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A246200 Self-inverse 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 bit-reverse" (A057889), the result should still be divisible by 3. Moreover, because the reversing operation is itself a self-inverse involution, and the prime factorization of any natural number is unique, we get a self-inverse permutation of nonnegative integers when we divide the bit-reversed 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 xrange(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

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Last modified October 23 05:56 EDT 2019. Contains 328335 sequences. (Running on oeis4.)