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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 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.)