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Consider the set $E_b = \left\{1, 1.2, 1.5, 1.8, 2.2, 2.7, 3.3, 3.9, 4.7, 5.6, 6.8, 8.2\right\}$. This is our base set. Let's define the set $E$ as follows: $$ E = \left\{ 10^k e \mid k=0,1,2,\ldots, \text{for every} e \in E_b \right\}$$ and $\Omega$ as the class of all subsets of $E$. We are intereted in defining a mapping $f$ from $\mathbb{N} \setminus \left\{1 \right\}$ to $\Omega$ such that $f(n)$ is mapped to a set $E_k \in \Omega$ according to the following rule: $$ \frac{1}{n} = \sum_{e_k \in E_k}{\frac{1}{e_k}}$$ Here are some examples:

  • $f(4) = \left\{10, 12, 15\right\}$ as $\frac{1}{4} = \frac{1}{10} + \frac{1}{12} + \frac{1}{15}$
  • $f(12) = \left\{12\right\}$
  • $f(19) = $ I haven't found a set that could be mapped to $19$, it could be an empty set
  • $f(20) = \left\{22, 220 \right\}$ as $\frac{1}{20} = \frac{1}{22} + \frac{1}{220}$

This question/puzzle was brought to my attention by a colleague (an electrical engineer), the idea is to synthesize a desired resistor (an integer-valued in this case) with a given set of standard resistors (set $E$) in parallel (two parallel resistors result in $\frac1{\frac{1}{R_1}+\frac{1}{R_2}}$).

Now I'm interested to know if we can synthesize a resistor with a minimum numbers of standard resistors, in other words a lower bound on the cardinality of $f(n)$. More interestingly, I'd like to know if someone can come up with an algorithm to construct such a mapping for every $n \in \mathbb{N} \setminus \left\{1 \right\}$, not necessarily with minimum cardinality. Also, for what numbers $f(n)$ is an empty set?

I hope I've articulated the question in mathematical terms clear enough.

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    $\begingroup$ Does this also have the scoring system "physicists are two points, mathematicians are three"? $\endgroup$ – null Aug 25 '15 at 20:23
  • $\begingroup$ $f(n)$ is never the empty set, but it may happen that $f(n)$ is not defined. Also, is it on purpose that you map to sets? I.e., one is not allowed to use several identical resistors? $\endgroup$ – Hagen von Eitzen Aug 25 '15 at 20:34
  • $\begingroup$ @HagenvonEitzen: identical resistors were not allowed, that's why I defined the mapping this way. $\endgroup$ – Ali Aug 25 '15 at 20:38
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    $\begingroup$ I think the usual convention is that $\sum_{e_k \in \{\}}{\frac{1}{e_k}} = 0$, so the assertion that $f(n) = \{\}$ would be the same as saying that $\frac 1n = 0$. In any case, it's easier to just say that $f(n)$ is undefined than to say that $f(n) = \{\}$ and also have to interpret $\sum_{e_k \in \{\}}{\frac{1}{e_k}}$ in such a way as to avoid contradictions (and then justify why you are doing it that way). $\endgroup$ – David K Aug 25 '15 at 21:04
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    $\begingroup$ I wonder what your colleague has against 150-, 220-, and 390-ohm resistors to cause him to omit them from the list of 10% standard values. $\endgroup$ – David K Aug 25 '15 at 23:51
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This is a partial answer, identifying some $n$ that cannot be synthesized with a finite subset of values from $E$.

The LCM of $E_{\text{b}}$ is $N=3541509972=2^2\cdot 3^3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 41\cdot 47$. Thus $\frac{N}{e}$ is an integer for $e\in E_{\text{b}}$. For general $e\in E$, $\frac{N}{e}$ therefore has fractional part with finite decimal length.

Consequently, if $f(n)$ is a finite set, then the sum for $\frac{N}{n}$ necessarily has fractional part with finite decimal length.

This implies that if $n$ has prime divisors other than those of $5N$, or prime divisors other than $2$ and $5$ that occur in greater multiplicity than in $N$, no finite set $f(n)$ can be given because $\frac{N}{n}$ would have infinite decimal expansion then. For example $19$ and $81$ cannot be synthesized.

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