# How to find the nth term of tribonacci series [duplicate]

I want to know the nth term of tibonacci series, given by the recurrence relation $$a_{n + 3} = a_{n + 2} + a_{n + 1} + a_n$$ with $a_1 = 1, a_2 = 2, a_3 = 4$, so the first few terms are $$1,2,4,7,13,24,44, \ldots$$ I am more interested in derivation of how the nth term is calculated rather than direct comming up with black box formula.

## marked as duplicate by Steven Stadnicki, Claude Leibovici, Asaf Karagila♦, user63181, NorbertJun 9 '14 at 7:33

• The tribonacci numbers are generated by the homogeneous linear recurrence relation $T_n = T_{n-1}+T_{n-2}+T_{n-3}$. See the methods described in this wikipedia article for info on solving linear recurrence relations: en.wikipedia.org/wiki/Recurrence_relation – JimmyK4542 Jun 9 '14 at 6:05
• I added the recurrence definition to your question, so people don't have to look it up themselves. – Arthur Jun 9 '14 at 6:07
• Do you know some method for the Fibonacci series? Have you tried something similar? – poolpt Jun 9 '14 at 6:07
• Yes there is Binet's methods to calculate nth term of fibonacci series. – Ankit Zalani Jun 9 '14 at 6:08
• Use linear algebra to solve it. – DeepSea Jun 9 '14 at 6:15

The $n^{th}$ tribonacci number $T_n$ is given by the closest integer to $$\frac{3 b}{b^2-2 b+4} \Big(\frac{a_++a_-+1}{3}\Big)^n$$ where $$a_{\pm}=\sqrt[3]{19 \pm 3 \sqrt{33}}$$ $$b=\sqrt[3]{586 + 102 \sqrt{33}}$$
See OEIS sequence A000073. We can write $$a_n = \sum_r \frac { -4\,{r}^{2}-3\,r+5}{ 22\;{r}^{n+3}}$$ where the sum is over the three roots of $r^3 + r^2 + r - 1$ (one real and two complex).