A048735 a(n) = (n AND floor(n/2)), where AND is bitwise and-operator (A004198).

0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 4, 6, 7, 0, 0, 0, 1, 0, 0, 2, 3, 8, 8, 8, 9, 12, 12, 14, 15, 0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 4, 6, 7, 16, 16, 16, 17, 16, 16, 18, 19, 24, 24, 24, 25, 28, 28, 30, 31, 0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 4, 6, 7, 0

Offset

0,7

Comments

To prove that (n AND floor(n/2)) = (3n-(n XOR 2n))/4 (= A048728(n)/4), we first multiply both sides by 4, to get 2*(n AND 2n) = (3n - (n XOR 2n)) and then rearrange terms: 3n = (n XOR 2n) + 2*(n AND 2n), which fits perfectly to the identity A+B = (A XOR B) + 2*(A AND B) (given by Schroeppel in HAKMEM link).

The number of 1's through 4*2^n appears to yield A000045(n+1). - Ben Burns, Jun 12 2017

Links

T. D. Noe, Table of n, a(n) for n = 0..1023

Beeler, M., Gosper, R. W. and Schroeppel, R., HAKMEM, ITEM 23 (Schroeppel)

Formula

a(n) = A048728(n)/4. (This was the original definition. AND-formula found Jan 01 2007).

Maple

seq(Bits:-And(n, floor(n/2)), n=0..200); # Robert Israel, Feb 29 2016

Mathematica

Table[BitAnd[n, Floor[n/2]], {n, 0, 127}] (* T. D. Noe, Aug 13 2012 *)

Prog

(PARI) a(n) = bitand(n, n\2); \\ Michel Marcus, Feb 29 2016
(Python)
def a(n): return n&int(n/2) # Indranil Ghosh, Jun 13 2017

Crossrefs

Cf. A003714 (positions of zeros), A003188, A050600.

Sequence in context: A024865 A025109 A348710 * A102037 A286530 A152857

Adjacent sequences: A048732 A048733 A048734 * A048736 A048737 A048738

Keyword

nonn,base

Author

Antti Karttunen, Apr 26 1999

Extensions

New formula and more terms added by Antti Karttunen, Jan 01 2007

Status

approved