A058301 Number of solutions to c(0)F(0) + ... + c(n)F(n) = 0, where c(i) = +-1 for i >= 0, number of (+1)'s >= number of (-1)'s, F(i) = A000045(i) = Fibonacci numbers.

1, 0, 2, 3, 0, 6, 4, 0, 8, 11, 0, 22, 16, 0, 32, 42, 0, 84, 64, 0, 128, 165, 0, 330, 256, 0, 512, 654, 0, 1308, 1024, 0, 2048, 2605, 0, 5210, 4096, 0, 8192, 10398, 0, 20796, 16384, 0, 32768, 41550, 0, 83100, 65536, 0, 131072, 166116, 0, 332232, 262144, 0

Offset: 0

Author

Naohiro Nomoto, Dec 08 2000

Keywords

• nonn

Examples

			a(3) = 3 because +0+1+1-2 = -0+1+1-2 = +0-1-1+2 = 0;
a(5) = 6 because +0+1-1-2-3+5 = +0-1+1-2-3+5 = +0+1-1+2+3-5 = -0+1-1+2+3-5 = +0-1+1+2+3-5 = -0-1+1+2+3-5 = 0.
		

Crossrefs

Cf. A000045, A002083, A047270, A047238.

Formula

a(3n+1) = 0, a(A047270(n)) = A002083(n+5), a(A047238(n)) = 2^n.

Extensions

More terms from Sean A. Irvine, Aug 02 2022