[seqfan] Re: Observations on some odd Fibonacci numbers

Vladimir Shevelev shevelev at bgu.ac.il
Wed Oct 6 21:53:46 CEST 2010

Thanks, Richard, for counter-examples.
Note  that F_{98} has divisor G_0=F_{49}. Therefore,  F_{98} is not a counter-example
in a wider sense (see my previous seqfan-massage). On the other hand, it is interesting that the other
first indices that are 7^2,11^2,13^2. Moreover ( since F_m|F_n iff m|n), the corresponding indices of G are 7,11,13. I think that  the sequence of counter-examples contains other values of prime^2;
moreover, it seems that, for odd prime p, the ratio F_{p^2}/F_p does not contain any Lucas number. 
Let us prove that, in order to avoid the counter-examples, it is sufficient to introduce a weak restriction on  G: indG>sqrt(ind F), such that  the corrected definition is:
"a(n) is the n-th odd Fibonacci number F with the property: F has at least one  proper Fibonacci divisor G for which ind G>sqrt(ind F) for which  F/G has not a proper Fibonacci divisor>3". 
Show that for a(n) there exists at least one of  such divisor G=G_0 for which the ratio F/G_0 is a Lucas number or product of some Lucas numbers.
Indeed, let a(n)=F_m. Since F_m is odd, then, as it is well-known, m is not multiple of 3. Furthermore, if m is even, then F_m has required representation ( see message by Tony). The case m=p^2, where p is prime, is impossible by the condition. The case m=p is impossible as well, since
F_p has not a proper Fibonacci divisor. Finally, if m=p*l or m=p^2*l, where l is odd and not multiple of 3, then, evidently, for every proper divisor d>1 of m, m/d is also a proprer divisor of m. Thus, F_d and F_{m/d} are  proper divisors of F_m, such that, for every G, a(n) contains also the proper divsor a(n)/G>3. It is impossible by the condition. This completes proof.
  The first terms of the sequence are:
21,55,377,6765,17711,121393,317811,5702887,39088169, 701408733, 1836311903,...
But now the sequence differs from sequence of {F} which are represented in the form 
 F=F_1*Prod {i=1,...,m}L_i (F_1>=3, m>=1) ( e.g., 987, 2178309, 102334155 are excluded). 

Best regards,

----- Original Message -----
From: Richard Mathar <mathar at strw.leidenuniv.nl>
Date: Tuesday, October 5, 2010 18:54
Subject: [seqfan] Re: Observations on some odd Fibonacci numbers
To: seqfan at seqfan.eu

> http://list.seqfan.eu/pipermail/seqfan/2010-October/006140.html
> vs>    I consider the following subsequence of 
> Fibonacci numbers:
> vs> 
> 21,55,377,987,6765,17711,121393,2178309,5702887,39088169,1836311903,...vs> with the definition: a(n) is the n-th odd Fibonacci number F with the 
> vs> property: F has a proper Fibonacci divisor G>1, but F/G has 
> not. 
> Note also that 6765 is not in my list because 6765 = 3*2255
> where 2255 has a proper Fibonacci divisor (that is, 55). So my 
> interpretationof the definition is that no odd F with two proper 
> divisors in A000045
> are in a(n). I am not sure whether to admit cases where F has 
> proper 
> Fibonacci divisors G of both types.
> vs> I noticed (without a proof) that F/G is a Lucas number or a 
> product of 
> vs> some Lucas numbers.
> The examples
> n= 49, F=7778742049, G=13, F/G=598364773, Lprod=false
> n= 98, F=135301852344706746049, G=13, F/G=10407834795746672773, 
> Lprod=falsen= 121, F=8670007398507948658051921, G=89, 
> F/G=97415813466381445596089, Lprod=false
> n= 169, F=93202207781383214849429075266681969, G=233, 
> F/G=400009475456580321242184872389193, Lprod=false
> are the first counter-examples according to my calculation which 
> are in the
> list of these a(n) where F/G is not a Lucas number or product of such.
> http://mersennus.net/fibonacci/f1000.txt
> J Brillhart et al, "Tables of Fibonacci and Lucas 
> Factorizations" Math Comp 50 (1988) 252, 
> http://www.jstor.org/stable/2007928
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 Shevelev Vladimir‎

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