An inequality related to
the sieve of Eratosthenes

The sieve of Eratosthenes removes the multiples of the primes $p\leq y$ from the set of positive integers $n\leq x$. Let $\Phi(x,y)$ denote the number of integers remaining. Answering a question of Ford, the first-named author [7] recently proved the following theorem.

If $y>\sqrt{x}$, then $\Phi(x,y)=\pi(x)-\pi(y)+1$ (where $\pi(t)$ is the number of primes in $[1,t]$), and so by the prime number theorem, Theorem A is essentially best possible when $x^{1-\epsilon}<y<\epsilon x$. When $y\leq\sqrt{x}$, there is a long history in estimating $\Phi(x,y)$, and in particular, we have the following theorem, essentially due to Buchstab (see [13, Theorem III.6.4]).

The Buchstab function $\omega(u)$ is defined as the unique continuous function on $[1,\infty)$ such that

Below is a graph of $\omega(u)$ for $u\in[1,8]$ generated by Mathematica.

It is known that $\lim_{u\to\infty}\omega(u)=e^{-\gamma}=0.561459483566885\dots$ and that $\omega(u)$ oscillates above and below its limiting value infinitely often. The minimum value of $\omega(u)$ on $[2,\infty)$ is $1/2$ at $u=2$ and the maximum value $M_{0}$ is $0.567143290409783\dots$, occurring at $u=2.76322283417162\dots$. In particular, it follows from Theorem B that if $c>M_{0}$ and $y\leq\sqrt{x}$ with $y$ sufficiently large depending on the choice of $c$, that $\Phi(x,y)<cx/\log y$. In addition, using an inclusion–exclusion argument plus the fact that the Mertens product $\prod_{p\leq y}(1-1/p)<M_{0}/\log y$ for all $y\geq 2$, the inequality $\Phi(x,y)<cx/\log y$ can be extended to all $2\leq y\leq\sqrt{x}$, but now with $x$ sufficiently large depending on $c$.

In light of Theorem A and given that $\Phi(x,y)$ is a fundamental (and ancient) function, it seems interesting to try and make these consequences of Theorem B numerically explicit. We prove the following theorem.

To prove this we use some numerically explicit estimates of primes due to Rosser–Schoenfeld, Büthe, and others. In addition we use a numerically explicit version of the upper bound in Selberg’s sieve.

Theorem B itself is also appealing. It provides a simple asymptotic formula for $\Phi(x,y)$ as $y\to\infty$ which is applicable in a wide range. Writing

one may attempt to establish numerically explicit lower and upper bounds for $\Delta(x,y)$ in the range $y\leq\sqrt{x}$ for suitably large $y\geq y_{0}$, where $y_{0}\geq 2$ is some numerically computable constant. More precisely, de Bruijn [3] essentially showed that for any given $\epsilon>0$, one has

for all $x\geq y\geq 2$, where

Recently, the first-named author [8] proved numerically explicit versions of this result applicable for $y$ in wide ranges.

Let $\pi(x)$ denote, as usual, the number of primes $p\leq x$. Let

where the principal value is taken for the singularity at $t=1$. There is a long history in trying to find the first point when $\pi(x)\geq{\rm li}(x)$, which we now know is beyond $10^{19}$. We prove a lemma based on this research.

After checking for $x\leq 10$, we remark that an immediate corollary of (2) is the inequality

For small values of $y\geq 2$ we can do a complete inclusion–exclusion to compute $\Phi(x,y)$. Let $P(y)$ denote the product of the primes $p\leq y$. We have

As a consequence, we have

We illustrate how this elementary inequality can be used in the case when $\pi(y)=5$, that is, $11\leq y<13$. Then the product in (4) is $16/77<.207793$. The remainder term in (4) is 16. And we have

when $x\geq 613$. There remains the problem of dealing with smaller values of $x$, which we address momentarily. We apply this method for $y<71$.

Pertaining to Table 1, for $x$ beyond the “$x$ bound” and $y$ in the given interval, we have $\Phi(x,y)<.6x/\log y$. The column “max” in Table 1 is the supremum of $\Phi(x,y)/(x/\log y)$ for $y$ in the given interval and $x\geq y^{2}$ with $x$ below the $x$ bound. The max statistic was computed by creating a table of the integers up to the $x$ bound with a prime factor $\leq y$, taking the complement of this set in the set of all integers up to the $x$ bound, and then computing $(j\log p)/n$ where $n$ is the $j$th member of the set and $p$ is the upper bound of the $y$ interval. The max of these numbers is recorded as the max statistic.

As one can see, for $y\geq 3$ the max statistic in Table 1 is below $.6$. However, for the interval $[2,3)$ it is above $.6$. One can compute that it is $<.6$ once $x\geq 10$.

This method can be extended to larger values of $y$, but the $x$ bound becomes prohibitively large. With a goal of keeping the $x$ bound smaller than $3\times 10^{7}$, we can extend a version of inclusion-exclusion to $y<241$ as follows.

First, we “pre-sieve” with the primes 2, 3, and 5. For any $x\geq 0$ the number of integers $n\leq x$ with $\gcd(n,30)=1$ is $(4/15)x+r$, where $|r|\leq 14/15$, as can be easily verified by looking at values of $x\in[0,30]$. We change the definition of $P(y)$ to be the product of the primes in $(5,y]$. Then for $y\geq 5$, we have

However, it is better to use the Bonferroni inequalities in the form

say, where $\nu(d)$ is the number of distinct prime factors of $d$. (We remark that the expression $b(y)$ could be replaced with $\frac{14}{15}b(y)$.) The inner sums in $s(y)$ can be computed easily using Newton’s identities, and we see that

We have verified that this $x$ bound is smaller than $30{,}000{,}000$ for $y<241$ and we have verified that $\Phi(x,y)<.6x/\log y$ for $x$ up to this bound and $y<241$.

This completes the proof of Theorem 1 for $y<241$.

In this section we prove Theorem 1 in the case that $u=\log x/\log y\geq 7.5$ and $y\geq 241$. Our principal tool is a numerically explicit form of Selberg’s sieve.

Let $\mathcal{A}$ be a set of positive integers $a\leq x$ and with $|\mathcal{A}|\approx X$. Let $\mathcal{P}=\mathcal{P}(y)$ be a set of primes $p\leq y$. For each $p\in\mathcal{P}$ we have a collection of $\alpha(p)$ residue classes mod $p$, where $\alpha(p)<p$. Let $P=P(y)$ denote the product of the members of $\mathcal{P}$. Let $g$ be the multiplicative function defined for numbers $d\mid P$ where $g(p)=\alpha(p)/p$ when $p\in\mathcal{P}$. We let

We define $r_{d}(\mathcal{A})$ via the equation

The thought is that $r_{d}(\mathcal{A})$ should be small. We are interested in $S(\mathcal{A},\mathcal{P})$, the number of those $a\in\mathcal{A}$ such that $a$ is coprime to $P$.

We will use Selberg’s sieve as given in [9, Theorem 7.1]. This involves an auxiliary parameter $D<X$ which can be freely chosen. Let $h$ be the multiplicative function supported on divisors of $P$ such that $h(p)=g(p)/(1-g(p))$. In particular if each $\alpha(p)=1$, then each $g(p)=1/p$ and $h(p)=1/(p-1)$, so $h(d)=1/\varphi(d)$ for $d\mid P$, where $\varphi$ is Euler’s function. Henceforth we will make this assumption (that each $\alpha(p)=1$). Let

where $\tau_{3}(d)$ is the number of ordered factorizations $d=abc$, where $a,b,c$ are positive integers. Selberg’s sieve gives in this situation that

Note that if $D\geq P^{2}$, then

so that $X/J=XV$. This is terrific, but if $D$ is so large, the remainder term $R$ in (5) is also large, making the estimate useless. So, the trick is to choose $D$ judiciously so that $R$ is under control with $J$ being near to $V^{-1}$.

Consider the case when each $|r_{d}(\mathcal{A})|\leq r$, for a constant $r$. In this situation the following lemma is useful.

To get a lower bound for $J$ in (5) we proceed as in [9, Section 7.4]. Recall that we are assuming each $\alpha(p)=1$ and so $h(d)=1/\varphi(d)$ for $d\mid P$.

Let

so that $I+J=V^{-1}$. Hence

so we want an upper bound for $IV$. Let $\varepsilon$ be arbitrary with $\varepsilon>0$. We have

and so, assuming each $\alpha(p)=1$,

In particular, if $y\geq 241$ and each $r_{d}(\mathcal{A})\leq r$, then

We shall choose $D$ so that the remainder term is small in comparison to $XV$, and once $D$ is chosen, we shall choose $\varepsilon$ so as to minimize $f(D,\mathcal{P},\varepsilon)$.

Our “target” for $\Phi(x,y)$ is $.6x/\log y$. We choose $D$ here so that our estimate for the remainder term is 1% of the target, namely $.006x/\log y$. Thus, in light of Lemma 2, we choose

We have verified that for every value of $y\leq 500{,}000$ and $x\geq y^{7.5}$ that the right side of (9) is smaller than $.6x/\log y$. Note that to verify this, if $p,q$ are consecutive primes with $241\leq p<q$, then $S(\mathcal{A},\mathcal{P})$ is constant for $p\leq y<q$, and so it suffices to show the right side of (9) is smaller than $.6x/\log q$. Further, it suffices to take $x=p^{7.5}$, since as $x$ increases beyond this point with $\mathcal{P}$ and $\varepsilon$ fixed, the expression $f(D,\mathcal{P},\varepsilon)$ decreases. For smaller values of $y$ in the range, we used Mathematica to choose the optimal choice of $\varepsilon$. For larger values, we let $\varepsilon$ be a judicious constant over a long interval. As an example, we chose $\varepsilon=.085$ in the top half of the range.

As in the discussion above we have a few choices to make, namely for the quantities $D$ and $\varepsilon$. First, we choose $x=y^{7.5}$, since the case $x\geq y^{7.5}$ follows from the proof of the case of equality. We choose $D$ as before, namely $.03x/(\log y)^{3}$. We also choose

Our goal is to prove a small upper bound for $f(D,\mathcal{P},\varepsilon)$ given in (7). We have

We treat the two sums separately. First, by Rosser–Schoenfeld [12, Theorems 9, 20], one can show that

for all $y\geq 2$, so that

for $y\geq 7$. For the second sum we have

At this point we use (2) , so that

and so

There are a few things to notice, but we will not need them. For example, ${\rm li}(y^{2\varepsilon})={\rm li}(e^{2})$ and ${\rm li}(11^{2\varepsilon})\approx\log(11^{2\varepsilon}-1)+\gamma$.

Let $S(y)$ be the sum of the right side of (10) and $1+\beta_{0}$ times the right side of (11). Then

The expression $XV$ in (9) is

We know from [10] that this product is $<e^{-\gamma}/\log y$ for $y\leq 2\times 10^{9}$, and for larger values of $y$, it follows from [6, Theorem 5.9] (which proof follows from [6, Theorem 4.2] or [2, Corollary 11.2]) that it is $<(1+2.1\times 10^{-5})e^{-\gamma}/\log y$. We have

We have verified that $(1-D^{-\varepsilon}e^{S(y)})^{-1}$ is decreasing in $y$, and that at $y=500{,}000$ it is smaller than $1.057$. Thus, (12) implies that

This concludes the case of $u\geq 7.5$.

In this section we prove that $\Phi(x,y)<.57163x/\log y$ when $u\in[2,3)$, that is, when $y^{2}\leq x<y^{3}$.

For small values of $y$, we calculate the maximum of $\Phi(x,y)/(x/\log y)$ for $y^{2}\leq x<y^{3}$ directly, as we did in Section 3 when we checked below the $x$ bounds in Table 1 and the bound $3\times 10^{7}$. We have done this for $241\leq y\leq 1100$, and in this range we have

Suppose now that $y>1100$ and $y^{2}\leq x<y^{3}$. We have

Indeed, if $n$ is counted by $\Phi(x,y)$, then $n$ has at most 2 prime factors (counted with multiplicity), so $n=1$, $n$ is a prime in $(y,x]$ or $n=pq$, where $p,q$ are primes with $y<p\leq q\leq x/p$.

Let $p_{j}$ denote the $j$th prime. Note that

Thus,

and so

where

We use Lemma 1 on various terms in (14). In particular, we have (assuming $y\geq 5$)

Via partial summation, we have

For $1100\leq t\leq 10^{4}$ we have checked numerically that

where $B=.261497\dots$ is the Meissel–Mertens constant. Further, for $10^{4}\leq t\leq 10^{6}$,

(The lower bounds here follow as well from [12, Theorem 20].) It thus follows for $1100\leq y\leq 10^{4}$ that