A162910 - OEIS

%I #49 Apr 24 2024 12:31:00

%S 1,2,1,3,3,1,2,5,4,4,5,2,1,3,3,8,7,5,7,7,5,7,8,3,3,1,2,5,4,4,5,13,11,

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

%U 21,18,14,19,16,11,17,19,14,13,7,11,17,13,15,18,18,15,13,17,11,7,13,14,19

%N Denominators of Bird tree fractions.

%C The Bird tree is an infinite binary tree labeled with rational numbers. The root is labeled with 1. The tree enjoys the following fractal property: it can be transformed into its left subtree by first incrementing and then reciprocalizing the elements; for the right subtree interchange the order of the two steps: the elements are first reciprocalized and then incremented. Like the Stern-Brocot tree, the Bird tree enumerates all the positive rationals (A162909(n)/A162910(n)).

%C From _Yosu Yurramendi_, Jul 11 2014: (Start)

%C If the terms (n>0) are written as an array (left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...

%C    1,

%C    2, 1,

%C    3, 3,1, 2,

%C    5, 4,4, 5,2,1, 3, 3,

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

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

%C then the sum of the m-th row is 3^m (m = 0,1,2,), each column k is a Fibonacci sequence.

%C If the rows are written in a right-aligned fashion:

%C                                                                         1,

%C                                                                       2,1,

%C                                                                   3,3,1,2,

%C                                                           5,4,4,5,2,1,3,3,

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

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

%C then each column k also is a Fibonacci sequence.

%C The Fibonacci sequences of both triangles are equal except the first terms of second triangle.

%C If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are the reverses of blocks of A162909 ( a(2^m+k) = A162909(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).

%C (End)

%H Ralf Hinze, <a href="http://www.cs.ox.ac.uk/ralf.hinze/publications/Bird.pdf">Functional pearls: the bird tree</a>, J. Funct. Programming 19 (2009), no. 5, 491-508.

%H <a href="/index/Fo#fraction_trees">Index entries for fraction trees</a>

%F a(2^m+k) = a(2^m-k-1) + a(2^(m-1)+k), a(2^m+2^(m-1)+k) = a(2^m-k-1), a(1) = 1, m=0,1,2,3,..., k=0,1,...,2^(m-1)-1. - _Yosu Yurramendi_, Jul 11 2014

%F If k is odd a(A080675(n)*2^m+k) = A268087(2^m+k), if k is even a(A136412(2^m+k+1)*2^m+k) = A268087(2^m+k), m >= 0, 0 <= k < 2^m, n > 0. a(A081254(n)) = 1, n > 0. - _Yosu Yurramendi_, Feb 21 2017

%F a(n) = A002487(1+A258996(A059893(n))) = A002487(1+A059893(A258746(n))), n > 0. - _Yosu Yurramendi_, Jul 14 2021

%e The first four levels of the Bird tree: [1/1] [1/2, 2/1] [2/3, 1/3, 3/1, 3/2], [3/5, 3/4, 1/4, 2/5, 5/2, 4/1, 4/3, 5/3].

%o (Haskell)

%o import Ratio; bird :: [Rational]; bird = branch (recip . succ) (succ . recip) 1; branch f g a = a : branch f g (f a) \/ branch f g (g a); (a : as) \/ bs = a : (bs \/ as); a162909 = map numerator bird; a162910 = map denominator bird

%o (R)

%o blocklevel <- 6 # arbitrary

%o a <- 1

%o for(m in 1:blocklevel) for(k in 0:(2^(m-1)-1)){

%o a[2^m+k]         = a[2^m-k-1] + a[2^(m-1)+k]

%o a[2^m+2^(m-1)+k] = a[2^m-k-1]

%o }

%o a

%o # _Yosu Yurramendi_, Jul 11 2014

%Y This sequence is the composition of A162912 and A059893: a(n) = A162912(A059893(n)). This sequence is a permutation of A002487(n+2).

%K easy,frac,nonn

%O 1,2

%A Ralf Hinze (ralf.hinze(AT)comlab.ox.ac.uk), Aug 05 2009