4
$\begingroup$

Every Fibonacci term $F(3n)$ is divisible by two

$F(3) = 2$

$F(6) = 8$

$F(9) = 34$

$...$

After seeking Fibonacci tables factorization until $F(10000)$, for every term $\frac{F(3n)}{2}$, it appears there is just one result which is prime

$\frac{F(9)}{2} = \frac{34}{2} = 17$

all other $\frac{F(3n)}{2}$ = composite numbers

I can't see some reason for this to happen, but since there is no more primes in so huge quantity of terms, I suppose there is not more any prime for $\frac{F(3n)}{2}$

http://mersennus.net/fibonacci

Miguel Velilla

$\endgroup$
2
$\begingroup$

Recall the well-known fact: $n \mid m \iff F_n \mid F_m$.

Thus $F_{3n}$ is divisible by $F_3=2$ and $F_n$. So if $n > 3$ and $\displaystyle \frac{F_{3n}}{2}$ is equal to prime number $p$ then $F_n \mid F_{3n}=2p$ so we have $F_{n}\in\{p, 2p\}$ (because $F_n>F_3=2$). In particular $F_{3n} \le 2F_n$ which is a contradiction as $2F_n < F_n+F_{n+1}=F_{n+2}<F_{3n}$.

$\endgroup$
0
$\begingroup$

It is well known that if $a$ divides $b$ then $Fib(a)$ divides $Fib(b)$.

Ignoring for the moment the cases where $a=1,2,3$ (in which case Fib(a)=$1,1,2$), this means that $Fib(3a)$ is divisible by $Fib(a) \gt 2$ so $\frac{Fib(3a)}{2}$ is divisible by $\frac{Fib(a)}{2} \gt 1$ if $Fib(a)$ is even or by $Fib(a) \gt 1$ if it is odd. So $\frac{Fib(3a)}{2}$ cannot be prime.

That leaves the early cases $\frac{Fib(3)}{2},\frac{Fib(6)}{2},\frac{Fib(9)}{2}$ which are $1,4,17$, and only the last of these is prime.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.