A229742 a(n) = A071585(n) - A071766(n).

0, 1, 2, 1, 3, 3, 1, 2, 4, 5, 4, 5, 1, 2, 3, 3, 5, 7, 7, 8, 5, 7, 7, 8, 1, 2, 3, 3, 4, 5, 4, 5, 6, 9, 10, 11, 9, 12, 11, 13, 6, 9, 10, 11, 9, 12, 11, 13, 1, 2, 3, 3, 4, 5, 4, 5, 5, 7, 7, 8, 5, 7, 7, 8, 7, 11, 13, 14, 13, 17, 15, 18, 11, 16, 17, 19, 14

Offset: 0

Author

N. J. A. Sloane, Oct 05 2013, at the suggestion of Kevin Ryde

Keywords

• nonn
• frac
• changed

Comments

From Yosu Yurramendi, Jun 30 2014: (Start)

If the terms (n>0) are written as an array (left-aligned fashion):

1,

2,1,

3,3, 1, 2,

4,5, 4, 5,1, 2, 3, 3,

5,7, 7, 8,5, 7, 7, 8,1,2, 3, 3,4, 5, 4, 5,

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

then the sum of the k-th row is 3^(k-1) and each column is an arithmetic sequence. The differences of the arithmetic sequences gives the sequence A071585 (a(2^(p+1)+k) - a(2^p+k) = A071585(k), p = 0,1,2,..., k = 0,1,2,...,2^p-1).

The first terms of each column give A071766. The second terms of each column give A086593. So, A086593(n) = A071585(n) + A071766(n).

If the rows (n>0) are written in a right-aligned fashion:

1,

2,1,

3,3,1,2,

4,5,4,5,1,2,3,3,

5,7,7,8,5,7,7,8,1,2,3,3,4,5,4,5,

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

then each column is a Fibonacci sequence (a(2^(p+2)+k) = a(2^(p+1)+k) + a(2^p+k) p = 0,1,2,..., k = 0,1,2,...,2^p-1, with a_k(1) = A071585(k) and a_k(2) = A071766(k) being the first two terms of each column sequence). (End)

Examples

			A229742/A071766 = 0, 1, 2, 1/2, 3, 3/2, 1/3, 2/3, 4, 5/2, 4/3, 5/3, 1/4, 2/5, 3/4, 3/5, 5, 7/2, 7/3, 8/3, 5/4, 7/5, 7/4, 8/5, ... (this is the HCS form of the Stern-Brocot tree).
		

Links

• Index entries for fraction trees

Crossrefs

Cf. A071585, A071766, A086593, A238837.

Programs

R

blocklevel <- 6 # arbitrary
a <- 1
for(m in 0:blocklevel) for(k in 0:(2^(m-1)-1)){
  a[2^(m+1)+k]             <- a[2^m+k] + a[2^m+2^(m-1)+k]
  a[2^(m+1)+2^(m-1)+k]     <- a[2^(m+1)+k]
  a[2^(m+1)+2^m+k]         <- a[2^m+2^(m-1)+k]
  a[2^(m+1)+2^m+2^(m-1)+k] <- a[2^m+k]
}
a
# Yosu Yurramendi, Jul 11 2014

Formula

From Yosu Yurramendi, May 26 2019: (Start)

a(2^(m+1)+2^m+k) = A071585( k)

a(2^(m+1) +k) = A071585(2^m+k), m >= 0, 0 <= k < 2^m. (End)

a(n) = A002487(A059893(A006068(n))) = A002487(1+A059893(A233279(n))), n > 0. - Yosu Yurramendi, Sep 29 2021