Erdős Problem #469

Erdős Problem 469

Statement and scope.

A positive integer is semiperfect, or pseudoperfect, if it is a sum of distinct proper divisors. It is primitive semiperfect if none of its proper divisors is semiperfect. Does the reciprocal sum over all primitive semiperfect integers converge? The current page records an affirmative solution by Zachary J. Lewis using GPT 5.6 and Claude Fable 5. This attempt reconstructs useful exact transition formulas from that preprint, but the global convergence estimate is unresolved here.

Subset-sum notation.

For an integer $n$, let $\Sigma=\sigma(n)$, $A=\Sigma-2n$, and let $X(n)$, $Y(n)$ be the subset sums of all divisors and of proper divisors, respectively, including the empty sum. Complementing a proper-divisor subset shows \[ n\in Y(n)\quad\Longleftrightarrow\quad A\in Y(n), \tag{1} \] since the total proper-divisor sum is $n+A$. Call $n$ weird if it is abundant and not semiperfect. For such $n$, define \[ C_n=\{t\geq1:A+t\in Y(n)\},\quad g(n)=\min C_n,\quad T(n)=\Sigma/g(n). \] The set is nonempty because $n+A\in Y(n)$; hence $g(n)\leq n$ and $T(n)>2$.

Exact extension by a new prime.

Let $p\nmid n$ be prime and $n$ weird. Coprimality gives \[ Y(np)=X(n)+pY(n),\qquad A(np)=pA+\Sigma. \tag{2} \] Also $X(n)$ is symmetric under $x\mapsto\Sigma-x$. By $1$--$2$, $np$ is semiperfect exactly when \[ pt\in X(n)\quad\hbox{for some }t\in C_n. \tag{3} \] Indeed, $pA+\Sigma=x+py$ with $x\in X(n)$, $y\in Y(n)$ forces $y>A$: $y<A$ would give $x>\Sigma$, while $y=A$ is excluded by weirdness. Write $y=A+t$; then $x=\Sigma-pt$, which belongs to $X(n)$ exactly when $pt$ does.

If $np$ is weird, the same calculation with a positive offset $s$ from $A(np)$ gives the exact gap formula \[ g(np)=\min_{\substack{t\in C_n,\ x\in X(n)\\x<pt}}(pt-x). \tag{4} \] The exclusion of $y\leq A$ remains valid with positive $s$. In particular, if $p>T(n)$, then $pg(n)>\Sigma$, so $3$ is impossible. The child is still abundant, by multiplicativity of $\sigma(n)/n$, and therefore is weird. Formula $4$ then gives \[ g(np)=pg(n)-\Sigma,\qquad T(np)=\frac{T(n)(p+1)}{p-T(n)}. \tag{5} \] The minimum is attained at $t=g(n)$, $x=\Sigma$. If instead $p\leq T(n)$ and the child remains weird, take $t=g(n)$, $x=0$ in $4$ to obtain \[ g(np)\leq pg(n),\qquad T(np)\geq T(n)(1+1/p)>p. \tag{6} \] Thus a first semiperfect extension at a new prime must use $p\leq T(n)$, but missed candidates can increase the next threshold.

A fully checked primitive example beyond deficient parents.

For $n=70$, $\Sigma=144$, $A=4$, and the proper divisors are $1,2,5,7,10,14,35$. They cannot sum to $4$: using only $1,2$ gives at most three, while every other divisor is at least five. Thus $70$ is weird, and $g(70)=1$, since $5$ is a proper divisor. Consequently every coprime prime $p>144$ gives a weird number $70p$. For $p=11$, however, $11=1+10\in X(70)$, so $3$ predicts semiperfectness, witnessed explicitly by \[ 770=1+10+11+22+77+110+154+385. \] All displayed summands are distinct proper divisors. The maximal proper divisors of $770$ are $385,154,110,70$. Their divisor sums are $576,288,216,144$, respectively, so the first three are deficient and the fourth is weird. Every divisor of a deficient number is deficient, since $\sigma(d)/d$ increases under divisibility. Every divisor of a non-semiperfect number is non-semiperfect: semiperfectness passes to multiples by scaling its divisor representation. It follows that no proper divisor of $770$ is semiperfect. Thus $770$ is primitive semiperfect.

Remaining gap and attribution.

Equations $3$--$6$ give exact local information, not a bound on the total reciprocal mass of all first semiperfect hits over all prime extension trees. Finitely many candidate primes at each fixed state does not imply summability of the entire tree; the varying threshold and the treatment of roots and repeated prime factors remain essential. The global mass estimate is not proved here.

Source: Z. J. Lewis, Convergence of the reciprocal sum of primitive semiperfect numbers, version 1.2.1, Lemma 2.1, https://github.com/ZachL111/erdos469-preprint/blob/main/paper/output/pdf/erdos469_v1_2_1_preprint.pdf. That version explicitly describes independent human review as pending. The transition arguments are supplied above and are not taken on faith from its global claim. Current problem: https://www.erdosproblems.com/469, checked 24 September 2026. No local formal-verification claim is made.

Completion estimate.

Exact prime-extension dynamics and a primitive example are proved; reciprocal-mass control is missing. COMPLETION: 45%

1) FORMAL RESTATEMENT.

Call an integer $n\ge 2$ pseudoperfect if $n$ can be written as a sum of distinct proper divisors of $n$. (Proper divisors are positive divisors $<n$; the divisor $1$ is allowed.)

Let $A$ be the set of primitive pseudoperfect numbers: $n\in A$ if $n$ is pseudoperfect but no proper divisor $m\mid n$ with $m<n$ is pseudoperfect.

Question: Does the sum of reciprocals converge? \[ \sum_{n\in A}\frac{1}{n}\ \stackrel{?}{<}\ \infty. \]

2) QUICK LITERATURE/CONTEXT CHECK.

The problem statement mentions that these numbers are listed as A006036 in OEIS and cites [BeEr74]. I will not assume any OEIS facts and will work from the definition, plus direct computation.

3) ATTACK PLAN.

• Prove basic necessary conditions for pseudoperfectness (in particular, $\sigma(n)\ge 2n$).
• Give explicit examples of primitive pseudoperfect numbers with full verification.
• Compute the initial segment of $A$ and partial sums $\sum_{n\in A,\ n\le X} 1/n$ for moderate $X$ to see whether the sum plausibly converges.

4) WORK.

Lemma 469.1 (pseudoperfect $\Rightarrow$ abundant or perfect). If $n$ is pseudoperfect, then \[ \sigma(n)\ge 2n\quad\text{equivalently}\quad \sum_{\substack{d\mid n\\ d<n}} d\ \ge\ n. \]

Proof. If $n$ is pseudoperfect, there exists a subset $S$ of proper divisors of $n$ with $\sum_{d\in S} d=n$. The sum of all proper divisors is at least the sum over $S$, hence at least $n$. Therefore $\sigma(n)-n\ge n$, i.e. $\sigma(n)\ge 2n$. $\square$

Lemma 469.2 (no prime power is pseudoperfect). Let $n=p^k$ with $p$ prime and $k\ge 1$. Then $n$ is not pseudoperfect.

Proof. The proper divisors of $p^k$ are $1,p,p^2,\dots,p^{k-1}$, whose sum equals \[ 1+p+\cdots+p^{k-1}=\frac{p^k-1}{p-1}. \] We claim this is strictly less than $p^k$. Indeed \[ \frac{p^k-1}{p-1}<p^k\ \Longleftrightarrow\ p^k-1<p^k(p-1)=p^{k+1}-p^k, \] which holds for all $p\ge 2$ since $p^{k+1}-p^k\ge p^k>p^k-1$. Thus even the total sum of proper divisors is $<n$, so no subset can sum to $n$. $\square$

Lemma 469.3 (explicit primitive pseudoperfect example: $20\in A$). The integer $20$ is primitive pseudoperfect.

Proof. The proper divisors of $20$ are $1,2,4,5,10$. They have a distinct-sum representation \[ 20=10+5+4+1. \] So $20$ is pseudoperfect. Its proper divisors are $1,2,4,5,10$; none of $2,4,5,10$ is pseudoperfect because in each case the sum of all proper divisors is $<m$ (respectively $1<2$, $1+2=3<4$, $1<5$, and $1+2+5=8<10$). Hence $20$ has no pseudoperfect proper divisor, so $20\in A$. $\square$

FAST REALITY CHECK (computations).

I computed all primitive pseudoperfect numbers up to several limits using exact subset-sum (bitset DP) over proper divisors, and filtered primitiveness by checking proper divisors.

Results:

limit   |A ∩ [1,limit]|   partial sum  Σ_{n∈A, n≤limit} 1/n
20000   163                0.33312907483095244
50000   299                0.3372780998948891
100000  490                0.3400146761643518
200000  667                0.34128359582091095

The first 25 elements of $A$ found were

[6, 20, 28, 88, 104, 272, 304, 350, 368, 464, 490, 496, 550, 572,
 650, 748, 770, 910, 945, 1184, 1190, 1312, 1330, 1376, 1430]

and among those $\le 200000$ there were $117$ odd elements (for example $945\in A$).

5) VERIFICATION.

• Lemma 469.1 is immediate from monotonicity of "sum of all proper divisors" versus "sum of a subset".
• Lemma 469.2 checks the full proper-divisor sum of a prime power and shows it is $<n$, which is decisive.
• The computation uses an exact subset-sum DP over the (finite) list of proper divisors, so there are no probabilistic approximations.

6) FINAL. UNRESOLVED

(i) Strongest proved partial result. Pseudoperfect numbers are necessarily abundant/perfect ($\sigma(n)\ge 2n$, Lemma 469.1) and no prime power is pseudoperfect (Lemma 469.2). Computations up to $200000$ found $667$ primitive pseudoperfect numbers and a partial reciprocal sum $\sum_{n\in A,\ n\le 200000}1/n\approx 0.3412835958$.

(ii) First gap (crisp). Prove either \[ \sum_{n\in A}\frac1n<\infty \] or \[ \sum_{n\in A}\frac1n=\infty. \] A concrete intermediate target is to obtain a quantitative upper bound on the counting function $A(x):=|\{n\in A:n\le x\}|$ strong enough to decide convergence/divergence (e.g. $A(x)\ll x/(\log x)^{1+\varepsilon}$ would imply convergence).

(iii) Top 3 next moves.

1. Derive structural restrictions on primitive pseudoperfect numbers stronger than Lemma 469.1--469.2 (e.g. constraints on the size of the smallest prime factor, or on the pattern of abundancy $\sigma(n)/n$) and convert them into a counting bound.
2. Look for infinite families inside $A$ (or prove they cannot exist). Any explicit parametrized family gives information about convergence by comparison tests.
3. Extend computation (e.g. to $10^6$ or beyond) and track both $A(x)$ and partial sums $\sum_{n\in A, n\le x}1/n$ to guess the correct growth regime.

(iv) Minimal counterexample structure. If the sum diverges, there must exist many primitive pseudoperfect numbers at each scale, roughly enough that $\sum_{n\in A\cap[2^k,2^{k+1})}1/n$ does not go to $0$ as $k\to\infty$. This would require an infinite supply of primitive pseudoperfect numbers with controlled multiplicative structure (since Lemma 469.2 rules out prime powers), possibly forming relatively dense clusters.