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# Problem 618

## Numbers with a given prime factor sum

Consider the numbers 15, 16 and 18:
$15=3\times5$ and $3+5=8$.
$16=2\times2\times2\times2$ and $2+2+2+2=8$.
$18=2\times3\times3$ and $2+3+3=8$.
15, 16 and 18 are the only numbers that have 8 as sum of the prime factors (counted with multiplicity).

We define $S(k)$ to be the sum of all numbers $n$ where the sum of the prime factors (with multiplicity) of $n$ is $k$.
Hence $S(8)=15+16+18=49$.
Other examples: $S(1)=0$, $S(2)=2$, $S(3)=3$, $S(5)=5+6=11$.

The Fibonacci sequence is $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3$, $F_5=5$, $\ldots$
Find the last nine digits of $\sum_{k=2}^{24}S(F_k)$.

## 给定质因数和的整数

$15=3\times5$，而$3+5=8$。
$16=2\times2\times2\times2$，而$2+2+2+2=8$。
$18=2\times3\times3$，而$2+3+3=8$。
15，16和18是仅有的质因数和（包括重复）为8的整数。