the least odd prime leaving prime remainder when divided by the first $n$ odd primes What is the least odd prime giving a prime remainder when divided by $3$?  It is $5$. What is the least odd prime giving prime remainders for both $3$ and $5$?  It is $17$.  For division by $3,5,7$ it is also $17$.  For division by $3,5,7,11$ it is $47$.  The sequence is $$5,17,17,47,863,887,887,887,887,887...$$
Note that $887$ leaves prime remainder under division by all the primes from $3$ to $31$ (but not $37$).  

If the sequence is continued, will any other prime occur five (or more)
  times in a row?

 A: Here's a very rough heuristic argument that you can expect repeats of length $5$ (or more) to occur infinitely often.
If $P_n$ is the least (odd) prime that gives a prime remainder when divided by each of the primes $3,5,7,\ldots,p_n$, the probability that its remainders under division by the next $4$ primes are also prime is approximately
$$\left(n\over p_n\right)^4$$
(A more "precise" approximation would be ${n\over p_{n+1}}\cdot{n+1\over p_{n+2}}\cdot{n+2\over p_{n+3}}\cdot{n+3\over p_{n+4}}$, but for large $n$, the displayed expression is good enough. The basic heuristic premise is that when you divide a "random" number by $q$, the remainder is equally likely to be any number smaller than $q$.)  So the probability that the next $4$ remainders are not all prime is approximately
$$1-\left(n\over p_n\right)^4$$
and so the probability that a repeat of length $5$ never happens again past some point $N$ is approximately
$$\prod_{n\gt N}\left(1-\left(n\over p_n\right)^4\right)$$
Now the prime number theorem implies $p_n\approx n\ln n$, so the approximated probability becomes
$$\prod_{n\gt N}\left(1-\left(1\over\ln n\right)^4\right)$$
But that infinite product "condiverges" to $0$ (i.e., its logarithm is the divergent series $\sum\ln(1-({1\over\ln n})^4)\approx-\sum({1\over\ln n})^4=-\infty)$, which we can interpret, heustically, as saying there is no chance that a repeat of length $5$ won't happens again.  Note, there is nothing special about the number $5$; the argument works the same for any fixed repeat length.
