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

Author

David W. Wilson

Keywords

• nonn
• easy
• frac
• nice

Comments

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

Examples

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

Links

• 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

Crossrefs

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

Cf. A020650, A086592.

Programs

Haskell

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

Maple

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

Mathematica

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 *)

R

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

Formula

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

Extensions

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