Smallest Value After Replacing With Sum of Prime Factors
<p>You are given a positive integer <code>n</code>.</p>

<p>Continuously replace <code>n</code> with the sum of its <strong>prime factors</strong>.</p>

<ul>
	<li>Note that if a prime factor divides <code>n</code> multiple times, it should be included in the sum as many times as it divides <code>n</code>.</li>
</ul>

<p>Return <em>the smallest value </em><code>n</code><em> will take on.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 15
<strong>Output:</strong> 5
<strong>Explanation:</strong> Initially, n = 15.
15 = 3 * 5, so replace n with 3 + 5 = 8.
8 = 2 * 2 * 2, so replace n with 2 + 2 + 2 = 6.
6 = 2 * 3, so replace n with 2 + 3 = 5.
5 is the smallest value n will take on.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Initially, n = 3.
3 is the smallest value n will take on.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
</ul>
Math; Number Theory; Simulation
class Solution {
    public int smallestValue(int n) {
        while (true) {
            int t = n, s = 0;
            for (int i = 2; i <= n / i; ++i) {
                while (n % i == 0) {
                    s += i;
                    n /= i;
                }
            }
            if (n > 1) {
                s += n;
            }
            if (s == t) {
                return s;
            }
            n = s;
        }
    }
}