[4*n+3] <- a[2*n] + a[2*n+1]
}
a
# Yosu Yurramendi, Jul 13 2014

a(1) = 1, a(2n) = a(n), a(2n+1) = A020650(2n). - Antti Karttunen, May 26 2004
a(2n) = A020650(2n+1). - Yosu Yurramendi, Jul 17 2014
a(2^m + k) = A093873(2^(m+1) + k) = A093873(2^(m+1) + 2^m + k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, May 18 2016
a(2^m + 2^r + k) = A093873(2^r + k)*(m-(r-1)) + A093873(k), m >= 0, r <= m-1, 0 <= k < 2^r. For k=0 A093873(0) = 0 is needed. - Yosu Yurramendi, Jul 30 2016
a((2n+1)*2^m) = A086592(n), m >= 0, n > 0. For n = 0 A086592(0) = 1 is needed. - Yosu Yurramendi, Feb 14 2017
a(4n+2) = a(4n+1) - a(4n) = a(2n+1) = a(4n+1) - a(n), n > 0. - Yosu Yurramendi, May 08 2018
a(1) = 1, a(n+1) = 2*floor(1/a(n))+1-1/a(n). - Jan Malý, Jul 30 2019
a(n) = A002487(A231551(n)), n > 0. - Yosu Yurramendi, Jul 15 2021

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

A020650 Numerators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = f(n)+1, f(2n+1) = 1/(f(n)+1)).

Original entry on oeis.org

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

Offset: 1

David W. Wilson

The fractions are given in their reduced form, thus gcd(a(n), A020651(n)) = 1 for all n. - Antti Karttunen, May 26 2004
From Yosu Yurramendi, Jul 13 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,
   4,1,4,3,5,2,5,3,
   5,1,5,4,7,3,7,4, 7,2, 7,5, 8,3, 8,5,
   6,1,6,5,9,4,9,5,10,3,10,7,11,4,11,7,9,2,9,7,12,5,12,7,11,3,11,8,13,5,13,8,
then the sum of the m-th row is 3^m (m = 0,1,2,), and each column is an arithmetic sequence.
If the rows are written in a right-aligned fashion:
                                                                          1,
                                                                        2,1,
                                                                   3,1, 3,2,
                                                         4,1, 4,3, 5,2, 5,3,
                                      5,1,5,4, 7,3, 7,4, 7,2, 7,5, 8,3, 8,5,
  6,1,6,5,9,4,9,5,10,3,10,7,11,4,11,7,9,2,9,7,12,5,12,7,11,3,11,8,13,5,13,8,
each column k is a Fibonacci sequence. (End)
a(n2^m+1) = a(2n+1), n > 0, m > 0. - Yosu Yurramendi, Jun 04 2016

			1, 2, 1/2, 3, 1/3, 3/2, 2/3, 4, 1/4, 4/3, ...

T. D. Noe, Table of n, a(n) for n = 1..10000
D. N. Andreev, On a Wonderful Numbering of Positive Rational Numbers, Matematicheskoe Prosveshchenie, Series 3, volume 1, 1997, pages 126-134 (in Russian). a(n) = numerator of r(n).
Shen Yu-Ting, A Natural Enumeration of Non-Negative Rational Numbers -- An Informal Discussion, American Mathematical Monthly, volume 87, number 1, January 1980, pages 25-29. a(n) = numerator of gamma_n.
Index entries for fraction trees

Cf. A020651.

Bisection: A086592.

import Data.List (transpose); import Data.Ratio (numerator)
a020650_list = map numerator ks where
   ks = 1 : concat (transpose [map (+ 1) ks, map (recip . (+ 1)) ks])
-- Reinhard Zumkeller, Feb 22 2014

A020650 := n -> `if`((n < 2),n, `if`(type(n,even), A020650(n/2)+A020651(n/2), A020651(n-1)));

f[1] = 1; f[n_?EvenQ] := f[n] = f[n/2]+1; f[n_?OddQ] := f[n] = 1/(f[(n-1)/2]+1); a[n_] := Numerator[f[n]]; Table[a[n], {n, 1, 94}] (* Jean-François Alcover, Nov 22 2011 *)
a[1]=1; a[2]=2; a[3]=1; a[n_] := a[n] = Switch[Mod[n, 4], 0, a[n/2+1] + a[n/2], 1, a[(n-1)/2+1], 2, a[(n-2)/2+1] + a[(n-2)/2], 3, a[(n-3)/2]]; Array[a, 100] (* Jean-François Alcover, Jan 22 2016, after Yosu Yurramendi *)

N <- 25 # arbitrary
a <- c(1,2,1)
for(n in 1:N){
  a[4*n]   <- a[2*n] + a[2*n+1]
  a[4*n+1] <-          a[2*n+1]
  a[4*n+2] <- a[2*n] + a[2*n+1]
  a[4*n+3] <- a[2*n]
}
a
# Yosu Yurramendi, Jul 13 2014

a(1) = 1, a(2n) = a(n) + A020651(n), a(2n+1) = A020651(2n) = A020651(n). - Antti Karttunen, May 26 2004
a(2n) = A020651(2n+1). - Yosu Yurramendi, Jul 17 2014
a((2*n+1)*2^m + 1) = A086592(n), n > 0, m > 0. For n = 0, A086592(0) = 1 is needed. For m = 0, a(2*(n+1)) = A086592(n+1). - Yosu Yurramendi, Feb 19 2017
a(n) = A002487(1+A231551(n)), n > 0. - Yosu Yurramendi, Aug 07 2021

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

A284459 Permutation of the positive integers: this permutation transforms the enumeration system of positive irreducible fractions A002487/A002487' (Calkin-Wilf) into the enumeration system A245327/A245328, and A162911/A162912 (Drib) into A020651/A020650 (Yu-Ting inverted).

Original entry on oeis.org

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

Offset: 1

Yosu Yurramendi, Mar 27 2017

nonn

The inverse permutation is A284460.

Yosu Yurramendi, Table of n, a(n) for n = 1..16383
Index entries for sequences that are permutations of the natural numbers

Cf. A002487, A020650, A020651, A162911, A162912, A245327, A245328, A284460.

maxrow <- 12 # by choice
a <- 1
b01 <- 1
for(m in 0:maxrow){
  b01 <- c(b01, c(1-b01[2^m:(2^(m+1)-1)], b01[2^m:(2^(m+1)-1)]) )
  for(k in 0:(2^m-1)){
    a[2^(m+1) +       k] <- a[2^m + k] + 2^(m + b01[2^(m+1) +       k])
    a[2^(m+1) + 2^m + k] <- a[2^m + k] + 2^(m + b01[2^(m+1) + 2^m + k])
}}
a
# Yosu Yurramendi, Mar 27 2017

maxblock <- 7 # by choice
a <- 1:3
for(n in 4:2^maxblock){
ones <- which(as.integer(intToBits(n)) == 1)
nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
anbit <- nbit
for(i in 2:(length(anbit) - 1))
   anbit[i] <- 1 - bitwXor(anbit[i], anbit[i-1])
a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
a
# Yosu Yurramendi, Apr 25 2021

a(n) = A258996(A231551(n)) = A231551(A092569(n)), n > 0 . - Yosu Yurramendi, Apr 10 2017

A153154 Permutation of natural numbers: A059893-conjugate of A006068.

Original entry on oeis.org

0, 1, 3, 2, 7, 4, 5, 6, 15, 8, 9, 14, 11, 12, 13, 10, 31, 16, 17, 30, 19, 28, 29, 18, 23, 24, 25, 22, 27, 20, 21, 26, 63, 32, 33, 62, 35, 60, 61, 34, 39, 56, 57, 38, 59, 36, 37, 58, 47, 48, 49, 46, 51, 44, 45, 50, 55, 40, 41, 54, 43, 52, 53, 42, 127, 64, 65, 126, 67, 124

Offset: 0

Antti Karttunen, Dec 20 2008

A002487(1+a(n)) = A020651(n) and A002487(a(n)) = A020650(n). So, it generates the enumeration system of positive rationals based on Stern's sequence A002487. - Yosu Yurramendi, Feb 26 2020

Inverse: A153153. a(n) = A059893(A006068(A059893(n))).

maxn <- 63 # by choice
a <- c(1,3,2)
#
for(n in 2:maxn){
  a[2*n] <- 2*a[n] + 1
  if(n%%2==0) a[2*n+1] <- 2*a[n+1]
  else        a[2*n+1] <- 2*a[n-1]
}
(a <- c(0,a))
# Yosu Yurramendi, Feb 26 2020

# Given n, compute a(n) by taking into account the binary representation of n
maxblock <- 8 # by choice
a <- c(1, 3, 2)
for(n in 4:2^maxblock){
  ones <- which(as.integer(intToBits(n)) == 1)
  nbit <- as.integer(intToBits(n))[1:tail(ones, n = 1)]
  anbit <- nbit
  for(i in 2:(length(anbit) - 1))
    anbit[i] <- bitwXor(anbit[i], anbit[i - 1])  # ?bitwXor
  anbit[0:(length(anbit) - 1)] <- 1 - anbit[0:(length(anbit) - 1)]
  a <- c(a, sum(anbit*2^(0:(length(anbit) - 1))))
}
(a <- c(0, a))
# Yosu Yurramendi, Oct 04 2021

From Yosu Yurramendi, Feb 26 2020: (Start)
a(1) = 1, for all n > 0 a(2*n) = 2*a(n) + 1, a(2*n+1) = 2*a(A065190(n)).
a(1) = 1, a(2) = 3, a(3) = 2, for all n > 1 a(2*n) = 2*a(n) + 1, and if n even a(2*n+1) = 2*a(n+1), else a(2*n+1) = 2*a(n-1).
a(n) = A054429(A231551(n)) = A231551(A065190(n)) = A284459(A054429(n)) =
       A332769(A284459(n)) = A258996(A154437(n)). (End)

A231550 Permutation of nonnegative integers: for each bit[i] in the binary representation, except the most and the least significant bits, set bit[i] = bit[i] XOR bit[i-1], where bit[i-1] is the less significant bit, XOR is the binary logical exclusive or operator.

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 6, 5, 8, 11, 14, 13, 12, 15, 10, 9, 16, 19, 22, 21, 28, 31, 26, 25, 24, 27, 30, 29, 20, 23, 18, 17, 32, 35, 38, 37, 44, 47, 42, 41, 56, 59, 62, 61, 52, 55, 50, 49, 48, 51, 54, 53, 60, 63, 58, 57, 40, 43, 46, 45, 36, 39, 34, 33, 64, 67, 70, 69, 76

Offset: 0

Alex Ratushnyak, Nov 10 2013

This permutation transforms the enumeration system of positive irreducible fractions A020651/A020650 into the enumeration system A002487/A002487' (Calkin-Wilf), and enumeration system A245327/A245326 into A162911/A162912 (Drib). - Yosu Yurramendi, Jun 16 2015

Cf. A003188, A006068, A231551.

Join[{0, 1}, Table[d = IntegerDigits[n, 2]; FromDigits[Join[{d[[1]]}, BitXor[Most@Rest@d, Rest@Rest@d], {d[[-1]]}], 2], {n, 2, 68}]] (* Ivan Neretin, Dec 28 2016 *)

a(n) = bitxor(n, if(n>3, bitand(n<<1, bitneg(0,logint(n,2))))); \\ Kevin Ryde, Jul 17 2021

for n in range(99):
  bits = [0]*64
  orig = [0]*64
  l = int.bit_length(int(n))
  t = n
  for i in range(l):
    bits[i] = orig[i] = t&1
    t>>=1
  for i in range(1, l-1):  bits[i] ^= orig[i-1]   # A231550
  #for i in range(1, l-1):  bits[i] ^= bits[i-1]   # A231551
  #for i in range(l-1):  bits[i] ^= orig[i+1]      # A003188
  #for i in range(1,l):  bits[l-1-i] ^= bits[l-i]  # A006068
  t = 0
  for i in range(l):  t += bits[i]<

R

a <- 1 maxlevel <- 8 # by choice # for(m in 0:maxlevel) for(k in 0:(2^m-1)){ a[2^(m+1) +2*k] <- 2*a[2^m+k] a[2^(m+2)-1-2*k] <- 2*a[2^m+k] + 1 } (a <- c(0,a)) # Yosu Yurramendi, Apr 10 2017

Formula

a(A231551(n)) = A231551(a(n)) = n.

a(n) = A284460(A258996(n)) = A092569(A284460(n)), n > 0. - Yosu Yurramendi, Apr 10 2017

cp's OEIS Frontend

A020651 Denominators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = f(n)+1, f(2n+1) = 1/(f(n)+1)).

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A020650 Numerators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = f(n)+1, f(2n+1) = 1/(f(n)+1)).

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

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A284459 Permutation of the positive integers: this permutation transforms the enumeration system of positive irreducible fractions A002487/A002487' (Calkin-Wilf) into the enumeration system A245327/A245328, and A162911/A162912 (Drib) into A020651/A020650 (Yu-Ting inverted).

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A153154 Permutation of natural numbers: A059893-conjugate of A006068.

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A231550 Permutation of nonnegative integers: for each bit[i] in the binary representation, except the most and the least significant bits, set bit[i] = bit[i] XOR bit[i-1], where bit[i-1] is the less significant bit, XOR is the binary logical exclusive or operator.

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A332769 Permutation of the positive integers: a(n) = A258996(A054429(n)) = A054429(A258996(n)).

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