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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
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.
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
KEYWORD
nonn,base,look
AUTHOR
Antti Karttunen, Aug 27 2014
STATUS
approved