In the late forties Mills [Mills47] proved that there was a real number A>1 for which [] is always a prime (n = 1,2,3,...). The purpose of this short note is to prove the following theorem (from an exercise of Ellison [EE85, exercise 1.23]), for which Mills' theorem is a special case. As a bonus, this theorem shows there are infinitely many c's which each have infinitely many A's for which is always prime (there are actually uncountably many choices of A [Wright1954]. We then approximate the least A for the sequence of primes (with c=3) after this proof.

Theorem: Let S={a_{n}} be any sequence of integers satisfying the following property:

(1) there exists real numbers x_{o} and w with 0 < w < 1, for which the open interval (x,x+x^{w}) contains an element of S for all reals x > x_{o}.

Then for every real number c > min(1/(1-w),2) there is a number A for which is a subsequence of S.

Proof: Use (1) to define a subsequence {b_{n}} of S recursively by

(2) b_{1} = least member of S for which b_{1}^{c}>x_{o}.
(3) b_{n+1} = least member of S satisfying b_{n}^{c} < b_{n+1 }< b_{n}^{c} + b_{n}^{wc}.

Because c> 1/(1-w), and c> 2 (see the lemma below) we get the last two inequalities in the following revision of (3):

For all positive integers n we can raise this to the c^{-(n+1)}th power to get

showing that the sequence converges. Call its limit A. Finally,

so = b_{n}, the chosen subsequence of S--completing the proof.

For the sequence of primes property (1) is known to hold with w=7/12 and x_{o} sufficiently large, so we can follow Mills and take c = 3. Note that it is not necessary to start with the minimal prime satisfying (2) (b_{o} > x_{o}) as long as we can find the necessary primes satisfying (3) until we find one greater than x_{o} (then we know all of the following primes exists).

The problem is, that even though surely there is a prime between consecutive cubes of integers (greater than one), we can not yet prove it! The best effective lower bound may be as large as the one in Cheng's draft paper [Cheng2003a]: 10^{6000000000000000000}. But it is trivial to prove, that if the Riemann Hypothesis hold, then there always are primes between consecutive integer cubes, starting with with b_{o} = 2 [CC2005].

Mills did not give an explicit A, but it is traditional for authors to start with b_{o} = 2 (even though the way Mills' wrote his result, starting with 2 would have been ruled out). Starting with b_{o} = 2 and taking the minimal prime at each step gives us the minimal sequence (given RH):

For now (until an appropriate effective bound is found), the above number is conjectured to be the least Mills' constant, often called the Mills' number. There are of course infinitely many others (to see this start with any other prime, or note that if A works, so does A^{3}, or ...)

Though amusing, this type of formula is useless for determining primes because we need to know the primes determined before we find A (and the subsequence of primes represented is so small!)

We end this note by proving an inequality used in the proof above.

Lemma: If x> 1 and c> 2, then 1 + x^{c} + x^{c-1}< (1 + x)^{c}.

Proof: Dividing by x^{c} and replacing x with 1/x we arrive at the equivalent inequality

0 < (1 + x)^{c} - (1 + x + x^{c}). (0 < x< 1)

The inequality clearly holds when c = 2 (because it reduces to x> 0) and when x = 0. Now if x>0, differentiate the right side with respect to c to get

(1 + x)^{c} log(1+x) - x^{c} log(x)

which is clearly positive, so the inequality above holds for all c > 2.