[seqfan] Re: Special Continued Fraction Expansion

Wouter Meeussen wouter.meeussen at telenet.be
Wed Jul 3 19:17:36 CEST 2013


Paul,

you didn't mention that r1  is produced by 3 <= r^2 <= 4
and r2 by both r^2 <3  and  r^2 >4

can you confirm this?
btw, W.R. Gosper 's continued fraction extensions from last year (on 
mathfun) didn't say anything about
ContinuedFractionK[1, x^k , {k, 0, \[Infinity]}]

so I guess you're on new territory.

Wouter

-----Original Message----- 
From: Paul D Hanna
Sent: Wednesday, July 03, 2013 1:27 PM
To: seqfan at list.seqfan.eu
Subject: [seqfan] Special Continued Fraction Expansion

SeqFans,
    Consider the constant(s) r such that the floor of the powers of r
equals the partial quotients of the continued fraction of r:

(*) r = [1; [r], [r^2], [r^3], [r^4], ..., floor(r^n), ...].

Suppose one arrives at a solution to r in (*) by recursion;
starting with some positive seed value r(0), then for n>=0,
(**) r(n+1) = [1; [r(n)], [r(n)^2], [r(n)^3], [r(n)^4], ...];

then it appears that there exists 2 fixed points:

r1 = 1.6915954196361070915206089538501262868270...
r2 = 1.7616596940944800771133433079549530812923...

depending on what value of r(0) you start with.

I wonder if it is easy to show that the following conjecture makes sense:

Starting with a positive seed value r(o),
if r(0) < sqrt(3), the recursion (**) restults in r1,
if r(0) > sqrt(3), the recursion (**) restults in r2.

The surprising thing (to me) is that the critical point would be r(0) = 
sqrt(3).

Below are two sequences associated with the 2 constants.
Are these interesting enough to submit to the OEIS?

Thanks,
    Paul (1)
The continued fraction expansion of the positive constant r < sqrt(3) that 
satisfies:
r = [1; [r], [r^2], [r^3], [r^4], ..., floor(r^n), ...].

r = 1.6915954196361070915206089538501262868270...

CF(r) = [1; 1, 2, 4, 8, 13, 23, 39, 67, 113, 191, 324, 548, 928, 1570, 2657, 
...].
(2)
The continued fraction expansion of the constant r > sqrt(3) that satisfies:
r = [1; [r], [r^2], [r^3], [r^4], ..., floor(r^n), ...].

r = 1.7616596940944800771133433079549530812923...

CF(r) = [1; 1, 3, 5, 9, 16, 29, 52, 92, 163, 287, 507, 893, 1573, 2772, 
...].

[END]

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