Here we address the following frequently asked question.

Is there a formula for producing a specific prime number? Let
me give you an example. The formula is given the number 52. In return,
the formula produces 239, the fifty-second prime number (I think).

Yes, there are many such formulas--but they are of recreational use only
because they are very inefficient. Most are either ways of encoding the
list of primes or very clever counting arguments. I will give a couple example
below, but for more information start with chapter three "Are There Functions
Defining Prime Numbers?" of Ribenboim's text [Ribenboim95
pp. 179-212].

Method One: Encoding Primes

Let p_{n} be the nth prime. In 1952 Sierpinski
suggested we define a constant A as follows:

A = = 0.02030005000000070...

Then using the floor function [x] (the greatest integer less than
or equal to x) we have

p_{n} =

Hardy and Wright [HW79
p345] give a variant of this: Let r be an integer greater than one
and define a constant B as follows:

B =

then

p_{n} =

This type formula would only be of value if the necessary constant could
be found without first finding the primes--this may be possible,
but it seems unlikely.

Method Two: Counting Primes with Wilson's Theorem

First use one of these three methods to define pi(x). Willans (1964) used

pi(n) = (sum from j=2 to n) sin^{2}(pi*(j-1)!^{2}/j)
/ sin^{2}(pi/j).

Minác (unpublished, proof in [Ribenboim95,
p181]) set

pi(n) = (sum from j=2 to n) [ ((j-1)!
+ 1)/j - [(j-1)!/j] ].

Hardy and Wright set pi(1) = 0, pi(2) = 1, and then [HW79
p414]

pi(n) = 1 + (sum from j=3 to n) ( (j-2)!
- j[(j-2)!/j] ).

(for all n>2). Then we have (still using the floor function [x]):

nth prime = 1 + (sum from m=1 to j=2^{n})
[ [ n/(1 + pi(m)) ]^{1/n} ]