# [seqfan] Re: Observations on some odd Fibonacci numbers

Thu Oct 7 23:45:48 CEST 2010

```Please ignore the previous email.

Now we consider sequence with the definition
"a(n) is the n-th odd Fibonacci number F with the property: (1) if F=F_m, then m is not divided by square of  any odd prime; (2) if G is the maximal proper Fibonacci divisor of F, then  F/G has not a
proper Fibonacci  divisor>3"
This allows to keep terms 987, 2178309, 102334155.
Number F=F_28=317811 is multiple of 3 (important example by Doug) has two proper Fibonacci divisors>3:
G_1= F_7=13 and G_2=F_{14}=377. In this case, we see that  the maximal proper Fibonacci divisor>3 is LCM (G_1, G_2) and coincides with one of G_i.
Show this in the general case when 3 is divisor of F. If there exist exactly r such divisors G_i of F/3 with indices in A000045 m_1,...m_r, i=1,...r, then LCM (m_1,...m_r) divides index of F. Thus F_{ LCM (m_1,...m_r)} divides F\3.  In both cases: all G_i are not multiple of 3 and and some of them are multiple of 3, we have: F_{ LCM (m_1,...m_r)}<F. Therefore, F_{ LCM (m_1,...m_r)}coincides with one of G_i ( in the oposite case, we have at least r+1 Fibonacci divisors>3). Thus, for some i=j, we have G_j=F_{ LCM (m_1,...m_r)}=  LCM (G_1,...G_r)=max(G_1,...G_r).
These arguments are quite relevant if  to teplace 3 by the minimal Fibonacci divisor of F.
We now should correct our proof from 6 Oct 2010. The conclusive part was based on statement that if F=F_m and d|m, then F has divisors G_1=F_d and G_2=F_{m/d} such that max (G_1,G_2)>3 and F/G_1 contains Fibonacci divisor G_2, while F/G_2 contains Fibonacci divisor G_1. This
is true if GCD(d, m/d)=2^v, v>=0, but, generally speaking, does not satisfy. Let us show that if G_j=F_l is, as in the above, max(G_1,...,G_r), then always GCD(m/l, l) =2^v, v>=0. Indeed, let odd d>1 such that d|(m/l) and d|l. Then m=adl and l=bd (with integer a,b). Thus m=abd^2. Let  p a prime divisor of d. Then p^2|m.  This contradicts to the condition (1). Therefore, now F/F_l
contains  Fibonacci divisor F_{m/l}, while  F/F_{m/l} contains  Fibonacci divisor F_l. Since max (F_l, F_{m/l})>3, then this contradicts to the condition (2). This completes our proof.

After that our previous proof with respect to the maximal Fibonacci divisor of F is valid and we conclude that always F/max(G_1,...,G_r) is Lucas number or products of some Lucas numbers.

Regards,

----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Thursday, October 7, 2010 9:24
Subject: [seqfan] Re: Observations on some odd Fibonacci numbers
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> Note that, condition ind G>sqrt(ind F) one can replace by weaker
> condition ind G is not equal to
> sqrt(ind F) ( with the quite same proof). This allows to keep
> terms 987, 2178309, 102334155.
> Now we conjecture that  sequence with the definition
> "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 not equals to sqrt(ind F) and  F/G has not a
> proper Fibonacci  divisor>3" 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, over
>
>
> ----- 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,
> >
> >
> >
> >
> > ----- 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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> > >
> >
> >
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> >
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> >
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