TITLE: Number Theory and d-Self-Contained Numbers
QUESTION [7 upvotes]: Given any natural number $N = a_{n}a_{n-1}\ldots a_{1}$, let us associate to it the set $S_{N} = \bigcup_{j=1}^{n}\{(a_{j},j)\}$. We're going to define a d-self-contained number as any natural number which satisfies the rule:
\begin{align*}
&\forall k\leq n,\exists\sigma\subset S_{N}\setminus\{(a_{k},k)\}; a_{k} = \sum_{s\in\sigma}(-1)^{e_{s}}\|p_{1}(s)\|^{d},\\
&\text{where}\,\,n\geq 3,\,\,e_{s}\in\{0,1\}\,\,\text{and}\,\,\#\sigma\geq 2
\end{align*}
In other words, a number is d-self-contained if each of its digits can be obtained from the others as a linear combination of their d-th power (where d is natural and fixed) whose coefficients belong to the set {-1,0,1}, where at least two of these coefficients are non-zero terms. It is worth saying that $p_{1}$ is the projection mapping on the first coordinate. Let us work it through examples.
The number 101 is 2-self-contained since we can rewrite its digits as: $1 = 0^{2} + 1^{2}$ and $0 = 1^{2} - 1^{2}$. Furthermore, the number 101 is d-self-contained for every d. Indeed, we have: 
$$1 = 0 + 1 = 0^{d} + 1^{d}\,\,\text{and}\,\,0 = 1 - 1 = 1^{d} - 1^{d}$$
On the other hand, the number 121 is not 2-self-contained given that:
$$1 \neq 2^{2} - 1^{2}\,\,\text{and}\,\,1\neq 2^{2} + 1^{2}$$
Although it is 1-self-contained. What about the 3-self-contained numbers? Here it comes an example which proves that they exist: 111111112. Undoubtedly, its digits satisfy the following relationships:
$$1 = 2^{3} - 1^{3} - 1^{3} - 1^{3} - 1^{3} - 1^{3} - 1^{3} - 1^{3}\,\,\text{and}\,\,2 = 1^{3} + 1^{3}$$
Even more, this number is also 1,2-self-contained. Really, we have:
$$1 = 2 - 1\,\,\text{and}\,\,2 = 1 + 1;\quad
1 = 2^{2} - 1^{2} - 1^{2} - 1^{2}\,\,\text{and}\,\,2 = 1^{2} + 1^{2}$$
Besides that, this example provides us with a way to build any d-self-contained number:
$$N = \overbrace{11\ldots 1}^{2^{d}}2\Rightarrow 2^{d} = \overbrace{1^{d} + 1^{d} + \ldots + 1^{d}}^{2^{d}}\,\,\text{and}\,\,1 = 2^{d} - \overbrace{1^{d} - 1^{d} - \ldots - 1^{d}}^{2^{d} - 1}$$
Since the definition has been made clear, I would like to make some questions. First of all, does any one can propose any criterion to identify them quickly? In second place, is there any formula which generates them all? And, finally, if we denote the set of d-self-contained numbers by $A_{d}$, does the next proposition hold: $A_{1}\supset A_{2}\supset\ldots\supset A_{k}\supset\ldots$? Thank you in advance for any contribution.

REPLY [2 votes]: Here's an easy answer to the last question: $10127$ is $3$-self-contained ($7 = 2^3 - 1^3$, $2^3 = 1^3+1^3$, $1^3 = 1^3+0^3$, $0=1^3-1^3$), but it's not $2$-self-contained because $7 > 1^2+0^2+1^2+2^2$.  Similarly I believe $10129$ and $111129$ are in $A_3 \setminus A_2$.  Thus $A_2 \not\supset A_3$, so the infinite descending chain proposition is false.
I suspect that, much like the OEIS sequence that Robert Soupe points to, the structure of $A_d$ for any fixed $d$ is quite simple and can described by a finite automaton.  But I hesitate to make any guesses as to how the complexity grows with $d$.