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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)). 6
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

COMMENTS

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.

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

LINKS

T. D. Noe, Table of n, a(n) for n=1..10000

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).

FORMULA

a(1) = 1, a(2n) = a(n), a(2n+1) = A020650(2n) - Antti Karttunen, May 26 2004

EXAMPLE

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

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-Fran├žois Alcover, Nov 22 2011 *)

PROG

(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

CROSSREFS

See A093873/A093875 for the full Kepler tree.

Cf. A020650, A086592.

Sequence in context: A038568 A071912 A070940 * A002487 A060162 A026730

Adjacent sequences:  A020648 A020649 A020650 * A020652 A020653 A020654

KEYWORD

nonn,easy,frac,nice

AUTHOR

David W. Wilson

EXTENSIONS

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

STATUS

approved

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Last modified April 21 06:23 EDT 2014. Contains 240824 sequences.