I'm reading this (very short, 1 page long) paper by W.H. Mills where he determines that there exists a real number $A$ such that $f(n) = \lfloor {A^3}^n \rfloor$ is a prime number for all positive integers $n$.

In the paper before Theorem 1, he has this sequence $\{P_n\}_{n=0}^{\infty}$ with the property

$$P_n^3 < P_{n+1} < (P_n + 1)^3 - 1. \hspace{20pt} (*)$$

He defines a new sequence $\{u_n\}_{n=0}^{\infty}$ by $u_n = P_n^{3-n}$ and proceeds to prove things about that sequence. In particular he states (without proof) that $u_{n+1} > u_n$ (which can also be written $P_{n+1}^{3-n-1} > P_n^{3-n}$).

I am trying to justify the step and it is turning out more difficult than I thought it would be. I tried using $(*)$ in the following way:

$$P_{n+1}^{3-n-1} > \left( P_n^3 \right)^{3-n-1} = P_n^{9-3n-3} = P_n^{6-3n}.$$

To finish, I would need the inequality $P_n^{6-3n} > P_n^{3-n}$. I can see how to get $P_n^{6-3n} > P_n^{3-3n}$, but the sentence $P_n^{3-3n} > P_n^{3-n}$ is clearly false, so this will not work.

This indicates to me that I should start from the other side of $(*)$. If you try it using the fact that for $x>0$, $(x+1)^c > x^c+x^{c-1}+1$, this is what happens:

$$\begin{array}{} (p_{n+1}+1)^{3-n} &> p_{n+1}^{3-n} + p_{n+1}^{3-n-1} + 1 \\ &> \left( p_n^{3-n} \right)^3 + \left( p_n^{3-n-1} \right)^3, \end{array}$$

then I would like to conclude $> p_n^{3-n}$, but it is not going to be true since $p_n^{3-n} < 1$ implies $\left( p_n^{3-n} \right)^3 < p_n^{3-n} < 1$.

What am I not seeing?


1 Answer 1


Unfortunate about the original typesetting. In various places where the exponent is written as $3-n,$ it should actually be $$3^{-n}.$$

So, from the basic $P_{n+1} > P_n^3,$ we get $P_n > P_0^{3^n}.$ So how do we get something of modest size? $$ P_n^{3^{-n}} > \left( P_0^{3^n} \right)^{3^{-n}} = P_0^{\;3^n \cdot 3^{-n}} = P_0. $$

Alright, just confirming, when it says exponent $3-n-1$ it should be $$ 3^{-n-1} $$

Your original problem with $u_{n+1}$ and $u_n$ is really $$ u_{n+1} = P_{n+1}^{3^{-n-1}} > \left( P_n^3 \right)^{3^{-n-1}} = P_n^{\; 3 \cdot 3^{-n-1} } = P_n^{3^{-n}} = u_n $$

  • $\begingroup$ Simple and sweet. Thanks! $\endgroup$
    – tomcuchta
    Jan 9, 2012 at 1:17
  • $\begingroup$ Aaargh, that's what was bugging me for a while... and in the end it turns out to be a typesetting problem $\endgroup$
    – Klangen
    May 31, 2017 at 13:45

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