%I #36 Apr 24 2024 22:17:10
%S 1,1,2,2,1,3,3,3,3,2,1,5,4,5,4,5,4,5,4,3,3,2,1,8,7,7,5,8,7,7,5,8,7,7,
%T 5,8,7,7,5,5,4,5,4,3,3,2,1,13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,
%U 13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,21,18,19,14,19
%N Numerators of an enumeration system of the reduced nonnegative rational numbers.
%C a(n)/A245326(n) enumerates all the reduced nonnegative rational numbers exactly once.
%C If the terms (n>0) are written as an array (in a left-aligned fashion) with rows of length 2^m, m = 0,1,2,3,...
%C 1,
%C 1,2,
%C 2,1,3,3,
%C 3,3,2,1,5,4,5,4,
%C 5,4,5,4,3,3,2,1,8,7,7,5,8,7,7,5,
%C 8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,
%C then the sum of the m-th row is 3^m (m = 0,1,2,), and each column k is a Fibonacci sequence.
%C If the rows are written in a right-aligned fashion:
%C 1,
%C 1,2,
%C 2, 1,3,3,
%C 3, 3, 2,1, 5, 4,5,4,
%C 5, 4, 5,4, 3, 3,2,1, 8, 7, 7,5, 8, 7,7,5,
%C 8,7,7,5,8,7,7,5,5,4,5,4,3,3,2,1,13,11,12,9,11,10,9,6,13,11,12,9,11,10,9,6,
%C then each column is an arithmetic sequence. The differences of the arithmetic sequences give the sequence A071585 (a(2^(m+1)-1-k) - a(2^m-1-k) = A071585(k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
%C If the sequence is considered by blocks of length 2^m, m = 0,1,2,..., the blocks of this sequence are permutations of terms of blocks from A002487 (Stern's diatomic series or the Stern-Brocot sequence), and, more precisely, the reverses of blocks of A229742 (a(2^m+k) = A229742(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1). Moreover, each block is the bit-reversed permutation of the corresponding block of A245327.
%H <a href="/index/Fo#fraction_trees">Index entries for fraction trees</a>
%F a(n) = A002487(A059893(A180200(n))) = A002487(1+A059893(A154435(n))). - _Yosu Yurramendi_, Sep 20 2021
%o (R)
%o blocklevel <- 6 # arbitrary
%o a <- 1
%o for(m in 0:blocklevel) for(k in 0:(2^(m-1)-1)){
%o a[2^(m+1)+k] <- a[2^m+2^(m-1)+k]
%o a[2^(m+1)+2^(m-1)+k] <- a[2^m+k]
%o a[2^(m+1)+2^m+k] <- a[2^(m+1)+k] + a[2^m+k]
%o a[2^(m+1)+2^m+2^(m-1)+k] <- a[2^(m+1)+2^m+k]
%o }
%o a
%Y Cf. A245326, A002487, A071585, A229742, A273494.
%K nonn,frac
%O 1,3
%A _Yosu Yurramendi_, Jul 18 2014