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

4

$begingroup$

This question already has an answer here:

• 
  Fibonacci, tribonacci and other similar sequences
  
  2 answers
  
  

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.

sequences-and-series

share|cite|improve this question

edited Jun 9 '14 at 6:05

Arthur

116k7116198

asked Jun 9 '14 at 5:58

Ankit ZalaniAnkit Zalani

12815

$endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

4

$begingroup$

This question already has an answer here:

• 
  Fibonacci, tribonacci and other similar sequences
  
  2 answers
  
  

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.

sequences-and-series

share|cite|improve this question

edited Jun 9 '14 at 6:05

Arthur

116k7116198

asked Jun 9 '14 at 5:58

Ankit ZalaniAnkit Zalani

12815

$endgroup$

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

4

4

4

1

$begingroup$

This question already has an answer here:

• 
  Fibonacci, tribonacci and other similar sequences
  
  2 answers
  
  

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.

sequences-and-series

share|cite|improve this question

edited Jun 9 '14 at 6:05

Arthur

116k7116198

asked Jun 9 '14 at 5:58

Ankit ZalaniAnkit Zalani

12815

$endgroup$

This question already has an answer here:

• 
  Fibonacci, tribonacci and other similar sequences
  
  2 answers
  
  

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.

This question already has an answer here:

• 
  Fibonacci, tribonacci and other similar sequences
  
  2 answers
  
  

sequences-and-series

sequences-and-series

share|cite|improve this question

edited Jun 9 '14 at 6:05

Arthur

116k7116198

asked Jun 9 '14 at 5:58

Ankit ZalaniAnkit Zalani

12815

share|cite|improve this question

edited Jun 9 '14 at 6:05

Arthur

116k7116198

asked Jun 9 '14 at 5:58

Ankit ZalaniAnkit Zalani

12815

share|cite|improve this question

share|cite|improve this question

edited Jun 9 '14 at 6:05

Arthur

116k7116198

edited Jun 9 '14 at 6:05

Arthur

116k7116198

edited Jun 9 '14 at 6:05

Arthur

116k7116198

116k7116198

asked Jun 9 '14 at 5:58

Ankit ZalaniAnkit Zalani

12815

asked Jun 9 '14 at 5:58

Ankit ZalaniAnkit Zalani

12815

asked Jun 9 '14 at 5:58

Ankit ZalaniAnkit Zalani

12815

12815

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

1

1

$begingroup$
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

$begingroup$
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

$begingroup$
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

$begingroup$
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

$begingroup$
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

$begingroup$
Use linear algebra to solve it.
$endgroup$
– DeepSea
Jun 9 '14 at 6:15

add a comment | 

2 Answers
2

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

answered Jun 9 '14 at 6:28

Claude LeiboviciClaude Leibovici

122k1157134

$endgroup$

add a comment | 

2

$begingroup$

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).

share|cite|improve this answer

answered Jun 9 '14 at 6:39

Robert IsraelRobert Israel

324k23214468

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

answered Jun 9 '14 at 6:28

Claude LeiboviciClaude Leibovici

122k1157134

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered Jun 9 '14 at 6:28

Claude LeiboviciClaude Leibovici

122k1157134

$endgroup$

add a comment | 

2

2

2

$begingroup$

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.

share|cite|improve this answer

answered Jun 9 '14 at 6:28

Claude LeiboviciClaude Leibovici

122k1157134

$endgroup$

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.

share|cite|improve this answer

answered Jun 9 '14 at 6:28

Claude LeiboviciClaude Leibovici

122k1157134

share|cite|improve this answer

share|cite|improve this answer

answered Jun 9 '14 at 6:28

Claude LeiboviciClaude Leibovici

122k1157134

answered Jun 9 '14 at 6:28

Claude LeiboviciClaude Leibovici

122k1157134

answered Jun 9 '14 at 6:28

Claude LeiboviciClaude Leibovici

122k1157134

122k1157134

add a comment | 

add a comment | 

2

$begingroup$

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).

share|cite|improve this answer

answered Jun 9 '14 at 6:39

Robert IsraelRobert Israel

324k23214468

$endgroup$

add a comment | 

2

$begingroup$

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).

share|cite|improve this answer

answered Jun 9 '14 at 6:39

Robert IsraelRobert Israel

324k23214468

$endgroup$

add a comment | 

2

2

2

$begingroup$

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).

share|cite|improve this answer

answered Jun 9 '14 at 6:39

Robert IsraelRobert Israel

324k23214468

$endgroup$

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).

share|cite|improve this answer

answered Jun 9 '14 at 6:39

Robert IsraelRobert Israel

324k23214468

share|cite|improve this answer

share|cite|improve this answer

answered Jun 9 '14 at 6:39

Robert IsraelRobert Israel

324k23214468

answered Jun 9 '14 at 6:39

Robert IsraelRobert Israel

324k23214468

answered Jun 9 '14 at 6:39

Robert IsraelRobert Israel

324k23214468

324k23214468

add a comment | 

add a comment | 

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