Given the following two continued fractions,with $|q|\lt1$
$F(q)= \cfrac{1}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^5}{1+\cfrac{q^8}{1+\cfrac{q^{12}}{1+\cfrac{q^{16}}{1+\cfrac{q^{24}}{1+\dots}}}}}}}}\tag1$
where the exponents are given by the Generating function
$f(z)=\frac{-2z^4-4z^3-3z^2-2z-1}{2z^2-1}$
and
$G(q)= 1-\cfrac{q}{1+\cfrac{q}{1-\cfrac{q}{1+\cfrac{q}{1-\cfrac{q^4}{1+\cfrac{q^{4}}{1-\cfrac{q^{4}}{1+\cfrac{q^{4}}{1-\dots}}}}}}}}\tag2$
in which case the exponents are given by the generating function
$g(z)=\frac{-2z^7-2z^6-2z^5-2z^4-z^3-z^2-z-1}{2z^4-1}$
and satisfy the recurrence relation $a(n+4)=2a(n)$
for $n\geq4$
How do we prove that $F(q)=G(q)$?
Note:by comparing coefficients of each continued fraction,we observe that they are equal.
Curiously though,by plotting each generating function on the complex plane yields beautiful images(f(z) and g(z)) of bugs
Let $\quad T_0(x, q):=1/(1+q/(1+q^2/(1+q^5/x))),$ $\; T_1(x, q):=1+q^2/(1+q^3/x),$ $\; T_2(x, q):=1-q/(1+q/(1-q/(1+q/x))),\;$ and $\; E(F, G, q):=F+q^4/(1-1/G).\;$ Then $E( T_1( -q^8/(1-1/y), q^4),\; T_2(y, q^4), q)= E( -q^4/(1-1/y),y,q)=0$ for all $y\ne 0.\;$ If for all $n>1,\;$ $F_{n-1}(q):=T_1(F_n(q),q^{2^n}), \; G_{n-1}(q):=T_2(G_n(q),q^{2^n}),$ then $E(F_{n+1}(q), G_{n+1}(q), q^{2^n})=0.$ Now, if $F(q):=T_0(F_1(q), q),\; G(q):=T_2(G_1(q), q),\;$ then $T_0(-q^4/(1-1/y), q)=T_2(y,q), q)$ for all $y\ne 0$, and if $y=G_1(q)$, then $F(q)=G(q)\;$ QED.
Please do not edit my continued fraction expressions. They are perfectly readable right now.