login
A265351
Permutation of nonnegative integers: a(n) = A263272(A263273(n)).
11
0, 1, 2, 3, 4, 11, 6, 5, 8, 9, 10, 29, 12, 13, 38, 33, 32, 35, 18, 7, 20, 15, 14, 17, 24, 23, 26, 27, 28, 83, 30, 31, 92, 87, 86, 89, 36, 37, 110, 39, 40, 119, 114, 113, 116, 99, 34, 101, 96, 95, 98, 105, 104, 107, 54, 19, 56, 21, 22, 65, 60, 59, 62, 45, 16, 47, 42, 41, 44, 51, 50, 53, 72, 25, 74, 69, 68, 71, 78, 77, 80, 81
OFFSET
0,3
COMMENTS
Composition of A263273 with the permutation obtained from its even bisection.
FORMULA
a(n) = A263272(A263273(n)).
As a composition of other related permutations:
a(n) = A264974(A265367(n)).
Other identities. For all n >= 0:
a(3*n) = 3*a(n).
a(n) = A265342(n)/2.
PROG
(Scheme) (define (A265351 n) (A263272 (A263273 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 a263272(n): return a263273(2*n)/2
def a(n): return a263272(a263273(n)) # Indranil Ghosh, May 25 2017
CROSSREFS
Inverse: A265352.
Sequence in context: A146027 A305619 A035358 * A065633 A160652 A131485
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Dec 07 2015
STATUS
approved