Hamlet to be or not to be in Primes? It is conjectured that, if you read $\pi$ long enough you'll find Hamlet. Since other numbers, like the Copeland–Erdős constant are known to be normal in base $10$, it should be true at least there. I wondered, if it also possible to find Hamlet or something else in the sequence of halved differences of subsequent primes:
$$
L_k = \frac{p_{k}-p_{k-1}}{2} \bmod 26
$$
($k>2$ for the nit-pickers ;-). This gives a sequence of number from $0$ to $25$, that one maps to the letters of the alphabet.
I checked the first million, but I couldn't even find my username. So before wasting more time, can one show Hamlet to be or not to be in Primes? 

This might boil down to the question, if we get enough 'A's, i.e. are there infinitely many twin primes, but maybe Hamlet or at least my username comes before the very last if there is one...

 A: The answer is NO.  Almost any contiguous fragment of the play will do, so let's use “The slings and arrows of outrageous fortune”, or numerically 20 8 5 19 12 9 14 7 19 1 14 4 1 18 18 15 23 19 15 6 15 21 20 18 1 7 5 15 21 19 6 15 18 20 21 14 5.  Doubling these determines the spacings modulo 52, and hence also modulo 13, which are:
$1,3,10,12,11,5,2,1,12,2,2,8,2,10,10,4,7,12,4,12,4,3,1,10,2,1,10,4,3,12,12,4,10,1,3,2,10.$
The partial sums of this table modulo 13 are
$ \color{red}1, \color{red}4, 1, \color{red}0, \color{red}{11}, \color{red}3, \color{red}5, \color{red}6, 5, \color{red}7, \color{red}9, 4, 6, 3, 0, 4, 11, \color{red}{10}, 1, 0, 4, 7, \color{red}8, 5, 7, 8, 5, 9, \color{red}{12}, 11, 10, 1, 11, 12, \color{red}2, 4, 1,$
which contains a complete system of residues mod $13$.  Aye, there's the rub.  For, in that sequence encoding this phrase, what remainders may occur, when we have cast off multiples of $13$, must give us zero.  This will not be prime unless it is very small (in the first few letters of the sequence).
As you can see, the sequence of prime gaps is not normal.  For similar reasons you won't find AAAAAAAAAAAA anywhere.  On the other hand, Shiu proved that there are arbitrarily long strings of consecutive primes all lying in the same congruence class mod $n$ (one can even prescribe the congruence class).  Therefore we can find strings of ZZZZZZZZZZZZ, as long as devoutly to be wished.  To sleep, perchance to dream...
