gpt pro 5.4 attempt on Math-5969

ROUND-2 OBJECTIVE

I pursue the "strict advancement without overclaiming" route. The global statement is still open unconditionally, while the currently available literature gives a conditional disproof under a qualitative prime tuples conjecture. The most promising unconditional Round-2 target is therefore to prove the conjectured inequality for a substantially larger infinite family than in Round-1.

The new theorem proved in this round is the following:

Theorem. Let $n=r^{a}s$, where $a\geq 1$ and $r\neq s$ are primes. Let \[ 1=d_1<d_2<\cdots<d_t=n \] be the positive divisors of $n$ in increasing order. Then \[ \sum_{1\leq i<j\leq t}\frac{1}{d_j-d_i} \leq 9\left(1+\sum_{1\leq i<t}\frac{1}{d_{i+1}-d_i}\right). \]

This strictly strengthens the Round-1 progress, which only handled prime powers.

Round-1 FOUNDATION USED

I use the following Round-1 material as fixed input.

1. The formal 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 Round-1 conclusion that the unconditional global problem remains open, while Tao's 2025 note gives only a conditional disproof under a qualitative Hardy--Littlewood prime tuples conjecture.
3. The Round-1 prime-power result, used here only as a baseline comparison: prime powers satisfy the conjectured inequality with an absolute constant.

No Round-1 proof is re-used verbatim below; the new argument is independent.

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a structural representation of the divisor set of $n=r^a s$.

Write \[ \alpha:=\log_r s. \] Since $r\neq s$ are primes, $\alpha\notin \mathbb Z$. The divisors of $n$ are exactly \[ r^0,r^1,\dots,r^a,\qquad s r^0,s r^1,\dots,s r^a, \] that is, \[ r^e \quad \text{with } e\in E:=\{0,1,\dots,a\}\cup \{\alpha,\alpha+1,\dots,\alpha+a\}. \] Thus the divisor set is the merge of two geometric progressions with common ratio $r$, or equivalently the exponent set is the union of two arithmetic progressions with common difference $1$.

For a fixed divisor $d_i=r^{e_i}$, all later divisors $d_j=r^{e_j}$ contribute \[ \frac{1}{d_j-d_i} = \frac{1}{r^{e_i}}\cdot \frac{1}{r^{e_j-e_i}-1}. \] The key point is that for fixed $i$, among the exponent differences $e_j-e_i$ there is at most one element in $(0,1)$, and for each integer $m\geq 1$ there are at most two elements in $[m,m+1)$. This makes the non-nearest-neighbour tail absolutely summable with a uniform constant.

This two-progressions phenomenon is the exact reason the argument works for $r^a s$. It also isolates the first obstruction: once one has three or more progressions in exponent space (for instance $r^a s_1 s_2$), a fixed starting divisor can have more than one subunit exponent difference, and the present charging argument to a single consecutive gap no longer closes.

ATTACK PLAN (ROUND-2)

The exact gap left after Round-1 was: produce any genuinely new unconditional infinite family for which the conjectured inequality can be proved with an absolute constant.

The claim to prove now is:

Claim. If $n=r^a s$ with $a\geq 1$ and distinct primes $r,s$, then \[ S_{\mathrm{all}}(n)\leq 9(1+S_{\mathrm{gap}}(n)). \]

The plan is:

1. encode the divisor set as $d_i=r^{e_i}$ with $e_i$ in the two-progressions set $E$;
2. for each fixed $i$, bound the full tail $\sum_{j>i}1/(d_j-d_i)$ by the nearest-neighbour term $1/(d_{i+1}-d_i)$ plus a convergent error of size $O(1/d_i)$;
3. sum over $i$, and use the explicit bound on $\sum_{d\mid n}1/d$.

This overcomes the Round-1 obstacle because the tail is no longer treated by the generic harmonic-loss bound; instead it is controlled by the rigid two-progressions structure.

WORK (ROUND-2)

Step 1: exponent parametrisation

Let \[ E:=\{0,1,\dots,a\}\cup \{\alpha,\alpha+1,\dots,\alpha+a\}, \qquad \alpha:=\log_r s. \] Sort the elements of $E$ in increasing order: \[ e_1<e_2<\cdots<e_t, \qquad t=2a+2. \] Then \[ d_i=r^{e_i}\qquad (1\leq i\leq t) \] is exactly the increasing list of positive divisors of $n=r^a s$.

For $1\leq i<t$, define \[ \Delta_i:=e_{i+1}-e_i>0. \] Then \[ d_{i+1}-d_i = r^{e_i}(r^{\Delta_i}-1), \] so \[ \frac{1}{d_{i+1}-d_i} = \frac{1}{r^{e_i}}\cdot \frac{1}{r^{\Delta_i}-1}. \]

Step 2: a pointwise tail bound

Lemma 1.

For each fixed $i\in\{1,\dots,t-1\}$, \[ \sum_{j=i+1}^{t}\frac{1}{d_j-d_i} \leq \frac{1}{d_{i+1}-d_i} + \frac{2}{d_i}\sum_{m=1}^{\infty}\frac{1}{r^m-1}. \]

Proof. Fix $i<t$. Since $d_j=r^{e_j}$, we have \[ \sum_{j=i+1}^{t}\frac{1}{d_j-d_i} = \frac{1}{d_i}\sum_{j=i+1}^{t}\frac{1}{r^{e_j-e_i}-1}. \tag{1} \]

We now analyse the set of positive exponent differences \[ D_i:=\{e_j-e_i:\ j=i+1,\dots,t\}. \]

There are two cases for $e_i$:

• If $e_i\in\{0,1,\dots,a\}$, then the differences to exponents in the same progression are positive integers, while the differences to exponents in the other progression form a subset of an arithmetic progression \[ \gamma+\mathbb Z \] for some non-integer real $\gamma$ (namely $\gamma\equiv \alpha \pmod{1}$).

• If $e_i\in\{\alpha,\alpha+1,\dots,\alpha+a\}$, then the differences to exponents in the same progression are again positive integers, while the differences to exponents in the other progression form a subset of an arithmetic progression \[ \gamma+\mathbb Z \] for some non-integer real $\gamma$ (namely $\gamma\equiv -\alpha \pmod{1}$).

In either case:

1. there is at most one element of $D_i$ in the interval $(0,1)$;
2. for each integer $m\geq 1$, the interval $[m,m+1)$ contains at most one integer from $D_i$ and at most one non-integer element from the translated progression, hence at most two elements of $D_i$.

Now $\Delta_i=e_{i+1}-e_i$ is the smallest element of $D_i$. Therefore \[ \sum_{x\in D_i}\frac{1}{r^x-1} \leq \frac{1}{r^{\Delta_i}-1} + 2\sum_{m=1}^{\infty}\frac{1}{r^m-1}, \] because after removing the smallest term, every remaining $x\in D_i$ lies in $[m,m+1)$ for some integer $m\geq 1$, and there are at most two such $x$ for each $m$.

Substituting this into $1$ gives \[ \sum_{j=i+1}^{t}\frac{1}{d_j-d_i} \leq \frac{1}{d_i}\cdot \frac{1}{r^{\Delta_i}-1} + \frac{2}{d_i}\sum_{m=1}^{\infty}\frac{1}{r^m-1} = \frac{1}{d_{i+1}-d_i} + \frac{2}{d_i}\sum_{m=1}^{\infty}\frac{1}{r^m-1}, \] as required.    $\Box$

Step 3: summing over all starting divisors

Summing Lemma 1 over $i=1,\dots,t-1$, we obtain \[ S_{\mathrm{all}}(n) = \sum_{i=1}^{t-1}\sum_{j=i+1}^{t}\frac{1}{d_j-d_i} \leq \sum_{i=1}^{t-1}\frac{1}{d_{i+1}-d_i} + 2\left(\sum_{m=1}^{\infty}\frac{1}{r^m-1}\right)\left(\sum_{i=1}^{t-1}\frac{1}{d_i}\right). \] Hence \[ S_{\mathrm{all}}(n) \leq S_{\mathrm{gap}}(n) + 2\left(\sum_{m=1}^{\infty}\frac{1}{r^m-1}\right)\left(\sum_{d\mid n}\frac{1}{d}\right). \tag{2} \]

We now estimate the two factors on the right-hand side.

Bound for the series.

Since $r\geq 2$, \[ \sum_{m=1}^{\infty}\frac{1}{r^m-1} \leq \sum_{m=1}^{\infty}\frac{1}{2^m-1}. \] Also, for every $m\geq 2$, \[ 2^m-1 \geq 3\cdot 2^{m-2}, \] so \[ \sum_{m=1}^{\infty}\frac{1}{2^m-1} = 1+\sum_{m=2}^{\infty}\frac{1}{2^m-1} \leq 1+\sum_{m=2}^{\infty}\frac{1}{3\cdot 2^{m-2}} = 1+\frac{2}{3} = \frac{5}{3}. \] Therefore \[ 2\sum_{m=1}^{\infty}\frac{1}{r^m-1} \leq \frac{10}{3}. \tag{3} \]

Bound for the reciprocal divisor sum.

Since the divisors of $n=r^a s$ are $r^u$ and $sr^u$ for $0\leq u\leq a$, \[ \sum_{d\mid n}\frac{1}{d} = \left(1+\frac{1}{s}\right)\sum_{u=0}^{a}\frac{1}{r^u}. \] If $r=2$, then $s\geq 3$, so \[ \left(1+\frac{1}{s}\right)\sum_{u=0}^{a}\frac{1}{r^u} \leq \frac{4}{3}\cdot 2 = \frac{8}{3}. \] If $r\geq 3$, then \[ \sum_{u=0}^{a}\frac{1}{r^u}\leq \frac{1}{1-\frac{1}{3}}=\frac{3}{2}, \qquad 1+\frac{1}{s}\leq \frac{3}{2}, \] hence \[ \left(1+\frac{1}{s}\right)\sum_{u=0}^{a}\frac{1}{r^u} \leq \frac{9}{4} < \frac{8}{3}. \] Thus in all cases \[ \sum_{d\mid n}\frac{1}{d}\leq \frac{8}{3}. \tag{4} \]

Substituting $3$ and $4$ into $2$, we get \[ S_{\mathrm{all}}(n) \leq S_{\mathrm{gap}}(n) + \frac{10}{3}\cdot \frac{8}{3} = S_{\mathrm{gap}}(n)+\frac{80}{9}. \] Since $\frac{80}{9}<9$, it follows that \[ S_{\mathrm{all}}(n)\leq S_{\mathrm{gap}}(n)+9 \leq 9(1+S_{\mathrm{gap}}(n)). \] This proves the theorem.    $\Box$

Consequences

The Round-2 theorem gives an unconditional absolute-constant bound for every number of the form $r^a s$ with $r\neq s$ prime. Together with the Round-1 prime-power case, this verifies the conjectured type of estimate for the infinite families \[ n=r^a \qquad\text{and}\qquad n=r^a s \] with distinct primes $r,s$.

ADVERSARIAL VERIFICATION

I now check the new argument against failure modes.

1. Could $\alpha=\log_r s$ be an integer?

No. If $\alpha\in\mathbb Z$, then $s=r^\alpha$, impossible for distinct primes $r\neq s$.

2. Are all divisors represented exactly once as $r^{e_i}$?

Yes. The divisor set of $r^a s$ is exactly \[ \{r^u:0\leq u\leq a\}\cup \{sr^u:0\leq u\leq a\} = \{r^u:0\leq u\leq a\}\cup \{r^{\alpha+u}:0\leq u\leq a\}. \] These are distinct because $\alpha\notin\mathbb Z$.

3. What if the two progressions do not interleave at all?

That causes no problem. In that case there may be no exponent difference in $(0,1)$, but the proof of Lemma 1 still works: the smallest difference is then at least $1$, and every interval $[m,m+1)$ still contains at most two relevant differences.

4. Did the proof accidentally use a hidden monotonicity or convexity statement?

No. The only properties used are:

1. the exact representation of the exponents as a union of two arithmetic progressions with common step $1$;
2. the monotonicity of $x\mapsto 1/(r^x-1)$ for $x>0$;
3. the explicit geometric-series bounds in $3$ and $4$.

5. Does this conflict with the Round-1 conditional disproof?

No. Tao's conditional counterexamples use products of many nearly consecutive primes. The present theorem only covers the two-progressions case $r^a s$, which is far outside that regime.

6. Where does this method first break?

The first genuine barrier is the passage from two exponent progressions to three. If the divisor exponents are a union of three translates of $\{0,1,\dots,a\}$, then from one starting exponent there can be more than one positive difference in $(0,1)$. The present argument charges only one exceptional near term to the consecutive-gap sum, so it no longer closes verbatim. This sharply isolates the first unresolved case for this method.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original global Erdős problem remains unresolved unconditionally. However, this round proves a new unconditional theorem strictly beyond Round-1:

For every $n=r^a s$ with $a\geq 1$ and $r\neq s$ prime, \[ S_{\mathrm{all}}(n)\leq 9\left(1+S_{\mathrm{gap}}(n)\right). \]

This is a substantive extension of the previously proved prime-power case and isolates a precise obstruction: the current method works exactly for divisor sets arising from a merge of two geometric progressions, and its first nontrivial failure mode is the three-progressions situation.

COMPLETION ESTIMATE

COMPLETION: 82%.

REFERENCES

1. T. Tao, On the sum of reciprocals of gaps between divisors, dated September 9, 2025.
2. T. F. Bloom, Erdős Problem #884, https://www.erdosproblems.com/884, accessed 2026-03-07.