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Math-759 · mathematics · Combinatorics · Mathematics

Erdős Problem #320

Status Solved · erdosproblems.comLLM status Solved from source status

Statement

For $N\in\mathbb{N}$ define \[ S(N):=\left|\left\{\sum_{n\in A}\frac{1}{n}\ :\ A\subseteq \{1,2,\dots,N\}\right\}\right|, \] the number of distinct subset sums of the reciprocals $1,1/2,\dots,1/N$. Estimate $S(N)$ as $N\to\infty$.

Restatement from Math-759-A3 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

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Math-759-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-759

Erdős Problem #320: counting reciprocal subset sums

Partial proof with explicit structural recursion, 24 September 2026.

1) Statement

Let \[S(N)=\left|\left\{\sum_{n\in A}\frac1n:A\subseteq\{1,\ldots,N\}\right\}\right|,\qquad s(N)=\log S(N). \] Estimate $S(N)$. The empty sum is included. This attempt proves \[c\frac N{\log N}\le s(N)\le C\frac{N\log\log N}{\log N}\] for all sufficiently large $N$, with absolute positive constants. The sharper iterated-logarithm order reported by the current source is not derived.

2) Prime denominators give distinct sums

All subsets of the primes at most $N$ have distinct reciprocal sums. Indeed, a collision would give $\sum_{p\le N}\epsilon_p/p=0$, with $\epsilon_p\in\{-1,0,1\}$ not all zero. Multiply by $P=\prod_{p\le N}p$ and reduce modulo a prime $q$ with $\epsilon_q\ne0$. Every term except $\epsilon_qP/q$ vanishes modulo $q$, and that remaining term is nonzero, a contradiction. Consequently \[S(N)\ge2^{\pi(N)}.\] The classical Chebyshev estimates $c_0x/\log x\le\pi(x)\le C_0x/\log x$, for sufficiently large $x$, imply the claimed lower bound. These standard prime-counting estimates are the only analytic number-theoretic input below.

3) Separate smooth denominators from large-prime denominators

Set $y=N/\log N$, taking $N$ large enough that $y>\sqrt N$. An integer $d\le N$ either has all prime factors at most $y$, or has exactly one prime factor $p>y$, with exponent one. In the latter case $d=pm$, with $m\le N/p<\log N<y$.

Every smooth denominator divides \[L_y=\prod_{p\le y}p^{\lfloor\log N/\log p\rfloor}. \] Thus its reciprocal subset sums lie on the lattice $L_y^{-1}\mathbb Z$ and in $[0,H_N]$, where $H_N=\sum_{j\le N}1/j\le1+\log N$. Their number is at most $\lfloor L_yH_N\rfloor+1$. Chebyshev's upper bound gives \[\log L_y\le\pi(y)\log N\ll\frac{N}{\log N}. \] The logarithm of the number of smooth subset sums is therefore $O(N/\log N)$.

For a fixed $p>y$, sums using denominators $pm$ are precisely $1/p$ times subset sums from $1,\ldots,\lfloor N/p\rfloor$, and number $S(\lfloor N/p\rfloor)$. Adding one choice from each denominator class can only identify choices, not create more than their product. We obtain the exact-form upper recursion \[s(N)\le C_1\frac N{\log N}+\sum_{y<p\le N}s(\lfloor N/p\rfloor).\tag{1}\]

4) A complete first bound from the recursion

Use $s(m)\le m\log2$. Partial summation and Chebyshev's upper bound yield \begin{align*} \sum_{y<p\le N}\frac1p &=\frac{\pi(N)}N-\frac{\pi(y)}y+\int_y^N\frac{\pi(t)}{t^2}\,dt\\ &\le\frac{C_2}{\log N}+C_2\log\!\left(\frac{\log N}{\log y}\right) \ll\frac{\log\log N}{\log N}. \end{align*} For the last step, $\log y=\log N-\log\log N$ and $-\log(1-u)\le2u$ when $0\le u\le1/2$. Substitution into $1$ proves \[s(N)\ll\frac{N\log\log N}{\log N}. \] All constants here are fixed; no iteration depth depending on $N$ has been absorbed into one of them.

5) Gap, sources and outcome

UNRESOLVED at the sharp scale in this attempt. The prime construction, the smooth-denominator counting lemma, recursion $1$, and the displayed coarse bounds are proved. Obtaining the current iterated-logarithm product requires a stronger lower construction and quantitative control through a number of recursive steps which itself grows with $N$. Iterating an unspecified constant in an $\ll$ estimate would not establish a uniform constant in that sharper result.

The decomposition follows the Bleicher--Erdős method explained at https://www.erdosproblems.com/320, checked 23 September 2026. That source credits a sharper lower bound to Bettin, Grenié, Molteni and Sanna, https://arxiv.org/abs/2509.10030, and the matching upper analysis to Young, Zhu, Luo and GPT 5.6 Sol. Their sharp results are context, not substituted for a proof in this partial attempt. No novelty or formal verification is claimed.

The estimate describes progress toward the sharp size estimate, not a probability.

6) COMPLETION ESTIMATE

COMPLETION: 30%

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Math-759-A3 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-759

Erdős problem 320

1) FORMAL RESTATEMENT

For $N\in\mathbb{N}$ define \[ S(N):=\left|\left\{\sum_{n\in A}\frac{1}{n}\ :\ A\subseteq \{1,2,\dots,N\}\right\}\right|, \] the number of distinct subset sums of the reciprocals $1,1/2,\dots,1/N$. Estimate $S(N)$ as $N\to\infty$.

2) QUICK LITERATURE/CONTEXT CHECK

Problem text records Bleicher--Erdős bounds: \[ \log S(N)\ge \frac{N}{\log N}\log_2\log N(1-o(1)), \quad \log S(N)\le \frac{N}{\log N}(\log_2\log N)^2(1+o(1)), \] with later improvements (Losonczy). I do not reproduce these proofs here.

3) ATTACK PLAN

Prove two clean self-contained lemmas: $1$ subset sums over reciprocals of distinct primes are unique $\Rightarrow$ explicit lower bound in terms of $\pi(N)$, $2$ every subset sum has denominator dividing $\mathrm{lcm}(1,\dots,N)$ $\Rightarrow$ simple upper bound $S(N)\le \mathrm{lcm}(1,\dots,N)\cdot H_N+1$. Also compute exact $S(N)$ for small $N$.

4) WORK

Lemma 4.1 (Prime-reciprocal subset sums are distinct).
Let $P$ be any finite set of distinct primes. Then the sums $\sum_{p\in A}1/p$ are all distinct as $A$ ranges over subsets of $P$. Proof. Suppose $\sum_{p\in A}1/p=\sum_{p\in B}1/p$. Let $C=A\triangle B$ be the symmetric difference, and cancel common terms to obtain \[ \sum_{p\in C_+}\frac1p=\sum_{p\in C_-}\frac1p \] with $C_+\cup C_-=C$ disjoint. Multiply by $Q=\prod_{p\in C} p$ to get an integer equality \[ \sum_{p\in C_+} \frac{Q}{p}=\sum_{p\in C_-}\frac{Q}{p}. \] Let $\ell$ be the largest prime in $C$. Then $Q/\ell$ is not divisible by $\ell$, whereas for $p\ne \ell$, $Q/p$ is divisible by $\ell$. Reducing mod $\ell$ forces the side containing $Q/\ell$ to be nonzero and the other side $0$, contradiction. Hence $C=\emptyset$ and $A=B$. $\square$
Corollary 4.2.
For every $N$, $S(N)\ge 2^{\pi(N)}$, where $\pi(N)$ is the number of primes $\le N$. Proof. Take $P=\{p\le N:\ p\text{ prime}\}$ and restrict subset sums to primes only; by Lemma 4.1 they are all distinct. $\square$
Lemma 4.3 (LCM upper bound).
Let $L_N=\mathrm{lcm}(1,2,\dots,N)$ and $H_N=\sum_{n=1}^N 1/n$. Then \[ S(N)\le L_N\cdot H_N+1. \] Proof. Every subset sum $s=\sum_{n\in A}1/n$ is a rational number whose denominator (in lowest terms) divides $L_N$ because each $1/n$ has denominator dividing $L_N$. Hence $s=m/L_N$ for some integer $m\ge 0$. Also $0\le s\le H_N$, so $0\le m\le L_N H_N$. Thus there are at most $L_NH_N+1$ possibilities. $\square$

FAST REALITY CHECK (computed exact values)

Brute enumeration (exact rational arithmetic) gives: \[ \begin{array}{c|cccccccccccc} N&1&2&3&4&5&6&7&8&9&10&11&12\\\hline S(N)&2&4&8&16&32&52&104&208&416&832&1664&1856 \end{array} \] and continuing: $S(13)=3712$, $S(14)=7424$, $S(15)=9664$, $S(16)=19328$, $S(18)=59264$, $S(20)=126976$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial results: $S(N)\ge 2^{\pi(N)}$ (Lemma 4.1/Cor. 4.2) and $S(N)\le L_NH_N+1$ (Lemma 4.3).

(ii) First gap: bridge the large gap between these elementary bounds and the refined asymptotics stated in the problem text (of order $\exp((N/\log N)\,\text{polylog}\log N)$).

(iii) Top 3 next moves: 1. Strengthen lower bound by selecting structured denominators (smooth numbers / prime products) with uniqueness mod a large prime factor. 2. Improve upper bound by bounding collisions via additive relations among reciprocals (Egyptian fraction structure). 3. Compute $S(N)$ for larger $N$ and analyze collision patterns to guide theory.

(iv) Minimal counterexample structure (to very large growth): many distinct subsets producing identical rational sums; minimal collisions likely arise from small-denominator identities (e.g. Egyptian fraction decompositions) propagating multiplicatively.

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Math-759-A2 · No claim recorded · LLM-generated · gpt 5.2 ·

gpt 5.2 attempt on Math-759

OUTPUT 1) FORMAL RESTATEMENT For $N\ge 1$, define \[S(N)=\left|\left\{\sum_{n\in A}\frac1n:\ A\subseteq\{1,\dots,N\}\right\}\right|,\] the number of distinct rational values attained by sums of reciprocals of subsets of $\{1,\dots,N\}$. Question: estimate the growth of $S(N)$.

2) QUICK LITERATURE/CONTEXT CHECK Browsing is not available. The problem statement lists sophisticated iterated-log bounds from the literature; I do not rederive them. I prove basic structural bounds and compute exact $S(N)$ for small $N$.

3) ATTACK PLAN Proof track: 1. Give easy upper bounds from the common denominator $L_N=\mathrm{lcm}(1,\dots,N)$ and the range $[0,H_N]$. 2. Give a clean exponential-in-$\pi(N)$ lower bound by restricting to subsets of primes and showing their reciprocal sums are all distinct. 3. Compute exact $S(N)$ for small $N$ to see collisions and growth.

Disproof track: 1. Not applicable; the task is to estimate growth, not a yes/no statement.

Chosen path: elementary bounds + computation.

4) WORK Lemma 1 (Prime-subset sums are all distinct). Let $P_N$ be the set of primes $\le N$. Then the map \[A\subseteq P_N\ \longmapsto\ \sum_{p\in A}\frac1p\] is injective. Consequently, \[S(N)\ge 2^{\pi(N)}.\]

Proof. Suppose $A,B\subseteq P_N$ and $\sum_{p\in A}1/p=\sum_{p\in B}1/p$. Let $p_0$ be the smallest prime in the symmetric difference $A\triangle B$ (so $p_0\in A\setminus B$ or vice versa). Let $D=\prod_{p\in A\cup B} p$. Multiply the equality by $D$ to get \[\sum_{p\in A}\frac{D}{p}=\sum_{p\in B}\frac{D}{p}.\] Reduce modulo $p_0$. For any prime $p\ne p_0$, $D/p$ is divisible by $p_0$ (since $p_0\mid D$ and $p\ne p_0$), hence contributes $0$ modulo $p_0$. The lone term $D/p_0$ appears on exactly one side (because $p_0$ is in exactly one of $A,B$), and $D/p_0$ is not divisible by $p_0$. Thus the two sides cannot be congruent modulo $p_0$, contradiction. Therefore $A=B$, proving injectivity and yielding $2^{\pi(N)}$ distinct sums. $\square$

Lemma 2 (A crude upper bound via $L_N$). Let $L_N=\mathrm{lcm}(1,2,\dots,N)$ and $H_N=\sum_{n=1}^N 1/n$. Then every subset sum $\sum_{n\in A}1/n$ can be written as $m/L_N$ with an integer $m\in[0,H_N L_N]$. Hence \[S(N)\le \lfloor H_N L_N\rfloor+1.\]

Proof. As in Problem #311 Lemma 1: \[\sum_{n\in A}\frac1n=\frac{1}{L_N}\sum_{n\in A}\frac{L_N}{n}=\frac{m}{L_N}\] with $m\in\mathbb{Z}_{\ge 0}$. Also $\sum_{n\in A}1/n\le H_N$, so $m\le H_N L_N$. There are at most $\lfloor H_N L_N\rfloor+1$ integers $m$ in this range. $\square$

FAST REALITY CHECK (computation). Exact values of $S(N)$ for $1\le N\le 15$: \[2,4,8,16,32,52,104,208,416,832,1664,1856,3712,7424,9664.\] Collisions begin at $N=6$ (since $S(6)=52<2^6$), reflecting identities like $1=1/2+1/3+1/6$.

5) VERIFICATION - Lemma 1 is a clean modular uniqueness argument using the smallest differing prime. - Lemma 2 is purely a denominator/range count and is very weak asymptotically but correct. - The computation is exact for $N\le 15$ and shows the first collisions at $N=6$.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: $S(N)\ge 2^{\pi(N)}$ by injectivity on prime subsets (Lemma 1), and $S(N)\le \lfloor H_N L_N\rfloor+1$ with $L_N=\mathrm{lcm}(1,\dots,N)$ (Lemma 2). Exact $S(N)$ for $N\le 15$ is listed in the reality check.

(ii) First gap: Bridge the gap between the elementary exponential-in-$\pi(N)$ lower bound and meaningful upper bounds, toward the iterated-log-type growth rates suggested by the literature bounds in the statement.

(iii) Top 3 next moves: 1. Prove stronger lower bounds by selecting structured subfamilies of denominators with provably distinct subset sums (beyond just primes). 2. Develop collision-counting upper bounds: quantify how many equalities $\sum_{n\in A}1/n=\sum_{n\in B}1/n$ can occur, perhaps via denominator size control. 3. Extend exact computation of $S(N)$ to larger $N$ using integer numerators mod $L_N$ to observe growth and collision structure.

(iv) Minimal counterexample structure (if a conjectured lower bound were false): Large families of distinct subsets $A\subseteq[1,N]$ producing the same rational sum, implying massive collision structure beyond those coming from classical Egyptian fraction identities.

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