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

Alexander P-sky apovolot at gmail.com
Tue Oct 5 16:23:19 CEST 2010

```FYI

---------- Forwarded message ----------
From: Alexander P-sky <apovolot at gmail.com>
Date: Tue, 5 Oct 2010 05:01:43 -0400
Subject: Re: [seqfan] Re: Observations on some odd Fibonacci numbers
To: "T. D. Noe" <noe at sspectra.com>, Vladimir Shevelev <shevelev at bgu.ac.il>

Through wayback machine
http://web.archive.org/web/20080430214712rn_1/mathforum.org/library/drmath/view/52699.html

A Fibonacci and Lucas Number Relation

Date: 08/24/98 at 04:59:12
From: David Ng
Subject: How do I prove a Fibonacci vs Lucas relation?

Hi,

This has been troubling me for quite some time now. How do I prove
that:

F(2n) = F(n) * L(n)

where F(x) is the xth Fibonacci number and L(y) is the yth Lucas
number?

I have only reached this:

F(2n) = (F(n+1))^2 - (F(n-1))^2

Date: 08/24/98 at 10:28:44
From: Doctor Floor
Subject: Re: How do I prove a Fibonacci vs Lucas relation?

Hi David,

Thank you for sending your question to Dr. Math.

To prove that F(2n) = F(n)L(n), I first like to use the explicit
formulae for F(n) and L(n). To make these formulae I first introduce
two very special members of the family of generalized Fibonacci
sequences (i.e. sequences such that X(n) = X(n-1) + X(n-2) ). These
special members are of the form:

g = g^1, g^2, g^3, g^4, ...  [g nonzero]

In such a formula we must have that:

g^n = g^(n-1) + g(n-2)
g^n - g^(n-1) - g(n-2) = 0
g^(n-2) * (g^2 - g - 1) = 0
g^2 - g - 1 = 0   [we can divide by g^(n-2) becuase g is nonzero]

So g must be one of the roots of g^2 - g - 1 = 0, so it must be one of
the following numbers:

P = (1+SQRT(5))/2  (a.k.a. the Golden Ratio)
Q = (1-SQRT(5))/2

So we have two very special generalized Fibonacci sequences:

P^1, P^2, P^3, P^4, ...
Q^1, Q^2, Q^3, Q^4, ...

I leave it for you to check that, when you have two generalized
Fibonacci sequences A(n) and B(n), and two numbers a and b, then
a*A(n) + b*B(n) is a generalized Fibonacci sequence. In fact, all
generalized Fibonacci sequences can be calculated in this way from
P^n and Q^n.

I also leave it for you to check that (check the first two, the rest
follows from being a generalized Fibonacci sequence):

F(n) = (P^n - Q^n)/SQRT(5)
L(n) = P^n + Q^n

Now that we know these formulae, we can conclude that:

F(n) * L(n) = (P^n - Q^n)/SQRT(5) * (P^n + Q^n)
= (P^n - Q^n)(P^n + Q^n)/SQRT(5)
= ((P^n)^2 - (Q^n)^2)/SQRT(5)
= (P^(2n) - Q^(2n))
= F(2n)

as desired.

If you have a math question again, please send it to Dr. Math.

Best regards,

- Doctor Floor, The Math Forum
Check out our web site! http://mathforum.org/dr.math/

Date: 08/24/98 at 10:42:58
From: Doctor Nick
Subject: Re: How do I prove a Fibonacci vs Lucas relation?

Hello David -

You've almost got it with your last statement.

Notice we can rewrite it as:

F(2n) = (F(n+1) - F(n-1))(F(n+1) + F(n-1))

and since F(n+1) - F(n-1) = F(n), we have:

F(2n) = F(n)(F(n+1) + F(n-1))

Now we use the fact that:

L(n) = F(n-1) + F(n+1)

which you probably already know, or you can prove it with induction.
This gives us:

F(2n) = F(n)L(n)

and that's it.

---------- Forwarded message ----------
From: "T. D. Noe" <noe at sspectra.com>
Date: Mon, 4 Oct 2010 19:07:17 -0700
Subject: [seqfan] Re: Observations on some odd Fibonacci numbers
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

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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