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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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Composition of A263273 with the permutation obtained from its even bisection.

LINKS

Antti Karttunen, Table of n, a(n) for n = 0..9841

Index entries for sequences that are permutations of the natural numbers

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.

Cf. A263272, A263273, A264974, A265367.

Cf. also A265342, A265353, A265355, A265356.

Sequence in context: A146027 A305619 A035358 * A065633 A160652 A131485

Adjacent sequences:  A265348 A265349 A265350 * A265352 A265353 A265354

KEYWORD

nonn,base

AUTHOR

Antti Karttunen, Dec 07 2015

STATUS

approved

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Last modified July 16 04:26 EDT 2019. Contains 325064 sequences. (Running on oeis4.)