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
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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)
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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
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So imagine you took *all* the cube roots instead of just one. You get a sequence which goes ∛2 ∛∛11 ∛∛∛1361 ∛∛∛∛2521008887 ...
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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
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This value A has the property that if you cube it, and round down, you get a prime if you cube it twice, and round down, you also get a prime if you cube it three times, and round down, you also get a prime ... In fact ⌊ A^3^n ⌋ is prime for all n > 0
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To repeat for emphasis: there is a real number A such that no matter how many times you cube it, if you round down, you always get a prime! It is called Mills' constant and it is one of the best numbers https://en.wikipedia.org/wiki/Mills%27_constant …
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It is not known for sure that Mills' constant is 1.3063778838630806904686144926... but if the Riemann Hypothesis is true then that's it
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Can we use Mills' constant to generate primes? Nope, other way around, we can use primes to construct Mills' constant
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End of conversation
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