Is this a valid formula for the tribonacci numbers?

Question

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.)

Answer 1

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 $$\frac 1{r_1}=-\frac{1}{3}+\frac{t}{3}-\frac{2}{3 t}$$ $$\frac 1{r_2}=-\frac{1}{3}-\frac{t}{6}+\frac{1}{3 t}+\frac i {\sqrt 3}\left(\frac 1 t +\frac t2\right)$$ $$\frac 1{r_3}=-\frac{1}{3}-\frac{t}{6}+\frac{1}{3 t}-\frac i {\sqrt 3}\left(\frac 1 t +\frac t2\right)$$ $$k=\frac 1{r_1\,r_2}=\frac{4}{9 t^2}-\frac{2}{9 t}+\frac{1}{3}+\frac{t}{9}+\frac{t^2}{9}$$