A206297 Position of n in the canonical bijection from the positive integers to the positive rational numbers.

1, 3, 5, 9, 13, 21, 25, 37, 45, 57, 65, 85, 93, 117, 129, 145, 161, 193, 205, 241, 257, 281, 301, 345, 361, 401, 425, 461, 485, 541, 557, 617, 649, 689, 721, 769, 793, 865, 901, 949, 981, 1061, 1085, 1169, 1209, 1257, 1301, 1393, 1425, 1509, 1549

Offset: 1

Author

Clark Kimberling, Feb 06 2012

Keywords

• nonn

Comments

The canonical bijection from the positive integers to the positive rational numbers is given by A038568(n)/A038569(n).

Appears to be a variant of A049691. - R. J. Mathar, Feb 11 2012

It appears that a(n) = 2*A005728(n) - 1. - Chris Boyd, Mar 21 2015

Examples

			The canonical bijection starts with 1/1, 1/2, 2/1, 1/3, 3/1, 2/3, 3/2, 1/4, 4/1, 3/4, 4/3, 1/5, 5/1, so that this sequence starts with 1,3,5,9,13 and A206350 starts with 1,2,4,8,12.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A038568, A038569, A206350.

A049691 is an essentially identical sequence. See also A018805.

Programs

Mathematica

a[n_] := Module[{s = 1, k = 2, j = 1},
  While[s <= n, s = s + 2*EulerPhi[k]; k = k + 1];
  s = s - 2*EulerPhi[k - 1];
  While[s <= n, If[GCD[j, k - 1] =
    = 1, s = s + 2]; j = j + 1];
  If[s > n + 1, j - 1, k - 1]];
t = Table[a[n], {n, 0, 3000}];   (* A038568 *)
ReplacePart[1 + Flatten[Position[t, 1]], 1, 1]
(* A206297 *)

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A206297(n):
    if n == 1:
        return 1
    c, j = 1, 2
    k1 = (n-1)//j
    while k1 > 1:
        j2 = (n-1)//k1 + 1
        c += (j2-j)*(A206297(k1+1)-2)
        j, k1 = j2, (n-1)//j2
    return (n-2)*(n-1)-c+j+2 # Chai Wah Wu, Aug 04 2024