gpt pro 5.2 attempt on Math-831

FORMAL RESTATEMENT

Let $A\subseteq\mathbb{N}$ be a finite nonempty set, and define \[ B := \{ n\ge 1 : \exists a\in A\ \text{such that } a\mid n\}, \] the set of positive integers divisible by at least one element of $A$. For $x\ge 1$ define the counting function \[F(x):=|B\cap[1,x]|\quad\text{and the density-initial-segment }\quad d(x):=\frac{F(x)}{x}. \] The problem asks whether the following inequality always holds: \[ \forall m>n\ge \max(A):\qquad \frac{F(m)}{m} < 2\frac{F(n)}{n}. \] It also notes that the constant $2$ would be best possible, since for $A=\{a\}$, $n=2a-1$, $m=2a$ makes the ratio approach $2$ from below.

QUICK LITERATURE/CONTEXT CHECK

The statement notes that an alternate version (with $a\mid n$ replaced by $a\nmid n$) likely contains a typo and has known counterexamples; the version stated here is the "multiples" version. Per the project integrity rule, I do not use results not explicitly present in the problem text.

ATTACK PLAN

Proof track:

• Use the structure of $B$ as a finite union of arithmetic progressions to obtain periodicity modulo $L=\mathrm{lcm}(A)$ and extract asymptotics $d(x)=\delta+O(1/x)$. This proves the inequality for sufficiently large $n$.
• Analyze small $n$ relative to $L$ by expressing $F(x)$ exactly in terms of $q=\lfloor x/L\rfloor$ and remainder $r=x\bmod L$.

Disproof track:

• Search computationally for a counterexample $(A,n,m)$ with small parameters.

I obtain (a) full proofs in some special regimes and (b) computational evidence for small parameters, but no complete proof or counterexample in general.

WORK

Lemma 1 (removing redundant divisors).

If $A$ contains two elements $a,b$ with $a\mid b$, then replacing $A$ by $A\setminus\{b\}$ does not change $B$. In particular, one may assume without loss of generality that $A$ is primitive: no element of $A$ divides another.

Proof.

If $a\mid b$, then every multiple of $b$ is automatically a multiple of $a$. Thus \[\{n: b\mid n\}\subseteq\{n: a\mid n\}. \] Therefore the union over $A$ of sets of multiples is unchanged if we remove $b$: \[ \bigcup_{t\in A}\{n: t\mid n\}=\left(\bigcup_{t\in A\setminus\{b\}}\{n: t\mid n\}\right)\cup\{n: b\mid n\}=\bigcup_{t\in A\setminus\{b\}}\{n: t\mid n\}. \] So $B$ is unchanged. $\square$

Lemma 2 (single divisor case; constant $2$ is sharp).

Let $A=\{a\}$ with $a\ge 1$. Then for every $m>n\ge a$, \[\frac{\lfloor m/a\rfloor}{m} < 2\frac{\lfloor n/a\rfloor}{n}. \] Moreover, taking $n=2a-1$ and $m=2a$ gives \[\frac{\lfloor m/a\rfloor/m}{\lfloor n/a\rfloor/n}=\frac{(2/a)}{(1/(2a-1))}=2-\frac{1}{a},\] showing the constant $2$ cannot be improved.

Proof.

Write $m=qa+r$ and $n=pa+s$ with integers $q,p\ge 1$ and remainders $0\le r,s\le a-1$. Then $\lfloor m/a\rfloor=q$ and $\lfloor n/a\rfloor=p$. We must show \[\frac{q}{qa+r} < 2\frac{p}{pa+s}. \] Cross-multiplying (all quantities positive) gives the equivalent inequality \[(pa+s)q < 2p(qa+r). \] Expand and cancel the common term $pqa$: \[s q < pqa + 2pr. \] Since $s\le a-1$, we have $sq\le (a-1)q < aq$. Since $p\ge 1$, $pqa \ge aq$. Thus $sq < pqa\le pqa+2pr$, proving the inequality. (Strictness holds because $sq<aq\le pqa$.)

For sharpness, if $n=2a-1$ then $\lfloor n/a\rfloor=1$, while for $m=2a$ we have $\lfloor m/a\rfloor=2$, yielding the stated ratio $2-1/a$. $\square$

Lemma 3 (periodicity and eventual validity for large $n$).

Let $A$ be finite and nonempty, and set $L:=\mathrm{lcm}(A)$. Then $B$ is periodic modulo $L$, and if we let \[\delta:=\frac{|B\cap[1,L]|}{L}>0, \] then for every integer $x\ge 1$, \[\left|d(x)-\delta\right|\le \frac{L}{x}. \] Consequently, for all $n\ge \max(A)$ with $n>\frac{3L}{\delta}$ and all $m>n$, the desired inequality $d(m)<2d(n)$ holds.

Proof.

Periodicity. If $n\equiv n'\pmod L$ then for every $a\in A$ (with $a\mid L$) we have $n\equiv n'\pmod a$, hence $a\mid n \Leftrightarrow a\mid n'$. Therefore membership in $B$ depends only on residue modulo $L$.

Density approximation. Write $x=qL+r$ with $q\ge 0$ and $0\le r<L$. Let $b:=|B\cap[1,L]|$ and $b(r):=|B\cap[1,r]|$. By periodicity, \[F(x)=qb + b(r). \] Thus \[d(x)=\frac{qb+b(r)}{qL+r}. \] Since $0\le b(r)\le L$ and $0\le r<L$, we can bound \[ \left|\frac{qb+b(r)}{qL+r} - \frac{b}{L}\right| = \left|\frac{L b(r) - b r}{L(qL+r)}\right| \le \frac{L\cdot L + b\cdot L}{L(qL+r)} \le \frac{2L}{x}. \] A slightly sharper bound is available because $b\le L$, giving \[\left|d(x)-\delta\right|\le \frac{L}{x}. \] (Indeed, $|L b(r) - b r|\le L\max\{b(r),b\}\le L^2$ and $L(qL+r)=Lx$.)

Eventual inequality. For $m>n$ we have \[d(m)\le \delta + \frac{L}{m} < \delta + \frac{L}{n},\qquad d(n)\ge \delta - \frac{L}{n}. \] If $n>\frac{3L}{\delta}$ then $\delta - L/n > \delta - \delta/3 = 2\delta/3>0$, and \[\delta + \frac{L}{n} < \delta + \frac{\delta}{3} = \frac{4\delta}{3} = 2\cdot\frac{2\delta}{3} \le 2\left(\delta - \frac{L}{n}\right) \le 2 d(n). \] Therefore $d(m)<2d(n)$ for all $m>n>3L/\delta$. $\square$

FAST REALITY CHECK (brute-force search for small counterexamples).

I searched over all sets $A\subseteq\{2,3,\dots,20\}$ with $|A|\le 5$ and all pairs $m>n\ge \max(A)$ with $m\le 200$. No counterexample to $F(m)/m < 2F(n)/n$ was found in this range.

VERIFICATION

• Lemma 2: verified the algebraic reduction and strictness. Also cross-checked against the explicit "sharpness" example $n=2a-1,m=2a$.
• Lemma 3: verified the periodic decomposition $F(qL+r)=qb+b(r)$ and the inequality chain in the "eventual" part; the constant $3L/\delta$ is a sufficient (not claimed optimal) threshold.
• Computational search: checked that the brute-force code tests all $(A,n,m)$ in the stated finite search box.

FINAL

UNRESOLVED

(i) Strongest proved partial result.

• If $A=\{a\}$ is a singleton, the inequality holds for all $m>n\ge a$, and the constant $2$ is best possible.
• For general finite $A$, writing $L=\mathrm{lcm}(A)$ and $\delta=|B\cap[1,L]|/L$, we have $d(x)=\delta+O(L/x)$ and therefore $d(m)<2d(n)$ holds for all $m>n>3L/\delta$.
• Brute-force search found no counterexample for $A\subseteq\{2,\dots,20\}$ with $|A|\le 5$ and $m,n\le 200$.

(ii) First gap (crisp). Prove or disprove the uniform inequality \[\forall\,\text{finite }A\subseteq\mathbb{N}\ \forall m>n\ge\max(A):\ \frac{|B\cap[1,m]|}{m}<2\frac{|B\cap[1,n]|}{n},\] where $B$ is the set of multiples of elements of $A$. My proofs only cover (a) singletons and (b) sufficiently large $n$ depending on $A$.

(iii) Top 3 next moves.

1. Use the exact periodic formula $F(qL+r)=qb+b(r)$ to reduce the inequality to finitely many inequalities in the variables $(q_n,r_n,q_m,r_m)$, perhaps showing the worst case occurs for small $q_n$.
2. Attempt to prove a monotonicity/convexity statement for the prefix averages of a "multiples" periodic set that would force the factor-$2$ bound.
3. Computational: extend exhaustive search to larger $\max(A)$ and larger $|A|$ (e.g. up to $\max(A)=50$) and record any near-tight examples to guess a general extremal configuration.

(iv) Minimal counterexample structure. A counterexample would consist of a finite set $A$ and a pair $m>n\ge\max(A)$ where the initial density $d(n)$ is unusually small compared to the eventual density $\delta$ of $B$, while $d(m)$ is already close to $\delta$. Such behaviour would likely require $L=\mathrm{lcm}(A)$ to be large and the set of "hit" residues modulo $L$ to be very unevenly distributed among initial segments.