A198192 Replace 2^k in the binary representation of n with n-k (i.e. if n = 2^a + 2^b + 2^c + ... then a(n) = (n-a) + (n-b) + (n-c) + ...).

0, 1, 1, 5, 2, 8, 9, 18, 5, 15, 16, 29, 19, 34, 36, 54, 12, 30, 31, 52, 34, 57, 59, 85, 41, 68, 70, 100, 75, 107, 110, 145, 27, 61, 62, 99, 65, 104, 106, 148, 72, 115, 117, 163, 122, 170, 173, 224, 87, 138, 140, 194, 145, 201, 204, 263, 156, 216, 219, 282, 226

Offset: 0

Author

Brian Reed, Oct 21 2011

Keywords

• nonn
• look
• base

Examples

			a(5) = (5-2) + (5-0) = 8 because 5 = 2^2 + 2^0.
a(7) = (7-2) + (7-1) + (7-0) = 18 because 7 = 2^2 + 2^1 + 2^0.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000120, A073642.

Programs

MATLAB

% n is number of terms to be computed:
function [B] = predAddition(n)
   for i = 0:n
      k = i;
      c = 0;
      s = 0;
      while(k ~= 0)
         if ((i - c) >= 0)
            s = s + mod(k,2)*(i-c);
         end
         c = c + 1;
         k = (k - mod(k,2))/2;
      end
      B(i+1) = s;
   end
end

Maple

b:= (n, k)-> `if`(n=0, 0, k*(n mod 2)+b(floor(n/2), k-1)):
a:= n-> b(n, n):
seq(a(n), n=0..100);  # Alois P. Heinz, Oct 25 2011

Formula

a(n) = n*A000120(n) - A073642(n). - Franklin T. Adams-Watters, Oct 22 2011

a(n) = b(n,n) with b(0,k) = 0, b(n,k) = k*(n mod 2) + b(floor(n/2),k-1) for n>0. - Alois P. Heinz, Oct 25 2011