gpt pro 5.4 attempt on Math-758

Round 2 continuation for Erdős problems #319, #359, #361

This file is written as a continuation of the Round-1 write-up. I explicitly indicate below whenever I use a Round-1 result as a black box.

---

Problem #319

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. Round 1 gave the lower bound \[ F(N)\ge \left(1-\frac1e+o(1)\right)N \] by using a singleton-sign construction \[ A=\{1\}\cup B,\qquad \sum_{b\in B}\frac1b=1. \] The most promising Round-2 step is therefore to determine whether this entire strategy can ever do better. I will prove that it cannot: within the class of solutions in which one sign class has size $1$, the constant $1-\frac1e$ is asymptotically best possible. Hence any future improvement for the full problem must use genuinely two-sided cancellation with at least two positive and two negative terms.

2) ROUND-1 FOUNDATION USED

I use the following Round-1 fact without re-proving it.

• Round 1 showed, using Croot's short-interval Egyptian-fraction theorem, that there exists \[ B\subseteq \left[\left(\frac1e-o(1)\right)N,N\right] \] with $\sum_{b\in B}1/b=1$, and hence \[ F(N)\ge \left(1-\frac1e+o(1)\right)N \] via the admissible set $A=\{1\}\cup B$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a sharp extremal principle for reciprocal sums:

Maximal-cardinality reciprocal-sum principle. For fixed $r>0$, among all $m$-element subsets of $\{1,\dots,N\}$, the reciprocal sum \[ \sum_{n\in B}\frac1n \] is minimized by the terminal interval \[ \{N-m+1,\dots,N\}. \]

This immediately turns the problem of maximizing $|B|$ subject to $\sum_{b\in B}1/b=r$ into a harmonic-tail calculation, and it gives a matching upper bound to the Round-1 lower-bound strategy.

4) ATTACK PLAN (ROUND-2)

The exact gap left by Round 1 was this: the construction \[ A=\{1\}\cup B,\qquad \sum_{b\in B}\frac1b=1 \] gave a lower bound, but there was no proof that this particular type of construction was already optimal.

So I will now prove two precise claims.

• (i) For every fixed rational $r>0$, if \[ M_r(N):=\max\left\{|B|:\ B\subseteq \{1,\dots,N\},\ \sum_{b\in B}\frac1b=r\right\}, \] then \[ M_r(N)=\left(1-e^{-r}+o(1)\right)N. \]
• (ii) Consequently, if an admissible solution to Problem #319 has one sign class of size $1$, then its total size is at most \[ \left(1-\frac1e+o(1)\right)N. \]

This closes the entire Round-1 "singleton-sign" avenue.

5) WORK (ROUND-2)

Proposition 319.1.

Fix a rational number $r>0$. Let \[ M_r(N):=\max\left\{|B|:\ B\subseteq \{1,\dots,N\},\ \sum_{b\in B}\frac1b=r\right\}. \] Then \[ M_r(N)=\left(1-e^{-r}+o(1)\right)N \qquad (N\to\infty). \]

Proof.

I first prove the upper bound.

Suppose $B\subseteq \{1,\dots,N\}$ has cardinality $m$ and \[ \sum_{b\in B}\frac1b=r. \] Since the function $n\mapsto 1/n$ is strictly decreasing, the reciprocal sum of an $m$-element subset is minimized by choosing the $m$ largest available integers. Therefore \[ r=\sum_{b\in B}\frac1b \ge \sum_{n=N-m+1}^{N}\frac1n = H_N-H_{N-m}, \] where $H_t$ denotes the $t$th harmonic number.

Using $H_t=\log t+\gamma+o(1)$ uniformly for $t\to\infty$, we get \[ r\ge \log\!\left(\frac{N}{N-m}\right)+o(1). \] Exponentiating, \[ \frac{N}{N-m}\le e^{\,r+o(1)}, \] so \[ 1-\frac{m}{N}\ge e^{-r+o(1)}=e^{-r}+o(1). \] Hence \[ m\le \left(1-e^{-r}+o(1)\right)N. \] This proves \[ M_r(N)\le \left(1-e^{-r}+o(1)\right)N. \]

For the lower bound I use Croot's theorem, exactly as in Round 1 but with $1$ replaced by $r$. Croot proved that for every fixed rational $r>0$ there are, for all sufficiently large $X$, distinct integers \[ X<n_1<\cdots<n_k<(e^r+o_r(1))X \] such that \[ \sum_{i=1}^k \frac1{n_i}=r. \] Choose $X=X(N)$ so that \[ (e^r+o_r(1))X\le N. \] Then there is a set \[ B\subseteq \left[(e^{-r}-o(1))N,\ N\right] \] with \[ \sum_{b\in B}\frac1b=r. \]

Let \[ I_N:=\left\{\left\lceil (e^{-r}-o(1))N\right\rceil,\dots,N\right\}. \] The reciprocal sum of the full interval is \[ \sum_{n\in I_N}\frac1n = \log(e^r)+o(1)=r+o(1). \] Since $B\subseteq I_N$ and $\sum_{b\in B}1/b=r$, it follows that \[ \sum_{n\in I_N\setminus B}\frac1n=o(1). \] Every omitted element of $I_N$ is at most $N$, so each contributes at least $1/N$. Hence \[ \frac{|I_N\setminus B|}{N}\le \sum_{n\in I_N\setminus B}\frac1n=o(1), \] that is, \[ |I_N\setminus B|=o(N). \] Therefore \[ |B|=|I_N|-o(N)=\left(1-e^{-r}+o(1)\right)N. \] So \[ M_r(N)\ge \left(1-e^{-r}+o(1)\right)N. \] Together with the upper bound, this proves the proposition. $\square$

Corollary 319.2 (singleton-sign solutions are completely understood).

Let $A\subseteq \{1,\dots,N\}$ be admissible for Problem #319, and write \[ A=P\sqcup Q,\qquad \sum_{p\in P}\frac1p=\sum_{q\in Q}\frac1q. \] If $\min\{|P|,|Q|\}=1$, then \[ |A|\le \left(1-\frac1e+o(1)\right)N. \]

Proof.

By symmetry it suffices to consider the case $P=\{p\}$ for some $p\in\{1,\dots,N\}$. Then \[ \sum_{q\in Q}\frac1q=\frac1p. \] Applying Proposition 319.1 with $r=1/p$ gives \[ |Q|\le \left(1-e^{-1/p}+o(1)\right)N. \] Since the function $p\mapsto 1-e^{-1/p}$ is decreasing for $p\ge 1$, its maximum occurs at $p=1$, giving \[ |Q|\le \left(1-\frac1e+o(1)\right)N. \] Hence \[ |A|=|Q|+1\le \left(1-\frac1e+o(1)\right)N. \] $\square$

Corollary 319.3 (exact asymptotic for the Round-1 class).

Let $F_{\mathrm{sing}}(N)$ be the maximum size of an admissible set in Problem #319 for which one sign class has size $1$. Then \[ F_{\mathrm{sing}}(N)=\left(1-\frac1e+o(1)\right)N. \]

Proof.

The upper bound is Corollary 319.2. The matching lower bound is exactly the Round-1 construction $A=\{1\}\cup B$ with $\sum_{b\in B}1/b=1$ and \[ |B|=\left(1-\frac1e+o(1)\right)N. \] $\square$

6) ADVERSARIAL VERIFICATION

I check the new argument against the obvious failure modes.

• Did I accidentally use minimality where only equality was needed? No. Proposition 319.1 is purely about subsets $B$ with a fixed reciprocal sum, and Corollary 319.2 uses only the equality \[ \sum_{p\in P}\frac1p=\sum_{q\in Q}\frac1q. \] So the upper bound for singleton-sign solutions is valid a fortiori.

• Could a non-terminal set of the same size have smaller reciprocal sum than the terminal interval? No. Replacing any denominator by a larger one strictly decreases its reciprocal. Repeating this replacement process pushes every $m$-element set to the terminal interval and can only decrease the reciprocal sum.

• Does the lower bound really give almost all elements of the interval? Yes. The missing reciprocal mass is $o(1)$, and every omitted denominator contributes at least $1/N$, so the number of omitted denominators is $o(N)$.

• Does the corollary really rule out every singleton-sign strategy, not just $P=\{1\}$? Yes. The proof treats an arbitrary singleton $\{p\}$ and shows that the constant is maximized at $p=1$.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 proves a genuine obstruction that was absent in Round 1:

• the entire singleton-sign class is now solved asymptotically, and
• the Round-1 constant $1-\frac1e$ is exactly the best possible constant obtainable from any construction with one positive term or, more generally, with one sign class of size $1$.

So any future improvement for the full problem must use at least two positive and two negative denominators. In other words, the Round-1 mechanism has now been exhausted, not merely exhibited.

8) COMPLETION ESTIMATE

COMPLETION: 60%

9) REFERENCES

• E. S. Croot III, On unit fractions with denominators in short intervals, Acta Arithmetica 99 $2001$, 99--114.
• T. F. Bloom, Erdős Problem #319, erdosproblems.com/319 (checked in this round).

---