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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.

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marked as duplicate by Steven Stadnicki, Claude Leibovici, Asaf Karagila, user63181, Norbert Jun 9 '14 at 7:33

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    $\begingroup$ 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 $\endgroup$ – JimmyK4542 Jun 9 '14 at 6:05
  • $\begingroup$ I added the recurrence definition to your question, so people don't have to look it up themselves. $\endgroup$ – Arthur Jun 9 '14 at 6:07
  • $\begingroup$ Do you know some method for the Fibonacci series? Have you tried something similar? $\endgroup$ – poolpt Jun 9 '14 at 6:07
  • $\begingroup$ Yes there is Binet's methods to calculate nth term of fibonacci series. $\endgroup$ – Ankit Zalani Jun 9 '14 at 6:08
  • $\begingroup$ Use linear algebra to solve it. $\endgroup$ – DeepSea Jun 9 '14 at 6:15
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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 $$a_{\pm}=\sqrt[3]{19 \pm 3 \sqrt{33}}$$ $$b=\sqrt[3]{586 + 102 \sqrt{33}}$$

You could find all required information in the links the comments and answers refer to.

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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).

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