A246209 Permutation of nonnegative integers: a(0) = 0, a(1) = 1, a(2n) = A117967(1+a(n)), a(2n+1) = A117968(a(n)).

0, 1, 5, 2, 15, 22, 3, 7, 52, 66, 35, 71, 4, 6, 11, 23, 137, 194, 148, 213, 36, 73, 99, 172, 17, 8, 16, 21, 12, 25, 33, 58, 462, 601, 447, 643, 431, 620, 304, 516, 37, 72, 104, 173, 127, 225, 419, 587, 45, 64, 9, 19, 47, 68, 49, 69, 13, 24, 29, 59, 43, 75, 152, 197, 1273, 1734, 1334, 1940, 1294, 1740, 899, 1556, 1404, 1837, 945, 1567, 389, 698, 1246, 1761, 41

Offset: 0

Author

Antti Karttunen, Aug 19 2014

Keywords

• nonn

Comments

This is an instance of entanglement permutation, where complementary pair A005843/A005408 (even and odd numbers respectively) is entangled with complementary pair A117967/A117968 (positive and negative part of inverse of balanced ternary enumeration of integers, respectively), with a(0) set to 0 and a(1) set to 1.

This implies that the even positions contain only terms of A117967 and apart from a(1) = 1, the odd positions contain only terms of A117968.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..2047
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Inverse: A246210.

Related permutations: A054429, A246207, A246211.

Cf. A117967, A117968.

Programs

Python

from sympy.ntheory.factor_ import digits
def a004488(n): return int("".join(str((3 - i)%3) for i in digits(n, 3)[1:]), 3)
def a117968(n):
    if n==1: return 2
    if n%3==0: return 3*a117968(n//3)
    elif n%3==1: return 3*a117968((n - 1)//3) + 2
    else: return 3*a117968((n + 1)//3) + 1
def a117967(n): return 0 if n==0 else a117968(-n) if n<0 else a004488(a117968(n))
def a(n): return n if n<2 else a117967(1 + a(n//2)) if n%2==0 else a117968(a((n - 1)//2))
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 07 2017

Formula

a(0) = 0, a(1) = 1, a(2n) = A117967(1+a(n)), a(2n+1) = A117968(a(n)).

As a composition of related permutations:

a(n) = A246207(A054429(n)).

a(n) = A246211(A246207(n)).