[seqfan] The Virtues of X_{n+1}=(4+3*X_{n})/(3+2*X_{n})

[Brad Klee]

Hey Seqfans,

It's Friday night somewhere,
maybe Saturday morning,
Let's do more iteration,
an easy-enough proof !

X_{n+1}=(4+3*X_{n})/(3+2*X_{n})
======================
4/3, 24/17, 140/99, 816/577, 4756/3363, 27720/19601, 161564/114243,
941664/665857, 5488420/3880899

List of Iterator Virtues
=============
* The rational iterator admits a fixed point at (+/-) sqrt(2), by setting
X_{n+1} = X_n and solving.

* With X = P/Q, the rational iterator decomposes to integer-only form :
( P_{n+1} , Q_{n+1} )=( 3*P_n + 4*Q_n , 2*P_n + 3*Q_n )

* By induction: If P_0 even, and Q_0 odd, then all P_n even all Q_n odd.
Vice versa also true.

* ( P_0 , Q_0 ) = ( 4 , 3 ) is a good enough initial condition, better than
( 0 , 1 ) in the sense that 0 does not provide any semblance of an
approximation to sqrt(2), ha! The iterated sequence seems to recover (
A005319, A001541 ); though, my reading of the entries did not rigorously
clarify this question.

* The transformation is invertible:
( P_n , Q_n ) = (3*P_{n+1} - 4*Q_{n+1},-2*P_{n+1} + 3*Q_{n+1} )

* All P_n coprime to Q_n, when P_0 coprime to Q_0. I stumbled over this
initially, making a stupid symmetry argument. Actually it's obvious after
calculating the inverse transformation. Let's work out the details
anyways.  Proof: Assume otherwise. With B coprime to C, and F a
hypothesized common factor, (P_{n+1},Q_{n+1})=F*( B , C ) is true if and
only if ( P_n , Q_n ) = F*(3*B - 4*C,-2*B + 3*C ), so ( P_n , Q_n ) also
have a common factor F. This sets up an induction with base case (4,3), two
coprime integers.

* Starting with (4/3), it is always possible to write: delta_n = xi - P_n /
Q_n, with xi = sqrt(2), and \delta_n > \delta_{n+1} > 0. That is, the
iterates approach increasingly close to sqrt(2). Or equivalently, X_{n+1} -
X_n > 0 whenever sqrt(2)>X>0, and 4/3 = 1.3... < sqrt(2) = 1.4...

* Applying the iteration, delta_{n+1} < (1/Q_{n+1})^2 leads to equivalent
condition delta_n*(1 + delta_n*(-6 + 4*xi)) < 1/Q_n^2. If delta_n <
(1/Q_n)^2 the condition at n+1 is satisfied by a narrow margin, for:
-6+4*xi = -0.3431457505... < 0 ( decimal expansion not in OEIS, ha! ). This
instantiates yet another induction, with base case xi-4/3 = 0.08... < 1/9 =
0.11... True.

* So, the iterates, also irreducible X_n = P_n/Q_n , satisfy Dirichlet's
irrationality criterion ( cf. Hardy & Wright, Introduction to the theory of
numbers, Ch. XI, Eq. 11.1.1 ). Then, when Hardy and Wright ask: "...how
/rapidly/ can we approximate to xi?" We are free to answer: The iterates
approach exponentially close to xi, for the recursion is derived as a
special case of the elliptic curve addition law on the Jacobi Quartic!
More precisely, in the limit around sqrt(2), the invariant differential
becomes dt = -dx/(x-sqrt(2)) , and the sequence X_n chooses discrete points
from the continuous time evolution x(t) at equal time intervals X_n =
x(n*t_0).

Caveat Emptor . . . I'm not much of an author, in math, even in physics,
nor in poetry. If you think I'm wrong, don't be too quiet.

Loose Ends
=======
* Does this iterator of many virtues produce A005319 / A001541 ?

* Did Jacobi or Weierstrass ever argue along these lines ?

* Does anyone else care about the picture with the parabolas intersecting
at (+/-) sqrt(2) ?

* Assuming the proof is already written, will we ever find it in the
library?

* And there is another secret identity, tending to infinity, and having to
do with Ramanujan's work.

I think these questions
and the last assertion,
will wait for another night,
maybe another morning.

Cheers,

Brad