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, 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

Offset: 0

Author

Alex Ratushnyak, Jul 29 2012

Keywords

• sign
• base
• easy

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

Crossrefs

Cf. A000045, A037861.

Programs

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
				

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 *)