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A214923 Total count of 1's in binary representation of Fibonacci(n) and previous Fibonacci numbers, minus total count of 0's. That is, partial sums of b(n) = -A037861(Fibonacci(n)).

%I A214923 #10 Jul 30 2012 15:12:23
%S A214923 -1,0,1,1,3,4,2,4,5,3,7,8,4,6,9,7,13,16,12,9,12,10,11,18,14,9,10,14,
%T A214923 17,22,18,19,15,19,20,18,18,21,15,13,18,24,24,27,33,32,43,37,28,31,33,
%U A214923 32,31,29,24,30,34,27,35,35,26,22,32,35,31,37,30,36,19,18
%N A214923 Total count of 1's in binary representation of Fibonacci(n) and previous Fibonacci numbers, minus total count of 0's. That is, partial sums of b(n) = -A037861(Fibonacci(n)).
%C A214923 b(n) = -A037861(Fibonacci(n)) begins: -1, 1, 1, 0, 2, 1, -2, 2, 1, -2, 4, 1, -4, 2, 3, -2, 6, 3, -4, -3, 3, -2, 1, 7, -4, -5, 1, 4, 3, 5, -4, 1, -4, 4, 1, -2, 0.  For example b(6) = -A037861(Fibonacci(6)) = -A037861(8) = -2.
%C A214923 Conjecture: a(n) contains infinitely many positive and infinitely many negative terms.
%H A214923 T. D. Noe, <a href="/A214923/b214923.txt">Table of n, a(n) for n = 0..10000</a>
%t A214923 Accumulate[Table[f = Fibonacci[n]; Count[IntegerDigits[f, 2], 1] - Count[IntegerDigits[f, 2], 0], {n, 0, 100}]] (* _T. D. Noe_, Jul 30 2012 *)
%o A214923 (Java)
%o A214923 import static java.lang.System.out;
%o A214923 import java.math.BigInteger;
%o A214923 public class A214923 {
%o A214923   public static void main (String[] args) {       // 51 minutes
%o A214923     BigInteger prpr = BigInteger.valueOf(0);
%o A214923     BigInteger prev = BigInteger.valueOf(1), curr;
%o A214923     long n, c0=1, c1, sum=0, count0=0, countPos=0, countNeg=0, max=0, min=0, maxAt=0, minAt=0;
%o A214923     for (n=0; n<10000000; ++n) {
%o A214923       c1 = prpr.bitCount();
%o A214923       if (n>0)
%o A214923         c0 = prpr.bitLength() - c1;
%o A214923       sum += c1-c0;
%o A214923       out.printf("%d, ", sum);
%o A214923       if (sum>0) ++countPos; else
%o A214923       if (sum<0) ++countNeg; else
%o A214923                  ++count0;
%o A214923       if (sum>max) { max=sum; maxAt=n; }
%o A214923       if (sum<min) { min=sum; minAt=n; }
%o A214923       curr = prev.add(prpr);
%o A214923       prpr = prev;
%o A214923       prev = curr;
%o A214923       //if ((n&65535)==0)
%o A214923       //  out.printf("%d  %d %d %d  %d %d  %d %d\n", n,
%o A214923       //      countPos, countNeg, count0, max, maxAt, min, minAt);
%o A214923     }
%o A214923     out.printf("\n\n%d  %d %d %d  %d %d  %d %d\n", n,
%o A214923             countPos, countNeg, count0, max, maxAt, min, minAt);
%o A214923     /// 10000000  6882307 3117686 7  3743769 5463976  -2088795 7963846
%o A214923   }
%o A214923 }
%Y A214923 Cf. A000045, A037861.
%K A214923 sign,base,easy
%O A214923 0,5
%A A214923 _Alex Ratushnyak_, Jul 29 2012