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ERDőS #252 · PARTIAL

Erdős problem #252 — wave 8o

Access date: 2026-07-28 (UTC).

Claim labels

primary source, or uses the explicitly named theorem.

below.

uniform theorem.

Step 0: live-page check

(d) I fetched the live problem page, its LaTeX view, and its discussion thread through the Bright Data browser path. At access time the page said OPEN, 0 claimed proofs, Currently working on this problem: None, and Interested in collaborating: None. Thus the mandatory stop condition did not fire. The page says it was last edited 22 January 2026.

The live page's verbatim statement is:

Let $k\geq 1$ and $\sigma_k(n)=\sum_{d\mid n}d^k$. Is\[\sum > \frac{\sigma_k(n)}{n!}\]irrational?

(d) The page lists the following known state: unconditional irrationality for \(1\leq k\leq4\); the cases \(k=1,2\) are attributed there to Erdős, the case \(k=3\) independently to Schlage-Puchta and to Friedlander--Luca--Stoiciu, and \(k=4\) to Pratt. It lists irrationality for every \(k\geq1\) conditional on either Schinzel's conjecture or Dickson's conjecture, and points to problem B14 in Guy's collection.

(d) There were three comments, none a claimed proof:

  1. Alfaiz (14 April 2026) added the Erdős--Kac Problem 4518 attribution.
  2. Dogmachine (4 January 2026) distinguished Dickson's conjecture from the

prime-tuples restriction; the page says it was updated in response.

  1. Quanyu Tang (5 September 2025) noted the two conditional all-\(k\)

results; the page says it was updated in response.

Primary-source audit

(b) Schlage-Puchta's paper, The irrationality of a number theoretical series, defines \(S_k=\sum_{n\geq1}\sigma_k(n)/n!\), proves \(S_3\) irrational, and proves the all-\(k\) assertion under Schinzel's Hypothesis H.

(b) Friedlander, Luca, and Stoiciu's paper, On the irrationality of a divisor function series, proves \(k=3\) unconditionally and proves the all-\(k\) assertion under Dickson's prime \(k\)-tuples conjecture.

(b) Pratt's paper, The irrationality of a divisor function series of Erdős and Kac, proves \(k=4\). Its introduction explicitly says that its proof pushes the sieve techniques to their limit and that new ideas seem necessary for \(k\geq5\).

(b) Deajim and Siksek's publisher record for On the \(\mathbb{Q}\)-linear independence of the sums gives a criterion conditional on Schinzel's conjecture and reports that the criterion was checked through the first 50 sums. It does not supply an unconditional \(k=5\) result.

(d) I searched the current arXiv API and web indexes using the exact series title and combinations of sigma_k(n), n!, irrationality, and k=5. The exact-title arXiv query returned only Pratt's 2022 paper. I found no post-2022 primary source claiming \(k=5\) or all \(k\). This is an honest search miss, not a proof that no unindexed literature exists.

Result

Write

\[ \alpha_5=\sum_{m\geq1}\frac{\sigma_5(m)}{m!}. \]

(d) Certified finite theorem. If \(\alpha_5=a/b\in\mathbb Q\) in lowest terms, then

\[ \boxed{b> 100000000000000000000000000000000000000000123456789.} \]

This does not prove \(\alpha_5\) irrational: a hypothetical rational denominator has no known a priori upper bound.

The same argument gives the following clean uniform target.

(a) Six-term sufficient criterion. For \(n>2\), put

\[ A_5(n)=\sum_{j=0}^{5} \frac{\sigma_5(n+j)}{n(n+1)\cdots(n+j)}. \]

If there are arbitrarily large \(n\) such that

\[ \tag{U5} \left\|A_5(n)\right\|> \frac{5\cdot7^5}{4n(n-2)}, \]

then \(\alpha_5\) is irrational. Here \(\|x\|\) is distance to the nearest integer. Thus (U5) is an exact statement whose missing word is “arbitrarily”; the certificate below proves it for one 51-digit \(n\).

Elementary reduction

(a) Suppose \(\alpha_5=a/b\) and \(b\mid(n-1)!\). Absolute convergence follows, for example, from \(\sigma_5(m)<\tfrac54m^5\). Therefore

\[ \begin{aligned} T_5(n) &:=(n-1)!\left(\alpha_5- \sum_{m=1}^{n-1}\frac{\sigma_5(m)}{m!}\right)\\ &=\sum_{j=0}^{\infty} \frac{\sigma_5(n+j)}{n(n+1)\cdots(n+j)} \end{aligned} \]

is an integer: both \((n-1)!\alpha_5\) and the removed finite sum are integers.

(a) For every \(m\geq1\),

\[ \sigma_5(m) =m^5\sum_{d\mid m}d^{-5} <m^5\left(1+\int_1^\infty x^{-5}\,dx\right) =\frac54m^5. \]

For \(j\geq6\), use \(n+j\leq n(j+1)\) and \(\prod_{i=0}^j(n+i)\geq n^{j+1}\) to get

\[ \frac{\sigma_5(n+j)}{n(n+1)\cdots(n+j)} <\frac54\frac{(j+1)^5}{n^{j-4}}. \]

The ratio of consecutive majorants is at most \((8/7)^5/n<2/n\). Hence the positive omitted tail \(R_5(n)\) satisfies

\[ \tag{1} 0<R_5(n)< \frac54\frac{7^5}{n^2}\frac1{1-2/n} =\frac{5\cdot7^5}{4n(n-2)}. \]

(a) If (U5) holds, (1) says that \(T_5(n)=A_5(n)+R_5(n)\) cannot be an integer. Given a rational denominator \(b\), any (U5) instance with \(n>b\) has \(b\mid(n-1)!\), contradicting the previous paragraph. This proves the six-term sufficient criterion.

The exact 51-digit certificate

Take

\[ N=100000000000000000000000000000000000000000123456789. \]

(d) The standalone checker verifies the following complete factorizations and proves every displayed factor prime:

N   = 431
      * 243684137338841
      * 952128292053325945036132006739059

N+1 = 2 * 5 * 17 * 613 * 51064608909191
      * 18791895729019492807343151467189

N+2 = 3 * 47 * 79 * 967 * 2767 * 9665209
      * 14618870706614971 * 23746143508304039

N+3 = 2^3 * 13 * 83 * 14724469
      * 786772055510269840721851286182582251649

N+4 = 7 * 31
      * 460829493087557603686635944700460829493088126529

N+5 = 2 * 3 * 11 * 601
      * 2521050773962587606514395199919326375233200309

(d) Direct divisor enumeration and exact rational arithmetic give

\[ \left\|A_5(N)\right\|=\frac rQ, \]

where

r =
3935951695477504858315043191776330373836460253236391717855622287851459140561276768250224310717661785747327574768822541853852920715289038225523748480863712279545120830391886345950066234039707520888940833206811291447505363739651950778697118941601651136280402144462103978956207613450062081607244532

Q =
63131313131313131313131313131313131313131780770674873737373737373737373737373737375180705116869425978535353535353535353535355911207684470956548095202967171717171717173917036694586712249892602297241492424242425328777329803743283291955138390353091422954768983989794532609986550246823135577492117441

The checker verifies the simpler exact comparison

\[ \tag{2}\frac rQ>\frac1{17}. \]

At this \(N\), the bound in (1), reduced to lowest terms, is

\[ \tag{3} \frac{84035}{ 40000000000000000000000000000000000000000098765430400000000000000000000000000000000060966314013107772} <10^{-95}. \]

Equations (2) and (3) prove (U5) at \(N\), so \(T_5(N)\notin\mathbb Z\).

(a) If \(b<N\), then \(b\mid(N-1)!\). If \(b=N\), the displayed factorization of \(N\) is a product of three distinct proper factors, each of which occurs in \((N-1)!\), so again \(b\mid(N-1)!\). The nonintegrality of \(T_5(N)\) therefore gives \(b>N\), proving the certified finite theorem.

Why the certificate itself is checkable

(a) The verifier uses the following elementary complete-\(p-1\) Lucas certificate. Suppose the complete prime factorization is \(p-1=\prod q^{e_q}\), and for every distinct \(q\) there is an \(a_q\) with

\[ a_q^{p-1}\equiv1\pmod p,\qquad \gcd\!\left(a_q^{(p-1)/q}-1,p\right)=1. \]

If a prime \(r\mid p\), the order of \(a_q\bmod r\) is divisible by \(q^{e_q}\); hence \(p-1\mid r-1\). A proper prime divisor has \(r<p\), an impossibility. Thus \(p\) is prime.

(d) The script stores complete \(p-1\) factorizations recursively, proves the leaves below \(10^6\) by trial division, and searches for (rather than trusts) every modular witness. It then:

  1. multiplies every factorization back to \(N+j\);
  2. enumerates all divisors and computes each \(\sigma_5(N+j)\) directly;
  3. independently checks the multiplicative divisor-sum formula;
  4. assembles \(A_5(N)\) both incrementally and over a common denominator; and
  5. performs all comparisons as integer cross-products through Python's exact

Fraction type.

Complete standard-library code is in erdos252_wave8o_reverify.py. Run it from the repository root with:

python runs/erdos252_wave8o_reverify.py

The successful run ends with:

human-scale bounds: distance > 1/17 and tail < 10^-95
exact comparison: distance > tail bound = True
CERTIFIED: if alpha_5 is rational in lowest terms a/b, then b > 100000000000000000000000000000000000000000123456789.

(d) I also recomputed all six factorizations, all six divisor sums, the reduced fraction \(r/Q\), and the exact tail comparison independently in PARI/GP; its output agreed digit-for-digit with the standard-library checker.

What remains

(a) No finite collection of (U5) instances proves irrationality, because a hypothetical reduced denominator may exceed every tested \(n\). The exact missing lemma for this route is:

Prove (U5) for an unbounded sequence of integers \(n\).

(b) Existing prime-pattern/sieve arguments provide the needed uniformity through \(k=4\), but Pratt explicitly reports that those techniques reach their limit there. The finite calculation above supplies no distribution theorem for the six arithmetic functions \(\sigma_5(n),\ldots,\sigma_5(n+5)\), so it does not bridge that analytic uniformity gap.

(d) Extending the computation to any finite height would only increase a conditional denominator lower bound. An exhaustive finite computation, regardless of core-hours, cannot establish the required unbounded sequence; the missing item is a theorem, not a larger search.

PARTIAL: For the smallest open case k=5, an exact six-term reduction and independently certified 51-digit computation prove that any rational value has reduced denominator greater than 100000000000000000000000000000000000000000123456789; irrationality still requires the uniform unbounded-(U5) lemma.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger