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%I #28 Aug 01 2017 06:21:29
%S 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,
%T 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,
%U 28,28,30,31,0,0,0,1,0,0,2,3,0,0,0,1,4,4,6,7,0
%N a(n) = (n AND floor(n/2)), where AND is bitwise and-operator (A004198).
%C 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).
%C The number of 1's through 4*2^n appears to yield A000045(n+1). - _Ben Burns_, Jun 12 2017
%H T. D. Noe, <a href="/A048735/b048735.txt">Table of n, a(n) for n = 0..1023</a>
%H Beeler, M., Gosper, R. W. and Schroeppel, R., <a href="http://www.inwap.com/pdp10/hbaker/hakmem/boolean.html#item23">HAKMEM, ITEM 23 (Schroeppel)</a>
%F a(n) = A048728(n)/4. (This was the original definition. AND-formula found Jan 01 2007).
%p seq(Bits:-And(n,floor(n/2)), n=0..200); # _Robert Israel_, Feb 29 2016
%t Table[BitAnd[n, Floor[n/2]], {n, 0, 127}] (* _T. D. Noe_, Aug 13 2012 *)
%o (PARI) a(n) = bitand(n, n\2); \\ _Michel Marcus_, Feb 29 2016
%o (Python)
%o def a(n): return n&int(n/2) # _Indranil Ghosh_, Jun 13 2017
%Y Cf. A003714 (positions of zeros), A003188, A050600.
%K nonn,base
%O 0,7
%A _Antti Karttunen_, Apr 26 1999
%E New formula and more terms added by _Antti Karttunen_, Jan 01 2007
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