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Problem 234

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Semidivisible numbers

For an integer n ≥ 4, we define the lower prime square root of n, denoted by lps(n), as the largest prime ≤ √n and the upper prime square root of n, ups(n), as the smallest prime ≥ √n.

So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37.
Let us call an integer n ≥ 4 semidivisible, if one of lps(n) and ups(n) divides n, but not both.

The sum of the semidivisible numbers not exceeding 15 is 30, the numbers are 8, 10 and 12. 15 is not semidivisible because it is a multiple of both lps(15) = 3 and ups(15) = 5.
As a further example, the sum of the 92 semidivisible numbers up to 1000 is 34825.

What is the sum of all semidivisible numbers not exceeding 999966663333 ?

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半可约整数

对于任意整数n ≥ 4，我们定义n的下素数平方根为≤ √n的最大素数，记作lps(n)，并定义n的上素数平方根为≥ √n的最小素数，记作ups(n)。

举例来说，lps(4) = 2 = ups(4)，lps(1000) = 31，ups(1000) = 37。
如果lps(n)和ups(n)中有且只有一个整除n，我们称整数n ≥ 4为半可约的。

所有不超过15的半可约整数之和为30，这些整数是8、10和12。15并不是半可约整数，因为它同时是lps(15) = 3和ups(15) = 5的倍数。
再举一例，不超过1000的92个半可约整数之和为34825。

不超过999966663333的所有半可约整数之和是多少？

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