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A023416
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Number of 0's in binary expansion of n.
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124
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1, 0, 1, 0, 2, 1, 1, 0, 3, 2, 2, 1, 2, 1, 1, 0, 4, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 0, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 4, 3, 3, 2, 3, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 0, 6, 5, 5, 4, 5, 4, 4, 3, 5, 4, 4, 3, 4, 3, 3, 2, 5, 4, 4, 3, 4, 3, 3, 2, 4, 3, 3, 2, 3, 2, 2, 1, 5, 4, 4, 3, 4, 3, 3, 2, 4
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,5
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COMMENTS
| Another version (A080791) has a(0) = 0.
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LINKS
| N. J. A. Sloane, Table of n, a(n) for n = 0..10000
R. Stephan, Some divide-and-conquer sequences ...
R. Stephan, Table of generating functions
R. Stephan, Divide-and-conquer generating functions. I. Elementary sequences
Index entries for sequences related to binary expansion of n
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FORMULA
| a(n) = 1, if n = 0; 0, if n = 1; a(n/2)+1 if n even; a((n-1)/2) if n odd.
a(n) = 1 - (n mod 2) + a(floor(n/2)) - Marc LeBrun (mlb(AT)well.com), Jul 12 2001
G.f.: 1 + 1/(1-x) * Sum(k>=0, x^(2^(k+1))/(1+x^2^k)). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Apr 15 2002
a(n) = A070939(n)-A000120(n).
a(n) = A008687(n+1) - 1.
a(n) = A000120(A035327(n)).
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MAPLE
| s1:=[];
for n from 0 to 200 do
t1:=convert(n, base, 2); t2:=subs(1=NULL, t1); s1:=[op(s1), nops(t2)]; od:
s1;
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MATHEMATICA
| Table[ Count[ IntegerDigits[n, 2], 0], {n, 0, 100} ]
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PROG
| (Haskell)
a023416 n = a023416_list !! n
a023416_list = 1 : c [0] where c (z:zs) = z : c (zs ++ [z+1, z])
-- More efficient variant:
a023416 0 = 1
a023416 1 = 0
a023416 n = a023416 n' + 1 - m where (n', m) = divMod n 2
-- Reinhard Zumkeller, Jun 16 2011, Mar 07 2011
(PARI) a(n)=n=binary(n); sum(i=1, #n, !n[i]) \\ Charles R Greathouse IV, Jun 10 2011
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CROSSREFS
| The basic sequences concerning the binary expansion of n are A000120, A000788, A000069, A001969, A023416, A059015.
With initial zero and shifted right, same as A080791.
Sequence in context: A116382 A050606 * A080791 A124748 A161225 A174980
Adjacent sequences: A023413 A023414 A023415 * A023417 A023418 A023419
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KEYWORD
| nonn,nice,easy,base
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AUTHOR
| David W. Wilson (davidwwilson(AT)comcast.net)
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