[seqfan] Re: Formula Needed for a Family of Continued Fractions

[Paul D Hanna]

SeqFans, 
      Of course, case m=0 is also interesting: 
1/(1 - x*(1-x)/(1 - x^2*(1-x^2)/(1 - x^3*(1-x^3)/(1 - x^4*(1-x^4)/(1 - ...))))) 
= 1 + x - x^5 - x^8 + x^16 + x^21 - x^33 - x^40 +...See: http://oeis.org/A143064 - Expansion of a Ramanujan false theta series. 
 
But in that case, P(x,0) and Q(x,0) are infinite q-series expansions, and so are outside the scope of my inquiry. 
   
---------- Original Message ----------
From: "Paul D Hanna" <pauldhanna at juno.com>
To: seqfan at list.seqfan.eu
Subject: [seqfan] Formula Needed for a Family of Continued Fractions
Date: Thu, 18 Jul 2013 17:42:32 GMT

SeqFans, 
   Wonder if there is a nice formula for polynomials P(x,m) and Q(x,m) for a certain family of continued fractions defined by: 
P(x,m)/Q(x,m) = 1/(1 - x*(1-x^(m+1))/(1 - x^2*(1-x^(m+2))/(1 - x^3*(1-x^(m+3))/(1 - x^4*(1-x^(m+4))/(1 - ...))))). 
 
It seems that the denominator polynomial Q(x,m) has a factor (1-x^m). 
But further patterns are not obvious. 
  
Here are some examples. 

CASE m=1: 
1/(1 - x*(1-x^2)/(1 - x^2*(1-x^3)/(1 - x^3*(1-x^4)/(1 - x^4*(1-x^5)/(1 - ...))))) 
= 1/(1-x).  

CASE m=2: 
1/(1 - x*(1-x^3)/(1 - x^2*(1-x^4)/(1 - x^3*(1-x^5)/(1 - x^4*(1-x^6)/(1 - ...))))) 
= P(x,2)/Q(x,2) UNKNOWN 
See: http://oeis.org/A227360 

CASE m=3. 
1/(1 - x*(1-x^4)/(1 - x^2*(1-x^5)/(1 - x^3*(1-x^6)/(1 - x^4*(1-x^7)/(1 - ...)))))
= (1-x^2-x^3)/((1-x^3)*(1-x-x^2))    (thanks to Joerg Arndt)
See: http://oeis.org/A173173  where A173173(n)=ceil(Fibonacci(n)/2). 

CASE m=4. 
1/(1 - x*(1-x^5)/(1 - x^2*(1-x^6)/(1 - x^3*(1-x^7)/(1 - x^4*(1-x^8)/(1 - ...)))))= P(x,2)/Q(x,2) UNKNOWN 
See: http://oeis.org/A227374

CASE m=5. 
1/(1 - x*(1-x^6)/(1 - x^2*(1-x^7)/(1 - x^3*(1-x^8)/(1 - x^4*(1-x^9)/(1 - ...))))).  
=  (1-x-x^4)*(1+x-x^3-x^4-x^5)/((1-x^5)*(1-x-x^2-x^3+x^5+x^6+x^7))  (thanks to R. J. Mathar)
See: http://oeis.org/A227375
 
Would be interesting to see a general formula for P(x,m)/Q(x,m). 
Thanks, 
     Paul

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