Can you find a second relationship between 3 consecutive Fibonacci numbers?

I literally cannot visualize any of this.

Do you mean that you do not see how the series is generated?
1+1 =2, giving 1 1 2
1+2 = 3 giving 1 1 2 3
2+3=5 1 1 2 3 5
and so on.

1,2,3 are ordinal numbers.

I won’t give it away. But here’s a hint. .Take two neighboring numbers in the series and divide the first by the second, for example: ⅔= .666 3/5= .60 5/8= .625 8/13= .615 As you work your way to infinity you will see they converge on a truly golden number. You can work magic with two golden ratios next to each other.

Ratio between them converges to phi.

f(n+2)*f(n) – f(n+1)*f(n+1) = (-1)exp(n)
That looks a bit clumsy, but if you multiply the largest by the smallest and take away the square of the middle number you will always get either +1 or -1.

That’s it!
For extra credit, you can do an inductive proof.

The formula can be derived from a more general result, which I learned about from the section on Fibonacci in Strange Curves, which is a really good popular math book.

Assume it is true for k then Fk+2Fk-F2k+1 = (-1)k
And Fk+2(Fk+2-Fk+1)-F2k+1= (-1)k
F2k+2 – (Fk+2.Fk+1)-F2k+1 = (-1)k
F2k+2—(Fk+2+Fk+1)Fk+1=(-1)k
F2k+2 –Fk+3Fk+1 =(-1)k
Reversing the signs Fk+3Fk+1-F2k+2 =(-1)k+1
Therefore it is true for k+1

Here is the way I did it.
Given that f(k+2)f(k) – f^2(k+1) = (-1)^k
We want to show that f(k+3)f(k+1) -f^2(k+2) = (-1)^(k+1)

f(k+3)f(k+1) -f^2(k+2) =
(f(k+2) + f(k+1))f(k+1) – f^2(k+1) =
f^2(k+1) + f(k+2)f(k+1) – f^2(k+1) = (using the inductive assumption)
f(k+2)f(k) – (-1)^k + f(k+2)f(k+1) – f^2(k+2) =
f(k+2)f(k) + (-1)^(k+1) + f(k+2)(f(k+1) – f(k+2)) =
f(k+2)f(k) + (-1)^(k+1) + f(k+2)(-f(k)) =
-1(k+1)