A264985 Self-inverse permutation of nonnegative integers: a(n) = (A264983(n)-1) / 2.

0, 1, 3, 2, 4, 9, 6, 10, 12, 5, 7, 11, 8, 13, 27, 18, 28, 36, 15, 19, 33, 24, 31, 30, 21, 37, 39, 14, 16, 32, 23, 22, 29, 20, 34, 38, 17, 25, 35, 26, 40, 81, 54, 82, 108, 45, 55, 99, 72, 85, 90, 63, 109, 117, 42, 46, 96, 69, 58, 87, 60, 100, 114, 51, 73, 105, 78, 94, 84, 57, 91, 111, 48, 64, 102, 75, 112, 93, 66, 118, 120, 41

Offset: 0

Author

Antti Karttunen, Dec 05 2015

Keywords

• nonn

Links

• Antti Karttunen, Table of n, a(n) for n = 0..9841
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Cf. A263273, A264983.

Cf. also A264989, A264991, A264992.

Programs

Mathematica

f[n_] := Block[{g, h}, g[x_] := x/3^IntegerExponent[x, 3]; h[x_] := x/g@ x; If[n == 0, 0, FromDigits[Reverse@ IntegerDigits[#, 3], 3] &@ g[n] h[n]]]; t = Select[f /@ Range@ 1000, OddQ]; Table[(t[[n + 1]] - 1)/2, {n, 0, 81}] (* Michael De Vlieger, Jan 04 2016, after Jean-François Alcover at A263273 *)

Python

from sympy import factorint
from sympy.ntheory.factor_ import digits
from operator import mul
def a030102(n): return 0 if n==0 else int(''.join(map(str, digits(n, 3)[1:][::-1])), 3)
def a038502(n):
    f=factorint(n)
    return 1 if n==1 else reduce(mul, [1 if i==3 else i**f[i] for i in f])
def a038500(n): return n/a038502(n)
def a263273(n): return 0 if n==0 else a030102(a038502(n))*a038500(n)
def a(n): return (a263273(2*n + 1) - 1)/2 # Indranil Ghosh, May 22 2017

Scheme

(define (A264985 n) (/ (- (A264983 n) 1) 2))

Formula

a(n) = (A264983(n)-1) / 2 = (1/2) * (A263273(2n + 1) - 1).