gpt pro 5.4 attempt on Math-5559

Erdős Problem #293

1) ROUND-2 OBJECTIVE

I pursue path (A): prove a strictly stronger theorem than the Round-1 upper bound. The main unresolved gap after Round 1 was that the bound \[ v(k)\le u_k+1=c_0^{(1+o(1))2^k} \] came from controlling the largest denominator in a single decomposition, whereas the definition of $v(k)$ depends on the union of all denominators occurring in all $k$-term decompositions. The most promising Round-2 move is therefore a counting argument.

2) Round-1 FOUNDATION USED

I use the following Round-1 ingredients without reproving them.

• The set \[ D_k=\{n>1:\ \mbox{$n$ occurs as some denominator in a $k$-term decomposition of $1$}\}. \]
• The definition \[ v(k):=\min\{n>1:\ n\notin D_k\}. \]
• The Round-1 upper bound $v(k)\le u_k+1$, where $u_1=1$ and $u_{j+1}=u_j(u_j+1)$.
• The exact values $v(1)=2$, $v(2)=2$, $v(3)=4$, $v(4)=11$, $v(5)=17$, $v(6)=103$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is to replace "largest denominator" control by "number of solutions" control.

Let $F(k)$ be the number of strictly increasing $k$-tuples \[ 1<n_1<\cdots<n_k,\qquad \frac1{n_1}+\cdots+\frac1{n_k}=1. \] Each such solution contributes at most $k$ denominators to $D_k$. Hence $D_k$ is controlled by $F(k)$, not by the largest possible denominator alone. This yields the elementary but asymptotically much stronger inequality \[ v(k)\le kF(k)+2. \]

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap is that the previous upper bound was of size $c_0^{(1+o(1))2^k}$, whereas the best currently available upper bound for $F(k)$ is much smaller, namely $c_0^{(\frac15+o(1))2^k}$.

So the plan is:

• prove the elementary inequality $v(k)\le kF(k)+2$ from the Round-1 definition of $D_k$;
• import the known upper bound for $F(k)$;
• compare the resulting bound with the Round-1 bound.

This overcomes the Round-1 obstacle because it uses the global combinatorics of all $k$-term solutions, not just a denominator bound inside one solution.

5) WORK (ROUND-2)

Lemma 1. For every $k\ge 1$, \[ |D_k|\le kF(k). \]

Proof. List all $F(k)$ many $k$-term decompositions of $1$: \[ 1=\frac1{n_1^{(j)}}+\cdots+\frac1{n_k^{(j)}} \qquad (1\le j\le F(k)). \] For each fixed $j$, the solution contributes at most $k$ denominators to $D_k$. Therefore \[ D_k\subseteq \bigcup_{j=1}^{F(k)} \{n_1^{(j)},\ldots,n_k^{(j)}\}, \] so by the union bound \[ |D_k|\le \sum_{j=1}^{F(k)} k = kF(k). \] This proves the claim.

Lemma 2. If $S\subseteq \{2,3,4,\ldots\}$ is finite, then the least integer $>1$ not contained in $S$ is at most $|S|+2$.

Proof. Among the $|S|+1$ integers \[ 2,3,\ldots,|S|+2 \] at least one is not in $S$, because $S$ contains only $|S|$ elements. Hence the least missing integer is at most $|S|+2$.

Corollary 3. For every $k\ge 1$, \[ v(k)\le |D_k|+2\le kF(k)+2. \]

Proof. Apply Lemma 2 with $S=D_k$, then use Lemma 1.

External input. Elsholtz and Planitzer proved the upper bound \[ F(k)\le c_0^{(\frac15+o(1))2^k}, \qquad c_0=1.264085\ldots, \] for the number of $k$-term Egyptian fraction decompositions of $1$.

Theorem 4 (Round-2 upper bound). As $k\to\infty$, \[ v(k)\le c_0^{(\frac15+o(1))2^k}. \]

Proof. By Corollary 3, \[ v(k)\le kF(k)+2. \] Insert the Elsholtz--Planitzer bound for $F(k)$. Since the prefactor $k$ and the additive term $2$ are both negligible on the logarithmic scale $2^k$, they are absorbed into the $o(1)$ term in the exponent. Hence \[ v(k)\le c_0^{(\frac15+o(1))2^k}. \]

Comparison with Round 1. Round 1 gave \[ v(k)\le u_k+1=c_0^{(1+o(1))2^k}. \] Round 2 improves the constant in the exponent from $1$ to $\frac15$.

6) ADVERSARIAL VERIFICATION

• Possible double-counting issue. The same denominator may occur in many solutions, but this only helps: the union of all denominator-sets is never larger than the sum of their sizes.
• Possible off-by-one issue in the passage from $|D_k|$ to $v(k)$. This is why Lemma 2 gives $|D_k|+2$, not $|D_k|+1$: the relevant integers start at $2$, not $1$.
• Compatibility with small cases. If $D_k=\emptyset$, Corollary 3 gives $v(k)\le 2$, which is sharp for $k=1,2$.
• Interaction with Round 1. The new bound does not replace the Round-1 Sylvester/Takenouchi bound pointwise for every small $k$; it improves the asymptotic upper bound. The two estimates are compatible and may be combined as \[ v(k)\le \min\{u_k+1,\ kF(k)+2\}. \]

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 proves a substantially stronger upper bound, \[ v(k)\le c_0^{(\frac15+o(1))2^k}, \] improving the Round-1 exponent constant from $1$ to $\frac15$. The lower bound is still only of the shape $\exp(ck^2)$, so the main asymptotic order of growth remains open.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 48%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #293, erdosproblems.com/293, accessed 2026-03-07.
2. W. van Doorn and Q. Tang, The smallest denominator not contained in a unit fraction decomposition of $1$ with fixed length, arXiv:2512.22083, 2025.
3. C. Elsholtz and S. Planitzer, The number of solutions of the Erdős--Straus equation and sums of $k$ unit fractions, Proc. Roy. Soc. Edinburgh Sect. A 150 $2020$, 1401--1427.