A056980 Number of blocks of {1, 1, 0} in binary expansion of n.

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 2

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• easy
• base

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Digit Block.

Crossrefs

Cf. A014082, A056974, A056975, A056976, A056977, A056978, A056979.

Cf. A213629.

Programs

Haskell

import Data.List (tails, isPrefixOf)
a056980 = sum . map (fromEnum . ([0,1,1] `isPrefixOf`)) .
                    tails . a030308_row
-- Reinhard Zumkeller, Jun 17 2012

Mathematica

a[1] = a[2] = 0; a[n_] := a[n] = If[OddQ[n], a[(n - 1)/2], a[n/2] + Boole[Mod[n/2, 4] == 3]]; Table[a[n], {n, 1, 102}] (* Jean-François Alcover, Oct 22 2012, after Ralf Stephan *)

PARI

a(n) = hammingweight(bitnegimply(bitand(n>>1, n>>2), n));
vector(102, i, a(i))  \\  Gheorghe Coserea, Sep 07 2015

Formula

a(2n) = a(n) + [n congruent to 3 mod 4], a(2n+1) = a(n). - Ralf Stephan, Aug 22 2003

a(n) = A213629(n,6) for n > 5. - Reinhard Zumkeller, Jun 17 2012