<s>[INST] Is this a valid formula for the tribonacci numbers? After about 5 pages of derivation, I derived a formula for the Tribonacci series:
$$
T(x) = \left( \frac{a-b-c + (b-c)r_1 + cr_1^2}{k(r_1-r_2)(r_1-r_3)} \right)r_1^n   
      + \left( \frac{a-b-c + (b-c)2_1 + cr_2^2}{k(r_2-r_1)(r_2-r_3)} \right)r_2^n   
      + \left(\frac{a-b-c + (b-c)r_3 + cr_3^2}{k(r_3-r_1)(r_3-r_2)} \right)r_3^n  
$$
Where $a = T_2,b=T_1,c=T_0$, and:
$$
1 - x - x^2 - x^3 = k(1-r_1)(1-r_2)(1-r_3)
$$
In this case: 
$$
r_1 = \frac{1}{\frac13\left(\sqrt[3]{3\sqrt{33}-17}-\dfrac2{\sqrt[3]{3\sqrt{33}-17}}-1\right)}
$$
and(approx)
$$ 
r_2 = \frac{1}{1.11514i - 0.771845}
$$
$$ 
r_3 = \frac{1}{1.11514i + 0.771845}
$$
$$
k = \frac{1}{r_2r_3}
$$
Is this a valid formula?
(I'll edit this answer to include the derivation, if it's really needed. It's a bit long.)
 [/INST] Just to make your formulae looking simpler.
I must precise that I did not check the calculations.
Define $$t=\sqrt[3]{17+3 \sqrt{33}}$$ this makes the roots to be such that
$$\frac 1{r_1}=-\frac{1}{3}+\frac{t}{3}-\frac{2}{3 t}$$
$$\frac 1{r_2}=-\frac{1}{3}-\frac{t}{6}+\frac{1}{3 t}+\frac i {\sqrt 3}\left(\frac 1 t +\frac  t2\right)$$
$$\frac 1{r_3}=-\frac{1}{3}-\frac{t}{6}+\frac{1}{3 t}-\frac i {\sqrt 3}\left(\frac 1 t +\frac  t2\right)$$
$$k=\frac 1{r_1\,r_2}=\frac{4}{9 t^2}-\frac{2}{9
   t}+\frac{1}{3}+\frac{t}{9}+\frac{t^2}{9}$$
</s>