A139352 Let the binary expansion of n be n = Sum_{k} 2^{r_k}, let e(n) be the number of r_k's that are even, o(n) the number that are odd; sequence gives o(n).

0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 3, 3, 1, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 3, 3, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2

Offset: 0

Author

Nadia Heninger and N. J. A. Sloane, Jun 07 2008

Keywords

• nonn
• base
• easy

Comments

e(n) + o(n) = A000120(n), the binary weight of n.

a(n) is also the number of 2's and 3's in the 4-ary representation of n. - Frank Ruskey, May 02 2009

Examples

			For n = 43 = 2^0 + 2^1 + 2^3 + 2^5, e(43)=1, o(43)=3. [Typo fixed by _Reinhard Zumkeller_, Apr 22 2011]
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Franklin T. Adams-Watters and Frank Ruskey, Generating Functions for the Digital Sum and Other Digit Counting Sequences, JIS 12 (2009), Article 09.5.6.

Crossrefs

Cf. A000035, A000120, A030308, A039004, A139351, A139353, A139354, A139355, A139370, A139371, A139372, A139373.

Programs

Fortran

c See link in A139351

Haskell

import Data.List (unfoldr)
a139352 = sum . map ((`div` 2) . (`mod` 4)) .
   unfoldr (\x -> if x == 0 then Nothing else Just (x, x `div` 4))
-- Reinhard Zumkeller, Apr 22 2011

Maple

A139352 := proc(n)
    local a,bdgs,r;
    a := 0 ;
    bdgs := convert(n,base,2) ;
    for r from 2 to nops(bdgs) by 2 do
        if op(r,bdgs) = 1 then
            a := a+1 ;
        end if;
    end do:
    a;
end proc: # R. J. Mathar, Jul 21 2016

Mathematica

a[n_] := Count[Position[Reverse@IntegerDigits[n, 2], 1]-1, {_?OddQ}];
Table[a[n], {n, 0, 99}] (* Jean-François Alcover, Mar 04 2023 *)
a[0] = 0; a[n_] := a[n] = a[Floor[n/4]] + If[Mod[n, 4] > 1, 1, 0]; Array[a, 100, 0] (* Amiram Eldar, Jul 18 2023 *)

PARI

a(n)=if(n>3,a(n\4))+n%4\2 \\ Charles R Greathouse IV, Apr 21 2016

Formula

G.f.: (1/(1-z))*Sum_{m>=0} (z^(2*4^m)/(1+(2*4^m))). - Frank Ruskey, May 03 2009

Recurrence relation: a(0)=0, a(4m) = a(4m+1) = a(m), a(4m+2) = a(4m+3) = 1+a(m). - Frank Ruskey, May 11 2009

a(n) = Sum_{k} A030308(n,k)*A000035(k). - Philippe Deléham, Oct 14 2011