--description--

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 $\frac{p}{q}\; (p>0, q>0)$ there exists at least one integer $n$ such that $\frac{f(n)}{f(n - 1)} = \frac{p}{q}$.

For instance, the smallest $n$ for which $\frac{f(n)}{f(n - 1)} = \frac{13}{17}$ 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

$$\frac{f(n)}{f(n - 1)} = \frac{123456789}{987654321}$$

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

--hints--

shortenedBinaryExpansionOfNumber() should return a string.

assert.isString(shortenedBinaryExpansionOfNumber());

shortenedBinaryExpansionOfNumber() should return the string 1,13717420,8.

assert.strictEqual(shortenedBinaryExpansionOfNumber(), '1,13717420,8');

--seed--

--seed-contents--

function shortenedBinaryExpansionOfNumber() {

  return true;
}

shortenedBinaryExpansionOfNumber();

--solutions--

// solution required