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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)). 1
 -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, 17, 22, 18, 19, 15, 19, 20, 18, 18, 21, 15, 13, 18, 24, 24, 27, 33, 32, 43, 37, 28, 31, 33, 32, 31, 29, 24, 30, 34, 27, 35, 35, 26, 22, 32, 35, 31, 37, 30, 36, 19, 18 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS 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. Conjecture: a(n) contains infinitely many positive and infinitely many negative terms. LINKS T. D. Noe, Table of n, a(n) for n = 0..10000 MATHEMATICA Accumulate[Table[f = Fibonacci[n]; Count[IntegerDigits[f, 2], 1] - Count[IntegerDigits[f, 2], 0], {n, 0, 100}]] (* T. D. Noe, Jul 30 2012 *) PROG (Java) import static java.lang.System.out; import java.math.BigInteger; public class A214923 {   public static void main (String[] args) {       // 51 minutes     BigInteger prpr = BigInteger.valueOf(0);     BigInteger prev = BigInteger.valueOf(1), curr;     long n, c0=1, c1, sum=0, count0=0, countPos=0, countNeg=0, max=0, min=0, maxAt=0, minAt=0;     for (n=0; n<10000000; ++n) {       c1 = prpr.bitCount();       if (n>0)         c0 = prpr.bitLength() - c1;       sum += c1-c0;       out.printf("%d, ", sum);       if (sum>0) ++countPos; else       if (sum<0) ++countNeg; else                  ++count0;       if (sum>max) { max=sum; maxAt=n; }       if (sum

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Last modified April 4 07:32 EDT 2020. Contains 333213 sequences. (Running on oeis4.)