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Mill's formula
Thu 02 April 2020 — download

One of my favourite mathematical formula is Mills' formula, published in 1947 in the Bulletin of the American Mathematical Society, Volume 53, Number 6, p604, under the title "A prime-representing function", by W. H. Mills, in 1946:

%% d_n = \lfloor{} A^{3^{n}} \rfloor{} %%

With $$A$$ being a constant, and $$d_{n}$$ being prime for all integers $$n \geq 1$$.

Something that I find fascinating about this formula, is that it's (currently) completely useless, because there is no known way to compute the constant $$A$$ without finding the primes in the first place.

The paper is only one page long and is pretty accessible if you have a background in mathematics. Otherwise, keep on reading for the hand-holding blog post.

The proof relies on a proof by Albert Edward Ingham, published in his 12 page article On the difference between consecutive primes, published in The Quarterly Journal of Mathematics, Oxford Ser. vol. 8 in 1937. As a side note, it's interesting to see that the Oxford University Press is charging EUR €27.00 to access an article written 83 years ago (local mirror). Anyway, here is the formula:

%% \tag{1} p_{n+1} - p_{n} < Kp^{5/8}_{n} %%

Namely, the gap between two primes numbers $$p_{n}$$ and $$p_{n+1}$$ (called the prime gap) is lesser than $$K·p_{n}^{5/8}$$, with $$K$$ being a fixed positive integer.

This boundary has since been reduced to $$p_{n+1} - p_n < K p_n^{1051/1920}$$, by Charles J. Mozzochi, in his paper "On the Difference Between Consecutive Primes", published in the Journal of Number Theory Volume 24, Issue 2, Pages 181-187.

The first lemma of the paper is the following proposition:

If $$N$$ is an integer greater than $$K^{8}$$ there exists a prime $$p$$ such that $$N^3 < p < (N+1)^3 - 1$$.

Let $$p_n$$ be the greatest prime less than $$N^3$$. Then:

%% \begin{aligned} N^3 &\lt p_{n+1} \\ &\lt p_n + Kp_{n}^{5/8} \\ &\lt N^3 + K(N^3)^{5/8} \\ &\lt N^3 + KN^{15/8} \\ &\lt N^3 + N^2 \\ \tag{2} &\lt (N + 1 )^3 - 1 \end{aligned} %%

Let $$P_0$$ be a prime greater than $$K^8$$. Then by the lemma we can construct an infinite sequence of primes, $$P_0, P_1, P_2, ...$$, such that $$P_n^3 < P_{n+1} < (P_n + 1)^3 - 1$$. Let

%% \tag{3} u_n = P^{3 - n}_n, v_n = (P_n + 1)^{3-n}. %%

Then

%% u_n > v_n u_{n+1} = P_{n+1}^{3 - (n+1)} = P_{n+1}^{3 - n - 1} \tag{4} \implies u_{n+1} > u_n, %%

and

%% u_{n+1} = (P_{n+1} + 1)^{3 - (n + 1)} = (P_{n+1} + 1)^{3 - n - 1} \tag{5} \implies v_{n+1} < v_n. %%

It follows at once that the $$u_n$$ form a bounded monotone increasing sequence. Let $$ A = \lim_{n \to \infty} u_n $$.

Theorem: $$ \lfloor{} A^{3^{n}} \rfloor{}$$ is a prime-representing function.

Proof: From $$(4)$$ and $$(5)$$ it follows that $$u_n < A < v_n $$, or $$ P_n < A^{3^n} < P_{n}+1 $$. Therefore $$ \lfloor A^{3^n} \rfloor = P_n $$, and $$ \lfloor{} A^{3^{n}} \rfloor{} $$ is a prime-representing function.

Interestingly, Làszlò Tòth proved in 2017, based on the work of Kuipers and Ansari who generalised this result to $$ \lfloor A^{3^{n}} \rfloor $$, where $$ c ∈ R $$ and $$ C \geq 2.106 $$, that $$ \lceil B^{c^{n}} \rceil $$ is also a prime-representing function.