[seqfan] Re: Observations on some odd Fibonacci numbers
Vladimir Shevelev
shevelev at bgu.ac.il
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
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