where the exponential and the gamma function are integral multiples of one another

Let $g(z) = \sin\left(\frac{\pi e^{z}}{\Gamma(z)}\right)\sin\left(\frac{\pi \Gamma(z)}{e^{z}}\right)$. $g(z)$ has a root in $\mathbb{C}$ (domain colored below) whenever there exists $n\in\mathbb{Z}$ such that $ne^{z}=\Gamma(z)$ or $n\Gamma(z)=e^{z}$

Do you see the double root at $z\approx 8.1636558707955126\ldots$ or so? That’s because at that point $2\exp(z)=\Gamma(z)$. There’s a whole necklace of double roots on the right side of the picture. The Rogers Ramanujan continued fraction

The Rogers-Ramanujan continued fraction $R(q), |q|<1$ is given by $R(q)=\sqrt{q}\frac{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}$ and looks like this on the unit disk: agm variants of ordinary trigonometric functions

The ordinary sine and cosine are averages of exponentials. Here, I play around with replacing the arithmetic average with the arithmetic geometric mean

Let $\mathrm{cas}(z)=\mathrm{agm}(e^{iz},e^{-iz})$

And since the $\mathrm{agm}(x,y)$ may be represented in terms of a complete elliptic integral of the first kind $K(m)$: $K(m)=\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-m\sin^2(\theta)}}$ $\mathrm{agm}(x,y) = \frac{\pi}{4} (x + y) \; / \; K\!\left[\left( \frac{x - y}{x + y} \right)^2 \right]$

Therefore $\mathrm{cas}(z) = \mathrm{agm}(e^{iz},e^{-iz}) = \frac{\pi}{4} (e^{iz} + e^{-iz}) \; / \; K\!\left[\left( \frac{e^{iz} - e^{-iz}}{e^{iz} + e^{-iz}} \right)^2 \right]$

Therefore: $\mathrm{cas}(z) = \frac{\pi\cos(z)}{2K(-tan^{2}(z))}$ Domain colored, with mpmath, it sort of looks like this: 