Erdoes #1053 - working report - Erdős Problem a Day

Erdős problem #1053 — prime-support reduction and exact checks

Accessed and computed on 2026-07-28 (UTC). In the labels below:

(a) means elementary-rigorous;
(b) means rigorous modulo the named theorem;
(c) means plausible/structural-unverified;
(d) means computational-only.

0. Mandatory live-page check

the Bright Data browser path, not by datacenter curl.

The verbatim live statement is:

> Call a number $k$-perfect if $\sigma(n)=kn$, where $\sigma(n)$ is the sum of the divisors of $n$. Must $k=o(\log\log n)$?

(d/page observation) The page status was OPEN. It displayed:

0 comments on this problem;
0 claimed proofs for this problem;
Currently working on this problem: None;
Interested in collaborating: None.

Thus the mandatory collision/claimed-proof stop condition did not apply.

(d/page observation) The accompanying text attributes the question to

Erdős via problem B2 of Guy's collection [Gu04]. It records Guy's

suggestion that there might be only finitely many $k$-perfect numbers for

$k\geq3$, says that the largest multiplicity currently represented by a

known example is $k=11$, calls these numbers multiply perfect, and links

OEIS A007539. The live page lists no further theorem about the question.

1. Literature check

The following are the directly relevant primary sources I could verify.

1. (d/source check) F. Luca and J. L. Varona,

Multiperfect numbers on lines of the Pascal triangle,

J. Number Theory 129 (2009), 1136–1148

(author-hosted PDF).

On pp. 1–2 they state this precise Erdős conjecture, say that no general

generation formula is known beyond the Euclid–Euler $k=2$ case, and

then prove finiteness only on specified lines of Pascal's triangle. This

is not a solution of the unrestricted question.

2. (d/source check) P. Erdős,

On perfect and multiply perfect numbers,

Ann. Mat. Pura Appl. (4) 42 (1956), 253–258,

proves that the counting function $P(x)$ of multiply-perfect numbers

satisfies $P(x)0$ and

sufficiently large $x$. This is a sparsity theorem, not a bound

improving the multiplicity of each individual number.

3. (d/source check) B. Hornfeck and E. Wirsing,

Über die Häufigkeit vollkommener Zahlen,

Math. Ann. 133 (1957), 431–438,

DOI 10.1007/BF01343756, sharpened

the counting result to $P(x)=o(x^\epsilon)$ for every

$\epsilon>0$. Again, this does not control $k$ on a selected

multiply-perfect number.

4. (d/source check) G. Robin, *Grandes valeurs de la fonction somme des

diviseurs et hypothèse de Riemann*, J. Math. Pures Appl. 63 (1984),

187–213, proves unconditionally, for $n\geq3$,

\[ \frac{\sigma(n)}n < e^\gamma\log\log n+\frac{0.6483}{\log\log n}. \]

Under RH the second term can be removed for $n>5040$. Either form

supplies only a constant multiple of $\log\log n$, not the desired

little-$o$.

5. (d/source/data check) A. Flammenkamp's current

Multiply Perfect Numbers page

supplies a gzip file of 5,932 known/claimed entries through

2022-12-31. Its current multiplicity counts agree with the live

tracker's assertion that examples are known only through $k=11$.

I audit the 5,880 proper entries below; I do not treat that list as a

completeness theorem.

Exact-phrase searches for the conjecture and searches on

sigma(n)=kn, multiplicity bounds, and multiply-perfect surveys also found

later papers about special shapes (Fibonacci numbers, binomial

coefficients, repdigits), odd multiperfect structure, radicals, and counts

with a bounded number of prime factors. (d/source-search result) I

found no primary source claiming a proof or counterexample to the

unrestricted asymptotic question. This search miss is not itself proof

that no such paper exists.

2. A prime-support criterion exactly equivalent to #1053

This is the main rigorous reduction.

For $n>1$, put

\[ I(n):=\frac{\sigma(n)}n,\qquad y:=\log n, \]

and define three nonnegative quantities:

\[ \begin{split} D(n)&:=\sum_{\substack{p\leq \log n\\p\nmid n}} \log\frac{p}{p-1},\\ E(n)&:=\sum_{p^a\parallel n} \log\frac{1}{1-p^{-(a+1)}},\\ T(n)&:=\sum_{\substack{p\mid n\\p>\log n}} \log\frac{p}{p-1}. \end{split} \]

Here and throughout, sums indexed by $p$ are over primes.

Proposition 1 (exact Euler-product identity)

(a) For every $n>1$,

\[ \boxed{\quad I(n)= \left(\prod_{p\leq y}\left(1-\frac1p\right)^{-1}\right) \exp\{-D(n)-E(n)+T(n)\}. \quad} \tag{1} \]

Proof. If $n=\prod p^{a_p}$, multiplicativity gives

\[ I(n)=\prod_{p^{a_p}\parallel n} \frac{1-p^{-(a_p+1)}}{1-p^{-1}}. \]

Insert the Euler factors for all $p\leq y$. A small prime not dividing

$n$ contributes the compensating factor $1-1/p$, every finite exponent

contributes $1-p^{-(a_p+1)}$, and every prime divisor above $y$

contributes its additional factor $(1-1/p)^{-1}$. Taking logs yields

(1). $\square$

Uniform bounds for the two error terms

(a) The exponent-truncation loss is uniformly bounded:

\[ 0\leq E(n) \leq\sum_p-\log(1-p^{-2}) =\log\zeta(2). \tag{2} \]

This uses only $a_p+1\geq2$.

(a) The large-prime tail tends uniformly to zero. If $r$ is the

number of prime divisors of $n$ exceeding $y$, then

\[ r\log y\leq \log\operatorname{rad}(n)\leq\log n=y. \]

Also $\log(p/(p-1))\leq1/(p-1)$. Therefore

\[ 0\leq T(n)\leq \frac{y}{(y-1)\log y} =O\!\left(\frac1{\log\log n}\right). \tag{3} \]

Mertens' prime-product theorem says

\[ \prod_{p\leq y}\left(1-\frac1p\right)^{-1} =e^\gamma\log y\,(1+o(1)). \]

Combining it with (1)–(3) gives the following uniform asymptotic.

Theorem 2 (support-defect formula)

(b), rigorous modulo Mertens' prime-product theorem. As $n\to\infty$,

\[ \boxed{\quad \frac{\sigma(n)/n}{e^\gamma\log\log n} =\exp\{-D(n)-E(n)+o(1)\}. \quad} \tag{4} \]

The $o(1)$ here is uniform in the prime factorisation of $n$.

Since $0\leq E(n)\leq\log\zeta(2)$, (4) has an exact qualitative

consequence:

\[ \boxed{\quad \frac{\sigma(n)}n=o(\log\log n) \quad\Longleftrightarrow\quad D(n)\longrightarrow\infty. \quad} \tag{5} \]

Indeed, $D\to\infty$ makes the right side of (4) tend to zero. If $D$

does not tend to infinity, some subsequence has $D\leq C$; on that

subsequence (2) and (4) give a positive lower bound

$\exp(-C-\log\zeta(2)+o(1))$ for the normalized ratio.

Finally,

\[ \log\frac{p}{p-1}=\frac1p+O\!\left(\frac1{p^2}\right), \]

and the sum of the $O(p^{-2})$ errors is uniformly bounded over every

subset of the primes. Hence (5) is equivalently

\[ \boxed{\quad \sum_{\substack{p\leq\log n\\p\nmid n}}\frac1p \longrightarrow\infty. \quad} \tag{6} \]

Exact reformulation of the open problem

Restrict (5) or (6) to integers satisfying $n\mid\sigma(n)$.

(b), rigorous modulo Mertens. Erdős problem #1053 is equivalent to:

> Must the reciprocal-prime mass of the primes $p\leq\log n$ which do

> not divide a multiply-perfect $n$ tend to infinity as

> $n\to\infty$?

Equally, a counterexample sequence is exactly a sequence of

multiply-perfect numbers whose omitted primes below $\log n$ have

bounded Euler cost $D(n)$. This cleanly separates the generic maximal

order of $\sigma(n)/n$ from the special divisibility equation

$n\mid\sigma(n)$.

3. A second exact reduction: the least number of multiplicity $k$

Let

\[ N(k):=\min\{n:\sigma(n)=kn\}, \]

with $N(k)=\infty$ when no such number exists, and let

$\mathcal K=\{k:N(k)<\infty\}$.

(a) Problem #1053 is also equivalent to

\[ \frac{\log\log N(k)}{k}\longrightarrow\infty \qquad(k\to\infty,\ k\in\mathcal K). \tag{7} \]

For a fixed $k$, $k/\log\log n\to0$ automatically as $n\to\infty$.

For unbounded $k$, every $k$-perfect $n$ satisfies $n\geq N(k)$;

conversely the numbers $N(k)$ themselves test the worst ratio for each

$k$.

Thus (7) asks for a lower bound of the form

\[ N(k)\geq \exp\!\exp\{kL(k)\} \quad\text{with }L(k)\to\infty \]

along attainable multiplicities. **(b), rigorous modulo Robin's stated

inequality**, the standard maximal-order machinery gives only

\[ \log\log N(k)\geq(e^{-\gamma}+o(1))k, \tag{8} \]

which is double-exponential growth with a constant coefficient in the

second logarithm. The missing step is precisely the divergence in (7),

not a better numerical value of the constant in (8).

4. Exact divisibility flow and a solved structural subcase

For a prime $\ell$, write $a_\ell=v_\ell(n)$, allowing

$a_\ell=0$. The equation $\sigma(n)=kn$ is equivalent, prime by

prime, to

\[ \boxed{\quad a_\ell+v_\ell(k) =\sum_{q^a\parallel n}v_\ell\!\left( \frac{q^{a+1}-1}{q-1}\right) \quad\text{for every prime }\ell. \quad} \tag{9} \]

(a) This is an exact valuation identity. It is the factor-chain

constraint absent from arbitrary integers having large

$\sigma(n)/n$.

Largest-prime source lemma

(a) If $n>1$ is $k$-perfect and $P$ is its largest prime factor,

then \(k \[ P^{v_P(n)} \ \Bigm|\! \prod_{\substack{q^a\parallel n\\qTo see \(k \[ k< \prod_{p\mid n}\left(1-\frac1p\right)^{-1} \leq\prod_{p\leq P}\left(1-\frac1p\right)^{-1}. \]

The last product is at most $P$: it equals $2$ and $3$ at

$P=2,3$; thereafter, if $Q$ is the preceding prime and the product

through $Q$ is at most $Q$, then the product through $P$ is at most

\(QP/(P-1)

Moreover $\sigma(P^a)\equiv1\pmod P$, so (9) forces all of

$P^{v_P(n)}$ to come from smaller prime powers, proving (10).

Complete squarefree classification

(a) Proposition 3. The only squarefree multiply-perfect positive

integers are

\[ n=1\quad(k=1),\qquad n=6\quad(k=2). \]

Proof. Let squarefree $n>1$ be $k$-perfect and let $P$ be its

largest prime factor. By the preceding lemma $P\nmid k$. Since

\[ \sigma(n)=\prod_{q\mid n}(q+1)=kn, \]

$P$ must divide $q+1$ for some prime \(q

$2\leq q+1\leq P$, so $q+1=P$. If $P>3$, then $P-1=q$ is an

even integer greater than $2$, not a prime. The case $P=2$ has no

smaller source. Hence $P=3$, $q=2$, and $n=2\cdot3=6$.

The separate case $n=1$ has $\sigma(1)=1$. $\square$

In particular, any hypothetical asymptotic obstruction to #1053 must use

nontrivial prime powers; squarefree near-extremizers cannot be

multiply-perfect.

5. From-scratch exhaustive computation through $10^8$

The standalone checker is

erdos1053_wave8b_reverify.py.

Its main search uses a linear Euler sieve. Along with $\sigma(i)$, it

stores the least-prime power $p^a\parallel i$ and

$\sigma(p^a)=1+p+\cdots+p^a$. If $i=mp^a$ with $(m,p)=1$, then

\[ \sigma(ip)=\frac{\sigma(i)}{\sigma(p^a)} \bigl(\sigma(p^a)+p^{a+1}\bigr); \]

if a new prime $q\nmid i$ is appended, then

$\sigma(iq)=\sigma(i)(q+1)$. (a) These identities prove the

recurrence computes the exact divisor sum of every integer visited.

As implementation cross-checks, the script:

compares every value through 20,000 with an independent

divisor-addition sieve;

trial-factors every reported hit and recomputes its geometric sigma

product independently;

uses the hard-coded expected table only as a final assertion, never to

guide the enumeration.

The command was

python runs/erdos1053_wave8b_reverify.py --limit 100000000

(d) It found exactly the following 14 pairs in

$1\leq n\leq10^8$:

| $n$ | $k$ | independently recovered factorisation |

|---:|---:|:---|

| 1 | 1 | $1$ |

| 6 | 2 | $2\cdot3$ |

| 28 | 2 | $2^2\cdot7$ |

| 120 | 3 | $2^3\cdot3\cdot5$ |

| 496 | 2 | $2^4\cdot31$ |

| 672 | 3 | $2^5\cdot3\cdot7$ |

| 8128 | 2 | $2^6\cdot127$ |

| 30240 | 4 | $2^5\cdot3^3\cdot5\cdot7$ |

| 32760 | 4 | $2^3\cdot3^2\cdot5\cdot7\cdot13$ |

| 523776 | 3 | $2^9\cdot3\cdot11\cdot31$ |

| 2178540 | 4 | $2^2\cdot3^2\cdot5\cdot7^2\cdot13\cdot19$ |

| 23569920 | 4 | $2^9\cdot3^3\cdot5\cdot11\cdot31$ |

| 33550336 | 2 | $2^{12}\cdot8191$ |

| 45532800 | 4 | $2^7\cdot3^3\cdot5^2\cdot17\cdot31$ |

(d) The exhaustive portion took 76.748 seconds. The combined

enumeration and database audit below took 105.929 seconds, with peak RSS

2,201,408 KiB. The bounded table is exact conditional only on correct

execution of the supplied integer program; it makes no assertion beyond

$10^8$.

6. Independent audit of all 5,880 published proper examples

For reproducibility the checker can download the source itself:

python runs/erdos1053_wave8b_reverify.py \
  --limit 0 --download-database --prove-primes

The exact downloaded file had SHA-256

acd1e1e65ff070ef74c5a4ca1ee261c00eb9dc90e614ff1f25943f6e5a47bc4f

For every $k\geq3$ row, the checker parses the claimed prime powers and

tests with Python big integers that

\[ \prod_{p^a\parallel n}\frac{p^{a+1}-1}{p-1} =k\prod_{p^a\parallel n}p^a. \]

It separately passes all distinct listed bases to PARI/GP's rigorous

isprime routine.

(d) Results:

5,880 proper entries checked;
5,496 distinct base factors proved prime by PARI/GP;
zero composite bases and zero sigma-identity failures;
counts by multiplicity

\[ \{3:6,\ 4:36,\ 5:65,\ 6:245,\ 7:516,\ 8:1136,\ 9:2164,\ 10:1710,\ 11:2\}. \]

The checker then recomputes $D(n)$ from Section 2 with 80-digit Decimal

logs. The exact log identity (1) had maximum numerical residual

$1.9\times10^{-78}$.

(d) The following table describes the examples in this particular

published list. The last two columns are the observed range of

$k/\log\log n$.

|---:|---:|---:|---:|---:|---:|

| 3 | 6 | 0.000000000000 | 0.685894379973 | 0.936042006869 | 1.915700702020 |

| 4 | 36 | 0.000000000000 | 0.732683172754 | 0.861101821750 | 1.713953687392 |

| 5 | 65 | 0.124494470244 | 0.681724642199 | 0.918787438110 | 1.586456161874 |

| 6 | 245 | 0.131634669161 | 0.604921528367 | 0.984838575377 | 1.562847859882 |

| 7 | 516 | 0.222800844840 | 0.527013744770 | 1.064270988441 | 1.439718299193 |

| 8 | 1136 | 0.260442906791 | 0.485737891954 | 1.103011285328 | 1.397682551629 |

| 9 | 2164 | 0.261012207980 | 0.452832697345 | 1.139923515284 | 1.386184883731 |

| 10 | 1710 | 0.269289606649 | 0.403550568333 | 1.194485657544 | 1.371104167698 |

| 11 | 2 | 0.299331608269 | 0.309769786696 | 1.311552817579 | 1.326130828944 |

Across all 5,880 rows, $0\leq D(n)\leq0.732683172754$.

(c) This finite data is compatible with a counterexample shape from

Theorem 2—known examples omit only bounded reciprocal-prime mass below

$\log n$. It is not evidence of a limiting counterexample: all examples

have $k\leq11$, the list is not a complete asymptotic family, and a

bounded finite range cannot decide whether $D(n)$ eventually diverges.

7. Exact wall

The standard machinery stalls at a sharply identifiable point.

1. (b) Mertens/Robin controls the full product of small-prime Euler

factors and yields only $k=O(\log\log n)$.

2. (a) The finite-exponent loss $E(n)$ is bounded by

$\log\zeta(2)$, and prime factors above $\log n$ contribute only

$o(1)$ on the logarithmic scale. Neither can produce the desired

little-$o$.

3. (a) Therefore the one missing lemma needed—and, by (5), a lemma

exactly equivalent to the problem—is

\[ n\mid\sigma(n),\ n\to\infty \quad\Longrightarrow\quad \sum_{\substack{p\leq\log n\\p\nmid n}}\frac1p\to\infty. \tag{11} \]

4. (a) The valuation-flow identity (9) is the available integrality

constraint. What is not known is a quantitative “factor-chain closure

cost” theorem deriving (11) from (9).

5. (a) The Hornfeck–Wirsing bound $P(x)=x^{o(1)}$ controls how many

multiply-perfect numbers there are, but sparsity of the set gives no

upper bound on the abundance or missing-prime defect of each member.

Brute-force extension cannot supply the missing uniformity. The linear

sieve already used 2.20 GB at $10^8$ and scales linearly; at $10^9$ it

would require roughly 22 GB and about 13 CPU-minutes at the observed rate.

Even that finite extension would not touch (11), and the published

factor-chain database already reaches vastly larger individual numbers.

PARTIAL: Proved that #1053 is exactly equivalent to divergence of the missing-prime sum in (11), classified all squarefree cases, and exhaustively verified the 14 multiply-perfect numbers through $10^8$; the unresolved step is a quantitative factor-chain lemma forcing that divergence.

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