[seqfan] Re: More (composite) terms for A233281
David Wilson
davidwwilson at comcast.net
Wed Feb 12 04:57:37 CET 2014
Let Fib(f(p)) > 0 be the smallest positive Fibonacci number divisible by p.
Empirically, it appears that
f(2) = 3, f(4) = 6, f(2^(k+3)) = 6*2^k.
For odd prime p, f(p) > 1, f(p^(k+1)) = f(p)*p^k.
If true (which I assume it is), then f(p^k) is composite for prime p and k >
1.
In other words, if k > 1, p^k | Fib(n) => n is composite.
This would imply that Fib(p) for prime p is squarefree.
As such Fib(p) is the product of k prime numbers.
This means it has 2^k divisors, k divisors are prime, 1 divisor is 1.
This leaves 2^k - k - 1 composite divisors.
2^k - k - 1 happens to be Eulerian number A(k, 1), I doubt there is a deeper
connection to Eulerian numbers.
We know that p divides Fib(f(p)) where f(p) = p + (0, -1, 1, 1, -1)
according to whether p mod 5 is (0, 1, 2, 3, 4).
Suppose p | F(q) for prime q.
Since m | Fib(n) => m | Fib(kn), and p | Fib(f(p)), we conclude q | f(p).
Hence f(p) = kq for some k >= 1.
|f(p) - p| <= 1, so p must be within 1 of some multiple of q.
So, with the exception of p = 5, all prime divisors of Fib(p) are of the
form kp +- 1.
For example, Fib(19) has prime divisors 37 = 2*19-1 and 113 = 4*19-1.
This implies that Fib(p) cannot be divisible by a prime smaller than 2p-1.
Not much of a lower bound, but a lower bound.
> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Hans
> Havermann
> Sent: Monday, February 10, 2014 12:13 PM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: More (composite) terms for A233281
>
> Attempting to understand this sequence better I have just posted a
> (hopefully) complete list of composites derived from Fibonacci numbers
with
> prime indices, up to index 997:
>
> http://chesswanks.com/num/CompositeDivisorsOfPrimeIndexFibonaccis.txt
>
> There appear to be an Eulerian number < http://oeis.org/A000295 > of
> solutions in each case. Antti's composites will of course be the sorted
> collection of all these solutions to infinity. My question is still how
one might
> know that smallish solutions (say, up to a given size) will not come from
> unsolved, future Fibonaccis.
>
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