Every integer greater than $1$ factors uniquely into primes. That is the uniqueness theorem from a first algebra course. Here is a stranger one, about the primes as a set.

Take any collection of integers greater than $1$ in which no element divides another. Weigh each integer $a$ by $1/(a\log a)$. Among all such collections, finite or infinite, the primes are the heaviest.

That is the Erdős primitive set conjecture, in print by 1974, proved by Jared Duker Lichtman in 2022. The paper is here. I first met the statement in locked-down Boston and was sure it had to be true. What follows is a walk from “what does divide mean?” to the $\pi/4$ that finishes the argument. Every term is defined before it is used. The published paper has extra numerical bounds for small primes and a separate estimate for the prime $2$; we will say exactly where those enter, and we will not pretend the napkin is the paper.

The calculus we need is substitution in integrals, the chain rule, and one named derivative: $\frac{d}{du}\arctan u = \frac{1}{1+u^2}$, with $\arctan 1 = \pi/4$. Throughout, $\log$ means the natural logarithm (what a calculus book calls $\ln$). So $\log e = 1$, and $\log 8 = 3\log 2$.

• Divisibility
• Primitive sets
• Infinite sums
• The scale $f(A)$
• A private territory for each $a$
• The territories do not overlap
• Mertens’ theorem, and a uniform bound
• Compete slot by slot
• How close to prime?
• Copies, and the $\sqrt{v}$ packing
• Buckets, then $\pi/4$
• Odd primes beat their own composites
• The prime $2$
• Why $f(\mathcal{P})$ is finite
• What is still open

---

Divisibility

We write $a \mid b$, read “$a$ divides $b$”, when $b$ is an integer multiple of $a$: $b = ka$ for some integer $k$. So $3 \mid 12$ because $12 = 4\cdot 3$, and $5 \nmid 12$ because $12/5$ is not an integer. Every integer divides itself. The integer $1$ divides everything.

The greatest common divisor $\gcd(a,b)$ is the largest positive integer that divides both $a$ and $b$. Two integers are coprime when $\gcd(a,b)=1$: they share no prime factor. So $\gcd(8,12)=4$ and $\gcd(8,15)=1$.

The fundamental theorem of arithmetic says that every integer $n > 1$ factors uniquely as a product of primes, up to order:

\[n \;=\; p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r}, \qquad p_1 < p_2 < \cdots < p_r.\]

For $60 = 2^2 \cdot 3 \cdot 5$, the primes are $2, 3, 5$ with exponents $2, 1, 1$. Uniqueness is the whole point: there is no other way to write $60$ as a product of primes.

Unique factorization gives a lemma we will need when two territories try to occupy the same integer.

Euclid’s lemma. If $a$ divides a product $bc$ and $\gcd(a,b)=1$, then $a$ divides $c$.

Proof. Write $bc = ak$. Every prime in the factorization of $a$ appears on the left, hence in $b$ or in $c$. None of those primes appear in $b$, because $a$ and $b$ are coprime. So they all appear in $c$, with at least the same exponents. Thus $a$ divides $c$.

Three bookkeeping functions come from the factorization.

• $p(n)$ is the smallest prime factor of $n$. So $p(60) = 2$, $p(35) = 5$, and $p(7) = 7$.
• $P(n)$ is the largest prime factor of $n$. So $P(60) = 5$, $P(35) = 7$, and $P(7) = 7$. For a prime $q$ one has $p(q) = P(q) = q$.
• $\Omega(n)$ is the number of prime factors of $n$, counted with repetition. So $\Omega(60) = 2+1+1 = 4$, $\Omega(12) = \Omega(2^2\cdot 3) = 3$, and $\Omega(7) = 1$. (The twin function $\omega(n)$ counts distinct primes and is not used here: $\omega(12)=2$.)

If $a$ divides $b$ and $a \neq b$, then $b = ac$ for some integer $c > 1$, so $\Omega(b) = \Omega(a) + \Omega(c) > \Omega(a)$. Dividing strictly increases the number of prime factors.

---

Primitive sets

Definition. A set $A$ of integers greater than $1$ is primitive if no element of $A$ divides another element of $A$.

That is the whole definition. Draw the integers with an arrow from $a$ to $b$ whenever $a$ divides $b$ and $a \neq b$. A primitive set is a selection of dots with no arrow between any two of them.

The primes ${2, 3, 5, 7, 11, \ldots}$. The only positive divisors of a prime $p$ are $1$ and $p$, and $1$ is excluded from every primitive set. So no prime divides another prime.

${6, 10, 15}$. Check every pair: $6 \nmid 10$, $6 \nmid 15$, $10 \nmid 15$, and the three reverse directions. Nothing divides anything else. (As factorizations: $2\cdot 3$, $2\cdot 5$, $3\cdot 5$. Each pair shares a prime, but neither number is a multiple of the other.)

${6, 12, 15}$ is not primitive, because $6 \mid 12$.

All the integers in an interval $(x, 2x]$, for any $x \ge 1$. If $a$ and $b$ both sit in $(x, 2x]$ and $a$ divides $b$ with $b \neq a$, then $b \ge 2a > 2x$, so $b$ has already left the interval.

All integers with $\Omega(n) = k$, for a fixed $k$. If $a$ divides $b$ and $a \neq b$, then $\Omega(b) > \Omega(a)$, so two numbers with the same $\Omega$ cannot divide one another. The primes are the case $k = 1$. The case $k = 2$ is the numbers with exactly two prime factors (repeats allowed): ${4, 6, 9, 10, 14, 15, \ldots}$.

Even perfect numbers ${6, 28, 496, 8128, \ldots}$. Each has the Euclid–Euler form $2^{p-1}(2^p-1)$ with $2^p-1$ itself prime. The odd parts $2^p-1$ and $2^q-1$ divide one another only when $p$ divides $q$; for distinct primes $p,q$ that never happens, so these numbers do not divide one another.

graph BT
    2 --> 4
    2 --> 6
    2 --> 10
    3 --> 6
    3 --> 9
    3 --> 15
    5 --> 10
    5 --> 15
    4 --> 12
    6 --> 12
    7["7"]
    style 2 fill:#38bdf8,color:#000
    style 3 fill:#38bdf8,color:#000
    style 5 fill:#38bdf8,color:#000
    style 7 fill:#38bdf8,color:#000
    style 6 fill:#a855f7,color:#fff
    style 10 fill:#a855f7,color:#fff
    style 15 fill:#a855f7,color:#fff

Blue nodes ${2,3,5,7}$: one primitive set. Purple nodes ${6,10,15}$: another. You cannot take both $6$ and $12$ — there is an arrow between them.

Why did anyone define this?

A number $n$ is abundant if its proper divisors (the positive divisors except $n$ itself) add up to more than $n$. For $12$, those divisors are $1, 2, 3, 4, 6$, summing to $16 > 12$, so $12$ is abundant. A number is perfect if they add up to exactly $n$ (so $6 = 1+2+3$), and deficient otherwise. Non-deficient means abundant or perfect.

Davenport proved in the 1930s that the abundant numbers occupy a positive proportion of the integers. The original proof was analytic and heavy. Erdős found a shortcut: it is enough to understand the primitive non-deficient numbers — those that are not multiples of a smaller non-deficient number. Every non-deficient number is a multiple of one of those, so the primitive ones control the whole set.

Once the definition was on the table, it was more interesting than the application. Classic Erdős. The question became: among all primitive sets, which one is the “largest”?

Largest in what sense is the next section but one. Counting how many elements you have, or what proportion of the integers they occupy, is the wrong scale, and it is worth seeing why before we change scale.

The natural density of a set $S$, when the limit exists, is

\[d(S) \;=\; \lim_{N \to \infty} \frac{\#\{ n \in S : n \le N \}}{N}.\]

The even numbers have density $1/2$. A single interval $(x,2x]$ is a finite set, so its density is $0$ — it occupies about half the integers up to $2x$, and then as $N\to\infty$ it is diluted to nothing.

What one can do is string infinitely many such intervals together, with huge gaps between them. Pick $x_1 = 2$ and take $(2,4]$. Skip far enough that those three integers are a negligible fraction of the next $x_2$, then take $(x_2, 2x_2]$. Repeat. Two running proportions oscillate:

• Upper density is the $\limsup$ of the proportion up to $N$. At each $N = 2x_i$ you have just swallowed an interval of length $x_i$, so the proportion is close to $1/2$.
• Lower density is the $\liminf$. At each $N = x_{i+1}$, just before the next interval, you have diluted the previous harvest down toward $0$.

Besicovitch showed that by placing the intervals farther and farther out you can make the upper density of a primitive set arbitrarily close to $1/2$. Behrend and Erdős showed that the lower density is always $0$. Density oscillates too wildly to pick a winner. We need a scale that notices small integers more than large ones, and that stays finite even for infinite sets.

---

Infinite sums

We will compare primitive sets by adding infinitely many positive numbers. Two facts from calculus, and two applications, are the entire analytic toolkit.

The harmonic series diverges. Group the terms of $\sum 1/n$ into blocks whose lengths are powers of two:

\[1 + \frac12 + \underbrace{\Bigl(\frac13+\frac14\Bigr)}_{\ge \frac12} + \underbrace{\Bigl(\frac15+\cdots+\frac18\Bigr)}_{\ge \frac12} + \underbrace{\Bigl(\frac19+\cdots+\frac1{16}\Bigr)}_{\ge \frac12} + \cdots.\]

There are infinitely many blocks, each at least $1/2$, so the sum is infinite. We write $\sum 1/n = \infty$.

The integral test. If $g(x)$ is positive and decreasing, the series $\sum_{n \ge N} g(n)$ and the integral $\int_N^\infty g(x)\,dx$ either both settle to a finite number or both run off to infinity. The picture is that the sum is a staircase of rectangles of height $g(n)$ and width $1$, and the integral is the area under the curve. The curve sits under the staircase starting at $N$ and over the staircase starting at $N+1$, so they sandwich each other.

Two integrals we will use. Substitute $u = \log x$, so $du = dx/x$:

\[\int_3^{\infty} \frac{dx}{x\log x} \;=\; \int_{\log 3}^{\infty} \frac{du}{u} \;=\; \infty,\] \[\int_3^{\infty} \frac{dx}{x(\log x)^2} \;=\; \int_{\log 3}^{\infty} \frac{du}{u^2} \;=\; \frac{1}{\log 3} \;<\; \infty.\]

So $\sum 1/(n\log n)$ diverges, and $\sum 1/(n(\log n)^2)$ converges. One extra $\log$ in the denominator is the difference between “infinite” and “finite.” That borderline is where the whole subject lives.

A word on notation: we write $g(x) \sim h(x)$ to mean $g(x)/h(x) \to 1$ as $x \to \infty$. The two sides become indistinguishable as a percentage, even if they still differ by a large number.

---

The scale $f(A)$

For a set $A$ of integers greater than $1$, define

\[f(A) \;=\; \sum_{a \in A} \frac{1}{a \log a}.\]

This is the Erdős sum. Each integer $a$ is given a positive weight that gets smaller as $a$ gets larger, but slowly. (Reminder: $\log$ is $\ln$, so $\log 6 \approx 1.792$, not $\log_{10} 6$.)

A small set, computed by hand. For $A = {6, 10, 15}$:

\(\log 6 \approx 1.792, \quad 6\log 6 \approx 10.75, \quad \frac{1}{6\log 6} \approx 0.093,\) \(\log 10 \approx 2.303, \quad \frac{1}{10\log 10} \approx 0.043, \qquad \log 15 \approx 2.708, \quad \frac{1}{15\log 15} \approx 0.025.\)

So $f({6,10,15}) \approx 0.161$. Tiny.

The primes, the first few terms. Write $\mathcal{P}$ for the set of all primes, and $f(p)$ as shorthand for $1/(p\log p)$.

$p$	$f(p) = 1/(p\log p)$	running sum
$2$	$0.7213$	$0.7213$
$3$	$0.3034$	$1.0248$
$5$	$0.1243$	$1.1490$
$7$	$0.0734$	$1.2224$
$11$	$0.0379$	$1.2604$
$13$	$0.0300$	$1.2903$
$17$	$0.0208$	$1.3111$
$19$	$0.0179$	$1.3290$

The first two primes already contribute more than $1$. The first eight already contribute $1.33$, against $0.161$ for ${6,10,15}$. Henri Cohen computed the full sum:

\[f(\mathcal{P}) \;=\; 1.6366\ldots\]

(That the infinite sum settles at all is not obvious. We will prove it at the end, using that the $n$th prime is about $n\log n$. For now, the table plus Cohen’s computation is the score to beat.)

Theorem (Lichtman, 2022). For every primitive set $A$, \(f(A) \;\le\; f(\mathcal{P}) \;=\; 1.6366\ldots\)

Why this weight, and not another?

• Weighing by $1/a$ makes the primes score $\sum 1/p = \infty$. The numbers with $\Omega(n) = k$ score infinity too (the $k=2$ case is $\sum_{p\le q} 1/(pq)$, which contains $\frac12\bigl((\sum 1/p)^2 - \sum 1/p^2\bigr)$). The scale cannot tell the contestants apart.
• Weighing by $1/a^2$ makes everything converge too easily. The primes score only $\sum_p 1/p^2 \approx 0.452$, and a pile of small composites can compete. The scale no longer sees the primes as special.
• Weighing by $1/(a(\log a)^2)$ still lets the primes converge, but for a bad reason: the integral test already says $\sum_n 1/(n(\log n)^2)$ converges, so every set of integers has a finite score. Primitivity is no longer the constraint that keeps the sum finite, and the contest stops being about the divisibility structure.

The weight $1/(a\log a)$ sits on the knife-edge from the last section: $\sum_n 1/(n\log n)$ diverges, while $\sum_p 1/(p\log p)$ converges. The scale is sensitive enough to see the primes as a finite budget, but not so harsh that composites become irrelevant. Primitivity is exactly what makes the sum finite; that is Erdős 1935, and it is where the later proof begins.

There is also a calculus identity behind the weight. For any $a \ge 2$,

\[\int_1^{\infty} a^{-t}\,dt \;=\; \frac{1}{a\log a}.\]

To see it, write $a^{-t} = e^{-t\log a}$ and substitute $u = t\log a$, so $du = \log a\, dt$:

\[\int_1^{\infty} e^{-t\log a}\,dt \;=\; \frac{1}{\log a} \int_{\log a}^{\infty} e^{-u}\,du \;=\; \frac{1}{\log a}\cdot e^{-\log a} \;=\; \frac{1}{a\log a}.\]

So $1/(a\log a)$ is the $t$-average of the scores $a^{-t}$. We will not need this identity for the proof. It is here because it is where the weight comes from, and because it explains a warning: if you ask for the stronger comparison $\sum_{a \in A} a^{-t} \le \sum_p p^{-t}$ at every $t > 1$, the answer is no. That holds only for $t \ge \tau \approx 1.1403$. Shift the original weight slightly to $1/(a(\log a+h))$ and the primes stop winning as soon as $h \ge 1.04$. The theorem, once proved, is only just true.

Before asking which primitive set is heaviest, you need to know that every primitive set has a finite score, bounded by a constant that does not depend on the set. That is the 1935 theorem, and it is the next three sections.

---

A private territory for each $a$

The idea is to give each $a$ in a primitive set a block of integers that nobody else in the set is allowed to claim, then see that the blocks cannot cover more than the whole number line.

For each integer $a \ge 2$, define the L-multiples of $a$ by

\[L_a \;=\; \bigl\{ a\cdot b \,:\, b \ge 1,\; \text{every prime factor of } b \text{ is at least } P(a) \bigr\}.\]

In words: start with $a$, and multiply by integers built only from primes $\ge$ the largest prime factor of $a$. The letter L alludes to lexicographic. Write the prime factorization of $n$ in nondecreasing order. Then $n$ is an L-multiple of $a$ when that factorization starts with the factorization of $a$, and continues with primes $\ge P(a)$.

Example: $a = 6$. Here $6 = 2\cdot 3$, so $P(6) = 3$. Then $L_6$ is $6$ times any integer built from primes $\ge 3$:

\[L_6 \;=\; \{6,\; 18,\; 30,\; 42,\; 54,\; 66,\; 78,\; 90,\; \ldots\}.\]

Walk through a few membership tests, writing factorizations in nondecreasing order.

• $18 = 2\cdot 3\cdot 3$. This starts with $2\cdot 3$, and the leftover $3$ is $\ge 3$. In $L_6$.
• $12 = 2\cdot 2\cdot 3$. This starts with $2\cdot 2$, not with $2\cdot 3$. Equivalently $12 = 6\cdot 2$ and $2 < 3$. Not in $L_6$.
• $36 = 2\cdot 2\cdot 3\cdot 3$. Starts with $2\cdot 2$, not $2\cdot 3$. And $36 = 6\cdot 6$ with $6 = 2\cdot 3$ carrying a $2 < 3$. Not in $L_6$.
• $90 = 2\cdot 3\cdot 3\cdot 5$. Starts with $2\cdot 3$; leftover $3,5 \ge 3$. In $L_6$.

Example: a prime $p$. Here $P(p) = p$, so $L_p$ is $p$ times any integer built from primes $\ge p$. Equivalently: $L_p$ is the set of integers whose smallest prime factor is $p$. Thus $L_2$ is the even numbers, $L_3$ is ${3, 9, 15, 21, 27, 33, \ldots}$, $L_5$ is ${5, 25, 35, 55, \ldots}$, and so on. Every integer $n \ge 2$ has a unique smallest prime factor, so the sets $L_p$, as $p$ runs over the primes, split the integers $\ge 2$ into disjoint pieces whose union is everything except $1$.

Density

We already defined natural density. The even numbers have density $1/2$. The multiples of $a$ have density $1/a$.

For $L_a$, a number $n$ lies in $L_a$ when $a$ divides $n$ and the cofactor $n/a$ is not divisible by any prime smaller than $P(a)$. So we need two things: “multiple of $a$”, which has density $1/a$, and “avoids the primes $< P(a)$.”

Why is the proportion of integers not divisible by any prime in a list $q_1, \ldots, q_k$ equal to $\prod_i (1-1/q_i)$? For one prime, the proportion not divisible by $2$ is $1-1/2$. For two primes, inclusion-exclusion:

\[1 - \frac12 - \frac13 + \frac16 \;=\; \frac13 \;=\; \Bigl(1-\frac12\Bigr)\Bigl(1-\frac13\Bigr).\]

The same pattern continues for any finite list (the sieve of Eratosthenes): the events “divisible by $q$” for distinct primes $q$ multiply, because a number is divisible by $q$ and $q’$ if and only if it is divisible by $qq’$. Hence

\[d(L_a) \;=\; \frac{1}{a} \prod_{q < P(a)}\Bigl(1 - \frac{1}{q}\Bigr).\]

Check: $a = 6$, by hand. The only prime $< P(6) = 3$ is $2$, so $d(L_6) = \frac16 \cdot \bigl(1-\frac12\bigr) = \frac1{12}$. Among $1, \ldots, 60$, first take the multiples of $6$:

\[6,12,18,24,30,36,42,48,54,60.\]

Now drop those whose cofactor $n/6$ is even (i.e. divisible by a prime $< 3$): $12=6\cdot 2$, $24=6\cdot 4$, $36=6\cdot 6$, $48=6\cdot 8$, $60=6\cdot 10$. What remains is ${6, 18, 30, 42, 54}$, five numbers, and $60/12 = 5$.

Check: a prime $p$. Then $d(L_p) = \frac1p \prod_{q < p}(1-1/q)$, which is the proportion of integers with smallest prime factor $p$. For $p = 2$, the product is empty. An empty product equals $1$ (the same way an empty sum equals $0$), so $d(L_2) = 1/2$, matching “the even numbers.”

---

The territories do not overlap

Disjointness lemma. If $A$ is primitive, then the sets $L_a$ for $a \in A$ are pairwise disjoint.

In other words: two $L$-sets can meet only if one of the two generators divides the other. Primitivity forbids that, so the territories of a primitive set never share an integer.

Before the proof, a picture of what the lemma is saying. Take $n = 90$. Is $90$ in two different $L$-sets of a primitive pair?

• $90 = 6 \cdot 15$ and $15 = 3\cdot 5$, all of whose primes are $\ge P(6) = 3$. So $90 \in L_6$.
• $90 = 10 \cdot 9$ and $9 = 3^2$, but $3 < P(10) = 5$. So $90 \notin L_{10}$.
• $90 = 18 \cdot 5$ and $5 \ge P(18) = 3$. So $90 \in L_{18}$.

Thus $90 \in L_6 \cap L_{18}$. But ${6, 18}$ is not primitive, because $6 \mid 18$. That is the lemma in one direction: if the territories meet, one generator divides the other.

The other direction, a primitive pair whose territories do not meet: ${6,10}$. Here $P(6)=3 < 5=P(10)$. Suppose some $n = 6s = 10 s’$ were in both, with every prime of $s$ at least $3$ and every prime of $s’$ at least $5$. Then every prime of $6$ is $\le 3 < 5$, so $6$ shares no prime with $s’$. Euclid’s lemma would force $6$ to divide $10$. It does not, so no such $n$ exists.

A same-largest-prime illustration, again not primitive: ${9,27}$. Both have $P=3$, and $27 = 9\cdot 3$ with cofactor $3 \ge 3$, so $27 \in L_9$. The smaller exponent of $3$ divides the larger. That is the second case of the proof.

Proof. Suppose $n \in L_a \cap L_{a’}$ with $a \neq a’$ both in $A$. Write $n = a\cdot s = a’\cdot s’$, where every prime factor of $s$ is $\ge P(a)$ and every prime factor of $s’$ is $\ge P(a’)$. Swap names if needed so that $P(a) \le P(a’)$.

If $s’ = 1$, then $n = a’$, so $a$ divides $a’$, contradicting primitivity.

Now suppose $s’ > 1$. Every prime factor of $a$ is $\le P(a) \le P(a’)$, and every prime factor of $s’$ is $\ge P(a’)$.

• If $P(a) < P(a’)$, these two ranges of primes do not overlap. So $a$ and $s’$ share no prime factors: $\gcd(a, s’) = 1$. But $a$ divides $a s = a’ s’$. Euclid’s lemma: an integer that divides a product and is coprime to one factor must divide the other, so $a$ divides $a’$. Contradiction.
• If $P(a) = P(a’) = p$, write $a = p^{\alpha} m$ and $a’ = p^{\beta} m’$ with $p$ dividing neither $m$ nor $m’$. Then $m$ and $m’$ are built from primes $< p$, while $s$ and $s’$ are built from primes $\ge p$. From $p^{\alpha} m s = p^{\beta} m’ s’$ we get that $m$ divides $m’ s’$. But $m$ shares no primes with $s’$, so Euclid again: $m$ divides $m’$. The same argument the other way gives $m’$ divides $m$. Thus $m = m’$, and $a, a’$ are $p^{\alpha} m$ and $p^{\beta} m$: the one with the smaller exponent divides the other. Contradiction.

So no such $n$ exists.

The argument used only finite arithmetic — unique factorization and Euclid’s lemma — not density, not primes being infinite, nothing. It is the engine of everything that follows.

Because the $L_a$ are disjoint, they cannot cover more than the whole number line. A catch: natural density is not a measure, so we cannot blindly add infinitely many densities. (The even numbers and the odd numbers each have density $1/2$, and their union has density $1=\tfrac12+\tfrac12$; that pair happens to add. An infinite family of densities need not.) We do not need to. For any finite subcollection of $A$, the corresponding $L_a$ are disjoint, the density of their union is the sum of their densities, and a subset of the integers cannot have density greater than $1$. Every finite sum of the $d(L_a)$ is therefore $\le 1$, and so the infinite sum is $\le 1$ as well:

\[\sum_{a \in A} d(L_a) \;=\; \sum_{a \in A} \frac{1}{a} \prod_{q < P(a)}\Bigl(1 - \frac{1}{q}\Bigr) \;\le\; 1.\]

This is a budget. Each $a \in A$ spends $d(L_a)$ of a total of at most $1$.

---

Mertens’ theorem, and a uniform bound

How large is that product $\prod_{q < P(a)}(1-1/q)$? A heuristic first, then the named theorem.

Taking logarithms, $\prod_{p \le x}(1-1/p) \approx \exp\bigl(-\sum_{p \le x} 1/p\bigr)$, because $\log(1-1/p) \approx -1/p$ for large $p$. The sum of reciprocals of primes up to $x$ grows like $\log\log x$ (one logarithm slower than the harmonic series, because the primes are about $1/\log t$ dense). So the product is about

\[\exp\bigl(-\log\log x\bigr) \;=\; \frac{1}{\log x}.\]

The constant that makes the $\sim$ true, rather than just “about,” is $e^{-\gamma}$. Here $\gamma = 0.57721\ldots$ is the Euler–Mascheroni constant

\[\gamma \;=\; \lim_{N \to \infty} \Bigl( \sum_{n=1}^{N} \frac{1}{n} - \log N \Bigr),\]

and $e^{\gamma} \approx 1.781$. Mertens proved in 1874 that

\[\prod_{q \le x} \Bigl(1 - \frac{1}{q}\Bigr) \;\sim\; \frac{e^{-\gamma}}{\log x}.\]

The harmonic sum and $\log N$ differ by $\gamma$ in the limit; Mertens is the prime-sieve version of that gap. We take the theorem as a named fact. Numerically it is already close by $x=100$:

$x$	$\prod_{p\le x}(1-1/p)$	$e^{-\gamma}/\log x$
$10$	$0.229$	$0.244$
$100$	$0.120$	$0.122$
$1000$	$0.0810$	$0.0813$

Apply this with $x = P(a)$. The off-by-one between “primes $< P(a)$” and “primes $\le P(a)$” is absorbed in the $\sim$, and we get

\[d(L_a) \;\sim\; \frac{e^{-\gamma}}{a \log P(a)},\]

or, rearranged,

\[\frac{1}{a \log P(a)} \;\sim\; e^{\gamma}\, d(L_a).\]

Always $P(a) \le a$, so $\log P(a) \le \log a$, so $1/(a\log a) \le 1/(a\log P(a))$. Combining,

\[\frac{1}{a\log a} \;\le\; \frac{1}{a\log P(a)} \;\approx\; e^{\gamma}\, d(L_a).\]

Sum over $A$ and use the budget:

\[f(A) \;\approx\text{at most}\; e^{\gamma} \sum_{a \in A} d(L_a) \;\le\; e^{\gamma} \cdot 1 \;=\; e^{\gamma} \approx 1.781.\]

Lichtman and Pomerance (2019) turn the $\approx$ into a strict inequality $f(A) < e^{\gamma}$. Two ingredients: explicit numerical bounds on Mertens’ product in place of the $\sim$, and the fact that a composite $a$ satisfies $a \ge 2P(a)$, hence $\log a \ge \log P(a) + \log 2$. (Equality holds for $a=4$ and for every $a=2p$; the strictness of $f(A)<e^{\gamma}$ comes from the explicit Mertens bounds as well as from this gap when it is present.)

Theorem (Erdős 1935; Lichtman–Pomerance 2019). For every primitive set $A$, $f(A) < e^{\gamma} = 1.781\ldots$

The primes score $1.6366$. The bound $1.781$ is about $9\%$ too large. That gap is the remaining problem.

Look back at the inequality $1/(a\log a) \le 1/(a\log P(a))$. This is an equality when $a$ is prime ($P(a) = a$), and a loss when $a$ is composite ($P(a) < a$). The 1935 argument treats every integer as if it were a prime. Lichtman’s idea is to charge composites for that loss — and to prove they cannot dodge the charge by clustering just below the primes.

---

Compete slot by slot

The weight is front-loaded:

\[f(2) = 0.721, \qquad f(3) = 0.303, \qquad f(5) = 0.124, \qquad f(6) = 0.093.\]

If you put $6$ in a primitive set, you cannot also put $2$ or $3$. Starting from the primes and making that trade produces the primitive set ${6} \cup {p : p \neq 2, 3}$, with score

\[f(\mathcal{P}) - f(2) - f(3) + f(6) \;=\; f(\mathcal{P}) - 0.932.\]

You sold the two heaviest elements in the store for pocket change. Any one-for-one substitution of a composite $a$ for the primes dividing $a$ looks like a bad deal. The conjecture says there is no clever infinite combination of such trades that comes out ahead.

The right way to organise the comparison is not globally but slot by slot. Every integer $a > 1$ has a smallest prime factor $p(a)$. Split any set $A$ by that prime:

\[A_p \;=\; \{ a \in A : p(a) = p \}, \qquad A \;=\; \text{the disjoint union of the } A_p.\]

More generally, for any integer $n\ge 2$ write $A_n := A \cap L_n$. When $n=p$ is prime this recovers the slot above, because $L_p$ is exactly the integers whose smallest prime factor is $p$. Then $f(A) = \sum_p f(A_p)$, because the blocks $A_p$ do not overlap. The conjecture would follow if, for every prime $p$ and every primitive $A$,

\[f(A_p) \;\le\; f(p) \;=\; \frac{1}{p\log p}.\]

Simply: in the slot of numbers whose smallest prime factor is $p$, the single best choice is to take ${p}$ itself, not a primitive bunch of composites all divisible by $p$. A prime $p$ with this property (for every primitive $A$) is called Erdős-strong. If every prime were Erdős-strong, the primes would win.

A concrete odd slot, before any asymptotic. The slot $p=5$ consists of integers whose smallest prime factor is $5$. The triple ${25, 35, 55}$ lives there (check: $5^2$, $5\cdot 7$, $5\cdot 11$, and none divides another):

\[f(25)+f(35)+f(55) \;\approx\; 0.0124+0.0080+0.0045 \;=\; 0.025,\]

against $f(5)\approx 0.124$. The slot $p=3$, skipping $3$ itself and taking ${9,15,21,33}$:

\[f(9)+f(15)+f(21)+f(33) \;\approx\; 0.0506+0.0246+0.0156+0.0087 \;=\; 0.099,\]

against $f(3)\approx 0.303$. Any finite bunch of composites in an odd slot is a rout. The work is to prove that no infinite primitive combination in that slot can catch up.

Here is a sufficient condition, from the 2019 bound. For $a \in A_p$ one has $p(a) = p$, so $a \in L_p$, and therefore $L_a \subset L_p$. By disjointness, $\sum_{a \in A_p} d(L_a) \le d(L_p)$. Combining with $f(a) \approx\text{at most } e^{\gamma}\, d(L_a)$,

\[f(A_p) \;\approx\text{at most}\; e^{\gamma}\, d(L_p) \;=\; e^{\gamma} \cdot \frac{1}{p} \prod_{q < p}\Bigl(1 - \frac{1}{q}\Bigr).\]

We want this $\le 1/(p\log p)$. Cancel $1/p$: it is enough that

\[e^{\gamma} \prod_{q < p}\Bigl(1 - \frac{1}{q}\Bigr) \;\le\; \frac{1}{\log p}. \tag{$\star$}\]

Mertens says the two sides of $(\star)$ are asymptotically equal. Direct computation shows that $(\star)$ holds for the first $10^8$ odd primes. For $p = 3$ it is already tight: the left side is $\approx 0.891$ and the right side is $1/\log 3 \approx 0.910$.

It fails for $p = 2$. The product over primes $q < 2$ is empty, hence equals $1$, and

\[e^{\gamma} \approx 1.781 \qquad\text{while}\qquad \frac{1}{\log 2} \approx 1.443.\]

Worse: even among odd primes, $(\star)$ is not a theorem. It fails for infinitely many primes, unconditionally. On the Riemann hypothesis it fails for a positive proportion of them (in log-density). The strategy “prove every prime is Erdős-strong via $(\star)$” cannot finish the conjecture. That is why the problem sat open.

The missing idea is to stop treating composites as if they were primes. Once we charge them a discount, $(\star)$ being tight stops mattering: a $21\%$ discount swallows the Mertens oscillation for every odd prime. (At $p=3$ the explicit error tracking inflates $\pi/4\approx 0.785$ up to $0.901$, still strictly less than $1$.) That is the next three sections.

---

How close to prime?

The loss in the 1935 argument, for a single $a$, is the ratio $\log P(a)/\log a \le 1$. Define

\[v(a) \;=\; \frac{\log a}{\log P(a)} - 1 \;\ge\; 0.\]

Then $v(a) = 0$ if and only if $P(a) = a$, if and only if $a$ is prime. Equivalently $a = P(a)^{1+v(a)}$: $v$ counts how many extra log-factors of $P(a)$ fit into $a$. For a prime power $p^k$ one has $v(p^k)=k-1$. Rearranging the definition,

\[\frac{1}{a\log a} \;=\; \frac{1}{1+v(a)}\cdot \frac{1}{a\log P(a)} \;\approx\; \frac{e^{\gamma}}{1+v(a)}\, d(L_a).\]

So $v(a)$ is the discount at which $a$ converts density into $f$-weight. Primes convert at full rate $e^{\gamma}$. A composite converts at rate $e^{\gamma}/(1+v(a))$.

Worked example: $a = 15 = 3\cdot 5$. Then $P(15) = 5$, and

\[v(15) \;=\; \frac{\log 15}{\log 5} - 1 \;=\; \frac{2.708}{1.609} - 1 \;\approx\; 0.683, \qquad \frac{1}{1+v(15)} \;\approx\; 0.594.\]

$a$	factorization	$P(a)$	$v(a)$	conversion $1/(1+v)$
$7$	$7$	$7$	$0$	$1$
$14$	$2\cdot 7$	$7$	$0.356$	$0.738$
$22$	$2\cdot 11$	$11$	$0.289$	$0.776$
$15$	$3\cdot 5$	$5$	$0.683$	$0.594$
$6$	$2\cdot 3$	$3$	$0.631$	$0.613$
$9$	$3^2$	$3$	$1$	$1/2$
$8$	$2^3$	$2$	$2$	$1/3$
$221$	$13\cdot 17$	$17$	$0.905$	$0.525$

Read the table against the 1935 loss. The dangerous composites are those with small $v$: a large prime times a small factor. For $a = 2p$,

\[v(2p) \;=\; \frac{\log 2}{\log p} \;\longrightarrow\; 0 \quad\text{as } p \to \infty.\]

So $2p$ looks almost like a prime to the 1935 bound: $P(2p) = p$ is a little more than half of $2p$, and the conversion rate tends to $1$. Numbers such as $9$ and $13\cdot 17$ are very composite in this sense ($P(a)^2 \approx a$, so $v \approx 1$) and convert at half rate or worse. They are no threat: even if they spent the whole density budget they would score at most $e^{\gamma}/2 \approx 0.89$, well below $f(\mathcal{P})$.

Lichtman’s outline says the same thing in the other direction: the 1935 argument already saves a lot when $P(a)^2 < a$. The critical case is a composite with $P(a)$ close to $a$ in size — a large prime with a small factor stuck on.

The remaining question: can a primitive set contain many dangerous (small-$v$) composites in the same slot? If it can fill the whole density of $L_p$ with $v \approx 0.01$ composites, the 1935 bound barely improves and $(\star)$ is still the bottleneck. If it cannot, we win.

---

Copies, and the $\sqrt{v}$ packing

A baby version of the idea. Suppose you have some disjoint subsets of the integers, of total density $D$, and suppose you can also fit a second family of disjoint copies of those subsets, still disjoint from the first family. Then $2D \le 1$, so $D \le 1/2$. More copies, smaller $D$.

Lichtman produces many copies of each $L_a$, indexed by extra multipliers $c$, provided every $a$ in the set is uniformly $v$-close to being prime — that is, $P(a)^{1+v} > a$, or equivalently $v(a) < v$. The copies live inside the ambient territory $L_n$ of the slot (for instance $L_p$ when we are competing with a prime $p$). That is the true statement, and it is local:

Proposition (Lichtman). Let $A$ be primitive, let $n \notin A$, and suppose $P(a)^{1+v} > a$ for every $a$ in the slot $A_n = A \cap L_n$. Then \(\sum_{a \in A_n} d(L_a) \;\le\; \sqrt{v}\cdot\bigl(1+o(1)\bigr)\cdot d(L_n).\)

For $v < 1$ this is strictly better than the trivial budget $\sum d(L_a) \le d(L_n)$. At $v = 1/4$ you may spend at most about half the slot; at $v = 1/100$, at most about a tenth. A primitive set cannot fill a slot with numbers that are all nearly prime.

Where does $\sqrt{v}$ come from? For each $a$, let $a^* = a/P(a)$ (so $a^$ is $a$ with its largest prime stripped off), and let $Q = P(a^)$ be the second-largest prime factor of $a$. The extra multipliers $c$ are the integers — including $c = 1$ — whose prime factors all lie in a window

\[\bigl[Q,\; Q^{1/\sqrt{v}}\bigr).\]

A fully numerical window, on a dangerous $2p$. Take $a=22=2\cdot 11$. Then $v(22)=\log 2/\log 11\approx 0.289$, so $\sqrt{v}\approx 0.538$ and $1/\sqrt{v}\approx 1.86$. Here $a^*=2$ and $Q=2$, and the window is

\[\bigl[2,\; 2^{1.86}\bigr) \;\approx\; [2,\; 3.62).\]

The primes in that window are $2$ and $3$. Already the three multipliers $c=1,2,3$ contribute extra copy-density $1+\frac12+\frac13\approx 1.83$, matching $1/\sqrt{v}\approx 1.86$. Those copies still have to sit inside the ambient slot, so the original $L_{22}$ is allowed to spend only about $1/1.86 \approx 0.54$ of the slot — the packing $\sqrt{v}$ on the nose, in the case the 1935 bound was most afraid of.

The one genuinely hard lemma in the paper, which we will not prove, is that the sets $L_{ac}$ ranging over $a \in A_n$ and over such $c$ remain pairwise disjoint, and all sit inside $L_n$, as long as $A$ is primitive and uniformly $v$-close. Primitivity is essential: the weaker “no one is an L-multiple of another” is enough for ordinary disjointness of the $L_a$, but not for the copies. (That is why the $e^{\gamma}$ bound is essentially sharp for that weaker class, while primitive sets can be pushed down to $f(\mathcal{P})$.) Everything after this lemma is Euler products, Mertens, and calculus.

Granting disjointness of the copies, their densities add, and cannot exceed $d(L_n)$. For each fixed $a$, the copy $L_{ac}$ is just $L_a$ scaled by $c$, so $d(L_{ac}) = d(L_a)/c$. The total copy-density attached to one $a$ is therefore $d(L_a)$ times $\sum 1/c$.

That sum is an Euler product. Unique factorization says every such $c$ is a product of primes from the window, so

\[\sum_c \frac{1}{c} \;=\; \prod_{Q \le q < Q^{1/\sqrt{v}}} \Bigl(1 + \frac{1}{q} + \frac{1}{q^2} + \cdots\Bigr) \;=\; \prod_{Q \le q < Q^{1/\sqrt{v}}} \Bigl(1 - \frac{1}{q}\Bigr)^{-1},\]

using the geometric series $1 + x + x^2 + \cdots = 1/(1-x)$ with $x = 1/q$. Mertens says $\prod_{q < x} (1-1/q)^{-1} \sim e^{\gamma} \log x$, so a product over a range is a ratio of logarithms (the $e^{\gamma}$ cancels):

\[\frac{\log\bigl(Q^{1/\sqrt{v}}\bigr)}{\log Q} \;=\; \frac{1}{\sqrt{v}}.\]

Each original $L_a$ therefore comes with about $1/\sqrt{v}$ times as much copy-density. All of those copies still sit inside $L_n$, so

\[\frac{1}{\sqrt{v}} \sum_{a \in A_n} d(L_a) \;\lesssim\; d(L_n) \qquad\Longrightarrow\qquad \sum_{a \in A_n} d(L_a) \;\lesssim\; \sqrt{v}\; d(L_n).\]

The exponent $1/\sqrt{v}$ on the window is the largest Lichtman could take and still prove the copies disjoint. A larger window would improve $\sqrt{v}$ to a smaller power of $v$, and would improve the $\pi/4$ in the next section.

When $v$ is small — the dangerous $2p$ numbers — the window $Q^{1/\sqrt{v}}$ is large, so there are many copies, so the original density is forced to be small. That is the packing constraint biting exactly on the critical case.

---

Buckets, then $\pi/4$

We now have two functions of $v$, inside a fixed slot $A_n$ of composites:

• an element at level $v$ converts density to $f$-weight at rate about $e^{\gamma}/(1+v)$;
• all the elements with $v(a) < v$ together spend at most about $\sqrt{v}$ of the ambient density $d(L_n)$.

To bound the total $f$, split the composites into buckets by $v$, which is how the paper actually proceeds, and only then pass to an integral.

Pick a partition $0 = v_0 < v_1 < \cdots < v_k = 1$ of the interval $[0,1]$. Put in bucket $i$ those $a \in A_n$ with $v_i \le v(a) < v_{i+1}$. (Anything with $v(a) \ge 1$, i.e. $P(a)^2 \le a$, converts at rate $\le 1/2$ and can be lumped into the last bucket.) In bucket $i$,

\[f(\text{bucket } i) \;\le\; \frac{e^{\gamma}}{1+v_i} \cdot d(\text{that bucket's $L$-sets}),\]

because every element there has conversion rate $\le 1/(1+v_i)$. Write $d_i$ for that bucket’s density. The packing constraint, applied to the union of the first $j$ buckets (the closer-to-prime elements), says

\[d_0 + d_1 + \cdots + d_{j-1} \;\le\; \sqrt{v_j}\, d(L_n).\]

The rates $1/(1+v_i)$ are decreasing in $i$: small $v$ is the high-value real estate. Given a decreasing sequence of rates and upper bounds on the cumulative densities, the most $f$ you can extract is by filling each cumulative budget as soon as it opens — spend as much as you are allowed at the highest remaining rate.

A three-bucket cash register, before the symbols. Rates $1 \ge 0.8 \ge 0.5$, and you may spend at most density $0.5$ at the top rate, at most $0.7$ in the first two buckets combined, at most $1$ in all three. Fill high-rate first: $0.5$ at rate $1$, then the newly unlocked $0.2$ at rate $0.8$, then the remaining $0.3$ at rate $0.5$. Revenue $0.50+0.16+0.15=0.81$. Any other allocation is worse. That is the paper’s rearrangement lemma, in English.

The worst case is therefore $d_0 + \cdots + d_{j-1} = \sqrt{v_j}\, d(L_n)$, so the density of bucket $i$ is at most the newly unlocked packing room

\[\bigl(\sqrt{v_{i+1}} - \sqrt{v_i}\bigr)\, d(L_n).\]

Hence

\[f(A_n) \;\le\; e^{\gamma}\, d(L_n) \sum_{i=0}^{k-1} \frac{\sqrt{v_{i+1}} - \sqrt{v_i}}{1+v_i}.\]

This is a Riemann sum for an integral. Taking a finer and finer partition ($v_i = i/k$, then $k \to \infty$),

\[\sum_{i} \frac{\sqrt{v_{i+1}} - \sqrt{v_i}}{1+v_i} \;\longrightarrow\; \int_0^1 \frac{1}{1+v}\cdot \frac{d}{dv}\bigl[\sqrt{v}\bigr]\, dv \;=\; \int_0^1 \frac{dv}{2\sqrt{v}\,(1+v)}.\]

The derivative of $\sqrt{v}$ is $\frac12 v^{-1/2} = 1/(2\sqrt{v})$: the extra packing room unlocked at level $v$ is $dv/(2\sqrt{v})$. Near $v=0$ the integrand blows up like $1/(2\sqrt{v})$, which is the white curve in the figure above. Do not panic: $\int_0^1 v^{-1/2}\,dv = 2$ is finite, so the singularity is integrable.

A coarse numerical check, four buckets, left endpoints:

$v_i$	$\sqrt{v_i}$	$\sqrt{v_{i+1}}-\sqrt{v_i}$	$1/(1+v_i)$	product
$0$	$0$	$0.500$	$1$	$0.500$
$0.25$	$0.500$	$0.207$	$0.800$	$0.166$
$0.50$	$0.707$	$0.159$	$0.667$	$0.106$
$0.75$	$0.866$	$0.134$	$0.571$	$0.076$
sum				$0.848$

The sum $0.848$ is a coarse overestimate of $\pi/4 \approx 0.785$, as a left Riemann sum on a decreasing integrand should be. Finer partitions close the gap.

Now the integral, by substitution. Set $u = \sqrt{v}$, so $v = u^2$ and $dv = 2u\, du$. The limits stay $0$ to $1$, and the $2u$ in the numerator cancels the $2\sqrt{v}=2u$ in the denominator:

\[\int_0^1 \frac{dv}{2\sqrt{v}\,(1+v)} \;=\; \int_0^1 \frac{2u\, du}{2u\,(1+u^2)} \;=\; \int_0^1 \frac{du}{1+u^2}.\]

The antiderivative of $1/(1+u^2)$ is $\arctan u$. And $\arctan 1 = \pi/4$, because that is the angle whose tangent is $1$: a $45^\circ$ angle, a quarter of a half-turn. Also $\arctan 0 = 0$. So

\[\boxed{\displaystyle \int_0^1 \frac{dv}{2\sqrt{v}\,(1+v)} \;=\; \frac{\pi}{4}}.\]

The left shaded region has area $\pi/4$. That same number is the angle on the right, in radians: $\mathrm{opp}/\mathrm{adj}=1/1=1$.

The factor $\pi/4 \approx 0.785$ is the proportion of a full-rate $e^{\gamma}$ that a slot of composites can realize under the $\sqrt{v}$ constraint. In the envelope, a slot $A_n$ of composites scores at most about

\[f(A_n) \;\le\; \frac{\pi}{4}\, e^{\gamma}\, d(L_n).\]

That is the true output of the $\pi/4$ argument: a bound on a slot of composites, in terms of the density of the ambient territory $L_n$. It is not a global bound “$f(\text{all composites}) \le 1.399$.” (That global-looking number $e^{\gamma}\pi/4 \approx 1.399$ does appear, as an upper bound on $f(A)$ for primitive sets of large elements. That is a different theorem, about tails.)

---

Odd primes beat their own composites

Specialize to $n = p$, an odd prime, with $p \notin A$. Then $A_p$ is a primitive set of composites all having smallest prime factor $p$, and the ambient territory is $L_p$, with

\[e^{\gamma}\, d(L_p) \;\approx\; f(p)\]

by Mertens (this is the same comparison that produced $(\star)$). The $\pi/4$ bound becomes

\[f(A_p) \;\le\; \frac{\pi}{4}\cdot\bigl(1+\text{a small Mertens error}\bigr)\cdot f(p).\]

The error is tracked in the paper with explicit bounds on Mertens’ product. The worst odd prime is $p = 3$, and even there the constant comes out strictly less than $0.901$. For larger odd primes it is better, and it tends to $\pi/4 \approx 0.785$ as $p \to \infty$. In all cases it is less than $1$:

Theorem (Lichtman, Theorem 1.3). For every primitive set $A$ and every odd prime $p$, \(f(A_p) \;\le\; f(p).\) If $p \notin A$, the inequality is strict: $f(A_p) < 0.901\, f(p)$.

Simply: you should never skip an odd prime in favour of composites with that smallest prime factor. Skipping $p$ and filling the slot with $2p$, $3p$, $p^2$, products of larger primes, whatever primitive combination you like, returns less than $90\%$ of what $p$ itself was worth.

That is the true reason the primes win on the odd slots. The $\pi/4$ is a local comparison of composites to their parent prime, not a global score for “all the composites in $A$.” The $21\%$ of slack is what swallows the Mertens oscillation that killed $(\star)$ for infinitely many large primes.

If $p \in A$, then primitivity forbids any other multiple of $p$ from joining $A$, so $A_p = {p}$ and $f(A_p) = f(p)$ on the nose.

---

The prime $2$

Every odd prime is Erdős-strong. The remaining slot is $p = 2$.

If $2 \in A$, then $A$ contains no other even number. The odd part of $A$ is a primitive set of odd numbers, and the odd-prime theorem gives $f(A_p) \le f(p)$ for every odd $p$. Adding $f(2)$ back, $f(A) \le f(\mathcal{P})$.

If $2 \notin A$, the criterion $(\star)$ fails and the constants around $\pi/4$ do not quite prove that $2$ is Erdős-strong. The idea of the paper’s substitute is a $2$-power split. Every even $a$ factors as $2^k$ times an odd integer. The odd part lives in an odd slot, where we already win; the $2^k$ costs a factor $\sim 2^{-k}$ in the weight, so a pile of even composites is cheaper, in bulk, than $2$ itself. Tracking the possible $k$ and applying the odd-prime bounds to the odd parts produces a global estimate $f(A) < 1.60 < f(\mathcal{P})$. We will not replay the constants. The conjecture does not need $2$ to be Erdős-strong: when $2$ is missing, this direct bound already loses to the primes.

Whether $2$ itself is Erdős-strong — whether some primitive set of even numbers, none of which is $2$, can score more than $f(2) \approx 0.721$ in the even slot — was left open in 2022.

graph TD
    A["primitive set A"] --> B{"2 in A?"}
    B -->|"yes"| C["no other evens"]
    C --> D["odd slots: each A_p scores at most f(p)"]
    D --> E["f(A) ≤ f(P)"]
    B -->|"no"| F["π/4 on each odd slot, plus a 2-power split"]
    F --> G["f(A) < 1.60 < f(P)"]
    style E fill:#38bdf8,color:#000
    style G fill:#38bdf8,color:#000

The napkin is the $\pi/4$ on a slot of composites. The paper is the Mertens-error tracking that makes $0.901 < 1$ for every odd prime, and the separate $1.60$ when $2$ is missing.

---

Why $f(\mathcal{P})$ is finite

We used $f(\mathcal{P}) = 1.6366\ldots$ as a finite number. Here is why the series converges.

The prime number theorem, taken as a named fact, says that the $n$th prime is $p_n \sim n\log n$. Then

\[\frac{1}{p_n \log p_n} \;\sim\; \frac{1}{n\log n \cdot \log(n\log n)}.\]

And $\log(n\log n) = \log n + \log\log n \sim \log n$, so

\[\frac{1}{p_n \log p_n} \;\sim\; \frac{1}{n(\log n)^2}.\]

We already know $\sum 1/(n(\log n)^2)$ converges, by the integral test in the infinite-sums section. By the limit comparison test, a series of positive terms that behaves like a convergent series converges. So $\sum_p 1/(p\log p)$ converges.

The same integral explains the tail. The amount contributed by primes larger than $x$ is about $1/\log x$:

primes up to	partial sum	gap to $1.6366$	$1/\log x$
$10^2$	$1.4216$	$0.215$	$0.217$
$10^4$	$1.5282$	$0.108$	$0.109$
$10^6$	$1.5642$	$0.072$	$0.072$
$10^8$	$1.5823$	$0.054$	$0.054$

Each extra two orders of magnitude in the cutoff trims the gap by a factor of about $2/3$, not by an order of magnitude. That is what a $1/\log x$ tail looks like. It is also why the conjecture cannot be checked by adding up primitive sets until $10^{12}$ and comparing: the tail is large enough to hide a counterexample sitting out at infinity.

Where the sum sits among its neighbours:

series	converges?	value / growth
$\sum_p 1/p^2$	yes	$0.4522\ldots$
$\sum_p 1/(p\log p)$	yes	$1.6366\ldots$
$\sum_p 1/p$	no	$\sim \log\log x$
$\sum_n 1/(n\log n)$	no	$\sim \log\log x$

The extra $\log$ in the denominator is not enough to make the sum over all integers converge. It is enough once you keep only the primes, because the primes themselves contribute another $1/\log n$ of sparsity. That second logarithm is a gift of the prime number theorem.

---

What is still open

Is $2$ Erdős-strong? Lichtman proved every odd prime is. A negative answer would not have damaged the 2022 theorem — the case $2 \notin A$ is already lost to $f(\mathcal{P})$ by the global bound $1.60$ — but it would have meant that some primitive set of even numbers, none of which is $2$, scores more than $f(2) \approx 0.721$ in the even slot.

The Erdős–Sárközy–Szemerédi tail conjecture (1968). Restrict to primitive sets of large elements:

\[\lim_{x \to \infty} \sup_{\substack{A \subset [x,\infty)\\ A \text{ primitive}}} f(A) \;\stackrel{?}{\le}\; 1.\]

This is where $e^{\gamma}\pi/4 \approx 1.399$ appears as a theorem: Lichtman’s method gives that upper bound on the lim-sup. The value $1$ is forced as a lower bound if the conjecture is true, because the set of integers with $\Omega(n) = k$ lives in $[2^k, \infty)$ and Lichtman had already proved that its Erdős sum tends to $1$ as $k \to \infty$.

A May 2026 preprint of Alexeev, Barreto, Li, Lichtman, Price, Shah, Tang, and Tao, arXiv:2605.00301, claims both that $2$ is Erdős-strong and that the tail conjecture is true, by a different method: a von Mangoldt-weighted Markov chain on the divisibility poset. It is marked preliminary. The 2022 $\pi/4$ argument remains the one this post unpacks.

---

The theorem says something simple enough to tell after one semester: among all the ways to pick integers so that none divides another, the primes are the heaviest, when each integer $a$ is weighed by $1/(a\log a)$. The reason, on a napkin, is local. In the slot of numbers with smallest prime factor $p$, a composite converts density at a discount $1/(1+v)$, the dangerous nearly-prime composites cannot spend more than density $\sqrt{v}$ of the slot, and the integral of that tradeoff is a quarter-circle. For every odd prime, that is strictly worse than taking $p$ itself.

Cheers.

---

Further reading

Jared Duker Lichtman, A proof of the Erdős primitive set conjecture. Forum of Mathematics, Pi (2023). The paper. The envelope in this post is the argument of its introduction; the actual inequalities are in sections 2–4.

Boris Alexeev, Kevin Barreto, Yanyang Li, Jared Duker Lichtman, Liam Price, Jibran Iqbal Shah, Quanyu Tang, and Terence Tao, Primitive sets and von Mangoldt chains. May 2026 preprint. A different method, claiming the tail conjecture and that $2$ is Erdős-strong.

Jared Duker Lichtman, Almost primes and the Banks–Martin conjecture. Proves $f(\mathbb{N}_k) \to 1$.

Paul Erdős and András Sárközy, On the divisibility of sequences of integers. Source of the 1968 tail conjecture.

Tsz Ho Chan, Jared Duker Lichtman, and Carl Pomerance, On the critical exponent for $k$-primitive sets. Where $\tau = 1.1403\ldots$ comes from, and why the $t$-pointwise analogue of the conjecture fails.

Jared Duker Lichtman and Carl Pomerance, The Erdős conjecture for primitive sets. The $e^{\gamma}$ bound.

Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory. Cambridge, 2007. Mertens’ theorems and the prime number theorem.

Quanta Magazine, Graduate Student’s Side Project Proves Prime Number Conjecture. The human story of the proof.