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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) + ...).

%I A198192 #21 Nov 14 2014 09:34:45
%S A198192 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,
%T A198192 68,70,100,75,107,110,145,27,61,62,99,65,104,106,148,72,115,117,163,
%U A198192 122,170,173,224,87,138,140,194,145,201,204,263,156,216,219,282,226
%N 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) + ...).
%H A198192 Alois P. Heinz, <a href="/A198192/b198192.txt">Table of n, a(n) for n = 0..10000</a>
%F A198192 a(n) = n*A000120(n) - A073642(n). - _Franklin T. Adams-Watters_, Oct 22 2011
%F A198192 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
%e A198192 a(5) = (5-2) + (5-0) = 8 because 5 = 2^2 + 2^0.
%e A198192 a(7) = (7-2) + (7-1) + (7-0) = 18 because 7 = 2^2 + 2^1 + 2^0.
%p A198192 b:= (n, k)-> `if`(n=0, 0, k*(n mod 2)+b(floor(n/2), k-1)):
%p A198192 a:= n-> b(n, n):
%p A198192 seq(a(n), n=0..100);  # _Alois P. Heinz_, Oct 25 2011
%o A198192 (MATLAB) % n is number of terms to be computed:
%o A198192 function [B] = predAddition(n)
%o A198192    for i = 0:n
%o A198192       k = i;
%o A198192       c = 0;
%o A198192       s = 0;
%o A198192       while(k ~= 0)
%o A198192          if ((i - c) >= 0)
%o A198192             s = s + mod(k,2)*(i-c);
%o A198192          end
%o A198192          c = c + 1;
%o A198192          k = (k - mod(k,2))/2;
%o A198192       end
%o A198192       B(i+1) = s;
%o A198192    end
%o A198192 end
%Y A198192 Cf. A000120, A073642.
%K A198192 nonn,look,base
%O A198192 0,4
%A A198192 _Brian Reed_, Oct 21 2011