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This is a copy of a conversation between ChatGPT & Anonymous.

don't search the internet. This is a test to see how well you can craft non-trivial, novel and creative proofs given a "number theory and primitive sets" math problem. Provide a full unconditional proof or disproof of the problem. Problem: "Is it true that, for any $x$, if $A\subset [x,\infty)$ is a primitive set of integers (so that no distinct elements of $A$ divide each other) then\[\sum_{a\in A}\frac{1}{a\log a}< 1+o(1),\]where the $o(1)$ term $\to 0$ as $x\to \infty$?" information you may or may not need to help with the above problem "It is proved that\[\sum_{a\in A}\frac{1}{a\log a}< e^{\gamma}\frac{\pi}{4}+o(1)\approx 1.399+o(1).\]" "It is proved that if $A$ is the set of all integers with exactly $k$ prime factors (so that $A\subset [2^k,\infty)$ and $A$ is a primitive set) then\[\sum_{a\in A}\frac{1}{a\log a}\geq 1+O(k^{-1/2+o(1)}),\]" "It is proved that\[\sum_{a\in A}\frac{1}{a\log a}= 1-(c+o(1))k^22^{-k}\]where $c\approx 0.0656$ is an explicit constant." REMEMBER - this unconditional argument may require non-trivial, creative and novel elements.

I’m treating this as a genuine extremal proof search, with two live paths: either compress primitive tails by prime-factor layers to force a universal $1+o(1)$ bound, or construct a counterexample beating 1 by a fixed amount.

I found a promising probabilistic route: use a “first-entry into $[x,\infty)$ ” random divisibility chain built from prime powers, weighted by $\Lambda(q)$ . Then $\sum_{q\mid n}\Lambda(q)=\log n$ converts the target sum into hitting probabilities, with only an $O(1/\log x)$ boundary loss.

A competing construction now looks much less viable: because unions of disjoint prime-support blocks stay primitive, any counterexample would need to add a second large-prime block to a near-extremal $\Omega(n)=k$ layer, but the scale mismatch makes that extra contribution too small to overcome the known deficit.

Yes.

In fact one can prove something slightly stronger.

Fix any constant $0<\delta<\log 2$ . Then for every primitive set $A\subset [x,\infty)$ ,

\sum_{a\in A}\frac1{a(\log a-\delta)}\le 1+O_\delta\!\left(\frac1{\log x}\right) \qquad (x\to\infty).

Since

\frac1{a\log a}\le \frac1{a(\log a-\delta)},

this immediately gives

\sum_{a\in A}\frac1{a\log a}\le 1+o(1).

So the statement in the problem is true.

Proof

Write $\Lambda$ for the von Mangoldt function. I will sum over all integers $q\ge1$ ; only prime powers contribute because $\Lambda(q)=0$ otherwise.

The key identity is the exact divisor sum

\sum_{q\mid n}\Lambda(q)=\log n.

The proof is a weighted “first-entry” decomposition on the divisibility poset, but with one important twist: I only allow large prime-power jumps. The finitely many small prime powers are absorbed into the initial mass, and that costs only $O(1/\log x)$ .

1. A tail estimate for prime powers

Let

M(t):=\sum_{q\le t}\frac{\Lambda(q)}{q}.

A standard unconditional estimate (Mertens’ first theorem) says

M(t)=\log t+O(1).

For $m\ge1$ , $y\ge2$ , set

T_\delta(m,y):=\sum_{q\ge y}\frac{\Lambda(q)}{q\,\log(mq)\,(\log(mq)-\delta)}.

I claim that, uniformly as $my\to\infty$ ,

T_\delta(m,y) = \frac1\delta\log\frac{\log(my)}{\log(my)-\delta} + O_\delta\!\left(\frac1{\log^2(my)}\right). \tag{1}

To prove this, write $L=\log m$ , $U=\log y$ , and

F(u):=\frac1{(L+u)(L+u-\delta)}.

Then

T_\delta(m,y)=\int_{U}^{\infty} F(u)\,dM(e^u).

Since $M(e^u)=u+E(u)$ with $E(u)=O(1)$ , we get

T_\delta(m,y)=\int_U^\infty F(u)\,du+\int_U^\infty F(u)\,dE(u).

On any finite interval, $E$ has bounded variation, so integration by parts gives

\int_U^\infty F(u)\,dE(u) = -F(U)E(U)-\int_U^\infty E(u)F'(u)\,du.

Now

F(U)\ll_\delta \frac1{(L+U)^2}, \qquad F'(u)\ll_\delta \frac1{(L+u)^3},

and $E(u)=O(1)$ , so

\int_U^\infty F(u)\,dE(u) = O_\delta\!\left(\frac1{(L+U)^2}\right).

Also

\int_U^\infty \frac{du}{(L+u)(L+u-\delta)} = \frac1\delta\log\frac{L+U}{L+U-\delta}.

This is exactly (1).

2. Choose a cutoff $Y$ and define the jump kernel

Fix $\delta\in(0,\log2)$ .

Using (1), choose once and for all an integer $Y\ge 2$ , large enough in terms of $\delta$ , so that for all sufficiently large $m$ ,

(\log m-\delta)\,T_\delta(m,Y)\le 1. \tag{2}

Why is this possible? By (1),

(\log m-\delta)T_\delta(m,Y) = (\log m-\delta)\left( \frac1\delta\log\frac{\log(mY)}{\log(mY)-\delta} +O_\delta\!\left(\frac1{\log^2 m}\right) \right).

Using $\log(1+t)\le t$ ,

\frac1\delta\log\frac{\log(mY)}{\log(mY)-\delta} \le \frac1{\log(mY)-\delta},

(\log m-\delta)T_\delta(m,Y) \le \frac{\log m-\delta}{\log(mY)-\delta} + O_\delta\!\left(\frac1{\log m}\right) = 1-\frac{\log Y}{\log(mY)-\delta}+O_\delta\!\left(\frac1{\log m}\right).

Since $\log Y$ is a fixed positive constant which we may choose as large as we like, the deficit from the main term dominates the error for large $m$ , giving (2).

Now fix $x$ so large that (2) holds for all $m\ge x$ .

Define

W(n):=\frac1{n(\log n-\delta)}.

For $m\ge x$ and $q\ge Y$ , define

P(m,mq):= \frac{(\log m-\delta)\Lambda(q)} {q\,\log(mq)\,(\log(mq)-\delta)}.

Because of (2), for every $m\ge x$ ,

\sum_{q\ge Y}P(m,mq)\le 1.

So this is a genuine sub-Markov jump kernel on $[x,\infty)$ : from $m$ , the chain either jumps to $mq$ with the above probability, or stops.

Every jump multiplies by an integer $q\ge Y>1$ , so every path is strictly increasing and each visited state divides every later one.

3. The entrance mass

Define the “source” mass on integers $n\ge x$ by

b_x(n):= \frac1{n\log n(\log n-\delta)} \left( \sum_{\substack{q\mid n\\ q<Y}}\Lambda(q) + \sum_{\substack{q\mid n\\ q\ge Y\\ n/q<x}}\Lambda(q) \right). \tag{3}

This means:

all small prime-power divisors $q<Y$ are inserted directly at $n$ , rather than represented by transitions;
among the large prime powers $q\ge Y$ , those whose parent $n/q$ lies below $x$ are also inserted directly at $n$ .

Let

B_x:=\sum_{n\ge x} b_x(n).

We will show

B_x=1+O_\delta\!\left(\frac1{\log x}\right). \tag{4}

First assume this and finish the main argument.

Start the sub-Markov chain with initial distribution

\mathbb P(N_0=n)=\frac{b_x(n)}{B_x}.

Let $v(n)$ be the probability that the chain ever visits $n$ .

Because the chain is increasing, the only ways to hit $n$ are:

start at $n$ , or
come from $n/q$ via one last jump with $q\ge Y$ .

v(n)=\frac{b_x(n)}{B_x} + \sum_{\substack{q\mid n\\ q\ge Y\\ n/q\ge x}} v(n/q)\,P(n/q,n). \tag{5}

Now I claim that

v(n)=\frac{W(n)}{B_x}\qquad(n\ge x). \tag{6}

This is proved by induction on $n$ . If it holds for all smaller integers, then from (5),

v(n)=\frac1{B_x}\left( b_x(n)+ \sum_{\substack{q\mid n\\ q\ge Y\\ n/q\ge x}} W(n/q)\,P(n/q,n) \right).

But

W(n/q)\,P(n/q,n) = \frac{1}{(n/q)(\log(n/q)-\delta)} \cdot \frac{(\log(n/q)-\delta)\Lambda(q)} {q\log n(\log n-\delta)} = \frac{\Lambda(q)}{n\log n(\log n-\delta)}.

Hence

b_x(n)+ \sum_{\substack{q\mid n\\ q\ge Y\\ n/q\ge x}} W(n/q)\,P(n/q,n) = \frac1{n\log n(\log n-\delta)} \sum_{q\mid n}\Lambda(q) = \frac1{n(\log n-\delta)} = W(n),

because the two sums in $b_x(n)$ together with the transition sum partition all divisors $q\mid n$ . Thus (6) holds.

Now let $A\subset[x,\infty)$ be primitive. A sample path cannot hit two distinct elements of $A$ , because along the path each earlier state divides each later state. Therefore

\sum_{a\in A} v(a)\le 1.

Using (6),

\sum_{a\in A}\frac1{a(\log a-\delta)} = \sum_{a\in A}W(a) = B_x\sum_{a\in A}v(a) \le B_x.

So by (4),

\sum_{a\in A}\frac1{a(\log a-\delta)} \le 1+O_\delta\!\left(\frac1{\log x}\right).

Finally,

\sum_{a\in A}\frac1{a\log a} \le \sum_{a\in A}\frac1{a(\log a-\delta)} \le 1+O_\delta\!\left(\frac1{\log x}\right) = 1+o(1).

This proves the theorem, assuming (4).

4. Estimating the total entrance mass $B_x$

Split $B_x$ according to the two terms in (3):

B_x=B_x^{\mathrm{small}}+B_x^{\mathrm{ent}}.

The small- $q$ part

We have

B_x^{\mathrm{small}} = \sum_{q<Y}\frac{\Lambda(q)}{q} \sum_{m\ge x/q} \frac1{m\log(mq)(\log(mq)-\delta)}.

For each fixed $q<Y$ , the inner summand is decreasing in $m$ , so by comparison with the integral,

\sum_{m\ge x/q} \frac1{m\log(mq)(\log(mq)-\delta)} \ll_\delta \int_{x/q}^{\infty} \frac{dt}{t\log(tq)(\log(tq)-\delta)}.

Substitute $u=\log(tq)$ , $dt/t=du$ . Since $t=x/q$ gives $u=\log x$ ,

\int_{x/q}^{\infty} \frac{dt}{t\log(tq)(\log(tq)-\delta)} = \frac1\delta\log\frac{\log x}{\log x-\delta} = O_\delta\!\left(\frac1{\log x}\right).

Because $Y$ is fixed,

B_x^{\mathrm{small}} = O_\delta\!\left(\frac1{\log x}\right). \tag{7}

The first-entry part

Now

B_x^{\mathrm{ent}} = \sum_{m<x}\frac1m \sum_{q\ge \max(Y,x/m)} \frac{\Lambda(q)}{q\log(mq)(\log(mq)-\delta)}.

Apply the tail estimate (1) with $y=\max(Y,x/m)$ . Then

m\,y=\max(mY,x),

so if we put

M_m:=\log(m\,y)=\max(\log(mY),\log x),

we have

M_m\in[\log x,\ \log x+\log Y].

Therefore, uniformly in $m<x$ ,

\sum_{q\ge \max(Y,x/m)} \frac{\Lambda(q)}{q\log(mq)(\log(mq)-\delta)} = \frac1\delta\log\frac{M_m}{M_m-\delta} + O_\delta\!\left(\frac1{M_m^2}\right) = \frac1{\log x}+O_\delta\!\left(\frac1{\log^2 x}\right).

Hence

B_x^{\mathrm{ent}} = \left(\frac1{\log x}+O_\delta\!\left(\frac1{\log^2 x}\right)\right) \sum_{m<x}\frac1m.

Since $\sum_{m<x}1/m=\log x+O(1)$ ,

B_x^{\mathrm{ent}} = 1+O_\delta\!\left(\frac1{\log x}\right). \tag{8}

From (7) and (8),

B_x=1+O_\delta\!\left(\frac1{\log x}\right),

which is exactly (4).

The proof is complete. ∎

Conclusion

So the answer to your problem is yes:

\sup_{\substack{A\subset [x,\infty)\\ A\ \text{primitive}}} \sum_{a\in A}\frac1{a\log a} \le 1+o(1).

The argument actually gives the stronger weighted form

\sum_{a\in A}\frac1{a(\log a-\delta)}\le 1+O_\delta\!\left(\frac1{\log x}\right)

for every fixed $0<\delta<\log2$ .

This is also sharp at the constant level, since the usual $k$ -almost-prime examples already show that the extremal value approaches $1$ .

no internet: Please write a full correct resolution to the problem formatted as a publishable maths research paper using amsart using a4paper, margin=1in. Keep the title brief and to the point. The abstract should only be at most 6 sentences. Use section headings sparingly. Do not add an author entry. Be rigorous and self-contained. Ensure to address any issues you raised. Give the LaTeX in a code markdown block.

I’m checking the argument carefully before I typeset it, especially the normalization and the tail estimate, so the paper version is fully watertight.

I’m also stress-testing the cutoff and sub-Markov construction, including boundary terms, to ensure the first-entry decomposition gives an exact $1+O(1/\log x)$ majorant and the weight recursion closes correctly.

I’ve now shaped the core argument into a paper draft, and I’m polishing the prime-power Mertens lemma so that the analytic input is stated cleanly and proved in a publishable form.

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Chat history

Proof

1. A tail estimate for prime powers

2. Choose a cutoff $Y$ and define the jump kernel

3. The entrance mass

4. Estimating the total entrance mass $B_x$

The small- $q$ part

The first-entry part

Conclusion

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Chat history

You said:

ChatGPT said:

Proof

1. A tail estimate for prime powers

2. Choose a cutoff YYY and define the jump kernel

3. The entrance mass

4. Estimating the total entrance mass BxB_xBx​

The small-qqq part

The first-entry part

Conclusion

You said:

ChatGPT said:

2. Choose a cutoff $Y$ and define the jump kernel

4. Estimating the total entrance mass $B_x$

The small- $q$ part