Each Fibonacci number is the sum of its two predecessors. My previous post looked at generalizing this to the so-called Tribonacci numbers, each being the sum of its three predecessors. One could keep going, defining the Tetrabonacci numbers and in general the n-Fibonacci numbers for any n at least 2.
For the definition to be complete, you have to specify the first n of the n-Fibonacci numbers. However, these starting values hardly matter for our purposes. We want to look at the limiting ratio of consecutive n-Fibonacci numbers, and this doesn’t depend on the initial conditions. (If you were determined, you could find starting values where this isn’t true. It’s enough to pick integer initial values, at least one of which is not zero.)
As shown in the previous post, the ratio is the largest eigenvalue of an n by n matrix with 1’s on the first row and 1’s immediately below the main diagonal. The characteristic polynomial of such a matrix is
λn – λn-1 – λn-2 – … -1
and so we look for the largest zero of this polynomial. We can sum the terms with negative coefficients as a geometric series and show that the eigenvalues satisfy
λn – 1/(2 – λ) = 0.
So the limiting ratio of consecutive n-Fibonacci numbers is the largest root of the above equation. You could verify that when n = 2, we get the golden ratio φ as we should, and when n = 3 we get around 1.8393 as in the previous post.
As n gets large, the limiting ratio approaches 2. You can see this by taking the log of the previous equation.
n = -log(2 – λ)/log(λ).
As n goes to infinity, λ must approach 2 so that the right side of the equation also goes to infinity.