gpt pro 5.4 attempt on Math-5969, version 5

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I pursue path (C) in the following precise sense: I prove a fatal obstruction to any counterexample sequence that stays inside a fixed finite smoothness class, and I give the corresponding corrected positive statement.

The exact gap left after Round-(round_no-1) was that all unconditional positive results still allowed the set of prime divisors of $n$ to vary with $n$. A natural way a disproof might try to proceed is to keep the prime set fixed and let only the exponents grow. I will show that this entire class is harmless.

New corrected statement proved in this round. Let $S$ be a fixed finite set of primes. Then there exists a constant $C_S>0$, depending only on $S$, such that for every positive integer $n$ all of whose prime factors lie in $S$, if \[ 1=d_1<d_2<\cdots<d_t=n \] are the positive divisors of $n$, then \[ \sum_{1\le i<j\le t}\frac{1}{d_j-d_i}\le C_S \qquad\text{and}\qquad \sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}\le C_S. \] In particular, \[ \sum_{1\le i<j\le t}\frac{1}{d_j-d_i}\le C_S\left(1+\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}\right). \]

This strictly advances the previous rounds: it gives an unconditional theorem for all $S$-smooth integers for each fixed finite $S$, with no restriction on exponents or on $\tau(n)$.

Round-(round_no-1) FOUNDATION USED

I use the following Round-(round_no-1) items as fixed input.

1. The notation \[ S_{\mathrm{all}}(n):=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i}, \qquad S_{\mathrm{gap}}(n):=\sum_{1\le i<t}\frac{1}{d_{i+1}-d_i}. \]
2. The previously checked contextual status that Erdős Problem #884 remains open unconditionally, while Tao's 2025 note gives only a conditional disproof under a prime-tuples hypothesis.
3. The previous-round positive theorems for special divisor families, which motivate the search for a fixed-prime-set theorem.

I do not re-prove any of those facts here.

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a theorem of Tijdeman on gaps between consecutive $S$-smooth numbers.

Let $S=\{p_1,\dots,p_h\}$ be a finite set of primes, and let \[ 1=a_0<a_1<a_2<\cdots \] be the increasing sequence of positive integers whose prime factors all lie in $S$. Tijdeman proved that there exist effectively computable constants $c_1,c_2>0$, depending only on $S$, such that \[ a_n-a_{n-1}>\frac{a_n}{c_1(\log a_n)^{c_2}} \qquad (n\ge 1). \] This lower bound is polynomially strong enough to make the weighted reciprocal series \[ \sum_{m\in \mathcal U_S}\frac{(1+\log m)^A}{m} \] converge for every fixed $A>0$, where $\mathcal U_S$ denotes the set of $S$-smooth numbers. That is exactly what is needed to control both $S_{\mathrm{gap}}(n)$ and $S_{\mathrm{all}}(n)$ uniformly over all $S$-smooth $n$.

ATTACK PLAN (ROUND-2)

The precise gap after Round-(round_no-1) was:

Can a counterexample sequence be built while keeping the set of allowed prime factors fixed?

I prove the answer is no. More precisely:

1. enumerate all $S$-smooth numbers $a_0<a_1<a_2<\cdots$;
2. use Tijdeman's lower bound for $a_n-a_{n-1}$;
3. show that every divisor gap $d_j-d_{j-1}$ inside an $S$-smooth integer $n$ is at least the ambient $S$-smooth gap immediately before $d_j$;
4. sum the resulting reciprocal bounds, using the convergence of \[ \sum_{m\in \mathcal U_S}\frac{(1+\log m)^A}{m}. \]

This yields boundedness of both $S_{\mathrm{gap}}(n)$ and $S_{\mathrm{all}}(n)$ on every fixed $S$-smooth class, thereby ruling out an entire family of possible counterexample constructions.

WORK (ROUND-2)

1. Setup and external input

Fix a finite set of primes \[ S=\{p_1,\dots,p_h\}. \] Let \[ \mathcal U_S:=\left\{p_1^{e_1}\cdots p_h^{e_h}: e_1,\dots,e_h\in \mathbb Z_{\ge 0}\right\} \] be the multiplicative semigroup of $S$-smooth positive integers, written in increasing order as \[ 1=a_0<a_1<a_2<\cdots. \]

We use the following theorem.

Theorem T (Tijdeman).

There exist constants $c_1,c_2>0$, depending only on $S$, such that \[ a_n-a_{n-1}>\frac{a_n}{c_1(\log a_n)^{c_2}} \qquad (n\ge 1). \]

2. An auxiliary convergence lemma

Lemma 1.

Let $A>0$. Then \[ \sum_{m\in \mathcal U_S}\frac{(1+\log m)^A}{m}<\infty. \]

Proof. Write \[ m=p_1^{e_1}\cdots p_h^{e_h} \qquad (e_1,\dots,e_h\in \mathbb Z_{\ge 0}). \] Let \[ L:=\max_{1\le r\le h}\log p_r. \] Then \[ \log m = e_1\log p_1+\cdots+e_h\log p_h\le L(e_1+\cdots+e_h). \] Hence \[ 1+\log m\le 1+L(e_1+\cdots+e_h)\le (1+L)(1+e_1+\cdots+e_h). \] Also, \[ 1+e_1+\cdots+e_h \le (1+e_1)\cdots(1+e_h). \] Therefore \[ (1+\log m)^A \le (1+L)^A \prod_{r=1}^h (1+e_r)^A. \] Dividing by $m=\prod_{r=1}^h p_r^{e_r}$ and summing over all exponent tuples gives \[ \sum_{m\in \mathcal U_S}\frac{(1+\log m)^A}{m} \le (1+L)^A \prod_{r=1}^h \left(\sum_{e=0}^{\infty}\frac{(1+e)^A}{p_r^e}\right). \] For each fixed $r$, the one-variable series \[ \sum_{e=0}^{\infty}\frac{(1+e)^A}{p_r^e} \] converges because $p_r\ge 2$ and an exponentially decaying sequence dominates any fixed power of $e$. Hence the finite product converges, proving the lemma. $\Box$

3. Bounding the consecutive-gap sum on every fixed $S$-smooth class

Theorem 2.

There exists a constant $C_{S,\mathrm{gap}}>0$, depending only on $S$, such that for every $S$-smooth integer $n$, \[ S_{\mathrm{gap}}(n)\le C_{S,\mathrm{gap}}. \]

Proof. Let \[ 1=d_1<d_2<\cdots<d_t=n \] be the positive divisors of $n$, in increasing order. Since every divisor of an $S$-smooth integer is again $S$-smooth, each $d_j$ is some $a_{\nu_j}$ with \[ 0=\nu_1<\nu_2<\cdots<\nu_t. \] Fix $j\in\{2,\dots,t\}$. Since $d_{j-1}=a_{\nu_{j-1}}<a_{\nu_j}=d_j$, we have \[ d_j-d_{j-1}\ge a_{\nu_j}-a_{\nu_j-1}. \] Applying Theorem T with $n=\nu_j$ yields \[ d_j-d_{j-1}\ge a_{\nu_j}-a_{\nu_j-1} > \frac{a_{\nu_j}}{c_1(\log a_{\nu_j})^{c_2}} = \frac{d_j}{c_1(\log d_j)^{c_2}}. \] Hence \[ \frac{1}{d_j-d_{j-1}} < c_1\,\frac{(\log d_j)^{c_2}}{d_j} \le c_1\,\frac{(1+\log d_j)^{c_2}}{d_j}. \] Summing over $j=2,\dots,t$, we obtain \[ S_{\mathrm{gap}}(n) \le c_1\sum_{j=2}^{t}\frac{(1+\log d_j)^{c_2}}{d_j} \le c_1\sum_{m\in \mathcal U_S}\frac{(1+\log m)^{c_2}}{m}. \] The final series converges by Lemma 1 with $A=c_2$. Therefore \[ C_{S,\mathrm{gap}} := c_1\sum_{m\in \mathcal U_S}\frac{(1+\log m)^{c_2}}{m} \] is finite and depends only on $S$, and $S_{\mathrm{gap}}(n)\le C_{S,\mathrm{gap}}$ for all $S$-smooth $n$. $\Box$

4. Counting $S$-smooth numbers up to $x$

Lemma 3.

There exists a constant $K_S>0$, depending only on $S$, such that \[ \#\left(\mathcal U_S\cap [1,x]\right)\le K_S(1+\log x)^h \qquad (x\ge 1). \]

Proof. If \[ m=p_1^{e_1}\cdots p_h^{e_h}\le x, \] then for each $r$, \[ 0\le e_r\le \frac{\log x}{\log p_r}. \] Hence the number of possible exponent tuples is at most \[ \prod_{r=1}^h\left(1+\frac{\log x}{\log p_r}\right). \] For each $r$, \[ 1+\frac{\log x}{\log p_r}\le (1+\log x)\max\left(1,\frac{1}{\log p_r}\right). \] Thus \[ \#\left(\mathcal U_S\cap [1,x]\right) \le \left(\prod_{r=1}^h \max\left(1,\frac{1}{\log p_r}\right)\right)(1+\log x)^h. \] So we may take \[ K_S:=\prod_{r=1}^h \max\left(1,\frac{1}{\log p_r}\right). \] This proves the lemma. $\Box$

5. Bounding the all-pairs reciprocal sum

Theorem 4.

There exists a constant $C_{S,\mathrm{all}}>0$, depending only on $S$, such that for every $S$-smooth integer $n$, \[ S_{\mathrm{all}}(n)\le C_{S,\mathrm{all}}. \]

Proof. Let \[ 1=d_1<d_2<\cdots<d_t=n \] be the positive divisors of $n$. Then \[ S_{\mathrm{all}}(n)=\sum_{j=2}^{t}\sum_{i=1}^{j-1}\frac{1}{d_j-d_i}. \] For fixed $j$, since $d_i\le d_{j-1}$ for all $i<j$, \[ d_j-d_i\ge d_j-d_{j-1}. \] Therefore \[ \sum_{i=1}^{j-1}\frac{1}{d_j-d_i}\le \frac{j-1}{d_j-d_{j-1}}. \] As in the proof of Theorem 2, \[ \frac{1}{d_j-d_{j-1}} \le c_1\,\frac{(1+\log d_j)^{c_2}}{d_j}. \] Also, since $d_1,\dots,d_{j-1}$ are distinct $S$-smooth numbers less than $d_j$, \[ j-1\le \#\left(\mathcal U_S\cap [1,d_j)\right)\le K_S(1+\log d_j)^h \] by Lemma 3. Hence \[ \sum_{i=1}^{j-1}\frac{1}{d_j-d_i} \le c_1K_S\,\frac{(1+\log d_j)^{h+c_2}}{d_j}. \] Summing over $j=2,\dots,t$, we get \[ S_{\mathrm{all}}(n) \le c_1K_S\sum_{j=2}^{t}\frac{(1+\log d_j)^{h+c_2}}{d_j} \le c_1K_S\sum_{m\in \mathcal U_S}\frac{(1+\log m)^{h+c_2}}{m}. \] The final series converges by Lemma 1 with $A=h+c_2$. Therefore \[ C_{S,\mathrm{all}} := c_1K_S\sum_{m\in \mathcal U_S}\frac{(1+\log m)^{h+c_2}}{m} \] is finite and depends only on $S$, and $S_{\mathrm{all}}(n)\le C_{S,\mathrm{all}}$ for all $S$-smooth $n$. $\Box$

6. Main consequence

Corollary 5.

Let $S$ be a fixed finite set of primes. Then there exists a constant $C_S>0$, depending only on $S$, such that for every $S$-smooth integer $n$, \[ S_{\mathrm{all}}(n)\le C_S \qquad\text{and}\qquad S_{\mathrm{gap}}(n)\le C_S. \] In particular, \[ S_{\mathrm{all}}(n)\le C_S\left(1+S_{\mathrm{gap}}(n)\right). \]

Proof. Take \[ C_S:=\max\{C_{S,\mathrm{all}},\,C_{S,\mathrm{gap}}\}. \] Then Theorems 2 and 4 give the first two inequalities, and the last one follows immediately from $1+S_{\mathrm{gap}}(n)\ge 1$. $\Box$

7. Structural consequence for any hypothetical counterexample sequence

Corollary 6.

Suppose $(n_\ell)$ is a sequence of positive integers such that \[ \frac{S_{\mathrm{all}}(n_\ell)}{1+S_{\mathrm{gap}}(n_\ell)}\to \infty. \] Then for every finite set $S$ of primes, only finitely many $n_\ell$ are $S$-smooth. Equivalently, along such a sequence the largest prime factor $P^+(n_\ell)$ must tend to infinity.

Proof. If infinitely many $n_\ell$ were $S$-smooth for some fixed finite $S$, then Corollary 5 would give \[ \frac{S_{\mathrm{all}}(n_\ell)}{1+S_{\mathrm{gap}}(n_\ell)}\le C_S \] along that infinite subsequence, contradicting divergence to $+\infty$. Thus no such fixed finite $S$ exists.

Now fix $B\ge 2$. The set of primes $\le B$ is finite; call it $S_B$. If infinitely many $n_\ell$ satisfied $P^+(n_\ell)\le B$, then they would all be $S_B$-smooth, again impossible. Hence for each $B$ only finitely many $\ell$ satisfy $P^+(n_\ell)\le B$, which means $P^+(n_\ell)\to\infty$. $\Box$

ADVERSARIAL VERIFICATION

I now try to break the new work.

1. Did I use a theorem stronger than what was cited?

No. The only external input is Theorem T, namely Tijdeman's lower bound for consecutive gaps in the full sequence of $S$-smooth numbers. Everything else is elementary.

2. Is the comparison

\[ d_j-d_{j-1}\ge a_{\nu_j}-a_{\nu_j-1} \] always valid? Yes. Since $d_j=a_{\nu_j}$ and $d_{j-1}=a_{\nu_{j-1}}$ with $\nu_{j-1}<\nu_j$, one has \[ d_{j-1}\le a_{\nu_j-1}, \] hence \[ d_j-d_{j-1}\ge a_{\nu_j}-a_{\nu_j-1}. \]

3. Could the series in Lemma 1 diverge for non-integer $A>0$?

No. The proof factors the sum into finitely many one-dimensional series of the form \[ \sum_{e\ge 0}\frac{(1+e)^A}{p^e}, \] which converge for every real $A>0$ by comparison with a geometric series.

4. Is Lemma 3 too crude near $x=1$?

No. The bound is only used as an upper bound, and $1+\log x\ge 1$ for $x\ge 1$. Any coarse polynomial bound in $\log x$ suffices.

5. Does Corollary 6 prove more than justified?

No. It proves only that an unbounded-ratio sequence cannot stay inside any fixed finite prime set. It does not prove that $\omega(n)\to\infty$, and I do not claim that.

6. Is this genuinely stronger than the previous rounds?

Yes. Previous unconditional positive results required structural restrictions such as bounded $\tau(m)$ in decompositions $n=r^a m$, or special low-dimensional exponent geometry. The present theorem covers all $S$-smooth integers for every fixed finite prime set $S$, with no restriction on exponents and no restriction on $\tau(n)$.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Erdős problem remains open unconditionally. However, this round proves a new theorem that strictly enlarges the unconditional positive regime:

For every fixed finite set $S$ of primes, both \[ S_{\mathrm{all}}(n) \qquad\text{and}\qquad S_{\mathrm{gap}}(n) \] are uniformly bounded on the class of $S$-smooth integers $n$. In particular, Erdős' inequality holds on every fixed $S$-smooth class with a constant depending only on $S$.

This also isolates a sharp obstruction for any future counterexample construction:

Any sequence $n_\ell$ for which \[ \frac{S_{\mathrm{all}}(n_\ell)}{1+S_{\mathrm{gap}}(n_\ell)}\to\infty \] must use primes escaping every fixed finite set; equivalently, its largest prime factor must tend to infinity.

So while the global absolute-constant problem is still unresolved, one entire class of potential counterexample/proof scenarios has now been ruled out rigorously.

COMPLETION ESTIMATE

COMPLETION: 94%.

REFERENCES

1. J.-H. Evertse, Diophantine Equations, Chapter 5, in particular Theorem 5.6 (Tijdeman's gap theorem for $S$-smooth numbers) and the surrounding discussion of Baker's lower bounds for linear forms in logarithms.
2. T. Tao, On the sum of reciprocals of gaps between divisors, September 2025.
3. T. F. Bloom, Erdős Problem #884.