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Given any natural number $N = a_{n}a_{n-1}\ldots a_{1}$, we're going to define its digits-partition as the next set $D_{N} = \bigcup_{j=1}^{n}\bigcup_{k=1}^{p(a_{j})}\{(P_{k},j)\}$, where each pair $(P_{k},j)$ indicates a partition $P_{k}$ corresponding to $a_{j}$ in the form of a multiset, for $1\leq k\leq p(a_{j})$ and $1\leq j \leq n$. It is worth mentioning here that the function $p(n)$, which gives the number of partitions of $n$, is well-defined and has an explicit expression.

Here it comes some examples of digits-partitions: \begin{align*} D_{5} & = \{(\{1,1,1,1,1\},1),(\{1,1,1,2\},1),(\{1,1,3\},1),(\{1,4\},1),(\{2,3\},1)\}\\\\ D_{14} & = \{(\{1,1,1,1\},1),(\{1,1,2\},1),(\{1,3\},1),(\{2,2\},1),(\{1\},2)\}\\\\ D_{15} & = \{(\{1,1,1,1,1\},1),(\{1,1,1,2\},1),(\{1,1,3\},1),(\{1,4\},1),(\{2,3\},1),(\{1\},2)\}\\ \end{align*} Once it has been understood, here it comes the utmost definition: we are going to consider N to be a p-power-partitioned number if it satisfies the following condition \begin{align*} &\exists\sigma = \bigcup_{j=1}^{n}\{(P_{k_{j}},j)\}\subset D_{N};\, N = \sum_{j=1}^{n}\sum_{s\in P_{k_{j}}}c_{s}s^{p},\\ &\text{where}\,\,c_{s}\in\{-1,0,1\}\,\,\text{and}\,\,p\,\text{is a fixed natural} \end{align*} Perhaps it is good to give some examples. For instance, the number 15 is 2-power-partitioned since we may choose $\sigma = \{(\{1,4\},1),(\{1\},2)\}$ to rewrite it as $15 = (-1^{2} + 4^{2}) + 0\cdot 1^{2}$. Similarly, the number 16 is also 2-power-partitioned. Indeed, if we set $\sigma = \{(\{1,1,4\},1),(\{1\},2)\}$, we can express it as follows: $16 = (-1^{2} + 1^{2} + 4^{2}) + 0\cdot 1^{2}$. The number 14 is 4-power-partitioned once, for $\sigma = \{(\{1,1,2\},1),(\{1\},2)\}$, we can express it as: $14 = (-1^{4} - 1^{4} + 2^{4}) + 0\cdot 1^{4}$.

Hopefully I have been clear enough and you have understood the definition. Thence I would like to make some questions. Could anyone provide me a proof there are finitely many p-power-partitioned numbers for each $p$? Besides this, I'd be extremely thankful as well if someone gave me a criterion to identify them. And, finally, I seek for an answer if there is a formula which generates at least some of them. Thank you in advance for any contribution.

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    $\begingroup$ For a fixed base b and fixed p you should be able to tell that there are only finitely many such numbers. It is unclear that for base 10 there exists even one such number given sufficiently large p. Can you motivate this any further? It is not clear why one wants to consider such numbers. Gerhard "Partitioning Digits Is Very Limiting" Paseman, 2016.07.02. $\endgroup$
    – Gerhard Paseman
    Jul 3, 2016 at 4:26
  • $\begingroup$ In fact, it is just recreation. Should I post elsewhere? $\endgroup$
    – user89024
    Jul 3, 2016 at 4:32
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    $\begingroup$ I don't know. However, I encourage you to provide more than just definitions and a few examples. In particular, find what interesting properties hold. I would hope that you yourself would have seen and posted that there are only finitely many such numbers given p, and perhaps given an upper bound for the largest possible number. Gerhard "Encourages You To Show Work" Paseman, 2016.07.02 $\endgroup$
    – Gerhard Paseman
    Jul 3, 2016 at 4:51
  • $\begingroup$ Indeed, my bad. I will try to work harder on the problem before posting it here. Thank you for the advice. $\endgroup$
    – user89024
    Jul 3, 2016 at 22:25

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