Golden ratio, $n$-bonacci numbers, and radicals of the form $\sqrt[n]{\frac{1}{n-1}+\sqrt[n]{\frac{1}{n-1}+\sqrt[n]{\frac{1}{n-1}+\cdots}}}$

The following infinite nested radical $$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$$ is known to converge to $\phi=\displaystyle\frac{\sqrt{5}+1}{2}$.

It is also known that the similar infinite nested radical $$\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\sqrt[3]{\frac{1}{2}+\cdots}}}}$$ converges to $\displaystyle\frac{1}{t_3-1}=1.19148...$, where $t_3=1.83929...$ is the tribonacci constant, corresponding to the ratio to which adjacent tribonacci numbers tend. Note the similarities with the previous case, since $\displaystyle\frac{1}{\phi-1}=\phi$.

Interestingly, assuming that the degree of the radicals is $n$ and that the constant term under the radicals is $1/(n-1)$, the relationship between this type of nested radicals and the n-bonacci constants is valid even for the degenerate 1-bonacci numbers (whose sequence reduces to a series of 1). In this case, the radicals are canceled out, the constant term becomes $1/0=\infty$, and the result is $\infty$, which is in accordance with $\frac{1}{1-1}=\infty$.

Based on these considerations, it could be hypothesized that similar nested radicals may exist for higher order n-bonacci numbers. For example, for $n=4$, we could expect that a nested radical giving $\displaystyle\frac{1}{t_4-1}=1.07809...$, where $t_4=1.92756...$ is the so-called tetranacci constant (limit of the ratio between adjacent 4-bonacci numbers), might be

$$\sqrt[4]{\frac{1}{3}+\sqrt[4]{\frac{1}{3}+\sqrt[4]{\frac{1}{3}+\sqrt[4]{\frac{1}{3}+\cdots}}}}$$

Unfortunately, in this case the radical seems not to work (it gives $1.09279...$ instead of the expected $1.07809...$). Also, I was not able to find any rational number that, inserted in this nested radical as the constant term, satisfies the equivalence (the numerical value that works is $k=0.27282...$). Similar results are obtained for higher values of $n$. Which is the reason for this? Is there any way to prove or disprove the existence of similar nested radicals linked to n-bonacci numbers for $n>3$?

Solution 1:

This is equivalent to $x^n=\dfrac1{n-1}+x$ , which means solving a polynomial of the n-th degree in x. The generalized Fibonacci numbers fulfill the general equation $x^{−n}+x=2$, so there is no reason to expect any universally valid connection between the two. See also strong law of small numbers.

Solution 2:

(Edit: Generalized previous answer.)

Actually, you are partially right. There is a general approach, but it works in radicals only for $n \leq 5$. Given,

$$x = \sqrt[n]{b+a\sqrt[n]{b+a\sqrt[n]{b+a\sqrt[n]{b+\dots}}}}$$

Raise to the $n$th power,

$$x^n = b+a\sqrt[n]{b+a\sqrt[n]{b+a\sqrt[n]{b+\dots}}}$$

Or,

$$x^n = b+ax\tag1$$

The general eqn of degree $n\leq5$ can be transformed, in radicals, to this form with the case $n=5$ called the Bring quintic. Hence, given the positive real root of $n$-nacci eqn, $y^{-n}+y = 2$, then,

$n = 2$: Fibonacci constant $y$:

$$y = \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dots}}}} = 1.618033\dots$$

$n = 3$: Tribonacci constant $y$:

$$y(3-y) = \sqrt[3]{14-2\sqrt[3]{14-2\sqrt[3]{14-2\sqrt[3]{14-\dots}}}} = 2.134884\dots$$

$n = 4$: Tetranacci constant $y$:

$$y(3-y) = \sqrt[4]{41-11\sqrt[4]{41-11\sqrt[4]{41-11\sqrt[4]{41-\dots}}}} = 2.06719\dots$$

$n = 5$: Pentanacci constant $y$:

$$\text{doable, but complicated}$$

You can still do so for $n=5$, but the LHS will be more complicated and $a,b$ will be algebraic numbers with square roots and cube roots. But if you are curious how to transform the general quintic into the form $(1)$, see this post.

P.S. Let me guess. You found that other nested radical for the tribonacci constant in this website?