[seqfan] Special Continued Fraction Expansion

Paul D Hanna pauldhanna at juno.com
Wed Jul 3 13:27:01 CEST 2013


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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