**Problem:**

Define f(0)=1 and f(n) to be the number of ways to write n as a sum of powers of 2 where no power occurs more than twice.

For example, f(10)=5 since there are five different ways to express 10:

10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1

It can be shown that for every fraction p/q (p[>]0, q[>]0) there exists at least one integer n such that

f(n)/f(n-1)=p/q.

For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241.

The binary expansion of 241 is 11110001.

Reading this binary number from the most significant bit to the least significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the string 4,3,1 the Shortened Binary Expansion of 241.

Find the Shortened Binary Expansion of the smallest n for which

f(n)/f(n-1)=123456789/987654321.

Give your answer as comma separated integers, without any whitespaces.

For example, f(10)=5 since there are five different ways to express 10:

10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1

It can be shown that for every fraction p/q (p[>]0, q[>]0) there exists at least one integer n such that

f(n)/f(n-1)=p/q.

For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241.

The binary expansion of 241 is 11110001.

Reading this binary number from the most significant bit to the least significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the string 4,3,1 the Shortened Binary Expansion of 241.

Find the Shortened Binary Expansion of the smallest n for which

f(n)/f(n-1)=123456789/987654321.

Give your answer as comma separated integers, without any whitespaces.

**Solution:**

4782

**Code:**

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

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

public interface EulerSolution{

public String run();

}

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