Matematické Fórum

Skrytý text:

We can see that several first    satisfy equation  . 

We can ask: Isn't this equation valid for every ?  First we'll find a necessary condition for such hypothese.


LEMMA. If is a sequence satisfying

(1)            ,

then

(2)              .


PROOF.  According to (1) we obtain
                       
(a)                              ,

(b)                              ,

(a) - (b)                ,

hence (2) . QED.


NEXT STEPS:  Resolving (2) with

(3)               

and showing

(4)                .

Characteristic polynomial of (2) : ,  its roots:  , general solution of (2):

(5)                                 .

From (3), (5) we obtain

(6)                .

It is not difficult to find that every  is a positive integer.  Positivity from (5), (6) , integrity from  (3), (2).

To prove (4), I let it to the kind reader :-) . With the Viete formulas for it is quite easy.  Therefore

                        ,

which corresponds with the recoursive formula for .

RESULT: 

            .

Skrytý text:
Let us define sequences and of positive integers in the following way:

and are coprime for each and



where the last fraction has levels. In other words, are defined by finite continued fractions. It can be proved by induction that



where is the Fibonacci sequence, i.e.



Thus and , .

Prove now by induction that



1.



2. Suppose that



Then



If is an odd number then , and



Thus is a sequence of positive integers.

An exact formula for can be derived from the one for the Fibonacci sequence taking .