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Problem 175

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Fractions involving the number of different ways a number can be expressed as a sum of powers of 2

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.

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与幂和表示有关的分数

记$f(0)=1$，$f(n)$为将$n$写成$2$的幂次的和且任意幂次出现不超过两次的方式数。

例如，$f(10)=5$，因为$10$恰好有$5$种不同的表示方式：
$10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1$

对于任意分数$p/q$（其中整数$p>0$，整数$q>0$），我们都能找到至少一个整数$n$，使得$f(n)/f(n-1)=p/q$。

例如，使得$f(n)/f(n-1)=13/17$的最小的$n$是$241$。

$241$的二进制表示为$11110001$。

从左往右读这个二进制串我们得到$4$个$1$，$3$个$0$，再$1$个$1$。因此，我们称数字串$4,3,1$是$241$的简式二进制表示。

找出满足下式的最小的$n$的简式二进制表示：
$f(n)/f(n-1)=123456789/987654321$.

你的答案应当用半角逗号“,”隔开各个整数，且没有任何空格。

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