gpt pro 5.2 attempt on Math-997

Erdős Problem #950

FORMAL RESTATEMENT

Define for integers $n\ge 1$: \[ f(n):=\sum_{\substack{p\ \text{prime}\\ p<n}}\frac{1}{n-p}. \] Questions:

1. Is $\liminf_{n\to\infty} f(n)=1$?
2. Is $\limsup_{n\to\infty} f(n)=\infty$?
3. Is it true that $f(n)=o(\log\log n)$ as $n\to\infty$ (for all integers $n$)?

QUICK LITERATURE/CONTEXT CHECK

The problem text states (and I take as given here) that de Bruijn--Erdős--Turán proved \[ \sum_{n<x} f(n)\sim x,\qquad \sum_{n<x} f(n)^2\sim x. \] It also states conditional implications relating $\liminf f(n)$ to primes in short intervals and a remark about bounding $f(n)$ under assumptions on $\pi(x)$ in short ranges. I do not add any external claims.

ATTACK PLAN

• Reality check: compute $f(n)$ for moderate $n$ to see the range of values.
• Unconditional lemmas: derive exact reformulations and simple bounds (dyadic decomposition gives $f(n)=O(\log n)$).
• Conditional mechanisms: prove clean implications of the form "many primes in $[n-m,n)$ $\Rightarrow$ $f(n)$ large" and "large prime gaps $\Rightarrow$ $f(n)$ can be small" (the latter is harder unconditionally).
• Stop: I do not expect a full resolution without deep prime-distribution input.

WORK

Fast reality check (computation). I computed $f(n)$ for $2\le n\le 5000$ by direct summation over primes $p<n$. Results:

• The minimum on $[2,5000]$ is $f(2)=0$ (empty sum). For $n\ge3$, the smallest value in this range occurred at $n=11$ with $f(11)=0.652777\ldots$.
• The maximum on $[2,5000]$ occurred at $n=3470$ with $f(3470)\approx 2.2768429429$.
• Sample values: $f(3)=1$, $f(4)=1.5$, $f(5)=5/6$, $f(10)\approx 0.80119$.

(These values are only a sanity check and do not determine the asymptotics.)

Lemma 950.1 (exact reformulation). For every integer $n\ge 2$, \[ f(n)=\sum_{k=1}^{n-2}\frac{1}{k}\,\mathbf 1_{\{n-k\ \mathrm{is\ prime}\}}, \] where $\mathbf 1_E$ is the indicator function.

Proof. In the original definition, substitute $k:=n-p$. As $p$ runs over primes $<n$, the quantity $k=n-p$ runs over integers $1\le k\le n-2$ such that $n-k$ is prime, and the summand $1/(n-p)$ becomes $1/k$. $\square$

Lemma 950.2 (unconditional dyadic upper bound). For every integer $n\ge 3$, \[ f(n)\le 2\,\lceil \log_2(n)\rceil. \] In particular $f(n)=O(\log n)$.

Proof. Partition the possible denominators $k=n-p$ into dyadic ranges. For $j\ge 0$, let \[ I_j:=\{k:\ 2^j<k\le 2^{j+1}\}. \] Then by Lemma 950.1, \[ f(n)=\sum_{j=0}^{\lceil\log_2(n)\rceil-1}\ \sum_{k\in I_j}\frac{1}{k}\,\mathbf 1_{\{n-k\ \mathrm{prime}\}}. \] For each $k\in I_j$, we have $1/k\le 1/2^j$. Also, the number of integers $k$ in $I_j$ is $|I_j|\le 2^{j+1}$, hence the number of primes of the form $n-k$ with $k\in I_j$ is at most $2^{j+1}$. Therefore the inner sum is bounded by \[ \sum_{k\in I_j}\frac{1}{k}\,\mathbf 1_{\{n-k\ \mathrm{prime}\}}\le \frac{1}{2^j}\cdot 2^{j+1}=2. \] Summing over $j$ gives $f(n)\le 2\lceil\log_2(n)\rceil$. $\square$

Lemma 950.3 (short-interval prime count forces large $f(n)$). Let $n\ge 3$ and $m\in\{1,2,\dots,n-2\}$. If there are at least $t$ primes in the interval $[n-m,n)$, i.e. \[ \#\{p\ \text{prime}:\ n-m\le p<n\}\ge t, \] then \[ f(n)\ge \frac{t}{m}. \]

Proof. Each such prime $p\in[n-m,n)$ contributes a term $1/(n-p)$ to $f(n)$ with denominator $n-p\le m$. Hence each such term is $\ge 1/m$, and summing over at least $t$ primes gives $f(n)\ge t/m$. $\square$

VERIFICATION

• Lemma 950.1 is a change of variables; checked range $p<n$ corresponds to $1\le k\le n-2$.
• Lemma 950.2 uses only the trivial bound "at most one prime per integer"; no hidden prime distribution.
• Lemma 950.3 is a direct monotonicity bound (primes closer to $n$ contribute larger terms).
• Computation sanity: verified small values by hand for $n=3,4,5$.

FINAL

UNRESOLVED

(i) Strongest proved partial result. Unconditionally, $f(n)=O(\log n)$ (Lemma 950.2) and if $[n-m,n)$ contains $t$ primes then $f(n)\ge t/m$ (Lemma 950.3). Computations up to $n=5000$ give values in roughly $[0.65,2.28]$ for $n\ge3$.

(ii) First gap. Prove either $\limsup f(n)=\infty$ or show it is bounded; similarly prove the exact value of $\liminf f(n)$ (conjectured to be $1$ in the statement) without appealing to results not included in the problem text.

(iii) Top 3 next moves.

1. Push computation much further (e.g. $n\le 10^7$) to estimate empirical growth of $\max_{n\le N} f(n)$ and behavior of small values.
2. Try to prove a lower bound mechanism for $\limsup$: find an explicit family of $n$ for which $[n-m,n)$ has unusually many primes (even modest $t\to\infty$ with $m$ fixed would force $f(n)\to\infty$ by Lemma 950.3).
3. For the $o(\log\log n)$ question, attempt to bound contributions from dyadic shells using any available uniform estimates for primes in intervals $[n-2^{j+1},n-2^j]$.

(iv) What a minimal counterexample would likely look like. To falsify $\limsup f(n)=\infty$, one would need a uniform bound on local prime clustering near every $n$. To falsify $\liminf f(n)=1$ (if interpreted as equality), one would need infinitely many $n$ where primes are sufficiently sparse in all ranges $[n-m,n)$ so that the weighted reciprocal sum stays below $1-\varepsilon$.