# What is a non-trivial upper bound on the $k$th prime below a given prime $p$?

Given a prime number $$p_0$$, by Bertrand's postulate we know that $$\begin{gather} p_1\ge\frac{p_0}{2}\\ p_2\ge\frac{p_1}{2}\ge\frac{p_0}{2^2}\\ \vdots\\ p_k\ge\frac{p_0}{2^k} \end{gather}$$ where $$p_1,p_2,\dots$$ are prime numbers immediately preceding $$p_0$$.

Question. Is there a similar nice upper bound for the $$p_i$$?

A trivial upper bound would be $$p_i\le p_0-2i$$, but that is trivial and bad in the sense that for large $$i$$ this doesn't give a good approximation (since one or two primes may differ by $$2$$, but $$10$$ consecutive primes can't all differ by $$2$$, there must be something better). I am not claiming that the above result due to Bertrand gives a good approximation, but that at least gives exponential bounds, unlike in this case.

There is no good upper bound for $$p_k$$ in the following sense: for every $$k$$, there exists a constant $$c_k>0$$ such that $$p_k>p_0-c_k$$ holds for infinitely many primes $$p_0$$. This was proved by Maynard (2013). You will find a concrete value for $$c_k$$ in Maynard's work, which was decreased by Polymath8b (2014). The best conjectured value would follow from the Hardy-Littlewood prime tuple conjecture.

Note also that the crude lower bound $$p_k>p_0/2^k$$ can be improved greatly. For example, Baker-Harman-Pintz (2000) proved that, for $$x$$ sufficiently large, there is always a prime in the interval $$[x-x^{21/40},x]$$. In particular, we have $$p_1>p_0-p_0^{21/40}$$ for all but finitely many primes $$p_0$$. This is still far from being optimal. The Riemann Hypothesis implies that the exponent can be replaced by any number exceeding $$1/2$$. Further, a conjecture of Cramér states that $$x^{21/40}$$ can be replaced by $$2(\log x)^2$$.

• Is the best (known or expected) upper bound of the form $p_k \leq p_1 -c k \log k$ for some explicit constant $c$? Commented Apr 11, 2021 at 15:09
• @WillSawin: Maynard (2013) gave $k^3e^{4k}$ for the constant, Polymath (2014) gave $ke^{(4-24/181)k}$. Assuming the prime $k$-tuple conjecture, one can take $(1+o(1))k\log k$ for the constant. Commented Apr 11, 2021 at 15:38
• Sure, but those are upper bounds, and the original question was about lower bounds. The lower bound for the size of an admissible $k$-tuple is something like $(1/2+ o(1) ) k \log k$, which also gives an (unconditional) lower bound for gaps. Commented Apr 11, 2021 at 15:57
• @WillSawin: I see your question now. It is clear that $p_k\leq p_1-h_k+k$, where $h_k$ is the smallest diameter of an admissible $k$-tuple, and even that $p_k\leq p_1-h_k$ when $p_1$ is large. The prime $k$-tuple conjecture implies that the last bound is sharp. Now the large sieve gives $h_k\geq(1/2+o(1))k\log k$, and a simple construction yields $h_k\leq(1+o(1))k\log k$. The upper bound is conjectured to be the actual size of $h_k$. Commented Apr 11, 2021 at 16:16