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%I #32 Dec 25 2023 17:17:26
%S 1,1,2,2,3,1,3,3,5,2,5,3,4,1,4,5,8,3,8,5,7,2,7,4,7,3,7,4,5,1,5,8,13,5,
%T 13,8,11,3,11,7,12,5,12,7,9,2,9,7,11,4,11,7,10,3,10,5,9,4,9,5,6,1,6,
%U 13,21,8,21,13,18,5,18,11,19,8,19,11,14,3,14,12,19,7,19,12,17,5,17,9,16,7,16,9,11,2,11,11,18,7,18,11
%N Numerators in recursive bijection from positive integers to positive rationals, where the bijection is f(1) = 1, f(2n) = 1/(f(n)+1), f(2n+1) = f(n)+1.
%C a(n)/A245328(n) enumerates all the reduced nonnegative rational numbers exactly once.
%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 1, 2,
%C 2, 3,1, 3,
%C 3, 5,2, 5,3, 4,1, 4,
%C 5, 8,3, 8,5, 7,2, 7,4, 7,3, 7,4,5,1,5,
%C 8,13,5,13,8,11,3,11,7,12,5,12,7,9,2,9,7,11,4,11,7,10,3,10,5,9,4,9,5,6,1,6,
%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 2,3,1,3,
%C 3,5,2,5,3,4,1,4,
%C 5, 8,3, 8,5, 7,2, 7,4,7,3,7,4,5,1,5,
%C 8,13,5,13,8,11,3,11,7,12,5,12,7,9,2,9,7,11,4,11,7,10,3,10,5,9,4,9,5,6,1,6,
%C then each column is an arithmetic sequence.
%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 Stern-Brocot sequence), and, more precisely, the reverses of blocks of A020650 ( a(2^m+k) = A020650(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
%C Moreover, each block is the bit-reversed permutation of the corresponding block of A245325.
%H Michael De Vlieger, <a href="/A245327/b245327.txt">Table of n, a(n) for n = 1..16383</a>, rows 1-14, flattened.
%H <a href="/index/Fo#fraction_trees">Index entries for fraction trees</a>
%F a(2n) = A245328(2n+1) , a(2n+1) = A245328(2n) , n=0,1,2,3,...
%F a((2*n+1)*2^m - 2) = A273493(n), n > 0, m > 0. For n = 0, m > 0, A273493(0) = 1 is needed. For n = 1, m = 0, A273493(0) = 1 is needed. For n > 1, m = 0, numerator((2*n-1) = num+den(n-1). - _Yosu Yurramendi_, Mar 02 2017
%F a(n) = A002487(A284459(n)). - _Yosu Yurramendi_, Aug 23 2021
%t f[n_] := Which[n == 1, 1, EvenQ@ n, 1/(f[n/2] + 1), True, f[(n - 1)/2] + 1]; Table[Numerator@ f@ k, {n, 7}, {k, 2^(n - 1), 2^n - 1}] // Flatten (* _Michael De Vlieger_, Mar 02 2017 *)
%o (R)
%o N <- 25 # arbitrary
%o a <- c(1,1,2)
%o for(n in 1:N){
%o a[4*n] <- a[2*n+1]
%o a[4*n+1] <- a[2*n] + a[2*n+1]
%o a[4*n+2] <- a[2*n]
%o a[4*n+3] <- a[2*n] + a[2*n+1]
%o }
%o a
%o (PARI) a(n) = my(L=logint(n, 2), A=0); for(i=0, L, if(bittest(n, L-i), A++, A=1/(A+1))); numerator(A) \\ _Mikhail Kurkov_, Mar 12 2023
%Y Cf. A002487, A020651, A245325, A245328, A273493.
%K nonn,frac
%O 1,3
%A _Yosu Yurramendi_, Jul 18 2014
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