Talk:Fibonacci number: Difference between revisions
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imported>David E. Volk (New page: {{subpages}}) |
imported>Wlodzimierz Holsztynski (→Fibonacci number and Primenumber: exception F_4 = 3) |
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== Fibonacci number and Primenumber == | |||
*If <math>\ F_p\ </math> is a prime number then <math>\ p</math> is prime. (The converse is false.) | |||
is not correct: | |||
{| border=1 cellspacing="0" | |||
!n !!F(n) | |||
|- | |||
|0 ||0 | |||
|- | |||
|1 ||1 | |||
|- | |||
|2 ||1 | |||
|- | |||
|3 ||2 | |||
|- | |||
|4 ||3 | |||
|- | |||
|5 ||5 | |||
|} | |||
The forth Fibonacci number is 3. 3 is a Primenumber, but 4 is not a Primenumber. --[[User:Karsten Meyer|arbol01]] 07:30, 30 December 2007 (CST) | |||
:True. This is the only exception, due to the fact that 4=2*2 while F_2=1 is not greater than 1. Thus for the composite number 4=n=a*b=2*2 we do not get a proper divisor of F_a | F_n, i.e. F_2 | F_4, because F_2 is not proper, it is equal 1. It was my fault to overlook this case; I was happy (too happy) to see that the general theorem works accidentally also for prime F_3=2, where the index is prime 3. Thank you, [[User:Wlodzimierz Holsztynski|Wlodzimierz Holsztynski]] 16:23, 30 December 2007 (CST) | |||
Latest revision as of 16:23, 30 December 2007
- If is a prime number then is prime. (The converse is false.)
is not correct:
| n | F(n) |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 1 |
| 3 | 2 |
| 4 | 3 |
| 5 | 5 |
The forth Fibonacci number is 3. 3 is a Primenumber, but 4 is not a Primenumber. --arbol01 07:30, 30 December 2007 (CST)
- True. This is the only exception, due to the fact that 4=2*2 while F_2=1 is not greater than 1. Thus for the composite number 4=n=a*b=2*2 we do not get a proper divisor of F_a | F_n, i.e. F_2 | F_4, because F_2 is not proper, it is equal 1. It was my fault to overlook this case; I was happy (too happy) to see that the general theorem works accidentally also for prime F_3=2, where the index is prime 3. Thank you, Wlodzimierz Holsztynski 16:23, 30 December 2007 (CST)