[seqfan] Formula Needed for a Family of Continued Fractions

[Paul D Hanna]

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