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What is known about this generalized "continued fraction" $$ b_0+\frac{a_1}{\left(b_1+\frac{a_2}{\left(b_2+\frac{a_3}{\left(b_3+\dotsb\right)^n}\right)^n}\right)^n} $$ when the integer $n\ge 2$?

Wikipedia and wolfram articles on generalized continued fraction doesn't mention any continued fractions of this kind.

Of course one can calculate explicitly periodic continued fractions, but they are not interesting.

Cave Johnson
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Not interesting! Explicitly periodic generalized continued fractions are not interesting? You do realize that if you take $n=2$, and you assume a period of $k$, then your continued fraction (if it converges) is the root of a degree $2k+1$ polynomial?

Explicitly: If $x = [a,a,a,a,...;b,b,b,b,...]_2$, then $x = a + b / x^2$, and $x^3 - ax^2 - b = 0$. If $x = [a, a', a, a',...; b,b',b,b',...]_2$, then $x = a + b / (a' + b'/x^2)^2 = a + bx^4 / (a'x^2 + b')^2$ or $a'^2 x^5 - (aa'^2 + b) x^4 + 2a'b' x^3 - 2aa'b' x^2 + b'^2 x - ab'^2 = 0$. Setting $c = b/a'^2$ and $d = (b'/a')$, we get $x^5 - (a+c)x^4 + 2dx^3 - 2adx^2 +d^2x - ad^2 = 0$.

What I find interesting is that in the latter case, you don't have the full space of quintic polynomials -- you only get solutions to a 3-dimensional subspace.

Craig
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    Polynomials and their roots are interesting, there is no doubt. I just meant they are not interesting in this context of generalized continued fractions. So that remark was intended to avoid the discussion of periodic case and focus on more interesting case of non periodic cfracs. – Cave Johnson Nov 01 '17 at 16:30
  • Well, ordinary continued fractions only give you quadratic irrationals in the periodic case... – Craig Nov 01 '17 at 16:45
  • Nevertheless, I wrote that remark to specifically avoid discussion of periodic case. – Cave Johnson Nov 01 '17 at 16:50