In [this post](https://math.stackexchange.com/a/1403979/4781), the following two continued fractions discussed by Nicco are given,

$$A(q)= \left(\frac{\vartheta\_2(0,q)}{2\,q^{1/4}}\right)^2= \cfrac{1}{1-q+\cfrac{q(1\color{red}-q)^2}{1-q^3+\cfrac{q^2(1\color{red}-q^2)^2}{1-q^5+\cfrac{q^3(1\color{red}-q^3)^2}{1-q^7+\ddots}}}}\tag1$$

$$B(q)=\frac{1}{2\,q^{1/2}}\frac{\vartheta\_2(0,q^2)}{\vartheta\_3(0,q^2)} =\cfrac{1}{1-q+\cfrac{q(1\color{blue}+q)^2}{1-q^3+\cfrac{q^2(1\color{blue}+q^2)^2}{1-q^5+\cfrac{q^3(1\color{blue}+q^3)^2}{1-q^7+\ddots}}}}\tag2$$

where $|q|<1$. One can see their beautiful affinity. On a hunch, I decided to investigate the similar cfrac,

$$C(q)=\cfrac{1}{1-q\color{red}-\cfrac{q(1\color{red}-q)^2}{1-q^3\color{red}-\cfrac{q^2(1\color{red}-q^2)^2}{1-q^5\color{red}-\cfrac{q^3(1\color{red}-q^3)^2}{1-q^7\color{red}-\ddots}}}}\tag3$$

**Q1:** Is it true that,

$$C(q) \overset{\color{red}?}= \frac{1}{q}\sum\_{n=1}^\infty \frac{q^n}{1-q^{2n-1}} =\sum\_{n=0}^\infty\sigma\_0(2n+1)\,q^n = 1+2q+2q^2+2q^3+3q^4+\dots\tag4$$

where $\sigma\_0(2n+1)$ is the *[divisor function](http://mathworld.wolfram.com/DivisorFunction.html)*? (I've checked it to hundreds of coefficients and it seems to hold. More info at [A099774](https://oeis.org/A099774).) Note that the equality below is known,

$$\sum\_{n=1}^\infty\sigma\_0(n)\,q^n = \sum\_{n=1}^\infty \frac{q^n}{1-q^n}$$

**Q2:** Can $C(q)$ be also expressed in terms of the *Jacobi theta functions* $\vartheta\_k(0,q)$?