Is the $n$-th prime $p_n$ expressible as the difference of coprime $A, B$ such that the set of prime divisors of $AB$ is $\{p_1, \dots, p_{n-1}\}$? We define recursively 
$p_1=2,p_2=3$
and 
$$p_{n}= \min_{(A,B)\in F_{n-1}}|A-B| $$
Where
$$
\begin{split}
F_n=\{(A,B) |&\gcd (A,B)=1,\quad |A-B| \not =1,
\\\
&\text{both $A$ and $B$ are products of powers of $p_i$ for $i\le n$},
\\\
&\text{for each $i\le n$, either $p_i |A$ or $p_i |B$}\}
\end{split}
$$
Is always $p_n$ the $n-th$ prime?
 A: Barry Cipra has already computed the first few values.
The next couple of numbers $p_n$ are
$13 = 5 \cdot 11 - 2 \cdot 3 \cdot 7$,
$17 = 2 \cdot 7 \cdot 13 - 3 \cdot 5 \cdot 11$,
$19 = 2^2 \cdot 3 \cdot 5 \cdot 17 - 7 \cdot 11 \cdot 13 $,
$23 = 7 \cdot 11 \cdot 17^2 - 2 \cdot 3^2 \cdot 5 \cdot 13 \cdot 19$,
$29 = 3 \cdot 11 \cdot 13^2 \cdot 19 \cdot 23 - 2^{12} \cdot 5 \cdot 7 \cdot 17$.
I don't find such expression for 31 in numbers $\leq 10^{12}$.
The smallest I find is $47 = 3 \cdot 7 \cdot 19^2 \cdot 23 \cdot 29 -
2^5 \cdot 5 \cdot 11 \cdot 13^2 \cdot 17$.
If the abc conjecture is true, there are at most finitely many ways to express the $n$-th
prime as a difference of coprime numbers which are divisible only by the first $n-1$ 
primes.
Addendum: For a more extensive table, see http://www.fermatquotient.com/DiverseMinimas/S=M-N_min (found by Google'ing for the numbers obtained above).
The data supports the assumption that the OP's assertion is unlikely to be true.
A: You can find some amusing (I might say amazing) papers in this area by searching for "primes at a glance" and " primes at a (somewhat lengthy) glance". 
In the first paper they give (along with many other interesting things) what they "believe to be a complete list" of all pairs of integers $B,L$ with


*

*$N=B+L$

*$B \ge |L|$

*$\gcd(B,L)=1$

*if $p \le\sqrt{N}$, then $p$ divides $BL.$

*if $Q | BL$ and $p \lt q$ then $p | BL$


For $N \gt 1$ this proves $N$ to be prime as it rules out any proper divisors. Such a presentation provides an at a glance proof that $p$ is prime. For $N=31$ the presentations range from 
$31=2^4+3\cdot5$ 
to 
$31=2^3\cdot7\cdot11\cdot17\cdot23-3\cdot5^2\cdot13^2\cdot19$
The first primes for which they give no solutions are $541,547$
the last for which they do is
$2521=19\cdot43\cdot37\cdot2\cdot3^2\cdot5\cdot29^2\cdot41\cdot47^2-7\cdot11^3\cdot13\cdot17^2\cdot23^2\cdot31$
Without condition 5 there are many solutions. $88711$ is the product of $7,19,23,29$ and $72930$ is the product of $2,3,5,11,13,17$ so we can certainly find positive coprime integers with $1=88711x-72930y.$ Then $31=88711s-72930t$ is a difference of coprime values for $s=31x+72930$ and $t=31y+88711$ You can always do that.But probably not with $st$ having all prime factors below 31 ( in which case the prime factors would split the same way, given that none of them divide $31$.)
