

A330081


If the binary expansion of n is (b(1), ..., b(w)), then the binary expansion of a(n) is (b(1), b(3), b(5), ..., b(6), b(4), b(2)).


5



0, 1, 2, 3, 4, 6, 5, 7, 8, 10, 12, 14, 9, 11, 13, 15, 16, 20, 18, 22, 24, 28, 26, 30, 17, 21, 19, 23, 25, 29, 27, 31, 32, 36, 40, 44, 34, 38, 42, 46, 48, 52, 56, 60, 50, 54, 58, 62, 33, 37, 41, 45, 35, 39, 43, 47, 49, 53, 57, 61, 51, 55, 59, 63, 64, 72, 68, 76
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

This sequence is a permutation of the nonnegative integers that preserves the binary length as well as the Hamming weight. See A330090 for the inverse.


LINKS



FORMULA

If n has w binary digits, then a^A003558(w1)(n) = n (where a^k denotes the kth iterate of the sequence).


EXAMPLE

For n = 1234:
 the binary expansion of 1234 is "10011010010",
 oddindexed bits are "101100",
 evenindexed bits are "01001", and in reverse order "10010",
 hence the binary expansion of a(1234) is "10110010010",
 so a(1234) = 1426.


PROG

(PARI) shuffle(v) = { my (w=vector(#v), o=0, e=#v+1); for (k=1, #v, w[if (k%2, o++, e)]=v[k]); w }
a(n) = fromdigits(shuffle(binary(n)), 2)


CROSSREFS

See A329303 for a similar sequence.


KEYWORD

nonn,base,easy


AUTHOR



STATUS

approved



