67 lines
1.6 KiB
Markdown
67 lines
1.6 KiB
Markdown
---
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id: 5900f4081000cf542c50ff1a
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challengeType: 5
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title: 'Problem 155: Counting Capacitor Circuits'
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forumTopicId: 301786
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---
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## Description
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<section id='description'>
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An electric circuit uses exclusively identical capacitors of the same value C.
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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.
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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 F each, we can obtain the following 7 distinct total capacitance values:
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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 ...
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Find D(18).
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Reminder : When connecting capacitors C1, C2 etc in parallel, the total capacitance is CT = C1 + C2 +...,
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whereas when connecting them in series, the overall capacitance is given by:
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</section>
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## Instructions
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<section id='instructions'>
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: <code>euler155()</code> should return 3857447.
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testString: assert.strictEqual(euler155(), 3857447);
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function euler155() {
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return true;
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}
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euler155();
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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// solution required
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```
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</section>
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