# Number Theory and d-Self-Contained Numbers

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 mentioning 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: $$N = \overbrace{11\ldots 1}^{2^{d}\,\text{times}}2\Rightarrow 2^{d} = \overbrace{1^{d} + 1^{d} + \ldots + 1^{d}}^{2^{d}}\,\,\text{and}\,\,1 = 2^{d} - \overbrace{1^{d} - 1^{d} - \ldots - 1^{d}}^{2^{d} - 1}$$ 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.

• As in a previous incarnation of this problem, such numbers can be identified with multi sets of digits, and such multisets of d-self-contained numbers are then closed under union (for a fixed d). There may be finitely many primitive solutions as in the earlier problem, but this is less clear. You can try computer enumeration to find such solutions, but you may also find some theory to aid in the search (Tarry-Escott comes to mind). I don't think a quick analytic solution is available. Gerhard "If Base Is Large Enough" Paseman, 2016.06.27. – Gerhard Paseman Jun 27 '16 at 20:24
• Also posted to m.se math.stackexchange.com/questions/1839834/… without notification to either site. – Gerry Myerson Jun 28 '16 at 0:01
• Sorry, I did not know that I should notify. How should I proceed? – APC89 Jun 28 '16 at 0:12
• All you have to do, apc89, is edit into your question at each site a link to the other site. No need to, now, as I've (essentially) done it for you, by putting links in my comments. – Gerry Myerson Jun 30 '16 at 2:54
• Also 1112 and 1113 are self-contained for all positive integers d. You might consider enumerating minimal digit multisets by hand and then forming larger multisets using conservative unions. Also, repeating a digit from a self-contained multiset preserves self-containment. It might be fruitful to look at minimal multisets associated with a particular digit set. Gerhard "Reductions Can Save Much Resources" Paseman, 2016.06.29. – Gerhard Paseman Jun 30 '16 at 5:12