[seqfan] Re: Q regarding A023172 (Numbers n such that n divides Fib(n).)
Max Alekseyev
maxale at gmail.com
Sun Nov 28 21:04:59 CET 2010
All terms of A023172 greater than 1 are multiples of 5 or 12.
Let n>1 divides Fibonacci number F(n) and let p a minimal prime divisor of n.
If p=2, then 3|n implying further that 4|n. Hence, 12|n.
If p is different from 2 and 5, then p divides either F(p+1) or F(p-1)
and thus p divides either F(gcd(n,p+1)) or F(gcd(n,p-1)). Minimality
of p implies that gcd(n,p-1)=1 and gcd(n,p+1)=1 (notice that p+1 being
prime implies p=2 which is not the case). Therefore, p divides F(1)=1,
a contradiction.
Therefore, any term of A023172 greater than 1 is a multiple 5 or 12.
Regards,
Max
On Sun, Nov 28, 2010 at 5:47 AM, Joerg Arndt <arndt at jjj.de> wrote:
> Can anyone answer the following?
>
> "Are there any a(n)>1 that are neither
> a multiple of 5 nor a multiple of 5?"
>
> See (the pink box near top in)
> http://oeis.org/history?seq=A023172
>
>
>
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