A056977 Number of blocks of {0, 1, 1} in binary expansion of n.
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1
Offset: 1
Links
- Gheorghe Coserea, Table of n, a(n) for n = 1..10000
- Eric Weisstein's World of Mathematics, Digit Block
Programs
-
Mathematica
a[n_, bits_] := (idn = IntegerDigits[n, 2]; ln = Length[idn]; lb = Length[bits]; For[cnt = 0; k = 1, k <= ln - lb + 1, k++, If[idn[[k ;; k + lb - 1]] == bits, cnt++]]; cnt); Table[ a[n, {0, 1, 1}], {n, 1, 102} ] (* Jean-François Alcover, Oct 23 2012 *) Table[SequenceCount[IntegerDigits[n,2],{0,1,1}],{n,120}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 03 2019 *)
-
PARI
a(n) = { if (n < 11, return(0)); my(k = logint(n,2) - 1); hammingweight(bitnegimply(bitand(n>>1, n), n>>2)) - bittest(n,k) }; vector(102, i, a(i)) \\ Gheorghe Coserea, Sep 17 2015
Formula
a(2n) = a(n), a(2n+1) = a(n) + [n>1 and n congruent to 1 mod 4]. - Ralf Stephan, Aug 22 2003