A245327 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.

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, 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, 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

Offset: 1

Author

Yosu Yurramendi, Jul 18 2014

Keywords

• nonn
• frac

Comments

a(n)/A245328(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,

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,

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

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,

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,

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

then each column is an arithmetic sequence.

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).

Moreover, each block is the bit-reversed permutation of the corresponding block of A245325.

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..16383, rows 1-14, flattened.
• Index entries for fraction trees

Crossrefs

Cf. A002487, A020651, A245325, A245328, A273493.

Programs

Mathematica

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 *)

PARI

a(n) = my(A=0); forstep(i=logint(n, 2), 0, -1, if(bittest(n, i), A++, A=1/(A+1))); numerator(A) \\ Mikhail Kurkov, Mar 12 2023

R

N  <- 25 # arbitrary
a <- c(1,1,2)
for(n in 1:N){
  a[4*n]   <-          a[2*n+1]
  a[4*n+1] <- a[2*n] + a[2*n+1]
  a[4*n+2] <- a[2*n]
  a[4*n+3] <- a[2*n] + a[2*n+1]
}
a

Formula

a(2n) = A245328(2n+1) , a(2n+1) = A245328(2n) , n=0,1,2,3,...

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

a(n) = A002487(A284459(n)). - Yosu Yurramendi, Aug 23 2021