login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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
OFFSET
0,3
COMMENTS
Composition of A263273 with the permutation obtained from its even bisection.
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