--description--

An electric circuit uses exclusively identical capacitors of the same value C.

The capacitors can be connected in series or in parallel to form sub-units, which can then be connected in series or in parallel with other capacitors or other sub-units to form larger sub-units, and so on up to a final circuit.

Using this simple procedure and up to n identical capacitors, we can make circuits having a range of different total capacitances. For example, using up to $n = 3$ capacitors of $60 μF$ each, we can obtain the following 7 distinct total capacitance values:

If we denote by $D(n)$ the number of distinct total capacitance values we can obtain when using up to $n$ equal-valued capacitors and the simple procedure described above, we have: $D(1) = 1, D(2) = 3, D(3)=7, \ldots$

Find $D(18)$.

Reminder: When connecting capacitors $C_1$, $C_2$ etc in parallel, the total capacitance is $C_T = C_1 + C_2 + \cdots$, whereas when connecting them in series, the overall capacitance is given by: $\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots$.

--hints--

capacitanceValues() should return 3857447.

assert.strictEqual(capacitanceValues(), 3857447);

--seed--

--seed-contents--

function capacitanceValues() {

  return true;
}

capacitanceValues();

--solutions--

// solution required