# Sum of reciprocals of integers minus primes

For any integer $m>2$, let $P_m$ be the set of primes less than $m$, and let $$f(m) = \sum\limits_{p \in P_m} \frac{1}{m-p}.$$ For example, $f(3)=\frac{1}{3-2}=1$, $f(4)=\frac{1}{4-2}+\frac{1}{4-3}=\frac{3}{2}$, and so on.

The question is to estimate $I=\inf\limits_{m>2} f(m)$.

A simple Mathematica calculation shows that $f(m)\geq f(223)\approx 0.60178$ for all $m$ up to $10,000$. It is true that $I>0$? Is $I>0.5$? Is $I=f(223)$?

• Likely yes. The sum seems not far from log(log m) - log (log (q)) where q is the distance between m and the next smallest prime. If you can prove it easily, it would establish a sub optimal but impressive upper bound on the size of prime gaps. Gerhard "Looking Backwards To Move Forward" Paseman, 2018.06.08. – Gerhard Paseman Jun 8 '18 at 15:46
• Upon reflection (pun intended), there are fewer terms with small denominator, so it is possible there is an upper bound to all the sums. This is going to require more thought. Gerhard "Backward May Be Wrong Direction" Paseman, 2018.06.08. – Gerhard Paseman Jun 8 '18 at 15:57
• Yes, $f(m)$ is in fact bounded. The contribution from primes less than $m/2$ is trivially at most $1$ (even if all such integers were prime). For primes between $m/2$ and $m$, if all integers in that range were prime then we'd get a contribution of $\log(m/2)$ or so; but at most a proportion $1/log(m/2)$ of those numbers are prime, so this contribution will be bounded as well (one can make this precise with partial summation). – Greg Martin Jun 8 '18 at 17:08
• @GregMartin: I think that $f(m)$ tends to infinity. Plausibly, any constant $c>0$ is admissible in my argument below, and $\liminf f(m)\geq\log(1/c)$ by that argument. The point is that the primes very close to $m$ give a bigger contribution than one would expect. – GH from MO Jun 8 '18 at 21:51

We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.
If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have $$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$ Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have $$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$ Here, $o(1)$ is meant as $m$ tends to infinity. Therefore, $$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$ As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.
Added. By a variant of the above argument, one can get an explicit lower bound for all $m$ (rather than for $m>m_0$) using an explicit prime number theorem in short intervals (in place of Huxley's result). For example, one can use the work of Dudek (as explained in this earlier MO post of mine) to get $$\sum_{x<p\leq x+3x^{2/3}}\log p>0.0006x^{2/3},\qquad x>\exp(8\times 10^{14}).$$
• : could this question lead to a solid numerical evidence for the truth of a weakened form of RH ? For example, would an effective value of $c$ less than $7/12$ follow from a proof that $I>f(223)$? – Sylvain JULIEN Jun 8 '18 at 19:55
• @SylvainJULIEN: I don't think so. It would be a fantastic achievement to bring down Huxley's $7/12$ to $1/2$, but I don't know any serious implication for the Riemann hypothesis. Likewise, Huxley's $7/12$ (the fact that it is less than $1$) does not seem to imply a quasi Riemann hypothesis (that the real parts of the Riemann zeta zeros are bounded away from $1$). – GH from MO Jun 8 '18 at 19:58
• @Bogdan: There is an explicit version of the quoted result for $c=2/3$. This yields an explicit (although very small) lower bound for all $m$ (upon noting that one can also get a simple lower bound from Bertrand's postulate). See the "Added" section in my post. I leave it to you to work out the details. – GH from MO Jun 11 '18 at 14:45