We use the following facts and heuristics. (1) The ordered partitions of $n$ into any positive integers (not just primes) is $2^{n-1}$. This can be guessed by checking small cases and finding a pattern, and is not difficult to prove. (2) The partitions of $\frac{2013}{n}$ into any positive integers equals the partitions of 2013 into integers from the set $\{n, 2 n, 3 n, \cdots\}$. (3) The small numbers matter more when considering partitions. (4) The set of primes $\{2,3,5,7, \cdots\}$ is close in size (near the small numbers) to $\{3,6,9, \cdots\}$ or $\{2,4,6, \cdots\}$. (5) The prime numbers get very sparse compared to the above two sets in the larger numbers. Thus, using these heuristics, the number of partitions of 2013 into primes is approximately $2^{\frac{2013}{3}}-1$ or $2^{\frac{2013}{2}-1}$, which, taking logarithms, give 670 and 1005.5, respectively. By (5), we should estimate something that is slightly less than these numbers.