Project Euler > Problem 155 > Counting Capacitor Circuits. (Java Solution)

Problem:

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 F 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 ...

Find D(18).

Reminder : When connecting capacitors C1, C2 etc in parallel, the total capacitance is CT = C1 + C2 +...,
whereas when connecting them in series, the overall capacitance is given by:

Solution:

232792560

Code:
The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}

We don't have code for that problem yet! If you solved that out using Java, feel free to contribute it to our website, using our "Upload" form.