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I conjecture a new generating function for the fibonacci numbers $F_{n}$. Given,the following conjectured q-continued fraction $$\chi(q)=\cfrac{1}{1+q-\cfrac{(1+q^2)}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}}$$

how do we show that, $\chi\Big(\frac{1}{q}\Big)= \sum_{n=1}^\infty (-1)^{n-1} F_{n}q^n$

is true?

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    $\begingroup$ Do you expect us to prove/disprove this ? How did you arrived to this result ? $\endgroup$
    – AlienRem
    Aug 17, 2015 at 13:42
  • $\begingroup$ Shouldn't the last numerator be equal to $q^4 (1-q)(1-q^3)(1-q^5)$? $\endgroup$
    – zhoraster
    Aug 17, 2015 at 13:58
  • $\begingroup$ Set up the recurrences for the convergents of $\chi(1/q)$ as $a_n / b_n$, and see what comes out. I'd not be too surprised if Fibonacci numbers fall out. $\endgroup$
    – vonbrand
    Aug 17, 2015 at 14:02
  • $\begingroup$ @vonbrand On the third convergent ,we have $$\frac{q+q^7}{1+q+q^2-q^6}$$ ,which looks very similar to the well known generating function for $F_{n}$ $\endgroup$
    – Nicco
    Aug 17, 2015 at 15:04
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    $\begingroup$ @vonbrand: It doesn't seem to be that simple. The generating function is different when $|q|<1$. Kindly see answer below. $\endgroup$ Aug 19, 2015 at 5:47

1 Answer 1

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(A partial answer.)

This is a special case of a conjectured equality discussed in this MO post. Let $|q|<1$, then,

$$\begin{aligned}U(q) &= \prod_{n=0}^\infty \frac{\big(1-a^2q^3(q^4)^n\big)\big(1-b^2q^3(q^4)^n\big)}{\big(1-a^2q(q^4)^n\big)\big(1-b^2q(q^4)^n\big)}\\ &= \dfrac{1} {1+ab-\dfrac{(a+bq)(b+aq)} {1+(ab)^3+\dfrac{(a-bq^2)(b-aq^2)q} {1+(ab)^5-\dfrac{(a+bq^3)(b+aq^3)q^2} {1+(ab)^7+\dfrac{(a-bq^4)(b-aq^4)q^3} {(1+(ab)^9-\ddots }}}}} \end{aligned}$$

If $a=q,\;b=1$, and $|q|<1$ then,

$$\begin{aligned}U(q) &=\prod_{n=0}^\infty \frac{\big(1-q^5(q^4)^n\big)\big(1-q^3(q^4)^n\big)}{\big(1-q^3(q^4)^n\big)\big(1-q(q^4)^n\big)} =\prod_{n=0}^\infty \frac{\big(1-q^5(q^4)^n\big)}{\big(1-q(q^4)^n\big)} = \frac{1}{1-q}\\ &=\cfrac{1}{1+q-\cfrac{\color{brown}{2q(1+q^2)}}{1+q^3+\cfrac{q^2(1-q)(1-q^3)}{1+q^5-\cfrac{q^3(1+q^2)(1+q^4)}{1+q^7+\cfrac{q^4(1-q^3)(1-q^5)}{1+q^9-\ddots}}}}} \end{aligned}$$

and it only takes a little algebraic manipulation to get the brown part of this cfrac to the form in the post. I get,

If $|q|<1$:

$$\chi(q) = \frac{1}{q}\tag1$$

If $|q|>1$:

$$\chi(q) = \sum_{n=1}^\infty (-1)^{n-1} \frac{F_{n}}{q^n} = \frac{q}{q^2+q-1}\tag2$$

where $(2)$ is a variant of the identity in this post. Thus, it should be specified that the generating function is only valid when $\color{blue}{|q|>1}$.

P.S. Such behavior is present in other cfracs. For example, for the Rogers-Ramanujan cfrac $R(q)$, if $|q|<1$, then $R(q) = R(q)$, but if $\color{blue}{|q|>1}$, then,

$$R(q) \to R(1/q^4),\quad \text{(even convergents)}$$

$$R(q) \to -1/R(-1/q),\quad \text{(odd convergents)}$$

See Section 2 of Berndt's paper.

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  • $\begingroup$ @Nicco: that kind of behaviour worried me, but now I know it's natural - Yeah, I know... $\endgroup$
    – Lucian
    Aug 19, 2015 at 16:45
  • $\begingroup$ @Lucian: While I understand such kind of humor, heaven forbid the recipient of your comment would be an evangelical Christian, or someone with a conservative ideology. $\endgroup$ Aug 19, 2015 at 16:58
  • $\begingroup$ @Tito piezasIII by letting the right handside be $n(1+q^2)$ ,we get generalized fibonacci numbers for every natural number $n$. $\endgroup$
    – Nicco
    Aug 20, 2015 at 7:03

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