# [seqfan] Formula Needed for a Family of Continued Fractions

Paul D Hanna pauldhanna at juno.com
Thu Jul 18 19:42:32 CEST 2013

```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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