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If denote the prime numbers, then there exists a constant

and a sequence

such that is the nth prime. It can be shown that this recurrence relation generates all the prime numbers -- however, the complexity of this problem (as is often the case with prime-generating constants) is determining the value of to a sufficiently high degree of accuracy. The proof of this result uses Bertrand's postulate.Here, I've used to denote the floor of x (you can think of that as 'rounding down x to the nearest whole number') and to denote the fractional part of x. So in other words, we'd have and .The exact value of can be represented as an infinite sum:You need about 25 terms in the series above to get all the primes less than 100, for example.

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