gpt pro 5.4 attempt on Math-732, version 2

Erdős Problem #252

1) ROUND-3 OBJECTIVE

Path (C): obstruction/correction.

Round 2 already extracted a correct conditional contradiction once two special prime-pattern families were available. The most promising continuation is therefore not to repeat that comparison, but to identify the exact structural consequence of the Round 2 mechanism.

The new goal is to prove a theorem strictly stronger than the Round 2 two-family criterion: for any fixed admissible congruence class, rationality of \[ \alpha_k=\sum_{n\ge 1}\frac{\sigma_k(n)}{n!} \] forces the fractional parts of \[ \frac{\sigma_k(q)}{q}=q^{k-1}\sigma_{-k}(q) \] to become rigid modulo $1$, with error $O_k(1/q)$, along every admissible sequence $q$ in that class. The previous prime-versus-semiprime contradiction will then become a corollary.

2) Round-2 FOUNDATION USED

I use the following Round 2 results without reproving them.

• Tail-integrality lemma. If $\alpha_k\in\mathbb{Q}$, then for every sufficiently large $n$, \[ \sum_{\nu\ge n}\frac{\sigma_k(\nu)}{(\nu)_{\nu-n+1}}\in\mathbb{Z}. \]
• Discarded-tail estimate. For fixed $k\ge 2$, truncating the above sum after its first $k$ terms leaves an error $O_k(1/n)$.
• Integer-polynomial approximation. For each $1\le i\le k$ there exists $P_{k,i}\in\mathbb{Z}[x]$ such that \[ \frac{(x+i-1)^k}{(x+i-1)_i}=P_{k,i}(x)+O_k\!\left(\frac1x\right), \] with $P_{k,1}(x)=x^{k-1}$.
• Congruence-stability lemma. If $D=(k!)^k$ and $q_1\equiv q_2\pmod D$, then for each $2\le i\le k$, \[ \sigma_{-k}(i)\left(P_{k,i}(q_1)-P_{k,i}(q_2)\right)\in\mathbb{Z}. \]

3) NEW INSIGHT / TOOL (ROUND-3)

For fixed $k\ge 3$ set \[ D:=(k!)^k. \] For a residue class $c\pmod D$, define the associated admissible class \[ \mathcal{A}_{k,c}:=\left\{q\in\mathbb{N}: q\equiv c\pmod D,\ \forall\,2\le i\le k,\ q+i-1=i\,r_i(q) \text{ with }r_i(q)\text{ prime and }r_i(q)>k\right\}. \] (The condition $r_i(q)>k$ ensures $\gcd(i,r_i(q))=1$.)

The new theorem is a mod-$1$ rigidity statement:

Rigidity theorem. Assume $\alpha_k\in\mathbb{Q}$ and $\mathcal{A}_{k,c}$ is infinite. Then there exists a class $\theta_{k,c}\in\mathbb{R}/\mathbb{Z}$ and a constant $C_k>0$ such that for every sufficiently large $q\in\mathcal{A}_{k,c}$, \[ \left\|\frac{\sigma_k(q)}{q}-\theta_{k,c}\right\|\le \frac{C_k}{q}. \] Equivalently, the values $\sigma_k(q)/q\pmod 1$ on any admissible class must collapse to a single phase at rate $O_k(1/q)$.

This is strictly stronger than the Round 2 two-family criterion: once the phase $\theta_{k,c}$ is known, any second subsequence with incompatible limiting fractional behaviour yields a contradiction automatically.

4) ATTACK PLAN (ROUND-3)

The exact gap left after Round 2 is that the contradiction was still packaged as a direct comparison between two specially chosen families. That did not yet isolate the universal consequence of the factorial-tail method.

The claims to prove now are:

1. For every admissible $q\in\mathcal{A}_{k,c}$, the Round 2 machinery yields a near-integer relation of the form \[ \left\|\frac{\sigma_k(q)}{q}+E_k(q)\right\|\ll_k \frac1q, \] where $E_k(q)=\sum_{i=2}^k\sigma_{-k}(i)P_{k,i}(q)$.
2. If $q,q'\in\mathcal{A}_{k,c}$, then $E_k(q)-E_k(q')\in\mathbb{Z}$; hence the normalized values $\sigma_k(q)/q$ satisfy a pairwise Cauchy estimate modulo $1$.
3. From this pairwise estimate, deduce existence of a unique phase $\theta_{k,c}$ on the whole admissible class.
4. Show that if an admissible class contains infinitely many primes and infinitely many numbers of the form $pr$ with fixed prime $p>k$ and $r$ prime, then the phase forced by the primes contradicts the phase forced by the semiprimes.

This overcomes the earlier obstacle by separating the algebraic part of the argument (which is now universal) from the prime-pattern existence part (which is the only remaining external input).

5) WORK (ROUND-3)

5.1 A universal near-integer relation on admissible classes

Fix $k\ge 3$ and assume, for contradiction, that $\alpha_k\in\mathbb{Q}$. For $n\ge 1$ write \[ S_k(n):=\sum_{i=1}^{k}\frac{\sigma_k(n+i-1)}{(n+i-1)_i}. \] By the Round 2 tail-integrality lemma together with the Round 2 discarded-tail estimate, there exists a constant $B_k>0$ such that \begin{equation}\label{wsref-0-65-71-3a-53-6b-2d-6e-65-61-72-2d-69-6e-74} \|S_k(n)\|\le \frac{B_k}{n} \end{equation} for every sufficiently large $n$.

Now fix a residue class $c\pmod D$ and a sufficiently large $q\in\mathcal{A}_{k,c}$. For each $2\le i\le k$ we may write \[ q+i-1=i\,r_i(q), \] where $r_i(q)$ is prime and $r_i(q)>k\ge i$, hence $\gcd(i,r_i(q))=1$. By multiplicativity of $\sigma_k$, \[ \sigma_k(q+i-1)=\sigma_k(i)\sigma_k(r_i(q))=\sigma_k(i)\left(1+r_i(q)^k\right). \] Since $r_i(q)^k=(q+i-1)^k/i^k$ and $\sigma_{-k}(i)=\sigma_k(i)/i^k$, this becomes \[ \sigma_k(q+i-1)=\sigma_{-k}(i)(q+i-1)^k+\sigma_k(i). \] Dividing by $(q+i-1)_i$ gives \[ \frac{\sigma_k(q+i-1)}{(q+i-1)_i} =\sigma_{-k}(i)\frac{(q+i-1)^k}{(q+i-1)_i}+\frac{\sigma_k(i)}{(q+i-1)_i}. \] By the Round 2 polynomial-division lemma, \[ \frac{(q+i-1)^k}{(q+i-1)_i}=P_{k,i}(q)+O_k\!\left(\frac1q\right), \] and because $i\ge 2$, \[ \frac{\sigma_k(i)}{(q+i-1)_i}=O_k\!\left(\frac1{q^2}\right). \] Hence for each $2\le i\le k$, \[ \frac{\sigma_k(q+i-1)}{(q+i-1)_i}=\sigma_{-k}(i)P_{k,i}(q)+O_k\!\left(\frac1q\right). \] Summing over $i=2,\dots,k$ yields \[ S_k(q)=\frac{\sigma_k(q)}{q}+E_k(q)+O_k\!\left(\frac1q\right), \] where \[ E_k(q):=\sum_{i=2}^{k}\sigma_{-k}(i)P_{k,i}(q). \] Combining this with (eq:Sk-near-int), and enlarging the constant if necessary, we obtain: \begin{equation}\label{wsref-0-65-71-3a-6d-61-73-74-65-72-2d-6e-65-61-72-2d-69-6e-74} \left\|\frac{\sigma_k(q)}{q}+E_k(q)\right\|\le \frac{C_k}{q} \end{equation} for every sufficiently large $q\in\mathcal{A}_{k,c}$, where $C_k>0$ depends only on $k$.

5.2 Pairwise rigidity modulo $1$

5.3 Existence of a unique phase on each admissible class

5.4 A concrete contradiction from incompatible subsequences

5.5 Recovery of the Round-2 two-family criterion

The previous round introduced, for each prime $p>k$, \[ \mathcal{Q}_k^{(0)}:=\left\{q\equiv1\pmod D: q\text{ prime and }(q+i-1)/i\text{ prime for }2\le i\le k\right\}, \] \[ \mathcal{Q}_{k,p}^{(1)}:=\left\{q\equiv1\pmod D: q=pr\text{ with }r\text{ prime},\ p^2\nmid q, \ (q+i-1)/i\text{ prime for }2\le i\le k\right\}. \] Discarding finitely many small elements if necessary, both sets are contained in $\mathcal{A}_{k,1}$. Therefore Corollary Corollary (prime and fixed-semiprime subsequences cannot coexist) immediately implies the Round 2 proposition:

If, for some prime $p>k$, both $\mathcal{Q}_k^{(0)}$ and $\mathcal{Q}_{k,p}^{(1)}$ are infinite, then $\alpha_k$ is irrational.

The new point is that this is now seen as a corollary of the stronger phase-rigidity theorem, not as the fundamental statement.

5.6 The sharpened obstruction

The exact obstruction isolated by this round is now the following.

Sharpened obstruction. To prove irrationality of $\alpha_k$ by the Round 2/Round 3 method, it is enough to exhibit a residue class $c\pmod D$ and an infinite admissible set $\mathcal{A}_{k,c}$ for which the fractional parts of $\sigma_k(q)/q$ do not approach a single phase at speed $O_k(1/q)$.

Equivalently: the entire remaining difficulty is no longer the factorial-tail analysis, but the production of admissible sequences whose normalized divisor sums have two incompatible mod-$1$ behaviours.

6) ADVERSARIAL VERIFICATION

• Use of Round 2 only where allowed. I did not reprove the Round 2 tail-integrality, tail-discard, polynomial-division, or congruence-stability lemmas; I used them explicitly as black boxes.
• Admissible-class definition. The added hypothesis $r_i(q)>k$ is deliberate: it guarantees $\gcd(i,r_i(q))=1$, which is exactly what multiplicativity of $\sigma_k$ needs. Discarding finitely many small $q$ does not affect infinitude arguments.
• Pairwise rigidity step. The only place where congruence information enters is the integrality of $E_k(q)-E_k(q')$, which follows directly from the Round 2 congruence-stability lemma because $q\equiv q'\pmod D$.
• Compactness argument. Proposition Proposition (phase rigidity) uses only compactness of $\mathbb{R}/\mathbb{Z}$ and Lemma Lemma (pairwise rigidity on one admissible class). No hidden uniform distribution input is used.
• Prime subsequence forces phase $0$. Along primes in an admissible class, $\sigma_k(q)/q\equiv1/q\pmod 1$, so the limiting phase is indeed $0$.
• Semiprime lower bound. For $q=pr$ with $r\ne p$, the number $r^{k-1}/p$ has exact denominator $p$ because $p$ is prime and $p\nmid r$. Hence its distance to the nearest integer is at least $1/p$.
• Interaction with Round 2. The new rigidity theorem strictly strengthens the previous round: it implies the Round 2 two-family proposition as a corollary and isolates the remaining obstruction more sharply.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The original problem remains unresolved in general, but this round proves a theorem strictly stronger than the previous round's criterion:

If $\alpha_k$ were rational, then on every infinite admissible congruence class $\mathcal{A}_{k,c}$ the normalized values $\sigma_k(q)/q$ would be rigid modulo $1$, converging to a single phase at rate $O_k(1/q)$.

This turns the earlier prime-versus-semiprime contradiction into a corollary and isolates a single sharp obstruction: one now only needs an admissible class with two incompatible mod-$1$ behaviours of $\sigma_k(q)/q$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 78%

9) REFERENCES

References

• [ScPu06] J.-C. Schlage-Puchta, The irrationality of a number theoretical series, Ramanujan Journal 12 $2006$, no. 3, 455--460.