A284571 Permutation of natural numbers: a(1) = 1, a(A005117(1+n)) = 2*a(n), a(A065642(1+n)) = 1 + 2*a(n).

1, 2, 4, 3, 8, 6, 16, 9, 5, 12, 32, 17, 18, 10, 24, 33, 64, 65, 34, 11, 36, 20, 48, 129, 7, 66, 19, 37, 128, 130, 68, 49, 22, 72, 40, 97, 96, 258, 14, 69, 132, 38, 74, 73, 21, 256, 260, 81, 13, 29, 136, 15, 98, 521, 44, 39, 144, 80, 194, 257, 192, 516, 23, 137, 28, 138, 264, 45, 76, 148, 146, 197, 42, 512, 147, 193, 520, 162, 26, 27

Offset: 1

Author

Antti Karttunen, Apr 17 2017

Keywords

• nonn

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Inverse: A284572.

Cf. A005117, A008683, A013928, A065642, A285328.

Similar or related permutations: A243343, A243345, A277695, A285111.

Programs

Python

from operator import mul
from sympy import primefactors
from sympy.ntheory.factor_ import core
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a285328(n):
    if core(n) == n: return 1
    k=n - 1
    while k>0:
        if a007947(k) == a007947(n): return k
        else: k-=1
def a013928(n): return sum(1 for i in range(1, n) if core(i) == i)
def a(n):
    if n==1: return 1
    if core(n)==n: return 2*a(a013928(n))
    else: return 1 + 2*a(a285328(n) - 1)
[a(n) for n in range(1, 121)] # Indranil Ghosh, Apr 17 2017

Formula

a(1) = 1, for n > 1, if A008683(n) <> 0 [when n is squarefree], a(n) = 2*a(A013928(n)), otherwise a(n) = 1 + 2*a(A285328(n)-1).