I have an integer $I$, and I want to express this in the following sum.

$$I\ =\ \sum_{i}^{m} {b_i}^{n_i}$$

where $b_i$, $n_i∈\mathbb{Z}$ are of two sets:

$$Base\ set:\ \{b\}_m\ =\ \{b_1,\ b_2,\ b_3,\ ...\ b_m \}$$

$$Power\ set:\ \{n\}_m\ =\ \{n_1,\ n_2,\ n_3,\ ...\ n_m \}$$



I could do this in a number of ways...

$$\{b\}_3=\{5,4,2\}$$ $$\{n\}_3=\{2,1,3\}$$

$$I = \sum^{3}{b_i}^{n_i}=5^2+4^1+2^3=37$$


$$\{b\}_2=\{6,1\}$$ $$\{n\}_2=\{2,1\}$$

$$I = \sum^{3}{b_i}^{n_i}=6^2+1^1=37$$

I was wondering, would it be possible to express each set as discrete functions?

$$\{b\}_m=B(I,m)$$ $$\{n\}_m=N(I,m)$$

  • $\begingroup$ I don't think so. For example, $4=2^2 = 4^1$, so $B(4,1)$ and $N(4,1)$ cannot be uniquely defined. You can construct less trivial examples, if you wish. $\endgroup$ – Crostul Apr 5 '16 at 16:43
  • $\begingroup$ Just a thought, but what if we treated B & N as 1xm matrices, and we were to plug those matrices into a multidimensional function (analogous to a plane) where it has multiple solutions. (I guess an eigenvalue/eigen-vector problem is what I'm getting at) $\endgroup$ – Ben Marconi Apr 5 '16 at 21:11

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