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A014081
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a(n) = number of occurrences of '11' in binary expansion of n.
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22
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0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 1, 2, 1, 1, 1, 2, 2, 2, 3, 4, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 3, 3, 3, 4, 5, 0, 0, 0, 1, 0, 0, 1, 2, 0, 0, 0, 1, 1, 1, 2, 3, 0, 0, 0, 1, 0, 0, 1, 2, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 1
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OFFSET
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0,8
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COMMENTS
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First occurrence of k: 0, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, ..., A000225[k-1]. [From Robert G. Wilson v, Apr 02 2009]
a(n) = A213629(n,3) for n > 2. - Reinhard Zumkeller, Jun 17 2012
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REFERENCES
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Prodinger, Helmut. Generalizing the sum of digits function. SIAM J. Algebraic Discrete Methods 3 (1982), no. 1, 35--42. MR0644955 (83f:10009). [See B_2(11,n) on p. 35. - N. J. A. Sloane, Apr 06 2014]
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LINKS
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Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
R. Stephan, Some divide-and-conquer sequences ...
R. Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Digit Block
Eric Weisstein's World of Mathematics, Rudin-Shapiro Sequence
Index entries for sequences related to binary expansion of n
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FORMULA
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a(4n) = a(4n+1) = a(n), a(4n+2) = a(2n+1), a(4n+3) = a(2n+1) + 1. - Ralf Stephan, Aug 21 2003
G.f.: 1/(1-x) * sum(k>=0, t^3/((1+t)*(1+t^2)), t=x^2^k). - Ralf Stephan, Sep 10 2003
a(n) = A000120(n) - A069010(n). - Ralf Stephan, Sep 10 2003
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EXAMPLE
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The binary expansion of 15 is 1111, which contains three occurrences of 11, so a(15)=3.
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MAPLE
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# To count occurrences of 11..1 (k times) in binary expansion of v:
cn := proc(v, k) local n, s, nn, i, j, som, kk;
som := 0;
kk := convert(cat(seq(1, j = 1 .. k)), string);
n := convert(v, binary);
s := convert(n, string);
nn := length(s);
for i to nn - k + 1 do
if substring(s, i .. i + k - 1) = kk then som := som + 1 fi od;
som; end; # This program no longer worked. Corrected by N. J. A. Sloane, Apr 06 2014.
[seq(cn(n, 2), n=0..300)];
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MATHEMATICA
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f[n_] := Count[ Partition[ IntegerDigits[n, 2], 2, 1], {1, 1}]; Table[ f@n, {n, 0, 104}] [From Robert G. Wilson v, Apr 02 2009]
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PROG
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(Haskell)
import Data.Bits ((.&.))
a014081 n = a000120 (n .&. div n 2) -- Reinhard Zumkeller, Jan 23 2012
(PARI) A014081(n)=sum(i=0, #binary(n)-2, bitand(n>>i, 3)==3) \\ - M. F. Hasler, Jun 06 2012
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CROSSREFS
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Cf. A014082, A033264, A037800, A056973, A000225, A213629, A000120, A069010.
First differences give A245194.
Sequence in context: A129753 A147693 A070936 * A091890 A029431 A091492
Adjacent sequences: A014078 A014079 A014080 * A014082 A014083 A014084
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KEYWORD
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nonn,easy,changed
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AUTHOR
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Simon Plouffe
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STATUS
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approved
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