This relates to another challenge Question about drawing Mandelbrot filaments.
It is possible to compute a formula for a continuous path inside the Mandelbrot Set connecting $c=i$ to $c=0$? Obviously, the part inside the cartoid or lobes is easy, but finding any in-Set curve that includes $i$ has eluded me. I know the Mandebrot boundary is infinitely long and detailed, but if a filament has any finite thickness there should be a finite-length path through it that doesn't follow the boundary.
I tried to start by finding a direction one could travel from $i$ for a very short distance and remain in-Set, but even that eluded me.
To show the topology we're up against, here is the $|z_{25}|==2$ contour in the vicinity of i.
So, can one derive a formula for the path I want?
As an aside, it's worth noting that $i$ is a Misiurewicz Point, meaning its orbit is not immediately periodic but becomes so after a finite number of steps, i.e., $z_3(i)=z_1(i)$. This property places i exactly on the boundary of the Mandelbrot set.
$$i\rightarrow-1+i\rightarrow-i\rightarrow-1+i\rightarrow-i\rightarrow-1+i\rightarrow-i...$$
Here's a zoom into the Mandelbrot set centered on $i$, with a factor of 10 magnification per frame. As you can extrapolate from the first frame, your path will have to go through an infinite number of bulbs before getting to any filaments. Then there will be an infinite number of tiny Mandelbrot islands to pass through, as they are dense in the filaments (they get very small very quickly). Finally you'll have to pass through an infinite number of 3-way Misiurewicz branch points, around which the Mandelbrot set is asymptotically self-similar.
All of this means I strongly doubt there is a simple formula for your desired path, if there is one at all. And if by "inside the Mandelbrot set" you mean "within the interior of the Mandelbrot set" then there is no such path possible, because you'll have to touch the boundary at the 3-way branching Misiurewicz points.
PS: this image is rendered using exterior distance estimation, which I explained in my answer to your other question.