$\color{brown}{\textbf{HINT}}$
Denote the target sequence $\{F_3(n)\}$ and let us try to estimate the probability $P(N)\}$ that natural number belongs to $\{F_3\}.$
Suppose
$$F_3(1)=1,\quad F_3(2)=2,\quad P(1)=P(2)=P(5) = P(6) = 1,\\ P(3)=P(4)=P(7)=P(8)=0,\tag1$$
$$V(N)=\sum\limits_{i=1}^{N}P(i),\quad F_3(N) = \left[V^{-1}(N)\right].\tag2$$
Let $P_a(N)$ is the probability that N does not belong to arithmetic progression and $P_g(N)$ is the similar probability for geometric progressions.
Suppose
$$P(N) = P_a(N)P_g(N)).\tag3$$
$\color{brown}{\textbf{Arithmetic Probability estimation.}}$
Suppose
$$P_a(N)=\prod\limits_{k=1}^{[N/2]}P_a(N,k),\tag4$$
where $P_a(N,k)$ is the probability that arithmetic progression $\{N-2k,N-k, N\}$ does not exist for any $j.$
Suppose
$$P_a(N,k) = \big(1-P(N-2k)\big)\big((1-P(N-k)\big).\tag5$$
$\color{brown}{\textbf{Geometric Probability estimation.}}$
Suppose
$$P_g(N)=\prod\limits_{k=1}^{\left[\,\sqrt Nn\,\right]}P_g(N,k),\tag6$$
where $P_g(N,k)$ is the probability that geometric progression $\left(\dfrac{N}{k^2}, \dfrac Nk, N\right\}$ with the denominator $k$ does not exist for all $i,j.$
Taking in account that the geometric progression can exist only if $k^2\,| \ N,$ suppose
$$P_g(N,k) = \left(1-\dfrac1{k^2}P\left(\dfrac{N}{k^2}\right)\right)\left(1-P\left(\dfrac{N}{k}\right)\right).\tag7$$
$\color{brown}{\textbf{Common model.}}$
Common model can be simplified to next one,
\begin{cases}
P(N) = 1-\prod\limits_{k=1}^{[N/2-1]}\Big(1-\big(1-P(N-2k)\big)\big(1-P(N-k)\big)\Big)\\
\times\prod\limits_{k=1}^{\left[\sqrt N\right]}\left(1-\left(1-\dfrac1{k^2}P\left(\dfrac{N}{k^2}\right)\right)\left(1-P\left(\dfrac{N}{k}\right)\right)\right)\\[4pt]
P(1)=P(2)=P(5) = P(6) = 1,\quad P(3)=P(4)=P(7)=P(8)=0\\
V(N)=\sum\limits_{i=1}^{N}P(i),\quad F_3(n) = \left[V^{-1}(n)\right].\tag8
\end{cases}
Looking the solution in the form of
$$\left\{\begin{align}
&P(N)=P_v(N),\quad\text{where}\quad v=\left[\dfrac{N-1 \mod 4}2\right],\\[4pt]
&P_0(N)=
\begin{cases}
1,\quad\text{if}\quad N<9\\
cN^{-s},\quad\text{otherwise}
\end{cases}\\[4pt]
&P_1(N)=
\begin{cases}
0,\quad\text{if}\quad N<9\\
dN^{-t},\quad\text{otherwise}
\end{cases}
\end{align}\right.\tag9$$
then
$$\begin{cases}
&P_0(N) =1-&\prod\limits_{k=1}^{[N/4-1]}\Big(1-\big(1-P_0(N-4k)\big)\big(1-P_1(N-2k)\big)\Big)\\
&&\times\Big(1-\big(1-P_1(N-4k-2)\big)\big(1-P_0(N-2k-1)\big)\Big)\\
&&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1}
\left(1-\left(1-\dfrac1{(2k-1)^2}P_0\left(\dfrac{N}{(2k-1)^2}\right)\right)
\left(1-P_1\left(\dfrac{N}{2k-1}\right)\right)\right)\\[4pt]
&&\times\left(1-\left(1-\dfrac1{4k^2}P_1\left(\dfrac{N}{4k^2}\right)\right)
\left(1-P_0\left(\dfrac{N}{2k}\right)\right)\right)\\[4pt]
&P_1(N) = 1- &\prod\limits_{k=1}^{[N/4-1]}\Big(1-\big(1-P_1(N-4k)\big)\big(1-P_0(N-2k)\big)\Big)\\
&&\Big(1-\big(1-P_0(N-4k-2)\big)\big(1-P_1(N-2k-1)\big)\Big)\\
&&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1}
\left(1-\left(1-\dfrac1{(2k-1)^2}P_1\left(\dfrac{N}{(2k-1)^2}\right)\right)
\left(1-P_0\left(\dfrac{N}{2k-1}\right)\right)\right)\\[4pt]
&&\times\left(1-\left(1-\dfrac1{4k^2}P_0\left(\dfrac{N}{4k^2}\right)\right)
\left(1-P_1\left(\dfrac{N}{2k}\right)\right)\right).
\end{cases}\tag{10}$$
Taking in account $(9),$ can be written
$$\begin{cases}
&P_0(N) =1-&\prod\limits_{k=[N/4-2]}^{[N/4-1]}\,c(N-2k-1)^{-s}\prod\limits_{k=1}^{[N/4-3]}\Big(1-\big(1-c(N-4k)^{-s}\big)\big(1-d(N-2k)^{-t}\big)\Big)\\
&&\times\Big(1-\big(1-d(N-4k-2)^{-t}\big)\big(1-c(N-2k-1)^{-s}\big)\Big)\\
&&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1}
\left(1-\left(1-\dfrac1{(2k-1)^2}c\left(\dfrac{N}{(2k-1)^2}\right)^{-s}\right)
\left(1-d\left(\dfrac{N}{2k-1}\right)^{-t}\right)\right)\\[4pt]
&&\times\left(1-\left(1-\dfrac1{4k^2}d\left(\dfrac{N}{4k^2}\right)^{-t}\right)
\left(1-c\left(\dfrac{N}{2k}\right)^{-s}\right)\right)\\[4pt]
&P_1(N) = 1- &\prod\limits_{k=[N/4-2]}^{[N/4-1]}\,c(N-2k)^{-s}\prod\limits_{k=1}^{[N/4-3]}\Big(1-\big(1-d(N-4k)^{-t}\big)\big(1-(N-2k)^{-s}\big)\Big)\\
&&\Big(1-\big(1-(N-4k-2)^{-s}\big)\big(1-d(N-2k-1)^{-t}\big)\Big)\\
&&\times\prod\limits_{k=1}^{\left[\sqrt N/2\right]-1}
\left(1-\left(1-\dfrac1{(2k-1)^2}d\left(\dfrac{N}{(2k-1)^2}\right)^{-t}\right)
\left(1-c\left(\dfrac{N}{2k-1}\right)^{-s}\right)\right)\\[4pt]
&&\times\left(1-\left(1-\dfrac1{4k^2}c\left(\dfrac{N}{4k^2}\right)^{-s}\right)
\left(1-d\left(\dfrac{N}{2k}\right)^{-t}\right)\right).
\end{cases}\tag{11}$$
Model $(11)$ should be checked theoretically and practically, but it gives the approach to the required estimations.
The next steps are estimation of parameters $$c,d,s,t$$ and using of obtained model.
Best Answer
I hope you’ve played with this, and noticed:
1.$\quad$It’s not a sum, it’s many sums, and each sum is finite.
2.$\quad$The base, $0.9$ in this case, plays no particular role, so that you can use any base $r$.
3.$\quad$The first few values are \begin{align} S_0&=0\\ S_1&=r\\ S_2&=r^3+r^2\\ S_3&=r^6+r^5+r^3\\ S^4&=r^{10}+r^9+r^7+r^4\\ S_5&=r^{15}+r^{14}+r^{12}+r^9+r^5\\ S_n&=r^n(S_{n-1}+1) \end{align} I see no way of getting a closed-form expression for $S_n$, a polynomial in $r$ of degree $\frac12(n^2+n)$, and most certainly not a numerical value once you evaluate $r$ to, in your case, $r=0.9\,$.
I do wonder where or how you came across this—without context, it seems a most unnatural problem.