A093875 Denominators in Kepler's tree of harmonic fractions.

1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5, 4, 4, 5, 5, 5, 5, 7, 7, 7, 7, 8, 8, 5, 5, 7, 7, 7, 7, 8, 8, 6, 6, 9, 9, 10, 10, 11, 11, 9, 9, 12, 12, 11, 11, 13, 13, 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

Offset: 1

Author

N. J. A. Sloane and Reinhard Zumkeller, May 24 2004

Keywords

• nonn
• easy
• frac

Comments

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).

It appears that A071585 is a bisection of this sequence, which itself is a bisection of A093873. - Yosu Yurramendi, Jan 09 2016

Examples

			The first few fractions are:
1 1 1 1 2 1 2 1 3 2 3 1 3 2 3 1 4 3 4 2 5 3 5 1 4 3 4 2 5 3 5
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ...
1 2 2 3 3 3 3 4 4 5 5 4 4 5 5 5 5 7 7 7 7 8 8 5 5 7 7 7 7 8 8
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Index entries for fraction trees

Crossrefs

The numerators are in A093873. Usually one only considers the left-hand half of the tree, which gives the fractions A020651/A086592. See A086592 for more information, references to Kepler, etc.

Cf. A071585, A093873

Programs

Mathematica

num[1] = num[2] = 1; den[1] = 1; den[2] = 2; num[n_?EvenQ] := num[n] = num[n/2]; den[n_?EvenQ] := den[n] = num[n/2] + den[n/2]; num[n_?OddQ] := num[n] = den[(n-1)/2]; den[n_?OddQ] := den[n] = num[(n-1)/2] + den[(n-1)/2]; A093875 = Table[den[n], {n, 1, 83}] (* Jean-François Alcover, Dec 16 2011 *)

Formula

a(n) = a([n/2]) + A093873([n/2]).

Conjecture of the comment in detail: a(2n+1) = a(2n), n > 0; a(2n+1) = A071585(n), n >= 0; a(2n) = A071585(n), n > 0. - Yosu Yurramendi, Jun 22 2016

a(2n) - A093873(2n) = a(n), n > 0; a(2n+1) - A093873(2n+1) = A093873(n), n > 0. - Yosu Yurramendi, Jul 23 2016

From Yosu Yurramendi, Jul 25 2016: (Start)

a(2^m) = m+1, m >= 0; a(2^m + 2) = 2m - 1, m >= 1; a(2^m - 1) = A000045(m+2), m >= 1.

a(2^(m+1) + k) - a(2^m + k) = a(k), m > 0, 0 <= k < 2^m. For k=0, a(0) = 1 is needed.

a(2^(m+2) - k - 1) = a(2^(m+1) - k - 1) + a(2^m - k - 1), m >= 0, 0 <= k < 2^m. (End)