The Ultimate Ratio

The sequence is called the ' Tribonacci numbers ', and each term is equal to the sum of the previous 3 terms (i.e. $T_n=T_{n-1}+T_{n-2}+T_{n-3}$ ).

(Note: the Brilliant wiki and the Wolfram Mathworld page for the Tribonacci numbers have the sequence starting with 1 zero while the Online Encyclopedia of Integer Sequences and Wikipedia have the sequence starting with 2 zeros. This does not affect the solution to this question.)

Assuming that the series approaches a common ratio $\displaystyle L=\lim_{n\to\infty}\dfrac{T_{n+1}}{T_n}$ , we can consider 4 very large consecutive terms in the series, $a,b,c$ and $d$ , so that: $a+b+c=d\qquad\text{and}\qquad\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=L$ Solving these equations, we can get a cubic equation for L: $\begin{aligned}\frac{c}{b}&=\frac{b}{a}&\frac{d}{c}&=\frac{b}{a}\\ c&=\frac{b^2}{a}&d&=\frac{bc}{a}\\ &&d&=\frac{b^3}{a^2}\\ a+b+c&=d\\ a+b+\frac{b^2}{a}&=\frac{b^3}{a^2}\\ \frac{b^3}{a^3}-\frac{b^2}{a^2}-\frac{b}{a}-1&=0\\ L^3-L^2-L-1&=0\end{aligned}$ To solve this, we will first remove the squared term by letting $L=y+\frac{1}{3}$ : $\left(y+\frac{1}{3}\right)^3-\left(y+\frac{1}{3}\right)^2-\left(y+\frac{1}{3}\right)-1=0\\ \left(y^3+y^2+\frac{1}{3}y+\frac{1}{27}\right)-\left(y^2+\frac{2}{3}y+\frac{1}{9}\right)-\left(y+\frac{1}{3}\right)-1=0\\ 3y^3-4y-\frac{38}{9}=0$ Now, instead of continuing with Cardano's method, we will let $y=\frac{4}{3}\cos\theta$ : $3\left(\frac{4}{3}\cos\theta\right)^3-4\left(\frac{4}{3}\cos\theta\right)-\frac{38}{9}=0\\ \frac{64}{9}\cos^3\theta-\frac{16}{3}\cos\theta-\frac{38}{9}=0\\ 4\cos^3\theta-3\cos\theta-\frac{19}{8}=0$ Using the trigonometric identity $\cos3\theta=4\cos^3\theta-3\cos\theta$ :* $\cos3\theta-\frac{19}{8}=0\\ \cos3\theta=\frac{19}{8}$ This has no real solutions, but there are complex ones, so we can still find the answer to the question with some rearranging: $\begin{aligned}L&=y+\frac{1}{3}\\ L-\frac{1}{3}&=\frac{4}{3}\cos\theta\\ \frac{3L-1}{4}&=\cos\theta\\ \cos^{-1}\frac{3L-1}{4}&=\theta\\ 3\cos^{-1}\frac{3L-1}{4}&=3\theta\\ \cos\left(3\cos^{-1}\frac{3L-1}{4}\right)&=\cos3\theta\\ &=\frac{19}{8}\\ 8\cos\left(3\cos^{-1}\frac{3L-1}{4}\right)&=\boxed{19}\end{aligned}$

---

*Concerning the trigonometric substitution

This is how one would determine what to let y equal for this trigonometric substitution to work:

We are going to let $y=k\cos\theta$ . The cubic expression equals zero, so we can multiply it by whatever we want, so it's simply a matter of finding a value of k such that the ratio between the coefficient of $y^3$ and the coefficient of $y$ is 4:3.

When we substitute in, the $3y^3$ will become $3k^3\cos^3\theta$ and the $4y$ will become $4k\cos^3\theta$ , so to make the ratio work: $\frac{3k^3}{4k}=\frac{4}{3}\\ k^2=\frac{16}{9}\\ k=\frac{4}{3}$ Now when we do the substitution we will be able to use the formula $\cos3\theta=4\cos^3\theta-3\cos\theta$ and solve the equation.