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This is an extension of "A question about the golden ratio and other numbers." Given $r$, suppose that $$c_0+c_1x+c_2x^2+ \cdots = \frac{1} {\lfloor{r}\rfloor+\lfloor{2r}\rfloor x+\lfloor{3r}\rfloor x^2+ \cdots}.$$ Let $L(r) = \lim_{i\to\infty} \frac{c_{i+1}}{c_i}$. Can someone prove that the limit $L(\phi)$ exists, where $\phi = \frac{1+ \sqrt{5}}{2}$? It appears that $$L(\phi) = -1.688924110769165206686359\ldots$$

$$(c_0,c_1,c_2,\ldots) = (1,−3,5,−9,17,−30,52,−90,154,−262,446,−758,1285,\ldots).$$

Also, it appears that $L(F_{k+1}/F_{k})$ exists, for $k \ge 5$, where $F_k$ denotes the $k$th Fibonacci number; e.g., $$\begin{eqnarray} L(8/5) &=&-1.69562076\ldots \newline L(13/8) &=& -1.76404686\ldots \newline L(21/13) &=& -1.68892398\ldots \newline L(34/21) &=& -1.68880982\ldots \end{eqnarray}$$

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    $\begingroup$ The rational ones are easy because everything is a rational function and hence the coefficients are solutions to a recurrence relation. For example $L(8/5)$ is just the smallest root of $x^8 + 3*x^7 + 4*x^6 + 6*x^5 + 8*x^4 + 7*x^3 + 5*x^2 + 4*x + 2$ (sorry for computer notation). $\endgroup$
    – Kevin Buzzard
    Commented Jan 17, 2017 at 17:42
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    $\begingroup$ One can simplify $L(8/5)$ to the largest root of $x^3+x^2+2$. $\endgroup$
    – Richard Stanley
    Commented Jan 17, 2017 at 19:11
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    $\begingroup$ What's going on in general presumably is that the ratio is simply the reciprocal of the first zero of the power series, which for the golden ratio will just be some random (negative, in this case) real (the power series clearly converges for $|x|<1$). For the rational numbers the zero will be some algebraic number satisfying some funky polynomial whose coefficients are related to floor(r),floor(2r),floor(3r)... . $\endgroup$
    – Kevin Buzzard
    Commented Jan 17, 2017 at 20:02
  • $\begingroup$ Clarification: by "the power series" I mean the one on the denominator, not the one with the $c_i$ in. $\endgroup$
    – Kevin Buzzard
    Commented Jan 17, 2017 at 22:11

1 Answer 1

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Some observations, but not a solution yet.

Let $t_k=\frac{F_{k+1}}{F_k}$ where $F_k$ is the $k^{th}$-Fibonacci number. The convention here for $F_k$ is that $F_3=2, F_4=3, F_5=5, \dots$. Denote the RHS in the above series by $$\Psi_k(x)=\left(\sum_{n=1}^{\infty}\lfloor nt_k\rfloor x^{n-1}\right)^{-1}.$$ Then, it seems that $$\Psi_k(x)=\frac{(1-x)(1-x^{F_k})}{P_k(x)}$$ for some polynomial $P_k(x)$ with the following rather curious properties:

(1) it has degree $F_k-1$;

(2) its coefficients are either $1$ or $2$ (although not clear which is which);

(3) $P_k(1)=F_k$;

(4) $L(t_k)=$ the smallest real root of $P_k(x)$;

(5) there might be a way to relate $(1-x)P_k(x)$ to other $(1-x)P_{\ell}(x)$ for $\ell<k$, recursively.

A few examples:

\begin{align} \Psi_4(x)&=\frac{(1-x)(1-x^3)}{1+2x+2x^3} \\ \Psi_5(x)&=\frac{(1-x)(1-x^5)}{1+2x+x^2+2x^3+2x^4} \\ \Psi_6(x)&=\frac{(1-x)(1-x^8)}{1+2x+x^2+2x^3+2x^4+x^5+2x^6+2x^7} \\ \Psi_7(x)&=\frac{(1-x)(1-x^{12})}{1+2x+x^2+2x^3+2x^4+x^5+2x^6+x^7+2x^8+2x^9+x^{10}+2x^{11}+2x^{12}}. \end{align}

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