Given a regular tetrahedron with its vertices $A,B,C,D$.
An attached tetrahedron can be constructed by a reflection of one point at the plane defined by the other 3 points.
For regular tetrahedron this can be simplified (as far as I know) to: $$A' = (B+C+D)\cdot2/3-A$$ (this will result in a regular tetrahedron attached to the BCD plane)
If we repeat this multiple times we can produce some 'straight' progression through space.
E.g. with given start $A_0,B_0,C_0,D_0$:
1.) Compute $A'$
2.) Set new tetrahedron to $A_{i+1}= B_i, B_{i+1}= C_i, C_{i+1}= D_i, D_{i+1}= A'$
3.) Go to 1.)
We will get something like this:
Question: Ist there any tetrahedron $A_k,B_k,C_k,D_k$ constructed that way with: $$A_k-A_0 = A_{2k}-A_k = \vec{v}$$ $$B_k-B_0 = B_{2k}-B_k = \vec{v}$$ $$C_k-C_0 = C_{2k}-C_k = \vec{v}$$ $$D_k-D_0 = D_{2k}-D_k = \vec{v}$$ And with this: $$A_{2k}-A_0 = 2\vec{v}$$ $$A_{n+k}-A_n = \vec{v} \text{ }\text{ }\text{ } \forall n$$ So the $(2k)$'th tetrahedron would be the $0$'th shifted by vector $2 \vec{v}$. And the $(n+k)$'th tetrahedron would be the $n$'th shifted by vector $\vec{v}$.
In tests (up to $k=1000$) the $k=254$ is very close to but not equal.
Is there any $k$ which does work out?
Or if not is there any other kind of attachment order which can produce such a vector ($\ne\vec{0}$, (plane attached to plane) )?
Edit: Thanks to Intelligenti pauca this is similar problem as for the Boerdijk–Coxeter Helix
with vertices coordinates: $$ (r \cos n\theta ,r\sin n\theta ,nh) $$ with: $$r=3\sqrt{3}/10, \theta=\pm\cos^{-1}(-2/3) \approx 131.81^\circ, h=1/\sqrt{10}$$ The problem is equal with finding integer $n_1$, $n_2$ with: $$\cos(n_1 \theta) = \cos(n_2 \theta)$$ $$\sin(n_1 \theta) = \sin(n_2 \theta)$$