GPT 6 Astra Ultra attempt on Math-5562

FORMAL RESTATEMENT

Erdős #297; a sharp entropy upper bound and a constructive exponential lower bound, 2026-09-24. Let \[ F(N)=\#\{E\subseteq[1,N]:\ \sum_{n\in E}1/n=1\}. \] Determine its growth, particularly its exponential rate.

QUICK LITERATURE AND CONTEXT CHECK

Liu--Sawhney and, independently, Conlon--Fox--He--Mubayi--Pham--Suk--Verstraëte determine $F(N)=\exp((\gamma_*+o(1))N)$, where $\gamma_*\approx0.631573$. The proof below establishes the matching upper exponent directly and proves a weaker positive exponential lower exponent using Martin's dense unit-fraction representations. It leaves a definite gap at the sharp lower bound. The upper-bound method is the standard exponential-weight argument underlying the known result.

OBJECTIVE AND ATTACK PLAN

Assign each subset a weight depending only on its reciprocal sum. Optimizing the resulting product gives the sharp upper exponent. For a lower bound, find linearly many terms that can be replaced independently by three unit fractions without introducing any repeated denominator.

WORK

The product bound. Fix $\lambda>0$. Expanding a product over subsets gives \[ \prod_{n=1}^N(1+e^{-\lambda N/n}) =\sum_{E\subseteq[1,N]}\exp\!\left(-\lambda N\sum_{n\in E}\frac1n\right). \] Each subset counted by $F(N)$ contributes exactly $e^{-\lambda N}$, and all other contributions are nonnegative. Hence \[ F(N)\leq e^{\lambda N}\prod_{n=1}^N(1+e^{-\lambda N/n}). \tag{1} \] The function $x\mapsto\log(1+e^{-\lambda/x})$, with value zero at $x=0$, is continuous on $[0,1]$. Its Riemann sums therefore imply \[ \limsup_{N\to\infty}\frac{\log F(N)}N \leq\Phi(\lambda):=\lambda+\int_0^1\log(1+e^{-\lambda/x})\,dx. \tag{2} \]

Optimization and the constant. Differentiating for $\lambda>0$ is justified by the exponential decay near $x=0$. We obtain \[ \Phi'(\lambda)=1-I(\lambda),\qquad I(\lambda)=\int_0^1\frac{dx}{x(1+e^{\lambda/x})} =\int_\lambda^\infty\frac{du}{u(1+e^u)}. \tag{3} \] The last expression shows that $I$ is continuous, strictly decreasing, tends to infinity as $\lambda\downarrow0$, and tends to zero as $\lambda\to\infty$. Thus there is a unique $\lambda_*>0$ with $I(\lambda_*)=1$. The derivative changes from negative to positive there, giving the unique global minimum \[ \gamma_*:=\Phi(\lambda_*),\qquad F(N)\leq\exp((\gamma_*+o(1))N). \tag{4} \] The limit $\Phi(0+)=\log2$ follows by dominated convergence, and $\Phi'<0$ near zero, so $\gamma_*<\log2$ strictly. Consequently $4$ already rules out $F(N)=2^{(1+o(1))N}$. The numerical values quoted by Liu--Sawhney are $\lambda_*\approx0.127191$ and $\gamma_*\approx0.631573$, corresponding to $\gamma_*/\log2\approx0.9112$ in base two. The exact integral definition, not the numerical approximation, is used in the proof.

A declared input for a positive exponential lower bound. Martin's Theorem 1, specialized to target $1$, gives for all sufficiently large $M$ a set $D\subseteq[1,M]$ with \[ \sum_{d\in D}\frac1d=1,\qquad |D|\geq\frac35M. \tag{5} \] Indeed the theorem gives $|D|>(1-e^{-1})M-O(M\log\log M/\log M)$, and $1-e^{-1}>3/5$. Its construction is imported here.

At least $M/10$ members of $D$ lie in $(M/2,M]$. On these members put an edge between $d$ and $e$ exactly when $2d=3e$ or $3d=2e$. Each vertex has at most two neighbors. Repeatedly selecting one vertex and deleting it and its neighbors therefore gives an independent set $T$ with \[ |T|\geq M/30. \tag{6} \] For each $d\in T$, independently either retain $1/d$ or use the identity \[ \frac1d=\frac1{2d}+\frac1{3d}+\frac1{6d}. \tag{7} \] All new denominators exceed $M$, so they avoid all original denominators. To check collisions between replacements for different $d,e\in T$, note that $d/e\in(1/2,2)$. Among ratios of elements of $\{2,3,6\}$, the only nontrivial possibilities in this interval are $3/2$ and $2/3$, exactly the edges we excluded. Thus every choice produces a representation with distinct denominators at most $6M$.

Different choices produce different sets: the original denominator $d\leq M$ is present precisely when its term was retained, and no replacement can produce it. There are consequently $2^{|T|}$ representations, giving \[ F(6M)\geq2^{M/30}. \] For arbitrary sufficiently large $N$, take $M=\lfloor N/6\rfloor$ and use monotonicity to obtain \[ F(N)\geq2^{N/180-O(1)}. \tag{8} \]

VERIFICATION AND REMAINING GAP

The argument proves a positive exponential lower rate and the exact known upper exponent. In particular it places $F(N)$ strictly below the full subset count on the exponential scale. It does not prove $F(N)\geq\exp((\gamma_*-o(1))N)$: that requires constructing exponentially many representations whose denominator distribution realizes the optimizing weights in $3$. The independent replacement construction loses most of that entropy. The published sharp lower bound is acknowledged but not used to conceal this gap.

SOURCES

• Current problem and independent resolutions: https://www.erdosproblems.com/297.
• Y. P. Liu and M. Sawhney, On further questions regarding unit fractions, Theorem 1.2: https://arxiv.org/html/2404.07113.
• G. Martin, Denser Egyptian Fractions, Theorem 1: https://arxiv.org/pdf/math/9811112.

CONCLUSION

Attempt status: unresolved within this reconstruction. The sharp upper exponent and a positive exponential lower bound are proved; matching the lower exponent remains a gap. Completion rate estimate: 75%; a heuristic estimate of the proof coverage.