Ivars Wrote:Thanks. is n-th iteration of , starting with , when n=0 .

Ok that means

where

and

.

I put some effort into investigating this sequence

. For the case of

and

that you describe one can see numerically that in the limit

for some constant

. By this observation we put up:

(1)

If

converges we can choose

so that

converges to 0. In this case the equation has to be satisfied in the limit too:

So that if

then

,

, where

has to be satisfied. Let

then all possible solutions are

and

for integer

. However only one of them is the actual value in

.

For example in our case

and

.

If vice versa

is given as above, what conditions are to be satisfied that

?

We put

into equation (1), first:

, for simplicity let

,

where we get

for solutions

and

for solutions

.

now into (1)

The interesting cases is (a) if

and

and (b) if

and

together with c)

. We assume in the following that

and only consider the case

.

If

then

and only positive values are subtracted from

, so

satisfies case (a).

If

then

and the condition

is equivalent to

.

In this case

and as both side are positive we can quadrate getting equivalently:

, let

so case (b) is occurs exactly for

So in the case

it is this satisified for all

otherwise for

.

As a last step we give conditions for c)

.

Let

and

. In the case that

we have to assure that the subtractions of

and

do not lead below

, i.e. at most

.

In the case

,

and

we showed already that

, so that the subtraction of

and the addition of

increases

. We shall ascertain that

.

The condition

is easy to check:

Now we know that

for

, hence the condition is satisfied if we demand:

.

The condition

, let

and

:

.

Similarly to the previous case the condition is satisfied for:

.

We know that

is striclty increasing for

and

(graph it!). Hence there is an

such that

(which is equivalent to

) for all

.

Now going backwards we get

Proposition.

Let , , , , and let be the solution of .

If there exists an and integer such that , then , particularly .
Verify the conditions for your choice:

,

,

,

,

.

The last condition that

, or even the stronger condition that

is quite probable for any sufficient chaotic choice of

and

.

So for your choice but also many other choices for

and

indeed

.

PS: By the length of the derivations I can not exclude that somewhere got an error into the computations, so dont take the result as granted.