Explicit Stability for Mills-Type Prime-Generating Constants

1. Introduction

In his classical paper [1], Mills showed:

There exists an absolute constant $A>1$ such that $\lfloor A^{3^n}\rfloor$ is prime for all $n\ge 0$.

Mills’ argument rests on an analytic statement about primes in short intervals (Ingham [2]); the existence of A is unconditional, but the proof is neither explicit nor computational. Later authors studied the computability of such a constant under explicit hypotheses on prime gaps, for example under the Riemann Hypothesis or using Baker–Harman–Pintz [3]. A representative modern reference is Caldwell–Cheng [4].

Our goal here is somewhat orthogonal: we assume one explicit hypothesis on the existence of primes between consecutive c-th powers and then push all the algebraic consequences as far as possible, writing down stability constants and a finite algorithm.

Hypothesis $\left(H_c\right)$

Fix $c>1$.

Definition 1  

(Prime-between-powers hypothesis). We say that $\left(H_c\right)$ holds if there exists $X_0\left(c\right)\ge 1$ such that for every real $x\ge X_0\left(c\right)$ the open interval

$$(x^c,{(x+1)}^c)$$

contains at least one prime number.

This is exactly the property one needs to iterate the Mills construction with exponent c. In particular, once $\left(H_c\right)$ is assumed, for every sufficiently large prime $p_n$ we can choose a prime

$$p_{n+1}\in \left(p_n^c,{(p_n+1)}^c\right).$$

We emphasize: this is where all the analytic depth lives. The rest of the paper is algebraic.

2. Prime Chains and a Telescoping Formula

Assume $\left(H_c\right)$. Fix a starting prime $p_1\ge 2$ and define inductively

$$p_{n+1}\in \left(p_n^c,{(p_n+1)}^c\right)\cap \mathbb{P}(n\ge 1),$$

(1)

which is non-empty for all sufficiently large n by $\left(H_c\right)$. We call ${\left(p_n\right)}_{n\ge 1}$ a Mills-type prime chain (with exponent c).

Lemma 1  

(Telescoping identity for $logA$). Let $\left(p_n\right)$ be a Mills-type prime chain with exponent $c>1$. Then there exists a real number $A>1$ such that

$$p_n=⌊A^{c^n}⌋(n\ge 1),$$

and $logA$ can be written as

$$logA={\displaystyle \frac{log{p}_1}{c}}+\sum _{n=1}^{\infty }{\displaystyle \frac{1}{c^{n+1}}}log\left({\displaystyle \frac{p_{n+1}}{p_n^c}}\right).$$

(2)

Moreover the series in (2) converges absolutely.

Proof. 

The proof is the classical one in Mills’ construction, but we sketch it for completeness. By (1),

$$1<{\displaystyle \frac{p_{n+1}}{p_n^c}}<{\left(1+{\displaystyle \frac{1}{p_n}}\right)}^c,$$

so the deviation from an exact c-th power is small. As in [1], there is a unique $A>1$ such that $A^{c^n}\in [p_n,p_n+1)$ for all n. Taking logs,

$$logA={\displaystyle \frac{1}{c^n}}log(p_n+{\theta }_n),0\le {\theta }_n<1.$$

Writing the same identity for $n+1$ and subtracting the two expressions produces the telescoping series (2). The upper bound ${(1+1/p_n)}^c=1+O(1/p_n)$ implies the summand is $O(1/p_n)$, and since $p_n\to \infty$ the series converges absolutely. □

Remark 1.  

Formula (2) is the key to everything that follows: it allows us tocomparetwo chains term by term, and it allows us toboundthe tail to get a finite algorithm.

3. A Double-Logarithmic Line

The growth of $\left(p_n\right)$ is very regular when seen through two logarithms.

Proposition 1  

(Double-log line). Let $\left(p_n\right)$ be as above. Then

$$loglog{p}_n=nlogc+loglogA+{\epsilon }_n,$$

(3)

with an error term satisfying

$$-{\displaystyle \frac{2}{p_{n}log{p}_n}}\le {\epsilon }_n\le {\displaystyle \frac{1}{p_{n}log{p}_n}}.$$

(4)

In particular ${\epsilon }_n\to 0$ as $n\to \infty$, so the graph of $loglog{p}_n$ is asymptotically a straight line of slope $logc$.

Proof. 

From the definition of A we have

$$p_n\le A^{c^n}<p_n+1.$$

Taking logs,

$$log{p}_n\le c^{n}logA\le log(p_n+1).$$

Now take logs again. For the right inequality we write

$$\begin{array}{c}loglog(p_n+1)=log\left(log{p}_n+log(1+1/p_n)\right)=log\left(log{p}_n(1+\frac{log(1+1/p_n)}{log{p}_n})\right)=loglog{p}_n+log\left(1+{\displaystyle \frac{log(1+1/p_n)}{log{p}_n}}\right).\end{array}$$

Since $log(1+1/p_n)\le 1/p_n$, the second term is $\le 1/(p_{n}log{p}_n)$. Thus

$$loglog(p_n+1)\le loglog{p}_n+{\displaystyle \frac{1}{p_{n}log{p}_n}}.$$

A similar Taylor expansion for $loglog{p}_n\ge loglog(p_n+1)-{\displaystyle \frac{2}{p_{n}log{p}_n}}$ (using the mean value theorem on $loglogx$ in $[p_n,p_n+1]$) gives the lower bound. Since the middle term

$$log(c^{n}logA)=nlogc+loglogA$$

is squeezed between these two, we obtain (3)–(4). □

4. Explicit Stability of the Constant

We now quantify the effect of choosing a different prime at some level.

Theorem 1  

(Explicit stability). Fix $c>1$ and assume $\left(H_c\right)$. Let ${\left(p_n\right)}_{n\ge 1}$ and ${\left(q_n\right)}_{n\ge 1}$ be two Mills-type chains with the same exponent c, and assume

$$p_k=q_k\mathrm{for}1\le k\le N.$$

Let $A_p$ and $A_q$ be the corresponding constants given by (2). Then

$$|log{A}_p-log{A}_q|\le {\displaystyle \frac{c}{c^{N+1}{p}_N}}+{\displaystyle \frac{c}{(c-1)c^{N+1}{p}_N^c}}.$$

(5)

In particular

$$|log{A}_p-log{A}_q|\ll {\displaystyle \frac{1}{c^{N+1}{p}_N}},$$

with an implied constant depending only on c.

Proof. 

Subtract the two series (2):

$$log{A}_p-log{A}_q=\sum _{n=N}^{\infty }{\displaystyle \frac{1}{c^{n+1}}}\left[log\left({\displaystyle \frac{p_{n+1}}{p_n^c}}\right)-log\left({\displaystyle \frac{q_{n+1}}{q_n^c}}\right)\right],$$

since all terms with $n<N$ cancel. For $n=N$ we use that

$${\displaystyle \frac{p_{N+1}}{p_N^c}},{\displaystyle \frac{q_{N+1}}{p_N^c}}\in \left(1,{\left(1+{\displaystyle \frac{1}{p_N}}\right)}^c\right),$$

so by the mean value theorem for log on that interval,

$$\left|log\left({\displaystyle \frac{p_{N+1}}{p_N^c}}\right)-log\left({\displaystyle \frac{q_{N+1}}{p_N^c}}\right)\right|\le clog\left(1+{\displaystyle \frac{1}{p_N}}\right)\le {\displaystyle \frac{c}{p_N}}.$$

This gives the first term

$${\displaystyle \frac{1}{c^{N+1}}}·{\displaystyle \frac{c}{p_N}}={\displaystyle \frac{c}{c^{N+1}{p}_N}}.$$

For $n>N$, write $n=N+j$ with $j\ge 1$. Both chains live, step by step, in intervals of the form

$$\left(p_{N+j-1}^c,{(p_{N+j-1}+1)}^c\right),\left(q_{N+j-1}^c,{(q_{N+j-1}+1)}^c\right).$$

Inductively, $p_{N+j-1}\ge p_N^{c^{j-1}}$ and likewise for $q_{N+j-1}$. Since $c>1$, we have $c^j\ge c$ for every $j\ge 1$, hence

$$p_N^{c^j}\ge p_N^c.$$

Therefore

$$\left|log\left({\displaystyle \frac{p_{N+j}}{p_{N+j-1}^c}}\right)-log\left({\displaystyle \frac{q_{N+j}}{q_{N+j-1}^c}}\right)\right|\le {\displaystyle \frac{c}{p_{N+j-1}}}\le {\displaystyle \frac{c}{p_N^{c^{j-1}}}}\le {\displaystyle \frac{c}{p_N^c}}.$$

Thus the tail is bounded by

$$\sum _{j=1}^{\infty }{\displaystyle \frac{1}{c^{N+1+j}}}·{\displaystyle \frac{c}{p_N^c}}={\displaystyle \frac{c}{c^{N+1}{p}_N^c}}\sum _{j=1}^{\infty }{\displaystyle \frac{1}{c^j}}={\displaystyle \frac{c}{c^{N+1}{p}_N^c}}·{\displaystyle \frac{1}{c-1}}.$$

Adding this to the $n=N$ term yields (5). □

Remark 2.  

This version works for every $c>1$, including $1<c<2$. The price we pay is that the second term in (5) now decays like $p_N^{-c}$ instead of $p_N^{-2}$, but this is still enough for the numerical procedure because $p_N$ grows doubly-exponentially along the chain.

5. A Four-Step Certifiable Procedure

We now restate the numerical part in the light of thm:stability.

Step 1. 

Fix $c>1$ and generate a chain. Under $\left(H_c\right)$ choose a prime $p_1$ and for $n=1,2,\dots$ choose $p_{n+1}$ to be any prime in $({\left(p_n\right)}^c,{(p_n+1)}^c)$.

Step 2. 

Form the partial sum

$$S_N:={\displaystyle \frac{log{p}_1}{c}}+\sum _{n=1}^N{\displaystyle \frac{1}{c^{n+1}}}log\left({\displaystyle \frac{p_{n+1}}{p_n^c}}\right).$$

Step 3. 

Bound the tail. By thm:stability,

$$0\le logA-S_N\le {\displaystyle \frac{c}{c^{N+1}{p}_N}}+{\displaystyle \frac{c}{(c-1)c^{N+1}{p}_N^c}}.$$

Thus, given a tolerance $\tau >0$, it suffices to choose N so that the right-hand side is $<\tau$.

Step 4. 

Exponentiate. Output $A\approx exp\left(S_N\right)$. Since we have a bound on $logA-S_N$, we also control the relative error in A itself.

6. What Is New and What Is Not

For the convenience of the reader (and a potential referee) we separate the ingredients.

• Not new. The general scheme “a prime in each short interval ⇒ a constant whose powers give primes” is Mills’ [1], with many conditional refinements (for example [4]). We do not claim novelty there.
• New here.

  (i)

  The telescoping identity (2) is written in a form tailored to compare two chains.

  (ii)

  The stability theorem thm:stability gives an explicit inequality with constants depending only on c; this justifies taking finite prefixes.

  (iii)

  The four-step algorithm is stated together with its tail bound, so the whole method is genuinely “fully explicit” once $\left(H_c\right)$ is granted.