Erdős Problem #252

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.

Erdős Problem #252

1) ROUND-2 OBJECTIVE

Path (C): keep the successful Round-1 factorial-tail framework, but sharpen it into an exact finite-dimensional reduction for the unresolved cases.

Round-1 already gave complete proofs for $k=1$ and $k=2$. The genuine remaining issue is the general case, where the naive prime-tail estimate is too crude. The most promising Round-2 move is therefore to determine exactly how many local terms one must keep, and exactly how small the remaining tail becomes.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without re-proving them.

• If \[ \alpha_k:=\sum_{n=1}^{\infty}\frac{\sigma_k(n)}{n!} \] is rational and $N$ is at least the denominator, then the factorial tail \[ R_{N,k}:=\sum_{n=N+1}^{\infty}\sigma_k(n)\frac{N!}{n!} \] is an integer.
• The cases $k=1$ and $k=2$ were completely settled in Round-1.

In this round I work only with $k\ge 2$ and with the prime choice $N=p-1$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is an exact truncation threshold for the prime-tail method.

For a prime $p$ define \[ R_{p-1,k}=\sum_{j=0}^{\infty}U_{j,k}(p), \qquad U_{j,k}(p):=\frac{\sigma_k(p+j)}{p(p+1)\cdots(p+j)}. \] Then:

• If one truncates after fewer than $k$ local terms, the omitted tail is still at least $1$ (in fact usually much larger).
• If one keeps exactly the first $k$ local terms, the remaining tail is already $O_k(1/p)$.

Thus the irrationality problem for fixed $k\ge 2$ is reduced to understanding the near-integrality of the single finite expression \[ Q_{p,k}:=\sum_{j=0}^{k-1}U_{j,k}(p) \] along large primes $p$.

4) ATTACK PLAN (ROUND-2)

Round-1 identified the obstruction qualitatively: for $k\ge 3$ the early terms of $R_{p-1,k}$ are too large for the $k=1,2$ argument to work directly.

The exact claims to prove now are:

• a lower-bound statement showing that truncating after $L+1$ terms with $L\le k-2$ can never produce a subunit error term;
• an upper-bound statement showing that truncating after the first $k$ terms always leaves an error of size $O_k(1/p)$.

This turns the vague Round-1 obstruction into a sharp theorem: the first unresolved term count is exactly $k$.

5) WORK (ROUND-2)

Fix $k\ge 2$ and let $p$ be a prime.

Claim 1 (lower bound on each local term).

For every $j$ with $0\le j\le k-1$, \[ U_{j,k}(p)\ge p^{k-j-1}. \]

Proof.

Since $p+j$ is a divisor of itself, \[ \sigma_k(p+j)=\sum_{d\mid (p+j)} d^k \ge (p+j)^k. \] Also, \[ p(p+1)\cdots(p+j)\le p\,(p+j)^j. \] Therefore \[ U_{j,k}(p) =\frac{\sigma_k(p+j)}{p(p+1)\cdots(p+j)} \ge \frac{(p+j)^k}{p\,(p+j)^j} =\frac{(p+j)^{k-j}}{p} \ge p^{k-j-1}. \] $\square$

Corollary 2 (fewer than $k$ terms can never give a small remainder).

Let $L$ be an integer with $0\le L\le k-2$. Then \[ \sum_{j=L+1}^{\infty}U_{j,k}(p)\ge p^{k-L-2}. \] In particular, for every sufficiently large prime $p$ the remainder after truncating after the first $L+1$ terms is larger than $1$.

Proof.

The omitted tail is positive, so it is at least its first term: \[ \sum_{j=L+1}^{\infty}U_{j,k}(p)\ge U_{L+1,k}(p). \] Apply Claim 1 with $j=L+1$. $\square$

This proves a genuine barrier: the Round-1 style argument cannot work after keeping fewer than $k$ local terms.

Claim 3 (uniform $O_k(1/p)$ tail after $k$ terms).

Define \[ Q_{p,k}:=\sum_{j=0}^{k-1}U_{j,k}(p), \qquad E_{p,k}:=\sum_{j=k}^{\infty}U_{j,k}(p). \] Then for every prime $p\ge 2^k$, \[ 0<E_{p,k}\le \frac{2^{k+1}\zeta(k)}{p}. \]

Proof.

For $k\ge 2$ one has the standard bound \[ \sigma_k(n)=n^k\sum_{d\mid n}\frac1{(n/d)^k} \le n^k\sum_{m=1}^{\infty}\frac1{m^k} =\zeta(k)n^k. \] Hence, with \[ \nu_{j,k}(p):=\zeta(k)\frac{(p+j)^k}{p(p+1)\cdots(p+j)}, \] we have $U_{j,k}(p)\le \nu_{j,k}(p)$ for every $j$.

First, \[ \nu_{k,k}(p) \le \zeta(k)\frac{(p+k)^k}{p^{k+1}} \le \zeta(k)\frac{(2p)^k}{p^{k+1}} =\frac{2^k\zeta(k)}{p}, \] because $p\ge k$.

Second, for every $j\ge k$, \[ \frac{\nu_{j+1,k}(p)}{\nu_{j,k}(p)} =\frac{(p+j+1)^k}{(p+j)^k}\cdot\frac1{p+j+1} =\frac{(p+j+1)^{k-1}}{(p+j)^k} \le \frac{2^{k-1}}{p+j} \le \frac12, \] since $p\ge 2^k$.

Therefore the majorant tail is geometric: \[ E_{p,k} \le \sum_{j=k}^{\infty}\nu_{j,k}(p) \le 2\nu_{k,k}(p) \le \frac{2^{k+1}\zeta(k)}{p}. \] The positivity $E_{p,k}>0$ is obvious. $\square$

Corollary 4 (finite-dimensional reduction).

Assume $\alpha_k\in\mathbf{Q}$. Then for every sufficiently large prime $p$, \[ R_{p-1,k}\in\mathbf{Z} \] and hence \[ \min_{m\in\mathbf{Z}} \left| \sum_{j=0}^{k-1}\frac{\sigma_k(p+j)}{p(p+1)\cdots(p+j)}-m \right| \le \frac{2^{k+1}\zeta(k)}{p}. \]

Proof.

The Round-1 factorial-tail lemma gives $R_{p-1,k}\in\mathbf{Z}$ for every sufficiently large prime $p$. Since \[ R_{p-1,k}=Q_{p,k}+E_{p,k}, \] Claim 3 shows that the distance from $Q_{p,k}$ to the nearest integer is at most $E_{p,k}\le 2^{k+1}\zeta(k)/p$. $\square$

Interpretation.

The open part of Problem #252 is now reduced to the following explicit task:

For fixed $k\ge 3$, show that for infinitely many large primes $p$ the finite sum \[ Q_{p,k}=\sum_{j=0}^{k-1}\frac{\sigma_k(p+j)}{p(p+1)\cdots(p+j)} \] is not within $O_k(1/p)$ of any integer.

This is a strict strengthening of the Round-1 obstruction, because it identifies the exact number of local terms that must be controlled.

6) ADVERSARIAL VERIFICATION

• Claim 1 only uses the trivial divisor $d=p+j$ and the elementary denominator bound \[ p(p+1)\cdots(p+j)\le p(p+j)^j. \] No hidden hypothesis is present.
• Corollary 2 is sharp in the sense relevant here: when $L=k-2$, the omitted tail is still at least $1$, so no interval of length $<1$ can arise from such a truncation.
• Claim 3 uses only the convergent Euler sum $\zeta(k)$ and a geometric-ratio estimate valid for every $p\ge 2^k$.
• The reduction in Corollary 4 is fully rigorous and depends only on the Round-1 integer-tail lemma.

Thus there is no logical gap in the new reduction: the unresolved issue is purely the arithmetic behavior of the explicit $k$-term local sum $Q_{p,k}$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict Round-2 advance is an exact threshold theorem for the prime-tail method.

• Truncating after fewer than $k$ local terms can never leave a subunit remainder.
• Truncating after exactly the first $k$ local terms always leaves an error $O_k(1/p)$.
• Therefore rationality of $\alpha_k$ would force the explicit $k$-term quantity \[ Q_{p,k}=\sum_{j=0}^{k-1}\frac{\sigma_k(p+j)}{p(p+1)\cdots(p+j)} \] to be $O_k(1/p)$-close to an integer for every sufficiently large prime $p$.

This is a genuine finite-dimensional reduction of the unresolved cases.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 55%

9) REFERENCES

Only references actually used or checked.

• T. F. Bloom, Erdős Problem #252, https://www.erdosproblems.com/252, accessed 2026-03-07.
• P. Erdős, Problem 4493, Amer. Math. Monthly 59 $1952$, 557--558.
• J.-C. Schlage-Puchta, The irrationality of a number theoretical series, Ramanujan J. 15 $2008$, 149--158; arXiv version checked at https://arxiv.org/abs/1105.1452.
• J. B. Friedlander, F. Luca, and M. Stoiciu, On the irrationality of a divisor function series, Integers 7 $2007$, A36.
• K. Pratt, The irrationality of a divisor function series of Erdős and Kac, arXiv:2209.11124.

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Problem 252: an explicit phase obstruction for the fourth divisor power

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. The target is irrationality of $S_k=\sum_{n\ge1}\sigma_k(n)/n!$ for every fixed positive integer $k$. I read 252.tex and 252_v2.tex fully, including their tail estimates and conditional prime patterns. A literature correction matters: Schlage-Puchta's cited paper already proves $k=3$ unconditionally, as well as the all-$k$ implication from Schinzel's hypothesis H. The following is a worked conditional obstruction for $k=4$, not a new proof of that known conditional theorem.

2) WORK

Put $D=24^4$. Suppose there are arbitrarily large $q\equiv1\pmod D$ for which all four numbers \[ q,\quad(q+1)/2,\quad(q+2)/3,\quad(q+3)/4 \] are prime. Assume toward contradiction that $S_4$ is rational. For large $q$, multiplication by $(q-1)!$ shows that \[ T_q=\sum_{i\ge1}\frac{\sigma_4(q+i-1)}{q(q+1)\cdots(q+i-1)} \] is an integer. The terms $i\ge5$ total $O(1/q)$: use $\sigma_4(m)\le\zeta(4)m^4$ and factorial decay. The first term is $q^3+1/q$. For $i=2,3,4$, the prime $r=(q+i-1)/i$ exceeds $i$, so multiplicativity gives $\sigma_4(ir)=\sigma_4(i)(r^4+1)$. Polynomial division consequently yields \[ T_q=q^3+\frac{17}{16}(q^2+3q+3) +\frac{82}{81}(q+5)+\frac{273}{256}+O(1/q). \] All three rational-polynomial values modulo one depend only on $q\bmod D$. At $q=1$ their sum is \[ \frac{119}{16}+\frac{164}{27}+\frac{273}{256} =\frac{100763}{6912}\equiv\frac{3995}{6912}\pmod1. \] This nonzero fixed phase contradicts integrality for sufficiently large $q$. Thus the specified prime-pattern infinitude would imply irrationality of $S_4$.

There is no elementary local obstruction to the pattern: writing $q=Dt+1$, the four forms are $(D/i)t+1$. Modulo $2$ or $3$ each is $1$; modulo every prime $\ell\ge5$ they exclude at most four residues, fewer than $\ell$.

3) ADVERSARIAL VERIFICATION

Local admissibility is not a proof of simultaneous primality. The argument's missing infinitude is exactly an unproved prime-pattern assertion. The polynomial remainders are $O(1/q)$, including the additional $+1$ in each prime divisor sum; they cannot cancel the fixed phase. Nothing here proves the $k=4$ case unconditionally, and the already established $k=3$ case must not be advertised as new progress.

4) FINAL

UNRESOLVED. An explicit rationality obstruction replaces an unspecified phase, but it still requires an unproved simultaneous-prime input. The full all-$k$ question is not settled.

5) SOURCES CHECKED

Both complete supplied versions; J.-C. Schlage-Puchta, The irrationality of a number theoretical series, https://arxiv.org/abs/1105.1452, checked 2026-09-23. The displayed divisions and phase are exact elementary calculations.

6) COMPLETION ESTIMATE

Subjective progress toward an unconditional solution for every $k$. COMPLETION: 15%