quarantine 'em

Okay, so once numbers get large enough, there's always a prime between any two consecutive cubes. For any n, there's always a prime number between n^3 and (n + 1)^3, if n is above some threshold

If the Riemann Hypothesis is true then this is true for *all* n, and we can more generally say that there is always a prime between any two cubes

So now imagine a sequence where a_1 = 2 a_{n + 1} = the next prime after (a_n)^3 This sequence goes 2, 11, 1361, 2521008887, ...

The interesting thing about this sequence is that because there's a prime between any two cubes, we know that a_{n + 1} is between (a_n)^3 and (a_n + 1)^3 a_{n + 1} can't be any larger than that

This means in turn that the cube root of a_{n + 1} is strictly between a_n and a_n + 1 (I'm going somewhere with this)

For example, the cube root of 11 is 2.2... < 3 the cube root of 1361 is 11.082... < 12 the cube root of 2521008887 is 1361.000001... < 1362 and so on forever

So imagine you took *all* the cube roots instead of just one. You get a sequence which goes ∛2 ∛∛11 ∛∛∛1361 ∛∛∛∛2521008887 ...

Or to put it another way 1.25992104989... 1.30529988079... 1.30637788382... 1.30637788386... ... We have essentially proved that this sequence converges to a value, which we'll call A