# Is the reverse of a previous prime always smaller?

For $$n \geq 2$$, let $$r(n)$$ be the previous prime to $$n$$; i.e., the largest prime strictly less than $$n$$. For example, $$r(3) = 2$$, $$r(10) = 7$$, and so on.

I have noticed that, usually, if you reverse the digits of $$r(n^3)$$ in base $$n$$, you obtain a smaller number. For example, $$r(5^3) = 113 = 423_5$$, and reversing the digits base $$5$$ gives $$324_5 < 423_5$$. This also seems to hold for powers other than $$3$$. Is this true in general? (For sufficiently large $$n$$, say.) If so, why? Is there a simple proof?

Here are three examples to show that, somehow, $$r(n^3)$$ is necessary: $$47$$ is a prime less than $$10^3$$ whose second digit is larger than its first; $$999$$ is closer to $$1000$$ than $$r(1000)$$, but reversing it in base $$10$$ gives the same number; $$414_5 = 109$$ is a prime less than $$5^3$$ which is fixed under reversal of base 5 digits.

Here's the start of an argument for squares: Let $$p = r(n^2) = d_1 n + d_0$$, where $$d_k \in \{0, 1, 2, \dots, n - 1\}$$. We can't have $$d_1 = d_0$$ since $$p$$ is prime (omitting some details). If we could prove $$d_1 = n - 1$$, then this would imply $$d_0 < d_1$$, so reversing the digits of $$p$$ would produce a smaller number. We have $$d_1 = \lfloor p / n \rfloor$$, so it would suffice to prove $$n - 1 \leq \frac{p}{n} < n,$$ or $$1 - \frac{1}{n} \leq \frac{r(n^2)}{n^2} < 1.$$ Based on my question here, this last thing seems difficult to prove.

• Searched $n$ up to 1000000 only 2,3 prevail. previous prime must fit $n-1$ mod $n$ or fail. This follows from weak application of Bertrand's postulate. – user645636 Nov 24 '19 at 2:49
• @RoddyMacPhee Thanks for checking more powers! Can you say more about why the previous prime must be $n - 1 \mod n$? I don't see why yet. – Robert D-B Nov 24 '19 at 2:53
• $p>{ n\over 2}$ sorry. – user645636 Nov 24 '19 at 2:56

• $$n-1\leq {p\over n}\leq n\implies$$ $$(n-1)^2< n^2-n\leq p\leq n^2$$ making this condition, a restricted form of Legendre's conjecture( so part of an unsolved problem, it's actually Oppermann's conjecture in disguise).
• Via Bertrand's postulate, for an $$m$$ base $$n>2$$ digit number, you get $$d_m\geq d_0\geq \lfloor {n\over 2}\rfloor$$ .
• The cube case, becomes problematic on the Bertrand's postulate front. $$5^3<2(4^3)$$ , However most stronger heuristics, I feel, are likely to support it.