[seqfan] Re: Observations on some odd Fibonacci numbers
Alonso Del Arte
alonso.delarte at gmail.com
Tue Oct 5 15:53:55 CEST 2010
At first blush F(2n) = F(n)L(n) would seem a much simpler way of
saying L(m)|F(n)
iff 2m|n, but that formulation gives up more Lucas divisors of F(n) when n
has more divisors. For example, F(12) = 144, divisible by the smaller
Fibonacci numbers 2, 3, 8. And we see that 144 is also divisible by 3, 4,
18, Lucas numbers indexed by 2, 3, 6.
Now if only I could remember the correlation between n and odd F(n) I'd be
halfway home to proving Shevelev's observation.
Al
On Mon, Oct 4, 2010 at 10:07 PM, T. D. Noe <noe at sspectra.com> wrote:
> 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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