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<min) { min=sum; minAt=n; }
curr = prev.add(prpr);
prpr = prev;
prev = curr;
//if ((n&65535)==0)
// out.printf("%d %d %d %d %d %d %d %d\n", n,
// countPos, countNeg, count0, max, maxAt, min, minAt);
}
out.printf("\n\n%d %d %d %d %d %d %d %d\n", n,
countPos, countNeg, count0, max, maxAt, min, minAt);
/// 10000000 6882307 3117686 7 3743769 5463976 -2088795 7963846
}
}
CROSSREFS
KEYWORD
sign,base,easy
AUTHOR
Alex Ratushnyak, Jul 29 2012
STATUS
approved