OFFSET
1,2
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
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
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)
EXAMPLE
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
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 *)
CROSSREFS
KEYWORD
nonn,easy,frac
AUTHOR
N. J. A. Sloane and Reinhard Zumkeller, May 24 2004
STATUS
approved