The reason for the near-periodic behavior of the choice $x = 10^5 + \frac{9}{250}$ has to do with the fact that $$\frac{x}{\pi} \approx 31831.00007753496977,$$ which is almost an integer with error $\epsilon \approx 0.000077 < 10^{-4}$. Moreover, $$\frac{1}{\epsilon} \approx 12897.4062021729,$$ and now you can see why this many terms are needed.
The above also suggests that if you can find some choice of $x$ such that $$\frac{x}{\pi} - \left\lfloor \frac{x}{\pi} \right\rfloor$$ is extremely tiny, you can make this phenomenon extend to as large of a value of $k$ as you please. It just so happens that the particular choice $10^5 + \frac{9}{250}$ is also close to a round number in base 10.
Found something! But it probably can go even faster.
As written in question we can project all members to a smaller sub-structure
$$\{\forall m: m^{p\cdot q} \bmod N\} \equiv (r_1) \times (r_2) \times (r_3) $$
If only one of those indices $i,j,k$ is $\not=0$ in
$$s_{ijk} \equiv s_0^{\beta^i\gamma^j\delta^k}\mod N$$
We can project $s_{ijk}$ and $s_0$ to all three sub-structures (with exponent $pq$,$pr$,$qr$). Finding a match there will be much faster. However this only works for small factors $p,q,r$. Assuming their prime factors are about the same size the giant-step requires computing the $\sqrt[\textstyle 18]{L_{all}}$ power of the target exponent $\beta, \gamma, \delta$ without knowing $\phi(N)$.
Let's assume only $i\not= 0$ depending on chosen projection $pq,pr,qr$ the sequence $x \mapsto x^\beta \bmod N$ projection $(x)^{\textstyle (\cdot)}$ would repeat after $r_1,q_1,p_1$ steps. Each of those have a length of about $\sqrt[\textstyle 9]{L_{all}}$ and with this the giant-step would be $\sqrt[\textstyle 18]{L_{all}}$ (as already mentioned above). The found indices can be used at the Chinese remainder theorem (indices with related mod $p_1,q_1,r1$).
With this we can compute $i$ if $j=k=0$. But that's not all! As those projections repeat after a short interval it does also work if they are $0$ modulo the related prime. E.g. if $i\not=0$ and the current value $v$ is projected $v^{pq}$ the index $j$ need to be $0 \equiv j \bmod r_2$ and $0\equiv k \bmod r_3$.
To find $i,j,k$ for every $s_{ijk}$ we can generate additional values around $s_0$ and search for matches of each of those.
The total area would be around $\sqrt[\textstyle 9]{L_{all}} \times \sqrt[\textstyle 9]{L_{all}} \times \sqrt[\textstyle 9]{L_{all}} $ size. (Maybe there is some better/faster way?)
Given this the complexity should be around
$$O((\sqrt[\textstyle 9]{L_{all}})^3 \cdot \sqrt[\textstyle 18]{L_{all}} \cdot (1 +\rho_{\not\phi}) \cdot 3 \cdot 3 :2) = O( {L_{all}}^{\textstyle \frac{7}{18}} \cdot \rho_{\not\phi})$$
with $\rho_{\not\phi}$ the steps needed to compute the $\sqrt[\textstyle 18]{L_{all}}$'th power without knowing $\phi(N)$. This does not include time needed for power, multiplications and other basic operators with numbers $<N$.
$\rightarrow $ If $L_{all}$ is small enough to be known (the assumption was made here, e.g. with cycle through) the giant-step can be done without knowing all factors of $\phi(N)$ and with this make it insecure. The security still remains in not leaking any factors of $L_{all}$ and with this it need to be bigger.
Update: At the alternative approach with $\lambda = \beta\cdot \gamma\cdot \delta$ and finding a solution for $m \equiv s_0^{\lambda^y} \bmod N$
Project those members to a shorter sequence (exponents $pq,pr,qr$). With this it will only take:
$$O(\sqrt[\textstyle 3]{L_{all}})$$
Can it be done even faster?
Best Answer
Calling
$$\delta(u,v)=\alpha(1-\gamma)\mbox{sign}(u)+\alpha\gamma \mbox{sign}(v) $$
we have
$$ \min_{u,v}\delta(u,v) \le \alpha(1-\gamma)\mbox{sign}(x_{n-2})+\alpha\gamma \mbox{sign}(x_{n-3})\le \max_{u,v}\delta(u,v) $$
and the recurrences
$$ x_n^i = \beta x_{n-1}^i+\min_{u,v}\delta(u,v)\\ x_n^s = \beta x_{n-1}^s+\max_{u,v}\delta(u,v)\\ $$
have the solutions
$$ x_n^i = C_0 \beta^{n-1}+\frac{(1-\beta^n)}{\beta-1}\min_{u,v}\delta(u,v)\\ x_n^s = C_0 \beta^{n-1}+\frac{(1-\beta^n)}{\beta-1}\max_{u,v}\delta(u,v)\\ $$
and after a little transient
$$ x_n^i \le x_n\le x_n^s $$
Attached the plot for $x_n^i, x_n^s$ in red and $x_n$ in blue. Note that the recurrences $x_n^i, x_n^s$ should begin at $x_3$. The values assumed are $\alpha=-3,\beta = 0.25, \gamma = 0.75$ and the initial conditions $x_1,x_2,x_3$ are random values in the range $(-10, 10)$
NOTE
Due to the structure
$$ x_n = \beta x_{n-1}+\cdots $$
there is an exponential component all along $n\to\infty$ which eliminates the periodic behavior possibility.
If instead we consider the recurrence
$$ x_n = \alpha(1-\gamma)\mbox{sign}(x_{n-2})+\alpha\gamma \mbox{sign}(x_{n-3}) $$
with $\beta = 0$ then periodic solutions can appear when the initial conditions $x_1, x_2, x_3$ have different signs as for instance $x_1=1,x_2 = -1, x_3 = 0$
or $x_1=1,x_2 = -1, x_3 = 1$
To solve
$$ z_n = \beta z_{n-1}+\alpha(1-\gamma)+\alpha\gamma $$
which is a linear recurrence we use the fact
$$ z_n = z_n^h+z_n^p\\ z_n^h -\beta z_{n-1}^h = 0\\ z_n^p -\beta z_{n-1}^p=\alpha(1-\gamma)+\alpha\gamma $$
then easily we find $z_n^h = C\beta^{n-1}$. Now considering $z_n^p = C_n\beta^{n-1}$ and substituting into the particular we can find also the recurrence for $C_n$ which is
$$ C_n-C_{n-1} = \alpha\beta^{1-n} $$
with solution
$$ C_n = \frac{\alpha \beta \left(1-\beta ^{-n}\right)}{\beta -1} $$
and finally
$$ z_n = C\beta^{n-1}+ \frac{\alpha \beta \left(1-\beta ^{-n}\right)}{\beta -1}\beta^{n-1}=C \beta^{n-1}+\frac{\alpha(1-\beta^n)}{\beta-1} $$
The general solution can be computed with the help of the following MATHEMATICA script