Representing the points of a doubly deleted Cantor set with a generic parameters $\alpha, \beta$ Let $0 < \alpha, \beta < 1$ such that $3\alpha + 2\beta = 1$ and consider the following construction of a modified Cantor set: At each step delete two subintervals of length $\beta \cdot \text{current interval}$ such that we are left with three subintervals of length $\alpha$. If our initial interval is the unit interval $[0,1]$, then after the first deletion we get the intervals $I_{1,1} = [0,\alpha], I_{1,2} = [\alpha + \beta, 2\alpha + \beta], I_{1,3} = [2\alpha + 2\beta, 1]$.
I am looking for a representation of the points of this set in the form of some infinite series $\sum_{k=1}^\infty h(k)$ for some $h$.
I had reasoned the following regarding the intervals and their representation:
From the construction we see that at $k$th step, length of the interval from which we delete stuff is $a^{k-1}$. Moreover, after the deletions we have three choices for the starting point of the interval: 1.) We can choose nothing so that we have an interval of the form $[0, a^k]$. 2.) We can choose to move by $\alpha + \beta$, giving an interval of the form $[\alpha^{k-1}(\alpha + \beta), \alpha^{k-1}(\alpha + \beta + \alpha)]$. 3.) We can choose to move by $2\alpha + 2\beta$ whence our interval is of the form $[\alpha^{k-1}2(\alpha + \beta), \alpha^{k-1}(2(\alpha + \beta) + \alpha)]$. By taking two binary functions $b_1,b_2\in\{0,1\}^{\mathbb{N}}$, my reasoning was that this choice process can be represented by
$$[\alpha^{k-1}(\alpha + \beta)(b_1(k) + b_1(k)b_2(k)), \alpha^{k-1}((\alpha + \beta)(b_1(k) + b_1(k)b_2(k)) + \alpha)]$$
so that the intervals at the $N$th step are given by
$$\left[\sum_{k=1}^\infty\alpha^{k-1}(\alpha + \beta)(b_1(k) + b_1(k)b_2(k)), \sum_{k=1}^\infty\alpha^{k-1}((\alpha + \beta)(b_1(k) + b_1(k)b_2(k)) + \alpha)\right]$$
where $b_1(k) = b_2(k) = 0, k > N$. If we take $b_1 \equiv 1 \equiv b_2$ then we see that $\sum_{k=1}^\infty\alpha^{k-1}(\alpha + \beta)(b_1(k) + b_1(k)b_2(k)) = 2(\alpha + \beta)\sum_{k=1}^\infty \alpha^{k-1} = \frac{2(\alpha + \beta)}{1 - \alpha} = \frac{1 - \alpha}{1 - \alpha} = 1$. This would lead me to believe that all points of this set are given in the form $\sum_{k=1}^\infty\alpha^{k-1}(\alpha + \beta)(b_1(k) + b_1(k)b_2(k))$ for some $b_1, b_2$, but I am not sure. To be specific, I have seen a similar representation in the classical middle Cantor set, but my sources have simply stated the representation is "easy to see", "trivial", "of immediate consequence" or something other along those lines. This is to say that I have not seen an analogue proof to which I could resort to.
 A: Here is an answer using the iterated function formalism I discussed as part of my answer to your other question Trouble understanding the "Modified Cantor sets" $C_{M,N}$ :
Fix $\alpha,\beta\in\mathbb{R}_{>0}$ with the property that $3\alpha+2\beta<1$ ($\dagger$). The construction involves shrinking $X=[0,1]$ into $3$ subintervals by a factor of $\alpha$ flushed left, and then spacing any two consecutive ones by a gap of exact length $\beta$. By the choices of $\alpha,\beta$ these intervals fit into $X$ in this configuration. Note that if $3\alpha+2\beta=1$ then there is no need to flush left (but then $\alpha,\beta$ seem to be "less generic"). As an IFS one can describe this construction as follows:
$$A_\bullet: \{0,1,2\}\to C^0(X;X), m\mapsto [x\mapsto \alpha x+m(\alpha+\beta)].$$
Using the notation in my earlier answer; the left endpoints of the subintervals constructed at step $k$ are parameterized as the images of the original left endpoint $0$ at time $k$ under the IFS:
$$\{A_{\epsilon_k\epsilon_{k-1}\cdots\epsilon_1}(0)\,|\, \epsilon_k,\epsilon_{k-1},...,\epsilon_1\in\{0,...,N-1\}\}.$$
As in my earlier answer, a straightforward induction gives
$$A_{\epsilon_k\epsilon_{k-1}\cdots\epsilon_1}(0)=(\alpha+\beta)\alpha^k \sum_{j=1}^k\alpha^{-j}\epsilon_j.$$
It's straightforward to verify convergence of any such series as $k\to\infty$, and via the change of variables $i=k+1-j$ one obtains power series expressions for any point in the attractor.

($\dagger$) One can of course consider more general situations, e.g. consider starting with an interval $I$ of length $L\in\mathbb{R}>0$, at  construct at the first step $M\in\mathbb{Z}_{\geq1}$ disjoint and consecutive subintervals $I_1,...,I_M$ (thus the maximum of $I_m$ is less than the minimum of $I_{m+1}$) of $I$ such that the gap between $I_m$ and $I_{m+1}$ is exactly $\beta_m\in\mathbb{R}>0$ (so that denoting by $\alpha_m$ the length of $I_m$ we have the constraining inequality $\sum_m \alpha_m+\sum_m \beta_m \leq L$); let us also center the subinterval $[\min(I_1),\max(I_M)]$ w/r/t the original interval $I$. It's straightforward to extract formulas for points using the IFS formalism.

Finally let me make one comment about your proposed solution. I haven't verified it thoroughly, but one perhaps negative aspect of it is that you are using two binary functions in your expression (with some possible cancellations of course). But from the dynamical point of view it's best to think of attractors like the Cantor set as the "limits" of $n$-ary trees; that is to say, one would want to be able to locate a point in the attractor with countably many yes/no (or more generally multiple but finite choice) questions. Roughly speaking such sets model the phase spaces of (most) dynamical systems from the measure theoretical point of view. For a discussion of this see Rudolph's book Fundamentals of Measurable Dynamics - Ergodic Theory on Lebesgue Spaces, p.9.
