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A245328 Denominators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = 1/(f(n)+1), f(2n+1) = f(n)+1).

%I A245328 #39 Apr 25 2024 12:14:40
%S A245328 1,2,1,3,2,3,1,5,3,5,2,4,3,4,1,8,5,8,3,7,5,7,2,7,4,7,3,5,4,5,1,13,8,
%T A245328 13,5,11,8,11,3,12,7,12,5,9,7,9,2,11,7,11,4,10,7,10,3,9,5,9,4,6,5,6,1,
%U A245328 21,13,21,8,18,13,18,5,19,11,19,8,14,11,14,3,19,12,19,7,17,12,17,5,16,9,16,7,11,9,11,2,18,11,18,7,15
%N A245328 Denominators in recursive bijection from positive integers to positive rationals (the bijection is f(1) = 1, f(2n) = 1/(f(n)+1), f(2n+1) = f(n)+1).
%C A245328 A245327(n)/a(n) enumerates all the reduced nonnegative rational numbers exactly once.
%C A245328 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 A245328    1,
%C A245328    2,1,
%C A245328    3,2, 3,1,
%C A245328    5,3, 5,2, 4,3, 4,1,
%C A245328    8,5, 8,3, 7,5, 7,2, 7,4, 7,3,5,4,5,1,
%C A245328    13,8,13,5,11,8,11,3,12,7,12,5,9,7,9,2,11,7,11,4,10,7,10,3,9,5,9,4,6,5,6,1,
%C A245328 then the sum of the m-th row is 3^m (m = 0,1,2,), and each column k is a Fibonacci sequence.
%C A245328 If the rows are written in a right-aligned fashion:
%C A245328                                                                         1,
%C A245328                                                                       2,1,
%C A245328                                                                   3,2,3,1,
%C A245328                                                           5,3,5,2,4,3,4,1,
%C A245328                                        8,5, 8,3, 7,5, 7,2,7,4,7,3,5,4,5,1,
%C A245328 13,8,13,5,11,8,11,3,12,7,12,5,9,7,9,2,11,7,11,4,10,7,10,3,9,5,9,4,6,5,6,1,
%C A245328 then each column is an arithmetic sequence. The differences of the arithmetic sequences, except the first on the right, give the sequence A093873 (Numerators in Kepler's tree of harmonic fractions) (a(2^(m+1)-1-k) - a(2^m-1-k) = A093873(k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
%C A245328 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 A020651 ( a(2^m+k) = A020651(2^(m+1)-1-k), m = 0,1,2,..., k = 0,1,2,...,2^m-1).
%C A245328 Moreover, each block is the bit-reversed permutation of the corresponding block of A245326.
%H A245328 Michael De Vlieger, <a href="/A245328/b245328.txt">Table of n, a(n) for n = 1..16383</a>, rows 1-14, flattened.
%H A245328 <a href="/index/Fo#fraction_trees">Index entries for fraction trees</a>
%F A245328 a(2n) = A245327(2n+1) , a(2n+1) = A245328(2n) , n=1,2,3,...
%F A245328 a((2*n+1)*2^m - 1) = A273493(n), n > 0, m >= 0. For n = 0 A273493(0) = 1 is needed. - _Yosu Yurramendi_, Mar 02 2017
%F A245328 a(n) = A002487(1+A284459(n)). - _Yosu Yurramendi_, Aug 23 2021
%t A245328 f[n_] := Which[n == 1, 1, EvenQ@ n, 1/(f[n/2] + 1), True, f[(n - 1)/2] + 1]; Table[Denominator@ f@ k, {n, 7}, {k, 2^(n - 1), 2^n - 1}] // Flatten (* _Michael De Vlieger_, Mar 02 2017 *)
%o A245328 (R)
%o A245328 N <- 25 # arbitrary
%o A245328 a <- c(1,2,1)
%o A245328 for(n in 1:N){
%o A245328   a[4*n]   <- a[2*n] + a[2*n+1]
%o A245328   a[4*n+1] <- a[2*n]
%o A245328   a[4*n+2] <- a[2*n] + a[2*n+1]
%o A245328   a[4*n+3] <-          a[2*n+1]
%o A245328 }
%o A245328 a
%o A245328 (PARI) a(n) = my(A=0); forstep(i=logint(n, 2), 0, -1, if(bittest(n, i), A++, A=1/(A+1))); denominator(A) \\ _Mikhail Kurkov_, Mar 12 2023
%Y A245328 Cf. A002487, A020651, A093873, A245326, A245327, A273493.
%K A245328 nonn,frac
%O A245328 1,2
%A A245328 _Yosu Yurramendi_, Jul 18 2014