[Alamo-Girl]

Thank you for your reply!

Another interesting thing is that you could concatenate the primes (for example) and divide by 13 (treating the string as a big number). The resulting quotient also generates Shakespeare.

Again, this formulation is more appealing because I do not see a high autocorrelation on first blush!

I don't know where. I could give an upper bound that would guarantee the you would find it before that, but it's a big bound.

Wouldn't that guarantee finding it by using a particular bit offset? IOW, the bit offset of the beginning bit of the number which is Shakespeare would obviously work - but can it be reduced?

[Doctor Stochastic]

That's why you only get an upper bound. Perhaps lower perhaps not. As these numbers all obey the "strong law of large numbers" they all have about the same behavior. The strong law (in this case) says that all bit sequences will appear with the proper frequency. That is half 1's, half 0's, a quarter each of 00,01,10,11, etc.