Erdős problem #252 — wave 8o
Access date: 2026-07-28 (UTC).
Claim labels
- (a) elementary-rigorous: a proof is given here from elementary facts.
- (b) rigorous-modulo-named-theorem: the claim is a theorem in the named
primary source, or uses the explicitly named theorem.
- (c) plausible/structural-unverified: no such claim is used in the result
below.
- (d) computational-only: a finite exact calculation/certificate, not a
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.
3. 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*](https://arxiv.org/abs/1105.1452), 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*](https://sites.williams.edu/mstoiciu/files/2012/08/irrationality.pdf),
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*](https://arxiv.org/abs/2209.11124), 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*](https://doi.org/10.1016/j.jnt.2010.11.009) 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}\(\prod_{i=0}^j(n+i)\geq n^{j+1}\) 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\(T_5(n)=A_5(n)+R_5(n)\) 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 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. (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 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 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 Complete standard-library code is in repository root with: The successful run ends with: (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. (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.Why the certificate itself is checkable
Fraction type.erdos252_wave8o_reverify.py. Run it from thepython runs/erdos252_wave8o_reverify.py
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.
What remains