The factorial-gap series has the unit circle as a natural boundary

The coefficients of $F$ are $1$ at the factorial indices and $0$ elsewhere. Hence their limsup root is $1$, so [L3] gives radius of convergence $1$. Since all coefficients are nonnegative, [L2] makes the boundary point $1$ singular.

Let $\omega$ be a root of unity. Choose $m\ge 1$ with ${\omega }^{m!}=1$. By [L4], ${\omega }^{n!}=1$ for every $n\ge m$, so $F(\omega z)-F(z)={\sum }_{n=0}^{m-1}({\omega }^{n!}-1)z^{n!},$ a polynomial.

Suppose $\omega$ were regular. Then some holomorphic function would extend $F$ across $\omega$, so after composing with $z\mapsto \omega z$ the function $F(\omega z)$ would extend holomorphically across $1$. Step 1.2 shows that $F$ differs from that extension by a polynomial, so $F$ itself would extend holomorphically across $1$, contradicting step 1.1. Therefore every root of unity on the unit circle is singular.

Roots of unity are dense on the unit circle. If some boundary point $\zeta$ were regular, a small extension disc around $\zeta$ would make every nearby boundary point regular as well, including some root of unity, contrary to step 2.1. Thus every point of the unit circle is singular, and [L1] makes the unit circle a natural boundary.