[seqfan] Re: Observations on some odd Fibonacci numbers

[Vladimir Shevelev]

I considered only odd Fibonacci numbers (A014437) for which the indices differ from A000045.
Therefore, I was surprised by my simple observation.
Now I think that in my terms (without indices) the following statement is true: an odd Fibonacci number F is represented by the product of the form F=F_1*Prod {i=1,...,m}L_i (F_1>=3, m>=1)
iff  F has at least one  proper Fibonacci divisor G>1 for which  F/G has not a proper Fibonacci divisor>3. 
 
In particular, we should prove that , at least, for 
one of  such divisors G=G_0, the ratio F/G_0 is a Lucas number or product of some Lucas numbers.
 E.g., F=317811 has two such divisors: 13 and 377. Here G_0=377 (not 13).

Best regards,
Vladimir


----- Original Message -----
From: "T. D. Noe" <noe at sspectra.com>
Date: Tuesday, October 5, 2010 10:26
Subject: [seqfan] Re: Observations on some odd Fibonacci numbers
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> At 5:49 PM +0000 10/2/10, Vladimir Shevelev wrote:
> > Dear SeqFans,
> >   I consider the following subsequence of Fibonacci 
> numbers:>21,55,377,987,6765,17711,121393,2178309,5702887,39088169,1836311903,...
> >with the definition: a(n) is the n-th odd Fibonacci number F 
> with the
> >property: F has a proper Fibonacci divisor G>1, but F/G has not.
> >   I noticed (without a proof) that F/G is a Lucas 
> number or a product of
> >some Lucas numbers.
> >   E.g., for F=6765, G=5 and F/G=1353=11*123; for 
> F=2178309, G=3 and
> >F/G=726103=7*47*2207; for F=1836311903, G=28657 and F/G=64079.
> >Could anyone verify (or disprove) this observation for further 
> terms of
> >the sequence?
> 
> 
> It seems that you have found the identity
> 
> F(2n) = F(n) * L(n).
> 
> Using this recursively gives
> 
> F(2^k n) = F(n) * L(2^(k-1) n) * L(2^(k-2) n) * ... * L(n).
> 
> Best regards,
> 
> Tony
> 
> 
> 
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> 

 Shevelev Vladimir‎