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Search: id:A020651
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| A020651 |
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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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5
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| 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; listen)
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OFFSET
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1,3
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COMMENT
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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 (Antti.Karttunen(AT)iki.fi), May 26 2004
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LINKS
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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).
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FORMULA
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a(1) = 1, a(2n) = a(n), a(2n+1) = A020650(2n) - Antti Karttunen (Antti.Karttunen(AT)iki.fi), May 26 2004
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EXAMPLE
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1, 2, 1/2, 3, 1/3, 3/2, 2/3, 4, 1/4, 4/3, ...
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MAPLE
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A020651 := n -> `if`((n < 2), n, `if`(type(n, even), A020651(n/2), A020650(n-1)));
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CROSSREFS
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See A093873/A093875 for the full Kepler tree.
Cf. A020650, A086592.
Sequence in context: A071912 A082909 A070940 this_sequence A002487 A060162 A026730
Adjacent sequences: A020648 A020649 A020650 this_sequence A020652 A020653 A020654
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KEYWORD
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nonn,easy,frac,nice
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AUTHOR
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David W. Wilson (davidwwilson(AT)comcast.net)
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EXTENSIONS
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Entry revised by njas, May 24 2004
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