When $x=2$ in the infinitely continued fraction $x+\frac{1}{x^2+\frac{1}{x^3+\ldots}}$, what algebraic value does it converge to?

Question

Say you have the infinitely continued fraction: $$x+\cfrac{1}{x^2+\cfrac{1}{x^3+\cfrac{1}{x^4+\ddots}}}$$

When $x=1$, you can see that it's $$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ddots}}}$$ which converges upon the golden ratio $\phi = 1.61803398875...$.

What if I were to plug in $x=2$? $$2+\cfrac{1}{4+\cfrac{1}{8+\cfrac{1}{16+\ddots}}}$$ Using a caluclator, it seems to converge upon the irrational number: $2.24248109286...$

Is there any way I can represent this irrational number algebraically?

Answer 1

Let $F(x)=x^0+\cfrac1{x^1+\cfrac1\cdots}~.$ Then $F(2)$ is OEIS A$214070$, for which no closed form is currently known.