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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 fiftysecond prime number (I think).
Yes, there are many such formulasbut 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. 179212].
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 primesthis 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*(j1)!^{2}/j)
/ sin^{2}(pi/j).
Minác (unpublished, proof in [Ribenboim95,
p181]) set
pi(n) = (sum from j=2 to n) [ ((j1)!
+ 1)/j  [(j1)!/j] ].
Hardy and Wright set pi(1) = 0, pi(2) = 1, and then [HW79
p414]
pi(n) = 1 + (sum from j=3 to n) ( (j2)!
 j[(j2)!/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} ]
