|
|
A048724
|
|
Write n and 2n in binary and add them mod 2.
|
|
60
|
|
|
0, 3, 6, 5, 12, 15, 10, 9, 24, 27, 30, 29, 20, 23, 18, 17, 48, 51, 54, 53, 60, 63, 58, 57, 40, 43, 46, 45, 36, 39, 34, 33, 96, 99, 102, 101, 108, 111, 106, 105, 120, 123, 126, 125, 116, 119, 114, 113, 80, 83, 86, 85, 92, 95, 90, 89, 72, 75, 78, 77, 68, 71, 66, 65, 192
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
Reversing binary representation of -n. Converting sum of powers of 2 in binary representation of a(n) to alternating sum gives -n. Note that the alternation is applied only to the nonzero bits and does not depend on the exponent of two. All integers have a unique reversing binary representation (see cited exercise for proof). Complement of A065621. - Marc LeBrun, Nov 07 2001
|
|
REFERENCES
|
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 2, p. 178, (exercise 4.1. Nr. 27)
|
|
LINKS
|
|
|
FORMULA
|
a(n) = Xmult(n, 3) (or n XOR (n<<1)).
a(2n) = 2a(n), a(2n+1) = 2a(n) + 2(-1)^n + 1.
G.f. 1/(1-x) * sum(k>=0, 2^k*(3t-t^3)/(1+t)/(1+t^2), t=x^2^k). - Ralf Stephan, Sep 08 2003
a(n) = sum(k=0, n, (1-(-1)^round(+n/2^k))/2*2^k). - Benoit Cloitre, Apr 27 2005
|
|
EXAMPLE
|
12 = 1100 in binary, 24=11000 and their sum is 10100=20, so a(12)=20.
a(4) = 12 = + 8 + 4 -> - 8 + 4 = -4.
|
|
MAPLE
|
a:= n-> Bits[Xor](n, n+n):
|
|
MATHEMATICA
|
|
|
PROG
|
(Haskell)
import Data.Bits (xor, shiftL)
a048724 n = n `xor` shiftL n 1 :: Integer
(Python)
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|