25585Re: Exact Recurrence Formula for Prime Numbers
- Jul 23, 2014---In email@example.com, <yuri1777@...> wrote :
> However, this formula cannot be used for directI like your use of a truncated Euler product and a truncated
> calculations due to rapidly growing huge precision
sum for zeta(4*prime(n)) to produce this impractical formula
for prime(n+1) given all of the first n primes. Your
choice of the multiplier 4 in the argument of zeta neatly
exploits Betrand's observation, proven by Chebyshev.
"Chebyshev said it, but I'll say it again;
There's always a prime between n and 2n."
Very neat, though utterly useless (as you acknowledge).
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