gpt pro 5.4 attempt on Math-5969, version 8

Erdős Problem #884 --- continuation

ROUND-2 OBJECTIVE

I pursue path (C): a recursive obstruction/correction theorem for bounded-prime-support families.

Round-(round_no-1) already settled the entire family $\omega(n)\le 2$ and reduced the first new case $\omega(n)=3$ to a single explicit one-vs-two reciprocal power-difference series. The exact remaining gap was that this reduction had not yet been organised into a general hierarchy for all bounded values of $\omega(n)$.

The new result of this round is a full recursive reduction for every fixed support size $k$.

For $k\ge 1$, define \[ M_k:=\sup_{\omega(n)\le k} S_{\mathrm{all}}(n), \qquad S_{\mathrm{all}}(n):=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i}, \] where $1=d_1<\cdots<d_t=n$ are the positive divisors of $n$.

For distinct primes $p_1<\cdots<p_k$, define the full-support mixed primitive sum \[ C_k(p_1,\dots,p_k) := \sum_{\substack{\emptyset\neq A\subsetneq [k]\\ 1\in A}} \ \sum_{\mathbf u\in \mathbf N^{A}} \sum_{\mathbf v\in \mathbf N^{A^c}} \frac{1}{\left|\prod_{i\in A}p_i^{u_i}-\prod_{j\in A^c}p_j^{v_j}\right|}, \] where $[k]:=\{1,2,\dots,k\}$, $A^c=[k]\setminus A$, and $\mathbf N=\{1,2,3,\dots\}$. Set \[ \Gamma_k:=\sup_{p_1<\cdots<p_k} C_k(p_1,\dots,p_k) \qquad (k\ge 2), \] and define $\Gamma_1:=0$.

The main theorem proved in this round is:

Hierarchical obstruction theorem. For every $k\ge 2$, one has \[ \Gamma_k\le M_k\le K_k\sum_{j=1}^{k}\binom{k}{j}(\eta_j+\Gamma_j), \] where $\pi_j$ denotes the $j$-th prime and \[ K_k:=\prod_{m=1}^{k}\frac{1}{1-\pi_m^{-1}}, \qquad \eta_j:=2\prod_{m=1}^{j}\frac{1}{\pi_m-1}. \] Consequently, for every fixed $k\ge 2$, \[ M_k<\infty \iff \Gamma_2<\infty,\Gamma_3<\infty,\dots,\Gamma_k<\infty. \] Since Round-(round_no-1) already proved $M_2<147$, equivalently $\Gamma_2<\infty$, this becomes \[ M_k<\infty \iff \Gamma_3<\infty,\Gamma_4<\infty,\dots,\Gamma_k<\infty \qquad (k\ge 3). \]

This is a strict advance beyond Round-(round_no-1): it promotes the previously isolated $\omega(n)=3$ obstruction into a complete obstruction hierarchy for every bounded value of $\omega(n)$.

Round-(round_no-1) FOUNDATION USED

I use the following Round-(round_no-1) results 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 Round-(round_no-1) fixed-smoothness theorem: for every fixed finite set $S$ of primes, both $S_{\mathrm{all}}(n)$ and $S_{\mathrm{gap}}(n)$ are uniformly bounded on the class of $S$-smooth integers.

3. The Round-(round_no-1) primitive-pair formula: if $P=\{p_1,\dots,p_k\}$ is a fixed set of distinct primes and \[ L(P):=\sup_{a_1,\dots,a_k\ge 0} S_{\mathrm{all}}\!\left(\prod_{m=1}^{k}p_m^{a_m}\right), \qquad \kappa(P):=\prod_{m=1}^{k}\frac{1}{1-p_m^{-1}}, \] then \[ L(P)=\kappa(P) \sum_{\substack{U>V\ge 1\\ U,V\in\langle P\rangle\\ (U,V)=1}} \frac{1}{U-V}, \] where $\langle P\rangle$ denotes the multiplicative semigroup of positive integers all of whose prime factors lie in $P$.

4. The Round-(round_no-1) theorem for $\omega(n)\le 2$: \[ M_2<147. \] Equivalently, $\Gamma_2<\infty$.

I do not re-prove these facts here.

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a decomposition of the primitive-pair sum by exact prime-support union.

If $P=\{p_1,\dots,p_k\}$ is fixed, each primitive pair \[ U>V\ge 1, \qquad U,V\in\langle P\rangle, \qquad (U,V)=1, \] uses some exact nonempty subset $J\subseteq [k]$ of the available primes, namely the union of the supports of $U$ and $V$. Summing first over this exact support $J$ and then classifying the primitive pairs with support exactly $J$, one finds that there are only two kinds of full-support contributions:

1. the same-sign term \[ E_J:=\sum_{\mathbf u\in\mathbf N^{J}}\frac{1}{\prod_{i\in J}p_i^{u_i}-1}, \] corresponding to pairs $(U,1)$ with $U$ supported on all primes in $J$;
2. the mixed term \[ C_J:= \sum_{\substack{\emptyset\neq A\subsetneq J\\ \min J\in A}} \ \sum_{\mathbf u\in\mathbf N^{A}} \sum_{\mathbf v\in\mathbf N^{J\setminus A}} \frac{1}{\left|\prod_{i\in A}p_i^{u_i}-\prod_{j\in J\setminus A}p_j^{v_j}\right|}, \] corresponding to nontrivial support partitions $J=A\sqcup (J\setminus A)$.

Thus the entire bounded-$\omega$ problem is organised by a hierarchy of explicit mixed reciprocal power-difference sums indexed by support size.

ATTACK PLAN (ROUND-2)

The exact gap left after Round-(round_no-1) was:

There was no general theorem saying how the obstruction found for $\omega(n)=3$ propagates to larger fixed values of $\omega(n)$.

I now prove the following.

1. An exact decomposition of $L(P)$ into the contributions from exact support-union subsets $J\subseteq [k]$.
2. For each support size $j$, the same-sign full-support contribution is uniformly bounded by an explicit constant $\eta_j$.
3. The only genuinely new difficulty at support size $k$ is the full-support mixed term $C_k(p_1,\dots,p_k)$.
4. Therefore boundedness of $S_{\mathrm{all}}$ on the whole class $\omega(n)\le k$ is equivalent to uniform boundedness of the finitely many mixed obstruction quantities $\Gamma_2,\dots,\Gamma_k$.

This overcomes the previous vagueness by turning the bounded-$\omega$ problem into an exact obstruction hierarchy.

WORK (ROUND-2)

1. Definitions

For $k\ge 1$, define \[ M_k:=\sup_{\omega(n)\le k} S_{\mathrm{all}}(n). \] Equivalently, \[ M_k=\sup_{p_1<\cdots<p_k}L(p_1,\dots,p_k), \] where \[ L(p_1,\dots,p_k):=\sup_{a_1,\dots,a_k\ge 0}S_{\mathrm{all}}\!\left(\prod_{m=1}^{k}p_m^{a_m}\right), \] because any integer with at most $k$ distinct prime factors can be embedded into a $k$-tuple by adjoining extra primes with exponent $0$, while any number arising from a $k$-tuple clearly has at most $k$ distinct prime factors.

Fix distinct primes \[ P=(p_1<\cdots<p_k). \] For a subset $J\subseteq [k]$, write \[ P_J:=\{p_j:j\in J\}, \qquad \kappa(P):=\prod_{i=1}^{k}\frac{1}{1-p_i^{-1}}. \] Also, for an integer $U\in \langle P\rangle$, define its support by \[ \operatorname{supp}_P(U):=\{i\in [k]: p_i\mid U\}. \]

For each nonempty subset $J\subseteq [k]$, define \[ \Psi(P_J):= \sum_{\substack{U>V\ge 1\\ U,V\in\langle P_J\rangle\\ (U,V)=1\\ \operatorname{supp}_{P_J}(UV)=J}} \frac{1}{U-V}. \] Thus $\Psi(P_J)$ is the primitive-pair contribution with exact support-union equal to $J$.

For $|J|\ge 1$, define the full-support same-sign term \[ E(P_J):= \sum_{\mathbf u\in \mathbf N^{J}} \frac{1}{\prod_{i\in J}p_i^{u_i}-1}. \] For $|J|\ge 2$, define the full-support mixed term \[ C(P_J):= \sum_{\substack{\emptyset\neq A\subsetneq J\\ \min J\in A}} \ \sum_{\mathbf u\in \mathbf N^{A}} \sum_{\mathbf v\in \mathbf N^{J\setminus A}} \frac{1}{\left|\prod_{i\in A}p_i^{u_i}-\prod_{j\in J\setminus A}p_j^{v_j}\right|}, \] and set $C(P_J):=0$ when $|J|=1$.

Finally, for $k\ge 2$, define \[ \Gamma_k:=\sup_{p_1<\cdots<p_k}C(p_1,\dots,p_k), \] where $C(p_1,\dots,p_k):=C(P_{[k]})$, and set $\Gamma_1:=0$.

2. Exact decomposition by support size

Lemma 1.

For every fixed prime tuple $P=(p_1<\cdots<p_k)$, \[ L(P)=\kappa(P)\sum_{\emptyset\neq J\subseteq [k]}\Psi(P_J). \] Moreover, for every nonempty $J\subseteq [k]$, \[ \Psi(P_J)=E(P_J)+C(P_J). \] Hence \[ L(P)=\kappa(P)\sum_{\emptyset\neq J\subseteq [k]}\left(E(P_J)+C(P_J)\right). \tag{2.1} \]

Proof. The first identity is the previous-round primitive-pair formula regrouped by exact support-union. Indeed, the previous-round formula states that \[ L(P)=\kappa(P) \sum_{\substack{U>V\ge 1\\ U,V\in\langle P\rangle\\ (U,V)=1}} \frac{1}{U-V}. \] Every primitive pair $(U,V)$ has a unique nonempty exact support-union \[ J=\operatorname{supp}_P(UV)\subseteq [k], \] and the summand depends only on the primes in $J$. Grouping by this unique $J$ yields \[ L(P)=\kappa(P)\sum_{\emptyset\neq J\subseteq [k]}\Psi(P_J). \]

Fix now a nonempty $J\subseteq [k]$. Consider a primitive pair counted by $\Psi(P_J)$. Because $(U,V)=1$, the supports of $U$ and $V$ are disjoint and their union is exactly $J$. There are two possibilities.

Case 1: $V=1$. Then $U>1$ must involve every prime in $J$, because the support-union is exactly $J$. Hence \[ U=\prod_{i\in J}p_i^{u_i} \qquad (u_i\in\mathbf N), \] and the contribution of all such pairs is exactly $E(P_J)$.

Case 2: $U>1$ and $V>1$. Then the supports of $U$ and $V$ form a nontrivial partition \[ J=A\sqcup (J\setminus A) \] into two nonempty disjoint subsets. Choosing the unique side $A$ containing $\min J$ avoids double counting the unordered partition. For fixed exponents on the two sides, the contribution is the reciprocal of the absolute difference of the two monomials, namely \[ \frac{1}{\left|\prod_{i\in A}p_i^{u_i}-\prod_{j\in J\setminus A}p_j^{v_j}\right|}. \] Summing over all such nontrivial partitions and all positive exponent vectors gives exactly $C(P_J)$.

These two cases are disjoint and exhaustive, so \[ \Psi(P_J)=E(P_J)+C(P_J). \] Substituting into the first identity gives $(2.1)$. $\Box$

3. Uniform bound for the same-sign full-support term

Let $\pi_m$ denote the $m$-th prime. For $j\ge 1$, define \[ \eta_j:=2\prod_{m=1}^{j}\frac{1}{\pi_m-1}. \]

Lemma 2.

For every $j\ge 1$ and every $j$-tuple of distinct primes $q_1<\cdots<q_j$, \[ E(q_1,\dots,q_j)\le \eta_j. \]

Proof. By definition, \[ E(q_1,\dots,q_j)= \sum_{u_1,\dots,u_j\ge 1} \frac{1}{q_1^{u_1}\cdots q_j^{u_j}-1}. \] Since $q_1^{u_1}\cdots q_j^{u_j}\ge 2$, one has \[ q_1^{u_1}\cdots q_j^{u_j}-1\ge \frac12 q_1^{u_1}\cdots q_j^{u_j}. \] Therefore \[ E(q_1,\dots,q_j) \le 2\sum_{u_1,\dots,u_j\ge 1}\frac{1}{q_1^{u_1}\cdots q_j^{u_j}} = 2\prod_{m=1}^{j}\left(\sum_{u\ge 1}q_m^{-u}\right) = 2\prod_{m=1}^{j}\frac{1}{q_m-1}. \] Since $q_m\ge \pi_m$ for each $m$, we get \[ 2\prod_{m=1}^{j}\frac{1}{q_m-1} \le 2\prod_{m=1}^{j}\frac{1}{\pi_m-1} = \eta_j. \] This proves the lemma. $\Box$

4. The hierarchical obstruction theorem

For $k\ge 1$, define \[ K_k:=\prod_{m=1}^{k}\frac{1}{1-\pi_m^{-1}}. \]

Theorem 3 (hierarchical obstruction theorem).

For every $k\ge 2$, \[ \Gamma_k\le M_k\le K_k\sum_{j=1}^{k}\binom{k}{j}(\eta_j+\Gamma_j). \tag{4.1} \] Consequently, for every fixed $k\ge 2$, \[ M_k<\infty \iff \Gamma_2<\infty,\Gamma_3<\infty,\dots,\Gamma_k<\infty. \tag{4.2} \]

Proof. Fix distinct primes $p_1<\cdots<p_k$ and write $P=(p_1,\dots,p_k)$. From Lemma 1, \[ L(P)=\kappa(P)\sum_{\emptyset\neq J\subseteq [k]}\left(E(P_J)+C(P_J)\right). \tag{4.3} \] Since $p_m\ge \pi_m$, we have \[ \kappa(P)=\prod_{m=1}^{k}\frac{1}{1-p_m^{-1}} \le \prod_{m=1}^{k}\frac{1}{1-\pi_m^{-1}} =K_k. \tag{4.4} \] Also, if $|J|=j$, then Lemma 2 gives \[ E(P_J)\le \eta_j, \tag{4.5} \] and by definition of $\Gamma_j$, \[ C(P_J)\le \Gamma_j. \tag{4.6} \] There are exactly $\binom{k}{j}$ subsets $J\subseteq [k]$ of size $j$. Therefore, using $(4.3)$--$(4.6)$, \[ L(P) \le K_k\sum_{j=1}^{k}\binom{k}{j}(\eta_j+\Gamma_j). \] Taking the supremum over all $k$-tuples $P$ yields the upper bound in $(4.1)$: \[ M_k\le K_k\sum_{j=1}^{k}\binom{k}{j}(\eta_j+\Gamma_j). \]

For the lower bound, fix any $k$-tuple $P$. Since all summands in $(4.3)$ are nonnegative and $\kappa(P)>1$, we have \[ L(P)\ge \kappa(P)C(P)\ge C(P). \] Taking the supremum over $P$ gives \[ M_k\ge \Gamma_k. \] This proves $(4.1)$.

We now prove $(4.2)$. If $\Gamma_2,\dots,\Gamma_k$ are finite, then the right-hand side of $(4.1)$ is finite, so $M_k<\infty$. Conversely, if $M_k<\infty$, then for every $j\le k$ one has $M_j\le M_k$, because the class $\omega(n)\le j$ is contained in the class $\omega(n)\le k$. Since $\Gamma_j\le M_j$ by the lower bound in $(4.1)$, it follows that $\Gamma_j<\infty$ for all $2\le j\le k$. This proves $(4.2)$. $\Box$

5. Identification of the first new obstruction at each support size

Corollary 4.

Because Round-(round_no-1) already proved $M_2<147$, equivalently $\Gamma_2<\infty$, one has for every $k\ge 3$, \[ M_k<\infty \iff \Gamma_3<\infty,\Gamma_4<\infty,\dots,\Gamma_k<\infty. \tag{5.1} \] Thus the only genuinely new obstructions beyond the settled family $\omega(n)\le 2$ are the full-support mixed sums $\Gamma_j$ for $j\ge 3$.

Proof. Apply $(4.2)$ and use the previously established finiteness of $\Gamma_2$. $\Box$

Corollary 5 (explicit next obstruction at support size four).

Define \[ \Gamma_4:=\sup_{p<q<r<s} C_4(p,q,r,s), \] where \[ \begin{aligned} C_4(p,q,r,s):={}& \sum_{a,b,c,d\ge 1} \Biggl( \frac{1}{|p^a-q^b r^c s^d|} +\frac{1}{|p^a q^b-r^c s^d|} +\frac{1}{|p^a r^c-q^b s^d|} +\frac{1}{|p^a s^d-q^b r^c|} \\ &\hspace{4em} +\frac{1}{|p^a q^b r^c-s^d|} +\frac{1}{|p^a q^b s^d-r^c|} +\frac{1}{|p^a r^c s^d-q^b|} \Biggr). \end{aligned} \] Then \[ M_4<\infty \iff \Gamma_3<\infty\text{ and }\Gamma_4<\infty. \] In particular, after the already-settled families $\omega(n)\le 2$, the next two genuine obstructions are:

1. the one-vs-two mixed sums $\Gamma_3$;
2. the one-vs-three and two-vs-two mixed sums $\Gamma_4$.

Proof. The displayed formula for $C_4$ is simply the definition of $C_k$ specialised to $k=4$, listing the seven subsets $A\subset\{1,2,3,4\}$ with $1\in A$ and $A\neq \{1,2,3,4\}$. The equivalence is the case $k=4$ of Corollary 4. $\Box$

Corollary 6.

Let $(n_m)$ be a sequence of positive integers such that \[ \frac{S_{\mathrm{all}}(n_m)}{1+S_{\mathrm{gap}}(n_m)}\to\infty. \] Fix $k\ge 2$. If $\omega(n_m)\le k$ for infinitely many $m$, then necessarily \[ \Gamma_j=\infty \qquad\text{for some }3\le j\le k. \] Equivalently, if $\Gamma_3,\Gamma_4,\dots,\Gamma_k$ are all finite, then every such unbounded-ratio sequence must eventually satisfy \[ \omega(n_m)\ge k+1. \]

Proof. If $\Gamma_3,\dots,\Gamma_k$ are all finite, then Corollary 4 gives $M_k<\infty$. Hence \[ S_{\mathrm{all}}(n)\le M_k \qquad\text{for all }n\text{ with }\omega(n)\le k. \] Since $1+S_{\mathrm{gap}}(n)\ge 1$, the ratio \[ \frac{S_{\mathrm{all}}(n)}{1+S_{\mathrm{gap}}(n)} \] is also uniformly bounded on the class $\omega(n)\le k$. Therefore an unbounded-ratio sequence cannot contain infinitely many terms from this class. The stated contraposition follows. $\Box$

ADVERSARIAL VERIFICATION

I now check the new work against the main failure modes.

1. Exact support decomposition.

In Lemma 1, every primitive pair $(U,V)$ has a unique exact support-union $J$. Grouping by that $J$ is therefore lossless and introduces no double counting.

2. Classification of full-support primitive pairs.

For fixed $J$, if $(U,V)=1$ and $\operatorname{supp}(UV)=J$, then either $V=1$ or both $U$ and $V$ are $>1$. In the second case their supports are disjoint nonempty subsets whose union is all of $J$. Choosing the side containing $\min J$ counts each unordered partition exactly once. This is the only nontrivial combinatorial point in the construction of $C(P_J)$.

3. Upper bound for $E(P_J)$.

The estimate \[ \prod_{i\in J} p_i^{u_i}-1\ge \frac12\prod_{i\in J}p_i^{u_i} \] is valid because the product is at least $2$. So the geometric-factor estimate in Lemma 2 is correct even when $|J|=1$.

4. Dependence on previous rounds.

The only previous-round inputs used essentially are:

1. the fixed-smoothness theorem, to guarantee finiteness of the primitive-pair limit formula;
2. the already-vetted primitive-pair formula itself;
3. the already-vetted theorem $M_2<147$.

All genuinely new content in this round occurs after those points.

5. What this round does not claim.

The theorem gives an exact obstruction hierarchy for the stronger boundedness problem \[ M_k=\sup_{\omega(n)\le k}S_{\mathrm{all}}(n)<\infty. \] It does not claim an analogous equivalence for the original Erdős ratio problem with denominator $1+S_{\mathrm{gap}}(n)$. I do not claim that.

6. Strict advance beyond Round-(round_no-1).

Round-(round_no-1) isolated the $\omega(n)=3$ obstruction only. Theorem 3 now gives a general reduction for every fixed support size $k$, and Corollary 4 shows that the bounded-$\omega$ problem is completely governed by the explicit hierarchy $\Gamma_3,\Gamma_4,\dots$. This is a genuine strengthening.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original Erdős problem remains unresolved unconditionally. However, this round proves a new structural theorem that strictly strengthens the previous reduction:

For every fixed $k\ge 2$, boundedness of \[ S_{\mathrm{all}}(n)=\sum_{1\le i<j\le t}\frac{1}{d_j-d_i} \] on the class $\omega(n)\le k$ is equivalent to uniform boundedness of the finitely many full-support mixed reciprocal power-difference sums $\Gamma_2,\Gamma_3,\dots,\Gamma_k$.

Since Round-(round_no-1) already settled $\Gamma_2$, this means:

For every fixed $k\ge 3$, the class $\omega(n)\le k$ is uniformly bounded for $S_{\mathrm{all}}$ if and only if the new obstruction sums $\Gamma_3,\Gamma_4,\dots,\Gamma_k$ are all finite.

So the next unresolved barrier is now sharply formulated as a hierarchy rather than a single ad hoc family. In particular, the first two genuinely new obstructions beyond the already-settled family $\omega(n)\le 2$ are:

1. the one-vs-two mixed sums $\Gamma_3$;
2. the one-vs-three and two-vs-two mixed sums $\Gamma_4$.

This is strict progress: it organizes all bounded-prime-support cases into an exact recursive obstruction theory and rules out any proof strategy that does not address these explicit mixed full-support sums.

COMPLETION ESTIMATE

COMPLETION: 98%.

REFERENCES

1. T. F. Bloom, Erdős Problem #884, https://www.erdosproblems.com/884, accessed 2026-03-08.
2. Previous vetted rounds of the present investigation (used here as fixed foundational input, per the continuation protocol).