Similar to Ramanujan's Step sum

We have $A \; = \; 1 + e^{\frac15\pi}R(e^{-\pi})$ where $R(q)$ is the Rogers-Ramanujan continued fraction. It is a remarkable fact that $R(e^{-\pi}) \; = \; \tfrac18(3 + \sqrt{5})(\sqrt[4]{5}-1)\left(\sqrt{10+2\sqrt{5}} - (3 + \sqrt[4]{5})(\sqrt[4]{5}-1)\right)$ and so we have an exact expression for $A$ . To $10$ significant figures, we have $A = \boxed{1.958650182}$ .

It can be proved that the function

$\rho(q)$ $=1+\frac{1}{1+\frac{q}{1+\frac{q^2}{1+\frac{q^3}{1+.....}}}}$ is continuous ,and of infinite class over the positive real numbers.( In complex field it has singularities.

So, $\rho(q)=\rho(0)+q\rho'(0)+\frac{q^2}{2!}\rho''(0)+\frac{q^3}{3!}\rho'''(0)+....$

We get, $\rho(0)=2,\rho'(0)=-1,\rho''(0)=2,\rho'''(0)=-4, ....$

So, $\rho(q)=2-q+q^2-q^4+q^5-q^6+q^7-q^9+q^{10}-q^{11}+q^{12}-q^{14}....$ .

Here $q^3,q^8,q^{13},q^{23},...$ terms are invalid. But what is the pattern? Yes. It is similar to Fibonacci sequence. See, the missing powers are $3,8,13,23...$ and $(3+8)+2=13;(8+13)+2=23....$ So, missing numbers will follow the Fibonacci sequence but $'2'$ ahead.

$\rho(e^{-\pi})≈(2-e^{-\pi}+e^{-2\pi}-e^{-4\pi}+q^{-5\pi})≈1.9586$ .

I know we need to solve it manually but is there any Python program to solve this, please share.