Let $q = e^{-2\pi}$. We are familiar with Ramanujan's beautiful continued fraction,

$$\cfrac{q^{1/5}}{1 + \cfrac{q} {1 + \cfrac{q^2} {1 + \cfrac{q^3} {1+\ddots}}}} = {\sqrt{5+\sqrt{5}\over 2}-{1+\sqrt{5}\over 2}} = 5^{1/4}\sqrt{\phi}-\phi$$

Let $q = e^{-\pi\sqrt{11}}$. Using Ramanujan's octic continued fraction, I found that,

$$\cfrac{\sqrt{2}\,q^{1/8}}{1 + \cfrac{q} {1 +q+ \cfrac{q^2} {1 + q^2+\cfrac{q^3} {1+q^3+\ddots}}}} = \left(\frac{1}{2}-\frac{1}{4}\sqrt{\frac{4v+7}{4v-7}}\right)^{1/8}$$

where $v = T+4$, and $T$ is the tribonacci constant. More simply,

Let $q = -e^{-\pi\sqrt{11}}$. Using the Heine continued fraction, then,

$$\cfrac{(-q)^{-1/24}} {1 + \cfrac{q} {1-q + \cfrac{q^3-q^2} {1 + \cfrac{q^5-q^3} {1 + \cfrac{q^7-q^4} {1+\ddots}}}}} = \frac{1}{T}+1$$

Q: Going higher, is there a $q$-continued fraction for the tetranacci constant, or the positive root of $x^4-x^3-x^2-x-1=0$?

  • $\begingroup$ Very interesting question. I put up over 7,000 -nacci constants and their generating functions. Looking for q-continued fractions for them too. Here is my link: genautica.com/math/naccis/naccis.html $\endgroup$ – Stefan Gruenwald Jul 18 '16 at 3:24
  • $\begingroup$ @StefanGruenwald: Your website is such a big page. Perhaps you can divide it into a dozen or more pages so it will load a dozen times faster? :) $\endgroup$ – Tito Piezas III Jul 18 '16 at 4:01
  • $\begingroup$ I have broken it down now: genautica.com/math/naccis/… $\endgroup$ – Stefan Gruenwald Aug 2 '16 at 6:12
  • $\begingroup$ @StefanGruenwald: That was a good move. I'll see if I can find q-continued fractions for some of them. $\endgroup$ – Tito Piezas III Aug 2 '16 at 15:42

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