## Problem:

Consider the infinite polynomial series AF(x) = xF1 + x2F2 + x3F3 + ..., where Fk is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... ; that is, Fk = Fk[−]1 + Fk[−]2, F1 = 1 and F2 = 1.

For this problem we shall be interested in values of x for which AF(x) is a positive integer.

Surprisingly AF(1/2) = (1/2).1 + (1/2)2.1 + (1/2)3.2 + (1/2)4.3 + (1/2)5.5 + ...
= 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ...
= 2

The corresponding values of x for the first five natural numbers are shown below.

x AF(x)
[√]2[−]1 1
1/2 2
([√]13[−]2)/3 3
([√]89[−]5)/8 4
([√]34[−]3)/5 5

We shall call AF(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690.

Find the 15th golden nugget.

748317

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 37 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p037 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p037().run());	}			public String run() {		long sum = 0;		for (int count = 0, n = 10; count < 11; n++) {			if (isTruncatablePrime(n)) {				sum += n;				count++;			}		}		return Long.toString(sum);	}			private static boolean isTruncatablePrime(int n) {		// Test if left-truncatable		for (long i = 10; i <= n; i *= 10) {			if (!Library.isPrime(n % (int)i))				return false;		}				// Test if right-truncatable		for (; n != 0; n /= 10) {			if (!Library.isPrime(n))				return false;		}				return true;	}	}