A290255 Number of 0's following directly the first 1 in the binary representation of n.

0, 1, 0, 2, 1, 0, 0, 3, 2, 1, 1, 0, 0, 0, 0, 4, 3, 2, 2, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 5, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Author

Emeric Deutsch, Sep 03 2017

Keywords

• nonn
• base

Examples

			a(19) = 2 because 19_2 = 10'0'11; a(21) = 1 because 21_2 = 10'101_2 (the counted 0's are marked).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Programs

Maple

a := proc (n) if type(log[2](n), integer) = true then log[2](n) else a(floor((1/2)*n)) end if end proc: seq(a(n), n = 1 .. 200);
# Alternate:
f := proc(n) option remember; local v;
v:= padic:-ordp(n,2);
  if n = 2^v then v else procname((n-2^v)/2^(v+1)) fi
end proc:
map(f, [$1..1000]); # Robert Israel, Sep 03 2017

Mathematica

sfo[n_]:=Module[{sidn2=Split[IntegerDigits[n,2]]},If[Length[ sidn2[[1]]]> 1, 0, Length[ sidn2[[2]]]]]; Join[{0},Array[sfo,110,2]] (* Harvey P. Dale, Mar 04 2018 *)

Formula

a(2^k)= k (k>=0); otherwise, a(n) = a(floor(n/2)).

G.f. q(x) + Sum_{j>=1} q(x^(2^j))*(x^(2^j)-x^(2^(j-1)))/(x-1) where q(z) = Sum_{j>=1} j*x^(2^j). - Robert Israel, Sep 03 2017