gpt pro 5.2 attempt on Math-5442

Erdős Problem #122

1) FORMAL RESTATEMENT

Let $f:\mathbb N\to\mathbb N$ be an arithmetic function. Consider any function $F:\mathbb N\to(0,\infty)$ such that \[ \frac{f(n)}{F(n)}\to 0\quad\text{for almost all }n \] (i.e. for all $\varepsilon>0$ the set $\{n: f(n)/F(n)>\varepsilon\}$ has asymptotic density $0$).

Question: for which functions $f$ is it true that there exist infinitely many $x\in\mathbb N$ such that \[ \frac{\#\{n\in\mathbb N:\ n+f(n)\in (x,x+F(x))\}}{F(x)}\to\infty\ ? \] (Here $(x,x+F(x))$ denotes the open interval in $\mathbb R$; equivalently we count those $n$ with $x<n+f(n)<x+F(x)$.)

2) QUICK LITERATURE/CONTEXT CHECK

From the problem text:

• Erdős--Pomerance--Sárközy proved the property holds for $f(n)=d(n)$ (divisor function) and for $f(n)=\omega(n)$ (number of distinct prime factors).
• Erdős believed it is false for $f(n)=\varphi(n)$ or $f(n)=\sigma(n)$.

No further sources are used here.

3) ATTACK PLAN

Try to identify simple necessary obstructions: classes of functions $f$ for which the conclusion cannot hold for some admissible choice of $F$. I give two such obstructions (bounded $f$, and eventually nondecreasing $f$), which already show the property is highly nontrivial and requires strong oscillation.

4) WORK

Lemma 122.1 (bounded $f$ cannot satisfy the conclusion).

Suppose $f$ is bounded: $f(n)\le B$ for all $n$. Then $f$ does not have the stated property.

Proof. Choose any function $F(n)\to\infty$ (e.g. $F(n)=n$). Then $f(n)/F(n)\to0$ for all $n$ and hence "for almost all $n$".

For any $x$, if $n+f(n)\in (x,x+F(x))$, then necessarily $n\in (x-B,\,x+F(x))$. Therefore \[ \#\{n:\ n+f(n)\in(x,x+F(x))\}\le \#\big(\mathbb N\cap(x-B,\,x+F(x))\big)\le F(x)+B+1. \] Dividing by $F(x)$ yields \[ \frac{\#\{n:\ n+f(n)\in(x,x+F(x))\}}{F(x)}\le 1+\frac{B+1}{F(x)}\to 1. \] In particular, this ratio cannot tend to $\infty$ along any sequence of $x$. Thus bounded $f$ fails the required property. $\square$

Lemma 122.2 (eventually nondecreasing $f$ cannot satisfy the conclusion).

Suppose $f$ is eventually nondecreasing: there exists $N_0$ such that $f(n+1)\ge f(n)$ for all $n\ge N_0$. Then $f$ does not have the stated property.

Proof. Define $g(n):=n+f(n)$. For $n\ge N_0$, we have \[ g(n+1)-g(n)=(n+1+f(n+1))-(n+f(n))=1+(f(n+1)-f(n))\ge 1. \] Hence $g$ is strictly increasing on $\{N_0,N_0+1,\dots\}$.

Fix any admissible $F$ with $f(n)/F(n)\to0$ for almost all $n$. For any $x$, the set $ \{n\ge N_0:\ g(n)\in(x,x+F(x))\} $ consists of integers $n$ whose strictly increasing images lie in an interval of length $F(x)$. Since $g$ increases by at least $1$ each step, there can be at most $\lceil F(x)\rceil$ such $n$ (more formally: the values $g(n)$ are distinct integers separated by $\ge1$, so an open interval of length $F(x)$ contains at most $\lceil F(x)\rceil$ of them). Thus for all sufficiently large $x$, \[ \#\{n:\ g(n)\in(x,x+F(x))\}\le F(x)+1. \] Dividing by $F(x)$ gives a uniform bound \[ \frac{\#\{n:\ n+f(n)\in(x,x+F(x))\}}{F(x)}\le 1+\frac{1}{F(x)}\le 2 \] for all large $x$. Hence the ratio cannot tend to $\infty$. Therefore any eventually nondecreasing $f$ fails the property. $\square$

5) VERIFICATION (FAST REALITY CHECK)

• Lemma 122.1 applies to $f(n)\equiv 1$ (bounded). Indeed $n+1$ hits any interval of length $F(x)$ in about $F(x)$ ways, ratio $\asymp 1$.
• Lemma 122.2 applies to $f(n)=n$ or $f(n)=n^2$ (eventually increasing). The map $n\mapsto n+f(n)$ is strictly increasing, so the count in any length-$F(x)$ interval is $O(F(x))$ and the normalized ratio is bounded.

6) FINAL

UNRESOLVED.

(i) Strongest fully proved partial result obtained here. Two rigorous obstruction lemmas: bounded $f$ cannot satisfy the property (Lemma 122.1), and eventually nondecreasing $f$ cannot satisfy it (Lemma 122.2).

(ii) Exact first gap. Beyond these obstructions, characterize which highly oscillatory arithmetic functions (such as $d(n)$ or $\omega(n)$) do satisfy the "local repetition" phenomenon for every admissible $F$.

(iii) Top 3 next moves.

1. Try to prove a sufficient condition in terms of distribution of values of $f(n)$ in short intervals (e.g. many repeats) that forces the normalized count to become large.
2. Test the conjectured negative cases $\varphi(n)$ and $\sigma(n)$ by attempting to construct an explicit admissible $F$ that keeps the normalized ratio bounded.
3. Explore probabilistic models: treat $f(n)$ as a random-like multiplicative function and estimate clustering of $n+f(n)$.

(iv) Minimal counterexample structure. If Erdős' belief is correct for $\varphi$ or $\sigma$, a minimal counterexample would be a specific admissible $F$ such that, along all sequences of $x\to\infty$, the interval $(x,x+F(x))$ contains only $O(F(x))$ values of $n+f(n)$ with a uniformly bounded proportionality constant.