---
id: 5900f4b91000cf542c50ffcc
title: '问题 333：特殊分区'
challengeType: 1
forumTopicId: 301991
dashedName: problem-333-special-partitions
---

# --description--

All positive integers can be partitioned in such a way that each and every term of the partition can be expressed as $2^i \times 3^j$, where $i, j ≥ 0$.

Let's consider only those such partitions where none of the terms can divide any of the other terms. For example, the partition of $17 = 2 + 6 + 9 = (2^1 \times 3^0 + 2^1 \times 3^1 + 2^0 \times 3^2)$ would not be valid since 2 can divide 6. Neither would the partition $17 = 16 + 1 = (2^4 \times 3^0 + 2^0 \times 3^0)$ since 1 can divide 16. The only valid partition of 17 would be $8 + 9 = (2^3 \times 3^0 + 2^0 \times 3^2)$.

Many integers have more than one valid partition, the first being 11 having the following two partitions.

$$\begin{align}   & 11 = 2 + 9 = (2^1 \times 3^0 + 2^0 \times 3^2) \\\\
  & 11 = 8 + 3 = (2^3 \times 3^0 + 2^0 \times 3^1) \end{align}$$

Let's define $P(n)$ as the number of valid partitions of $n$. For example, $P(11) = 2$.

Let's consider only the prime integers $q$ which would have a single valid partition such as $P(17)$.

The sum of the primes $q &lt;100$ such that $P(q) = 1$ equals 233.

Find the sum of the primes $q &lt; 1\\,000\\,000$ such that $P(q) = 1$.

# --hints--

`specialPartitions()` should return `3053105`.

```js
assert.strictEqual(specialPartitions(), 3053105);
```

# --seed--

## --seed-contents--

```js
function specialPartitions() {

  return true;
}

specialPartitions();
```

# --solutions--

```js
// solution required
```
