= Solution
{c}
\b[The <sequence> does not converge.] For $n=2^k$, the <largest prime factor> is $p(n)=2$, while the <sum of prime factors with multiplicity> is $s(n)=2k$. This <subsequence> therefore has ratio $1/k\to0$.
There are infinitely many <prime numbers>: if there were only $q_1,\ldots,q_r$, a <prime factor> of $q_1\cdots q_r+1$ would be absent from the list. Listing the <prime numbers> in increasing order gives a <subsequence> of indices $n=q_j$ for which $p(n)=s(n)=q_j$, so the ratio is constantly one. Every <subsequence> of a <convergent sequence> has the same <limit of a sequence>. The two distinct <subsequential limits> $\boxed{0\text{ and }1}$ rule out <convergence>.
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