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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). 8
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, 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, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

A.....(n)/a(n) enumerates all the reduced nonnegative rational numbers exactly once.

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

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,

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

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

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

                                                                        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,

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

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).

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).

Moreover, each block is the bit-reversed permutation of the corresponding block of A245326.

LINKS

Michael De Vlieger, Table of n, a(n) for n = 1..16383, rows 1-14, flattened.

Index entries for fraction trees

FORMULA

a(2n) = A245327(2n+1) , a(2n+1) = A245328(2n) , n=1,2,3,...

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

MATHEMATICA

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 *)

PROG

(R)

N <- 25 # arbitrary

a <- c(1, 2, 1)

for(n in 1:N){

  a[4*n]   <- a[2*n] + a[2*n+1]

  a[4*n+1] <- a[2*n]

  a[4*n+2] <- a[2*n] + a[2*n+1]

  a[4*n+3] <-          a[2*n+1]

}

a

CROSSREFS

Cf. A002487, A020651, A093873, A245326, A245327, A273493.

Sequence in context: A002487 A318509 A263017 * A060162 A026730 A318691

Adjacent sequences:  A245325 A245326 A245327 * A245329 A245330 A245331

KEYWORD

nonn,frac

AUTHOR

Yosu Yurramendi, Jul 18 2014

STATUS

approved

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Last modified January 17 10:30 EST 2019. Contains 319218 sequences. (Running on oeis4.)