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A265352 Permutation of nonnegative integers: a(n) = A263273(A263272(n)). 13
0, 1, 2, 3, 4, 7, 6, 19, 8, 9, 10, 5, 12, 13, 22, 21, 64, 23, 18, 55, 20, 57, 58, 25, 24, 73, 26, 27, 28, 11, 30, 31, 16, 15, 46, 17, 36, 37, 14, 39, 40, 67, 66, 199, 68, 63, 190, 65, 192, 193, 70, 69, 208, 71, 54, 163, 56, 165, 166, 61, 60, 181, 62, 171, 172, 59, 174, 175, 76, 75, 226, 77, 72, 217, 74, 219, 220, 79, 78, 235, 80, 81 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
Composition of A263273 with the permutation obtained from its even bisection.
LINKS
FORMULA
a(n) = A263273(A263272(n)).
As a composition of other related permutations:
a(n) = A265368(A264974(n)).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).
PROG
(Scheme) (define (A265352 n) (A263273 (A263272 n)))
(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(a263273(2*n)/2) # Indranil Ghosh, Jun 08 2017
CROSSREFS
Inverse: A265351.
Sequence in context: A064554 A290641 A340069 * A265368 A239972 A162425
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Dec 07 2015
STATUS
approved

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Last modified June 6 10:57 EDT 2024. Contains 373127 sequences. (Running on oeis4.)