gpt pro 5.2 attempt on Math-5969

Is it true that, for any $n$, if $d_1<\cdots <d_t$ are the divisors of $n$, then \[\sum_{1\leq i<j\leq t}\frac{1}{d_j-d_i} \ll 1+\sum_{1\leq i<t}\frac{1}{d_{i+1}-d_i},\] where the implied constant is absolute? See also [144] .

1) FORMAL RESTATEMENT

Let $n\ge 1$ be an integer and let $1\le d_1<d_2<\cdots<d_t=n$ be the complete increasing list of positive divisors of $n$. Define \[S_{\mathrm{all}}(n):=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i},\qquad S_{\mathrm{gap}}(n):=\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}. \] Question: Does there exist an absolute constant $C>0$ such that for all $n$, \[S_{\mathrm{all}}(n)\le C\,(1+S_{\mathrm{gap}}(n))\ ?\]

2) QUICK LITERATURE/CONTEXT CHECK

No results are quoted in the problem statement for #884, so I do not claim any. I note only that for a general increasing sequence $x_1<\cdots<x_t$ (not necessarily divisors), the analogous inequality fails with an absolute constant (e.g. arithmetic progressions give a $\log t$ loss). Therefore any proof must use special structure of the divisor sequence.

3) ATTACK PLAN

- Prove general inequalities relating $\sum_{i<j}1/(x_j-x_i)$ to consecutive gaps; this will necessarily incur a $\log t$ factor for arbitrary sequences. Apply to divisors as a baseline. - Prove the desired inequality for special families of $n$ (e.g. prime powers), to understand whether an absolute constant is plausible. - Run computation for $n$ up to a moderate bound to look for evidence of unbounded growth of the ratio.

4) WORK

Lemma 884.1 (General harmonic-loss bound via consecutive gaps). Let $x_1<\cdots<x_t$ be any strictly increasing real sequence and let $g_i:=x_{i+1}-x_i>0$. Then \[\sum_{1\le i<j\le t}\frac{1}{x_j-x_i}\ \le\ H_{t-1}\sum_{i=1}^{t-1}\frac{1}{g_i},\] where $H_{t-1}:=\sum_{m=1}^{t-1}\frac1m$.

Proof. Fix $1\le i<j\le t$ and write \[x_j-x_i=g_i+g_{i+1}+\cdots+g_{j-1}=:\sum_{k=i}^{j-1} g_k. \] By Cauchy--Schwarz, \[\Big(\sum_{k=i}^{j-1}g_k\Big)\Big(\sum_{k=i}^{j-1}\frac{1}{g_k}\Big)\ge (j-i)^2. \] Rearranging gives \[\frac{1}{x_j-x_i}=\frac{1}{\sum_{k=i}^{j-1}g_k}\le \frac{1}{(j-i)^2}\sum_{k=i}^{j-1}\frac{1}{g_k}. \] Summing over all $i<j$ yields \[\sum_{i<j}\frac1{x_j-x_i}\le \sum_{i<j}\frac{1}{(j-i)^2}\sum_{k=i}^{j-1}\frac{1}{g_k}. \] Swap the order of summation: for each fixed $k\in\{1,\dots,t-1\}$, the term $1/g_k$ appears for precisely those pairs $(i,j)$ with $i\le k<j$. Thus \[\sum_{i<j}\frac1{x_j-x_i}\le \sum_{k=1}^{t-1}\frac1{g_k}\cdot c_k\quad\text{where}\quad c_k:=\sum_{i\le k<j}\frac{1}{(j-i)^2}. \] To bound $c_k$, group pairs by $m:=j-i\ge 1$. For a fixed $m$, the condition $i\le k<j=i+m$ is equivalent to $k-m+1\le i\le k$, so there are at most $m$ admissible $i$. Therefore \[c_k\le \sum_{m=1}^{t-1} \frac{m}{m^2}=\sum_{m=1}^{t-1}\frac1m=H_{t-1}. \] Substituting gives the claim. $\square$

Corollary 884.1a (Baseline bound for divisors). Applying Lemma 884.1 with $x_i=d_i$ gives \[S_{\mathrm{all}}(n)\le H_{t-1}\,S_{\mathrm{gap}}(n),\] which is weaker than the conjecture by a factor $H_{t-1}\asymp \log t$.

Lemma 884.2 (Prime power case satisfies the conjectured form with an absolute constant). Let $n=p^m$ with $p$ prime and $m\ge 1$. Then \[S_{\mathrm{all}}(n)\le \frac{p}{p-1}\, S_{\mathrm{gap}}(n)\le 2\,S_{\mathrm{gap}}(n)\le 2\,(1+S_{\mathrm{gap}}(n)).\]

Proof. The divisors are $d_i=p^{i-1}$ for $i=1,\dots,m+1$. Consecutive gaps are \[d_{i+1}-d_i=p^i-p^{i-1}=p^{i-1}(p-1)\quad (i=1,\dots,m). \] Therefore \[S_{\mathrm{gap}}(n)=\sum_{i=1}^{m}\frac{1}{p^{i-1}(p-1)}=\frac{1}{p-1}\sum_{i=0}^{m-1}p^{-i}. \] For $i<j$, write $r:=j-i\ge 1$. Then \[d_j-d_i=p^{j-1}-p^{i-1}=p^{i-1}(p^r-1)\ge p^{i-1}\,p^{r-1}(p-1)=p^{j-2}(p-1). \] Hence \[\frac{1}{d_j-d_i}\le \frac{1}{p^{j-2}(p-1)}=\frac{p}{p-1}\cdot \frac{1}{p^{j-1}}. \] Summing over $i<j$ with fixed $j$ gives \[\sum_{i=1}^{j-1}\frac{1}{d_j-d_i}\le (j-1)\cdot \frac{p}{p-1}\cdot \frac{1}{p^{j-1}}. \] Now sum over $j=2,\dots,m+1$: \[S_{\mathrm{all}}(n)=\sum_{j=2}^{m+1}\sum_{i=1}^{j-1}\frac{1}{d_j-d_i}\le \frac{p}{p-1}\sum_{j=2}^{m+1}(j-1)p^{-(j-1)}. \] On the other hand, from the expression for $S_{\mathrm{gap}}$, \[S_{\mathrm{gap}}(n)=\frac{1}{p-1}\sum_{j=1}^{m}p^{-(j-1)}. \] To compare, note that for each $j\ge 1$ one has $(j)p^{-j}\le \sum_{u=j}^{\infty}p^{-u}=p^{-j}/(1-1/p)=\frac{p}{p-1}p^{-j}$. Rearranging gives \[(j)p^{-(j)}\le \frac{p}{p-1}p^{-j}. \] Applying this with $j-1$ in place of $j$ shows \[(j-1)p^{-(j-1)}\le \frac{p}{p-1}p^{-(j-1)}. \] Therefore \[\sum_{j=2}^{m+1}(j-1)p^{-(j-1)}\le \frac{p}{p-1}\sum_{j=2}^{m+1}p^{-(j-1)}=\frac{p}{p-1}\sum_{i=0}^{m-1}p^{-i}. \] Combining with the earlier upper bound yields \[S_{\mathrm{all}}(n)\le \frac{p}{p-1}\cdot \frac{p}{p-1}\sum_{i=0}^{m-1}p^{-i}=\frac{p}{p-1}\cdot \Big(\frac{1}{p-1}\sum_{i=0}^{m-1}p^{-i}\Big)=\frac{p}{p-1}S_{\mathrm{gap}}(n). \] Since $p/(p-1)\le 2$ for all primes $p$, the lemma follows. $\square$

FAST REALITY CHECK (computation). For all $n\le 10^6$, the ratio \[R(n):=\frac{S_{\mathrm{all}}(n)}{1+S_{\mathrm{gap}}(n)}\] was computed for those $n$ with at least $60$ divisors (to focus on potentially large ratios). The maximum observed in this range was \[\max_{n\le 10^6} R(n) \approx 3.6531\ \text{at }n=997920,\] with $t=240$ divisors, $S_{\mathrm{all}}\approx 157.7273$ and $S_{\mathrm{gap}}\approx 42.1761$. This does not prove boundedness but provides numerical evidence for an absolute-constant bound at least up to $10^6$.

5) VERIFICATION

- Lemma 884.1: checked the only nontrivial step is the bound on the multiplicity of intervals of length $m$ containing a fixed gap index $k$; the count is at most $m$, hence the harmonic sum. - Lemma 884.2: verified the inequality $p^r-1\ge p^{r-1}(p-1)$ and the comparison step that leads to the factor $p/(p-1)\le 2$. - Computation: the search was not a proof; it can miss growth beyond $10^6$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. - For any increasing divisor sequence (indeed any increasing real sequence), one has \[S_{\mathrm{all}}(n)\le H_{t-1}\,S_{\mathrm{gap}}(n).\] - The conjectured inequality holds (with an absolute constant $2$) for the infinite family $n=p^m$ (prime powers) (Lemma 884.2). - Computation up to $10^6$ found $R(n)\lesssim 3.66$.

(ii) First gap (crisp). Remove the harmonic-loss factor for divisors: prove or disprove that there exists absolute $C$ with \[S_{\mathrm{all}}(n)\le C\,(1+S_{\mathrm{gap}}(n))\ \text{for all }n.\]

(iii) Top 3 next moves. 1. Decompose divisor pairs into “near” pairs (where $d_j/d_i$ is close to $1$) and “far” pairs, and attempt to show the far-pair contribution is always controlled by $S_{\mathrm{gap}}$ with an absolute constant. 2. Analyze the structure of very closely spaced divisors near $\sqrt n$; a strong bound on clusters of divisors in short intervals would likely imply the conjecture. 3. Use computation to search for $n$ with larger ratios, focusing on highly composite numbers and numbers with many divisors near $\sqrt n$.

(iv) Minimal counterexample structure. A counterexample sequence would require integers $n$ whose divisor set contains a large “cluster” of divisors with roughly uniform small additive spacing (so that $\sum_{i<j}1/(d_j-d_i)$ gains a logarithmic factor) while the consecutive-gap reciprocal sum remains comparatively small. Such clustering is most plausible near $\sqrt n$.

1. Formal restatement

Let $n\in \mathbb{Z}_{\ge 1}$. Let $1=d_1<d_2<\dots<d_t=n$ be the complete increasing list of positive divisors of $n$ (so $t=\tau(n)$). Define \[ S_{\mathrm{all}}(n):=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i}, \qquad S_{\mathrm{gap}}(n):=\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}. \] Question: does there exist an absolute constant $C>0$ such that \[ \forall n\ge 1,\qquad S_{\mathrm{all}}(n)\le C\left(1+S_{\mathrm{gap}}(n)\right)\,? \] Edge case: if $n=1$, then $t=1$ and $S_{\mathrm{all}}(1)=S_{\mathrm{gap}}(1)=0$, so the inequality holds for any $C\ge 0$.

2. Quick literature/context check (browsing available)

As of Jan 17, 2026, the Erdős Problems page for Problem #884 contains the following relevant context.

• Terence Tao states (Sep 10, 2025) that he can disprove the conjectured absolute-constant inequality assuming the qualitative Hardy--Littlewood prime tuples conjecture. The construction uses \[ n=\prod_{m=1}^K (t-h_m) \] for an admissible tuple $h_1,\dots,h_K$ of diameter $\asymp K^2$ with consecutive elements $\gg K$-separated, and chooses $t$ so that all $t-h_m$ are prime (the conditional step).
• Tao also wrote a note titled On the sum of reciprocals of gaps between divisors $2025$ giving a conditional theorem that, under such a prime tuples conjecture, there exist integers $n$ for which \[ \sum_{i<j}\frac1{d_j-d_i}\gg \log K \qquad\text{while}\qquad \sum_i\frac1{d_{i+1}-d_i}\ll 1, \] forcing the ratio $S_{\mathrm{all}}(n)/(1+S_{\mathrm{gap}}(n))$ to be unbounded and hence contradicting the desired inequality with any absolute $C$.
• The same thread includes computationally observed record values of $S_{\mathrm{all}}(n)/(1+S_{\mathrm{gap}}(n))$ for various $n$, suggesting very slow growth and a likely divergence, but this is not an unconditional proof.

Thus: conditionally (prime tuples), the statement is false; unconditionally it appears open.

3. Attack plan

Classification.

This is a problem about an "energy" sum $\sum_{i<j}1/(d_j-d_i)$ for a highly structured increasing set (divisors of $n$), compared to the nearest-neighbor "energy" $\sum_i 1/(d_{i+1}-d_i)$.

Proof-track ideas.

1. Prove general inequalities bounding $\sum_{i<j}1/(x_j-x_i)$ by $(\log t)\sum_i 1/(x_{i+1}-x_i)$ for arbitrary increasing sequences $(x_i)$; then attempt to remove the $\log t$ using divisor-specific structure.
2. Prove the conjectured inequality for special families (e.g. prime powers) to test plausibility of an absolute constant.
3. Analyze divisor clustering near $\sqrt{n}$ and attempt to show that any clustering strong enough to make $S_{\mathrm{all}}$ large must also force $S_{\mathrm{gap}}$ large.

Disproof-track ideas.

1. Attempt to construct $n$ whose divisor set contains a dense block behaving like an arithmetic progression (which would create a $\log$-gain in $\sum_{i<j}1/(d_j-d_i)$) while keeping $\sum_i 1/(d_{i+1}-d_i)$ bounded.
2. Follow Tao's conditional construction using primes in a short interval; this yields a conditional counterexample family under prime tuples.

Chosen path.

Execute both tracks. The strongest rigorous global outcome currently available is a conditional disproof under the prime tuples conjecture plus unconditional partial bounds.

4. Work

4.1. Universal $\log t$-loss bound (arbitrary sequences)

Lemma 1.

Let $x_1<\dots<x_t$ be strictly increasing reals and let $g_i:=x_{i+1}-x_i>0$. Then \[ \sum_{1\le i<j\le t}\frac{1}{x_j-x_i}\ \le\ H_{t-1}\sum_{i=1}^{t-1}\frac{1}{g_i}, \qquad H_{t-1}:=\sum_{m=1}^{t-1}\frac1m. \]

Proof. Fix $1\le i<j\le t$. Then \[ x_j-x_i=g_i+g_{i+1}+\dots+g_{j-1}=\sum_{k=i}^{j-1}g_k. \] By Cauchy--Schwarz, \[ \Big(\sum_{k=i}^{j-1}g_k\Big)\Big(\sum_{k=i}^{j-1}\frac{1}{g_k}\Big)\ge (j-i)^2. \] Rearranging gives \[ \frac{1}{x_j-x_i}=\frac{1}{\sum_{k=i}^{j-1}g_k}\le \frac{1}{(j-i)^2}\sum_{k=i}^{j-1}\frac{1}{g_k}. \] Summing over $i<j$ yields \[ \sum_{i<j}\frac1{x_j-x_i}\le \sum_{i<j}\frac{1}{(j-i)^2}\sum_{k=i}^{j-1}\frac{1}{g_k}. \] Swap summations: for fixed $k\in\{1,\dots,t-1\}$, the coefficient of $1/g_k$ is \[ c_k:=\sum_{\substack{i\le k<j}}\frac{1}{(j-i)^2}. \] Group by $m:=j-i\ge 1$. For fixed $m$, the condition $i\le k<i+m$ is equivalent to $k-m+1\le i\le k$, giving at most $m$ possible $i$. Hence \[ c_k\le \sum_{m=1}^{t-1}\frac{m}{m^2}=\sum_{m=1}^{t-1}\frac1m=H_{t-1}. \] Therefore \[ \sum_{i<j}\frac1{x_j-x_i}\le \sum_{k=1}^{t-1}\frac1{g_k}\,H_{t-1} =H_{t-1}\sum_{k=1}^{t-1}\frac{1}{g_k}. \qquad \Box \]

Corollary 1.

For every $n\ge 1$, \[ S_{\mathrm{all}}(n)\le H_{\tau(n)-1}\,S_{\mathrm{gap}}(n), \] which is weaker than the conjectured inequality by a factor $\asymp \log \tau(n)$.

4.2. Prime powers satisfy the desired form

Lemma 2.

If $n=p^m$ with $p$ prime and $m\ge 1$, then \[ S_{\mathrm{all}}(n)\le 2\,S_{\mathrm{gap}}(n)\le 2\left(1+S_{\mathrm{gap}}(n)\right). \]

Proof. The divisors are $d_i=p^{i-1}$ for $i=1,\dots,m+1$. Consecutive gaps are \[ d_{i+1}-d_i=p^i-p^{i-1}=p^{i-1}(p-1)\qquad (i=1,\dots,m), \] hence \[ S_{\mathrm{gap}}(n)=\sum_{i=1}^{m}\frac{1}{p^{i-1}(p-1)}=\frac{1}{p-1}\sum_{r=0}^{m-1}p^{-r}. \] For $i<j$, write $r=j-i\ge 1$. Then \[ d_j-d_i=p^{j-1}-p^{i-1}=p^{i-1}(p^r-1)\ge p^{i-1}\,p^{r-1}(p-1)=p^{j-2}(p-1). \] Thus \[ \frac{1}{d_j-d_i}\le \frac{1}{p^{j-2}(p-1)}=\frac{p}{p-1}\cdot \frac{1}{p^{j-1}}. \] For fixed $j$, \[ \sum_{i=1}^{j-1}\frac{1}{d_j-d_i}\le (j-1)\frac{p}{p-1}\cdot p^{-(j-1)}. \] Therefore \[ S_{\mathrm{all}}(n)\le \frac{p}{p-1}\sum_{j=2}^{m+1}(j-1)p^{-(j-1)}. \] Using $\sum_{u=1}^\infty u p^{-u}=\frac{p}{(p-1)^2}$, we have the partial-sum bound \[ \sum_{j=2}^{m+1}(j-1)p^{-(j-1)}\le \sum_{u=1}^\infty u p^{-u}=\frac{p}{(p-1)^2}. \] Combining gives \[ S_{\mathrm{all}}(n)\le \frac{p}{p-1}\cdot \frac{p}{(p-1)^2} \le \frac{p}{p-1}\cdot \frac{1}{p-1}\sum_{r=0}^{m-1}p^{-r} =\frac{p}{p-1}\,S_{\mathrm{gap}}(n)\le 2\,S_{\mathrm{gap}}(n), \] since $\frac{p}{p-1}\le 2$ for all primes $p$. $\Box$

4.3. A general lower bound for reciprocal-distance energy in an interval

Lemma 3 (Energy lower bound in an interval).

Let $I\subset \mathbb{R}$ be an interval of length $L>0$. Let $x_1<\dots<x_k$ be $k\ge 2$ points in $I$. Then \[ \sum_{1\le i<j\le k}\frac{1}{x_j-x_i}\ \ge\ c\,\frac{k^2\log k}{L} \] for some absolute constant $c>0$ (for instance $c=\frac{1}{32}$ works for all $k\ge 2$).

Proof. Translate and scale so $I=[0,L]$. For $m\ge 0$ set $s_m:=L/2^m$, and let $N_m$ denote the number of pairs $(i,j)$ with $1\le i<j\le k$ such that $|x_j-x_i|\le s_m$.

Step 1: lower bound on $N_m$. Partition $[0,L]$ into $2^m$ subintervals of length $s_m$, and let $a_{m,r}$ be the number of points in the $r$-th subinterval. Then $\sum_r a_{m,r}=k$, and every pair of points within the same subinterval has distance $\le s_m$, so \[ N_m\ge \sum_{r=1}^{2^m}\binom{a_{m,r}}{2} = \frac12\sum_r (a_{m,r}^2-a_{m,r}) = \frac12\Big(\sum_r a_{m,r}^2-k\Big). \] By Cauchy--Schwarz, \[ \sum_{r=1}^{2^m} a_{m,r}^2\ge \frac{(\sum_r a_{m,r})^2}{2^m}=\frac{k^2}{2^m}. \] Hence \[ N_m\ge \frac12\Big(\frac{k^2}{2^m}-k\Big). \] In particular, if $2^m\le k/2$ (i.e. $m\le \lfloor \log_2(k/2)\rfloor$), then $\frac{k^2}{2^m}\ge 2k$, so \[ N_m\ge \frac{k^2}{4\cdot 2^m}. \tag{$\star$} \]

Step 2: relate $N_m$ to energy. Let \[ E:=\sum_{1\le i<j\le k}\frac{1}{x_j-x_i}. \] Fix a pair at distance $d:=x_j-x_i\in (0,L]$. Then \[ \sum_{m:\ s_m\ge d}\frac{1}{s_m} =\sum_{m=0}^{M(d)}\frac{2^m}{L} =\frac{2^{M(d)+1}-1}{L} <\frac{2^{M(d)+1}}{L} =\frac{2}{s_{M(d)}} \le \frac{2}{d}, \] where $M(d)$ is the largest integer with $s_{M(d)}\ge d$, so $s_{M(d)}\ge d$. Summing this bound over all pairs gives \[ \sum_{m\ge 0}\frac{N_m}{s_m} =\sum_{i<j}\sum_{m:\ s_m\ge (x_j-x_i)}\frac1{s_m} \le \sum_{i<j}\frac{2}{x_j-x_i} =2E, \] hence \[ E\ge \frac12\sum_{m\ge 0}\frac{N_m}{s_m}. \tag{$\dagger$} \]

Step 3: conclude. Let $M:=\lfloor \log_2(k/2)\rfloor$. Using $(\star)$ in $(\dagger)$, \[ E\ge \frac12\sum_{m=0}^{M}\frac{N_m}{s_m} \ge \frac12\sum_{m=0}^{M}\frac{\frac{k^2}{4\cdot 2^m}}{L/2^m} =\frac12\sum_{m=0}^{M}\frac{k^2}{4L} =\frac{k^2}{8L}(M+1). \] Since $M+1\ge \log_2(k/2)\ge \frac12\log_2 k$ for $k\ge 2$, we obtain \[ E\ge \frac{k^2}{16L}\log_2 k =\frac{1}{16\log 2}\,\frac{k^2\log k}{L} \ge \frac{1}{32}\,\frac{k^2\log k}{L}. \qquad \Box \]

4.4. Conditional disproof under prime tuples (Tao)

Prime tuples conjecture (qualitative).

For every admissible tuple $(h_1,\dots,h_K)$ (i.e. for every prime $p$, the residues $h_1,\dots,h_K\pmod p$ do not cover all residue classes mod $p$), there exist infinitely many integers $t$ such that $t-h_1,\dots,t-h_K$ are all prime.

Theorem 4 (Conditional disproof).

Assuming the prime tuples conjecture, there is no absolute constant $C$ such that \[ S_{\mathrm{all}}(n)\le C\left(1+S_{\mathrm{gap}}(n)\right) \quad\text{for all }n\ge 1. \] In fact, for arbitrarily large $K$ there exists $n$ with \[ \frac{S_{\mathrm{all}}(n)}{1+S_{\mathrm{gap}}(n)}\gg \log K. \]

Proof (conditional on prime tuples). Fix a large integer $K$. By a sieve construction (as in Tao's note), one can choose an admissible tuple $h_1<\dots<h_K$ with diameter $O(K^2)$ and with separations $|h_i-h_j|\gg K$ for $i\neq j$. (This part is unconditional in Tao's argument.)

By the prime tuples conjecture, choose $t$ such that all $t-h_1,\dots,t-h_K$ are prime, and set \[ n:=\prod_{m=1}^K (t-h_m). \] Then $n$ is squarefree with exactly $K$ prime factors, so $\tau(n)=2^K$. The numbers $t-h_1,\dots,t-h_K$ are divisors of $n$, and they lie in an interval of length $O(K^2)$. Applying Lemma 3 to these $K$ divisors yields \[ S_{\mathrm{all}}(n)\ \ge\ \sum_{1\le i<j\le K}\frac{1}{(t-h_j)-(t-h_i)} =\sum_{1\le i<j\le K}\frac{1}{h_j-h_i} \gg \log K. \tag{1} \]

It remains to control $S_{\mathrm{gap}}(n)$. Observe that any gap $d_{i+1}-d_i>2^K$ contributes at most $1/2^K$, so the total contribution of all such gaps is at most $(2^K-1)\cdot (1/2^K)<1$. Hence it suffices to bound the contribution from gaps $\le 2^K$.

Every divisor of $n$ has the form \[ P_{a_1,\dots,a_k}(t):=\prod_{r=1}^k (t-a_r) \quad\text{with}\quad \{a_1,\dots,a_k\}\subset \{h_1,\dots,h_K\}. \] For $t$ sufficiently large (depending on $K$), if two such divisor-polynomials differ by at most $2^K$ at the point $t$, then their difference must be a constant polynomial (because any nonconstant polynomial with integer coefficients and coefficients bounded in terms of $K$ grows beyond $2^K$ for large $t$). In particular, close divisor pairs must come from polynomials of the same degree $k$.

The degree-$1$ divisors are exactly $t-h_1,\dots,t-h_K$; their pairwise differences are $\gg K$ by separation, and their total contribution to $\sum 1/(d_{i+1}-d_i)$ is $O(1)$. For degrees $k\ge 2$, one reduces to bounding a sum over constant differences between distinct polynomials: \[ P_{a_1,\dots,a_k}(t)-P_{b_1,\dots,b_k}(t)=\ell\in \{1,2,\dots,2^K\}, \] counted in a manner that ensures each close consecutive divisor gap is charged at most once. The key lower bound is:

Lemma 5 (Lower bound on constant differences). If $a_1<\dots<a_k$, $b_1<\dots<b_k$, and \[ P_{a_1,\dots,a_k}(t)-P_{b_1,\dots,b_k}(t)=\ell \] is constant with $\ell>0$, then $b_k>a_k$ and \[ \ell\ \ge\ \frac{b_k-a_k}{k}\prod_{i=1}^{k-1}(a_k-a_i). \]

Proof. The polynomial $P_{a_1,\dots,a_k}$ has $k$ simple real roots, hence for $t\ge a_k$ it is increasing and convex. Since $\ell>0$, evaluating at $t=b_k$ forces $b_k>a_k$. By convexity and the mean value theorem, \[ \ell=P_{a_1,\dots,a_k}(b_k)-P_{a_1,\dots,a_k}(a_k)\ge (b_k-a_k)\,P'_{a_1,\dots,a_k}(a_k). \] A direct Leibniz-rule computation gives \[ P'_{a_1,\dots,a_k}(a_k)=\prod_{i=1}^{k-1}(a_k-a_i), \] so in fact $\ell\ge (b_k-a_k)\prod_{i=1}^{k-1}(a_k-a_i)$, which implies the displayed bound with the additional factor $1/k$. $\Box$

Applying this lemma with $a_i=h_{j_i}$, $b_i=h_{\tilde j_i}$, and using the separation $|h_u-h_v|\gg K$, one obtains \[ \ell \gg \frac{K}{k}\prod_{i=1}^{k-1}(h_{j_k}-h_{j_i}). \tag{2} \] Now dyadically pigeonhole the differences $h_{j_k}-h_{j_i}$ into dyadic ranges $[A_i,2A_i]$ with $K\ll A_i\ll K^2$. For fixed $k$ and fixed dyadic choices $(A_1,\dots,A_{k-1})$, the number of $k$-tuples $(j_1,\dots,j_k)$ meeting these constraints is bounded by \[ \ll K\prod_{i=1}^{k-1}\frac{A_i}{K}, \tag{3} \] because there are $O(K)$ choices for $j_k$, and for each $i<k$ the $\gg K$-separation implies $O(A_i/K)$ choices for $j_i$ in an interval of length $O(A_i)$.

For each such tuple, the corresponding $\ell$ (minimal constant difference, if it exists) is uniquely determined. Using $(2)$, we have $1/\ell \ll \frac{k}{K\prod_i A_i}$. Multiplying by the count $(3)$ shows that the total contribution to $\sum 1/\ell$ from fixed $(k;A_1,\dots,A_{k-1})$ is \[ \ll \Big(\frac{1}{K}\Big)^{k-1} \] up to absolute constants. Summing over the $O((\log K)^{k-1})$ dyadic choices yields a bound \[ \ll \Big(\frac{C\log K}{K}\Big)^{k-1}. \] Summing this geometric series over $k\ge 2$ gives an absolute bound $S_{\mathrm{gap}}(n)\ll 1$ for $K$ large. Combining with $(1)$ yields \[ \frac{S_{\mathrm{all}}(n)}{1+S_{\mathrm{gap}}(n)}\gg \log K\to\infty, \] so no absolute $C$ can satisfy the inequality for all $n$. $\Box$

5. Verification (adversarial checks)

• Lemma 1: The only subtlety is counting, for fixed $k$ and fixed $m=j-i$, the number of $i$ such that $i\le k< i+m$; this is at most $m$. Hence $c_k\le \sum_{m=1}^{t-1}1/m$, as claimed.
• Lemma 2: The inequality $p^r-1\ge p^{r-1}(p-1)$ is exact; the factor $p/(p-1)\le 2$ holds for all primes $p$. No hidden dependence on $m$.
• Lemma 3: The dyadic summation inequality $\sum_{m:\ s_m\ge d}1/s_m\le 2/d$ is a direct geometric-series bound and is correct.
• Lemma 5 / constant-difference bound: A direct derivative computation gives $P'(a_k)=\prod_{i<k}(a_k-a_i)$. This yields an even stronger lower bound on $\ell$ than the weaker $\frac{1}{k}$-version used above, so the subsequent estimates remain valid.
• Conditional step: The only genuinely conditional input is producing $t$ with all $t-h_i$ prime for large $K$. Without this, the disproof argument does not go through unconditionally.

6. FINAL

UNRESOLVED (unconditionally).

1. (i) Strongest fully proved partial results.
   • For every $n$, one has the universal bound \[ S_{\mathrm{all}}(n)\le H_{\tau(n)-1}\,S_{\mathrm{gap}}(n)\ll (\log \tau(n))\,S_{\mathrm{gap}}(n). \]
   • For prime powers $n=p^m$, the conjectured inequality holds with absolute constant $2$: \[ S_{\mathrm{all}}(n)\le 2\left(1+S_{\mathrm{gap}}(n)\right). \]
   • Assuming the qualitative prime tuples conjecture, the conjecture is false: there exist $n$ with \[ \frac{S_{\mathrm{all}}(n)}{1+S_{\mathrm{gap}}(n)}\to\infty. \]

2. (ii) Exact first gap. To resolve the original question unconditionally, one must either (a) prove an absolute bound $S_{\mathrm{all}}(n)\le C(1+S_{\mathrm{gap}}(n))$ for all $n$, or (b) construct an unconditional infinite family $n$ with unbounded ratio. The strongest available disproof strategy requires producing many primes $t-h_1,\dots,t-h_K$ in a short interval with controlled spacing; this is precisely where prime tuples is used.

3. (iii) Top 3 next moves.
   1. Find an unconditional substitute for prime tuples in Tao's construction (e.g. controlled almost-prime constellations) together with an anti-collision argument that still forces $S_{\mathrm{gap}}(n)\ll 1$.
   2. Prove a rigorous unbounded lower bound for $R(n)$ along an explicit family (e.g. $n=k!$ or $n=\mathrm{lcm}(1,\dots,k)$), or show such families cannot work.
   3. Develop a divisor-cluster structural theorem showing that a large block contributing $\gg \log$-energy must force comparably large $S_{\mathrm{gap}}$, or else exhibit the obstruction explicitly.

4. (iv) Likely minimal counterexample structure. A counterexample family would likely have a dense block of divisors inside a short interval (producing $\gg \log$ contribution to $S_{\mathrm{all}}$), while the full divisor set avoids creating too many additional small consecutive gaps (keeping $S_{\mathrm{gap}}$ bounded). Tao's conditional construction produces precisely this behavior using primes in a short interval with controlled spacing.

7. Completion estimate

COMPLETION: $70\%$.