If $ b$ is a number in the sequence, and $ p$ is the smallest prime dividing $ b$, then the next number is $ b+\frac{b}{p}=b\cdot\frac{p+1}{p}$. From this, we immediately see that the list is increasing, so we any other $ a$ must be less than 2002. Say that we had an even number in the sequence. Let this number be $ 2^i \cdot m$, where $ m$ is odd. Clearly, for the next $ i$ steps in the list, 2 will be the smallest prime, so we will multiply by $ \frac{3}{2}$ each time. After these $ i$ steps, we will be left with $ 3^i \cdot m$. Then, 3 will be the smallest prime, so the next number will be $ 2^2\cdot3^{i-1} \cdot m$. Then, the next numbers will be $ 2\cdot 3^i \cdot m, 3^{i+1}\cdot m$, and from here, a pattern of the smallest prime being 3,2,2 will repeat forever. Each number in this sequence is divisible by $ 4$ or $ 3$, and since $ 2002=2\cdot7\cdot11\cdot13$ is not divisible by $ 4$ nor $ 3$, it follows that no even number can precede 2002. If the first number is even, then it must be $ 2002$, giving us one valid value of $ a$. Otherwise, if $ a$ is odd, then the next number in the list must be even, and thus must be 2002. In this case, we must have $ 2002=a\cdot\frac{p+1}{p}$, or $ 2002p=a(p+1)$. Since $ a$ is odd and $ \gcd(p,p+1)=1$, we see that $ p+1$ must be an even factor of 2002. Testing out all 8 even factors of 2002, we see that the only such factors that are 1 greater than a prime are $ 2\cdot7=14$ and $ 2\cdot7\cdot13=182$. The first case corresponds to $ a=11\cdot13^2=1859$, and the second corresponds to $ a=11\cdot 181 = 1991$. But in both these cases, $ 11$ is the smallest prime, and the next value should be $ \frac{12a}{11}$, which will not give a value of 2002. Thus, our only valid value of $ a$ is $ 2002$.