[seqfan] Re: Observations on some odd Fibonacci numbers

Fri Oct 8 16:38:04 CEST 2010

Consider, finally,  sequence which defined as: "a(n) is the n-th odd Fibonacci number F with the property: (1) F has at least one proper Fib. divisor (2) index of the maximal proper Fib. divisor G* differs from sqrt(ind F); (3)  F/G* has not a proper Fibonacci  divisor>3".
 Let us back to the last rows of our proof from 6 Oct 2010. I was confused by  example by Doug F_{28}=317811. This number has Fibonacci divisors 3,13,377. Divisor F_7=13 does not contradict to the last rows of our proof, but 317811/13 is not represented as a product of Lucas numbers. Nevertheless, proof is not true for 13  o n l y since F_{28/7}=F_4=3 and 3 is only Fibonacci divisor allowed for F/G, therefore, we have not a contradiction. But our proof is, evidently, true, if there exists divisor d>1 of m=ind(F) for which simultaneously F_d and F_{m/d} differ from 3, i.e. both of  d and m/d differ from 4. Show that this always satisfies for choice d=t, where t=ind G*. Recall that we consider m of the forms pl or p^2*l, where l is odd and not multiple of 3. Besides, we can suppose that p is odd ( otherwise, F_m has the required representation). Note that, G=G* is unique proper Fib. divisor of F only in case F_8=21=F_4*L_4. Let  F>21. Then there exists a proper Fib. divisor G=F_k of F such that 3<=G<G*. Thus, since F_t>F_k>=3, then  t>4. If now m/t=4 then m=4t which contradicts to the supposition. Furthermore, since m, by the supposition, is odd, then t and m/t are odd. Bisides m/t>=3 since G* is proper divisor. Thus G=F_{m/t}>1 is proper divisor of F. Then, as I proved yesterday, G|G*. This means that m/t divides t. Therefore, m has the form m=a*t^2 with , by condition (2), a>1. This yields that F_{t^2} is proper divisor of F. Since F_{t^2}>F_t=G*, then the latter conradicts to the maximality of G*. This completes proof.

 Now we conjecture that sequence {a_n}contains all {F} which are 
represented in the form   F=F_1*Prod {i=1,...,m}L_i 
(F_1>=3, m>=1), and only them.
 To prove this conjecture, we should prove that, for odd 
prime p, the ratio F_{p^2}/F_p  is not Lucas number or a product of Lucas numbers. 
Moreover,  simultaneously it will be proved that the sequence contains all odd Fibonacci 
numbers which in A000045 have indices of the form 6k+2 and 6k+4, and only them.

Regards,Vladimir

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Wednesday, October 6, 2010 21:57
Subject: [seqfan] Re: Observations on some odd Fibonacci numbers
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> 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,
> Vladimir
> 
> 
> 
> 
> ----- 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
> > 
> > _______________________________________________
> > 
> > Seqfan Mailing list - http://list.seqfan.eu/
> > 
> 
>  Shevelev Vladimir‎
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 

 Shevelev Vladimir‎