Formula for Feigenbaum’s constant?

Question

I have conjectured a formula to calculate the Feigenbaum constant $\delta \approx 4.66920$.

$\delta\stackrel{?}{=}$ $$4+\cfrac{1\times 2 -1}{1+\cfrac{2\times 3 -1}{2^2+\cfrac{3\times 4 -1}{1+\cfrac{4\times 5 -1}{3^2+\cfrac{5\times 6 - 1}{1+\ddots}}}}}$$ which after several iterations is $\approx 4.66919$ (I believe).

This discovery, whether exact or approximate, was completely accidental. Can this be numerically verified? Thanks.

Answer 1

Using Python and the sympy library, I defined the following function to compute the continued fraction to arbitrary depth and precision:

def F(n, prec=100):
    x = sympy.Float(n+1, dps=prec)**2
    for i in range(n,0,-1):
        x = 1 + (2*i*(2*i+1)-1)/x
        x = i*i + (2*i*(2*i-1)-1)/x
    return x+3

At 100 decimal precision, the continued fraction has stabilised at $n=300$ to $$ 4.669197341028614722501735076952905036174621190442756866058551504040545753015009110638731541739920174 $$ which is confirmed by increasing depth and precision further.

The first Feigenbaum constant is $4.669201609102990671853203821578...$, so while the values are close, they are not the same.

---

For those not familiar with Python, range(n,0,1) counts down from $n$ to $1$. Arithmetic operations on Float objects are handled by the Float class even when the other value is an integer, which is why I don't need to explicitly convert i to a Float before computing with it.

The way the function works is it starts at the $2n$th term, ie the one that reads $(n+1)^2+\text{fraction}$ which it approximates to $(n+1)^2$ as a starting point, and then moves up one step at a time: two steps in each iteration.