Prove that numbers of the form 131, 1331, 13331, ... are divisible by 19 when they have length 15+18k

Note that $$ 3 \cdot \overbrace{13\cdots31}^L = (10^{L+1} - 1) - 6(10^L + 1) = 10^{L+1} - 6\cdot 10^L - 7\\ = 10\cdot 10^{L} - 6\cdot 10^L - 7 = 4\cdot 10^L - 7. $$ So, $\overbrace{13\cdots31}^L$ is a multiple of $19$ if and only if $4\cdot 10^L - 7$ is a multiple of $19$, which is to say that $$ 4 \cdot 10^L \equiv 7 \pmod{19} $$
Multiplying both sides by $5$ (mult. inverse of $4$) yields $$ (20) \cdot 10^L \equiv 35 \pmod{19}\\ 10^L \equiv -3 \pmod{19}.\\ $$ From here, it suffices to note that $18$ is the smallest positive value of $L$ for which $10^L \equiv 1 \pmod{19}$, and $10^{14} \equiv -3 \pmod{19}$.

Sequence $\,f_k\,$ has $\,\overbrace{3f_4 = 4(10)^3\!-7}^{\textstyle 3(1331) = 3993\ },\,$ and $\,3f_k = 4(10)^{k-1}\!-7\,$ similarly. Working $\bmod 19\!:$

$ f_{\color{#c00}n+j}\!\equiv f_j\! \iff\ \! 3f_{n+j}+7 \equiv 3f_j+7 \!$ $\iff \!4(10)^{n+j-1}\!\equiv 4(10)^{j-1}\!\!\!\iff\! 10^{\color{#c00}n}\equiv 1\!\!\iff\!\! \color{#c00}{18\mid n}$

since Fermat $\Rightarrow 10^{18}\!\equiv 1,\, $ but $10^6,10^9\!\not\equiv 1,\,$ so $10$ has order $\,\color{#c00}{18}\,$ by the Order Test.

Thus $\!\!\!\!\!\!\!\underbrace{f_{15}\equiv f_{-3}}_{\large 15\ \equiv\ -3\pmod{\!\color{#c00}{18}}}\!\!\!\!\!\!\!\equiv (\color{#0a0}{4(10)^{-4}\!-7})/3 \equiv 0,\,$ by $\, \color{#0a0}{7(10^4)}\equiv 7(5^2)\equiv 7(6)\equiv \color{#0a0}4$