---In primenumbers@yahoogroups.com, <yuri1777@...> wrote :

http://scribd.com/doc/168629586/Exact-Recurrence-Formula-for-Prime-Numbers

> However, this formula cannot be used for direct

> calculations due to rapidly growing huge precision

> required.

I like your use of a truncated Euler product and a truncated

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).

David