Primitive Sets: I · II · III · IV · V

Conversion against packing. The best a slot of composites can do is an integral, the integral is an arctangent, and the arctangent is a quarter-turn: $\pi/4\approx 0.785$ of the prime it is trying to replace.

Vol. IV left us inside a single slot $A_n$ of composites with two facts in tension. An element at distance-from-prime $v$ converts density into score at rate about $e^\gamma/(1+v)$, and the elements closer to prime than $v$ can spend at most about $\sqrt v$ of the slot’s density $d(L_n)$. This volume finds the best a challenger can do under both constraints, and then finishes the theorem.

Contents

buckets, then π/4
odd primes beat their own composites
the prime 2
the argument on one page
what is still open
further reading

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 $d(L_n)$.

Conversion rate, packing cap, and the extra-room integrand

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.

Check yourself. Why is "any other allocation worse"? Try moving $0.1$ of density from bucket 1 to bucket 2.

Bucket 1 can't hold more than $0.5$, so the only moves are down: shift density from a higher-rate bucket to a lower-rate one. Moving $0.1$ from rate $0.8$ to rate $0.5$ loses $0.1\times(0.8-0.5)=0.03$. Every legal move trades a unit at a higher rate for a unit at a lower one. In general this is summation by parts: $\sum r_i d_i = \sum_j (r_{j-1}-r_j)\,D_j + r_{k-1}D_k$ with $D_j$ the cumulative sums, and every coefficient $r_{j-1}-r_j$ is $\ge 0$, so the sum is largest when every $D_j$ is as large as allowed.

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. Watch them close it, and flip the axis to see the substitution coming:

buckets k bucket i's contributionintegrand

Each blue bar has area $(\sqrt{v_{i+1}}-\sqrt{v_i})/(1+v_i)$, bucket $i$'s worst-case contribution. On the $v$ axis the integrand $1/(2\sqrt v(1+v))$ blows up at $0$ (clipped here). On the $u=\sqrt v$ axis the same bars become a left Riemann sum for the tame curve $1/(1+u^2)$: the singularity was an artefact of the coordinate.

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 $\arctan u$. (Chain rule: if $\theta = \arctan u$ then $u=\tan\theta$, $du/d\theta = \sec^2\theta = 1+\tan^2\theta = 1+u^2$.) 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 integral is the area under $1/(1+u^2)$, which is the $45^\circ$ angle$

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 $(\star)$ in Vol. IV). 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 $3p$, $p^2$, products of larger primes, whatever primitive combination you like, returns less than $90\%$ of what $p$ itself was worth. (In the slot of an odd prime $p$ every element is odd, since its smallest prime factor is $p$.)

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 $A_p = \lbrace p\rbrace$ 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 four numbers that run the argument

Student. So where exactly does the napkin stop and the paper start?

Teacher. Three places. First, the disjointness of the copies $L_{ac}$ in Vol. IV, which we stated and did not prove; it is the combinatorial heart of the paper and the only place primitivity is used beyond the 1935 lemma. Second, every "$\approx$" that came from Mertens is, in the paper, an explicit inequality with an explicit error; that is what turns $\pi/4$ into $0.901$ at $p=3$. Third, the global bound $1.60$ when $2\notin A$. The shape of the argument (budget, discount, packing, quarter-turn) is exactly the napkin.

the argument on one page

Objects (Vol. I). A primitive set has no element dividing another. The primes are one; so are $\lbrace 6,10,15\rbrace$, every $(x,2x]$, and every $\lbrace\Omega = k\rbrace$. Counting and density cannot rank them.
Ruler (Vol. II). $f(A)=\sum 1/(a\log a)$, on the knife-edge: infinite over all integers, $1.6366$ over the primes.
Budget (Vol. III). Territories $L_a$ are disjoint for primitive $A$, so $\sum d(L_a)\le 1$. Mertens converts density to score at rate $e^\gamma$, giving $f(A)\lt e^\gamma = 1.781$.
Slots and discount (Vol. IV). Compete prime by prime. A composite converts at the discounted rate $e^\gamma/(1+v)$, and nearly-prime composites can fill at most $\sqrt v$ of a slot.
Quarter-turn (this volume). Discount against packing integrates to $\int_0^1 \frac{du}{1+u^2}=\pi/4$. Every odd prime beats its composites by nearly $10\%$. The prime $2$ is handled by a direct bound, $1.60 \lt 1.6366$.

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.

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 series unpacks.

Cheers.

Julian Henry

polyglot / software engineer / author

Primitive SetsVThe Quarter-Turn

buckets, then $\pi/4$

odd primes beat their own composites

the prime $2$

the argument on one page

what is still open

further reading