

A246200


Selfinverse permutation of natural numbers: a(n) = A057889(3*n) / 3.


9



0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 19, 14, 15, 16, 17, 18, 13, 20, 21, 22, 27, 24, 35, 38, 23, 28, 39, 30, 31, 32, 33, 34, 25, 36, 41, 26, 29, 40, 37, 42, 43, 44, 75, 54, 59, 48, 67, 70, 51, 76, 83, 46, 55, 56, 71, 78, 47, 60, 79, 62, 63, 64, 65, 66, 49, 68, 81, 50, 57, 72, 73, 82, 45, 52, 77, 58, 61, 80, 69
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

In binary system, 3 ("11" in binary), has a similar shortcut rule for divisibility as eleven has in decimal system. This rule doesn't depend on which end of the number representation it is applied from, thus, if we reverse the number 3*n with "balanced bitreverse" (A057889), the result should still be divisible by 3. Moreover, because the reversing operation is itself a selfinverse involution, and the prime factorization of any natural number is unique, we get a selfinverse permutation of nonnegative integers when we divide the bitreversed result with 3.


LINKS

Antti Karttunen, Table of n, a(n) for n = 0..10921
Index entries for sequences that are permutations of the natural numbers


FORMULA

a(n) = A057889(3*n) / 3.


PROG

(Scheme) (define (A246200 n) (/ (A057889 (* 3 n)) 3))
(Python)
def a057889(n):
x=bin(n)[2:]
y=x[::1]
return int(str(int(y))+(len(x)  len(str(int(y))))*'0', 2)
def a(n): return a057889(3*n)/3
print [a(n) for n in range(101)] # Indranil Ghosh, Jun 11 2017


CROSSREFS

Cf. A036215, A057889, A003714, A048724, A083822, A083824.
Sequence in context: A161950 A111470 A227508 * A262356 A277861 A173902
Adjacent sequences: A246197 A246198 A246199 * A246201 A246202 A246203


KEYWORD

nonn,base,look


AUTHOR

Antti Karttunen, Aug 27 2014


STATUS

approved



