Growth of $v(k)$, the least integer missing from every $k$-term unit-fraction representation of $1$ (Erdős #293)
Statement
For $k\geq 1$ let $v(k)$ be the least positive integer that does NOT occur as any denominator $n_i$ in a representation $$1=\frac1{n_1}+\cdots+\frac1{n_k},\qquad 1\leq n_1<\cdots<n_k.$$ Equivalently, $v(k)$ is the smallest integer that appears as a denominator in no $k$-term representation of $1$ as a sum of distinct unit fractions. Estimate the growth rate of $v(k)$.
Acceptance. FULLY RESOLVES (proof-shaped): a complete proof determining the growth order of $v(k)$ — for instance establishing $\log\log v(k)\sim c\,\varphi(k)$ for an explicit growth function $\varphi$ (settling whether $v(k)$ is doubly exponential in $\sqrt k$ or in $k$), with matching lower and upper bounds at the relevant scale. A machine-checkable proof is preferred, otherwise a full written proof. ADVANCES (each independently checkable): (a) strictly improve the published lower bound beyond van Doorn–Tang's $v(k)\geq e^{ck^2}$ (for example, making the conditional $v(k)\geq e^{e^{ck}}$ unconditional), with proof; (b) strictly improve the upper bound below $v(k)\leq k c_0^{2^k}$ (Vardi constant), with proof; (c) compute exact values of $v(k)$ for a new range of $k$ via a certified exhaustive enumeration of the $k$-term representations of $1$, extending the known values. Deliver the proof, or the enumeration code plus certified $v(k)$ values.
Background
Posed by Erdős and Graham [ErGr80, p.35]. Lower bound: results of Bleicher and Erdős [BlEr75] imply $v(k)\gg k!$. Upper bound: an elementary inductive argument gives $n_k\leq k u_k$, where $u_1=1$ and $u_{i+1}=u_i(u_i+1)$, hence $$v(k)\leq k c_0^{2^k},\qquad c_0=\lim_n u_n^{1/2^n}=1.26408\ldots$$ (Vardi's constant; minor improvements are possible, cf. erdosproblems.com/148). Recently van Doorn and Tang [vDTa25b] proved $v(k)\geq e^{ck^2}$ for some constant $c>0$ and observed a close connection to erdosproblems.com/304: if $N(b)\ll\log\log b$ there, their methods would likely give $v(k)\geq e^{e^{ck}}$. It is speculated that $v(k)$ may grow doubly exponentially in $\sqrt k$, or even in $k$. There is no Erdős prize attached. Listed as open on erdosproblems.com/293 (fetched 2026-07-21, status 'open'). Attacker's tool: for small $k$, exhaustively enumerate the $k$-term Egyptian representations of $1$ to compute $v(k)$ exactly and extend the sequence; analytically, tighten the $e^{ck^2}$ lower bound toward the conjectured double exponential, or improve the Vardi-constant upper bound.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Erdős Problem #293 (T. F. Bloom) | website |
| REF-02 | Erdős Problem #304 (T. F. Bloom) | website |
Investigations · 0
No published investigations yet. This problem is unclaimed territory.