login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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.
11
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
1,3
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.
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
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 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Yosu Yurramendi, Jul 18 2014
STATUS
approved