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FAQ: Is there a formula for the *n*th Prime?

### By Chris Caldwell

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 [Ribenboim95pp. 179-212].

### Method One: Encoding Primes

Let *p*_{n} be the *n*th 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 [HW79p345] 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 π(x). Willans (1964) used

π(n) = (sum fromj=2 ton) sin^{2}(pi*(j-1)!^{2}/j) / sin^{2}(pi/j).

Minác (unpublished, proof in [Ribenboim95, p181]) set

π(n) = (sum fromj=2 ton) [ ((j-1)! + 1)/j- [(j-1)!/j] ].

Hardy and Wright set π(1) = 0, π(2) = 1, and then [HW79p414]

π(n) = 1 + (sum fromj=3 ton) ( (j-2)! -j[(j-2)!/j] ).

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

nth prime = 1 + (sum fromm=1 toj=2^{n}) [ [n/(1 + π(m)) ]^{1/n}]