Reused writeup on Math-837 (gpt pro 5.2)

Erdős Problem 517 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/517

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/517.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/517.tex.

FORMAL RESTATEMENT

Let $(n_k)_{k\ge 1}$ be a strictly increasing sequence of nonnegative integers and let $(a_k)_{k\ge 1}$ be nonzero complex numbers such that the power series \[ f(z)=\sum_{k=1}^{\infty} a_k z^{n_k} \] converges for all $z\in\mathbb{C}$ (so $f$ is entire). Assume the gap condition \[ \frac{n_k}{k}\to\infty\quad\text{as }k\to\infty. \] Question: Must it follow that $f$ assumes every complex value infinitely often (i.e., for every $w\in\mathbb{C}$, the equation $f(z)=w$ has infinitely many solutions in $\mathbb{C}$)?

Edge cases: $f$ is transcendental because infinitely many terms are present. By classical Picard theory, there is at most one value $w$ that can fail to be assumed infinitely often; the conjecture asks that there is no such exceptional value.

QUICK LITERATURE/CONTEXT CHECK

The problem statement records: Fejér proved that if $\sum 1/n_k<\infty$ then $f$ assumes every value at least once; Biernacki proved under the same hypothesis every value is assumed infinitely often. Pólya proved that if $f$ has finite order then, assuming $\limsup (n_{k+1}-n_k)=\infty$, $f$ assumes every value infinitely often.

ATTACK PLAN

• Proof-track idea: show that the gap condition prevents $f$ from having an exceptional value $w$ with only finitely many preimages, by ruling out representations $f-w = P e^g$.
• Disproof-track idea: attempt to explicitly construct a lacunary entire function with $n_k/k\to\infty$ that omits a value or has only finitely many preimages for some value (i.e., build $f-w=P e^g$ with lacunary Maclaurin support).

I did not complete either track.

WORK

Fast reality check (the gap condition forces unbounded successive gaps). A quick check shows $n_k/k\to\infty$ implies $n_{k+1}-n_k$ cannot be bounded.

Additional computational sanity check (toy lacunary series). Consider the lacunary entire function $f(z)=\sum_{k\ge 1} \frac{z^{2^k}}{(2^k)!}$ (here $n_k=2^k$, so $n_k/k\to\infty$). Truncating at $k=4$ gives \[ P(z)=\frac{z^2}{2!}+\frac{z^4}{4!}+\frac{z^8}{8!}+\frac{z^{16}}{16!}. \] Using numpy.roots on the degree-$16$ coefficient array, I obtained:

• degree $16$; $16$ roots in total; $2$ roots at $0$ (numerically $|z|<10^{-8}$, consistent with the factor $z^2$);
• smallest nonzero root modulus $\approx 3.3422052103$ (approximately at $\pm 3.3422052103\, i$);
• largest root modulus $\approx 12.2987828606$.

This does not resolve the conjecture; it only checks that a representative lacunary pattern under the hypothesis behaves nontrivially in small truncations.

Lemma 1 (unbounded successive gaps). If $n_k/k\to\infty$ as $k\to\infty$, then \[ \limsup_{k\to\infty} (n_{k+1}-n_k)=\infty. \]

Proof. Suppose for contradiction that $n_{k+1}-n_k\le D$ for all $k$ for some fixed $D$. Then by telescoping, \[ n_k \le n_1 + D(k-1) \le (n_1+D)k. \] Dividing by $k$ gives $n_k/k\le n_1+D$, contradicting $n_k/k\to\infty$. $\square$

Lemma 2 (finite preimages imply an exponential-polynomial factorisation). Let $F$ be a nonconstant entire function and $w\in\mathbb{C}$. If the equation $F(z)=w$ has only finitely many solutions in $\mathbb{C}$ (counted without multiplicity), then there exist a nonzero polynomial $P$ and an entire function $g$ such that \[ F(z)-w = P(z) e^{g(z)}. \]

Proof. Let $G(z):=F(z)-w$. By assumption, $G$ is an entire function with only finitely many zeros (counted with multiplicity). Let those zeros be $\zeta_1,\dots,\zeta_m$ with multiplicities $m_1,\dots,m_m$. Define the polynomial \[ P(z):=\prod_{j=1}^m (z-\zeta_j)^{m_j}. \] Then $H(z):=G(z)/P(z)$ is entire and has no zeros: all zeros of $G$ have been divided out with the correct multiplicity. Since $H$ is holomorphic and nonvanishing on the simply connected domain $\mathbb{C}$, it admits an entire logarithm: define \[ g(z):=\int_0^z \frac{H'(\zeta)}{H(\zeta)}\,d\zeta. \] Because $H'/H$ is entire (holomorphic with no poles) and $\mathbb{C}$ is simply connected, the integral is path-independent and yields an entire function $g$ with $g'(z)=H'(z)/H(z)$. Then $(e^{-g}H)'=0$ (differentiate to check), so $H(z)=C e^{g(z)}$ for some constant $C\neq 0$. Absorbing $C$ into $P$ (replace $P$ by $CP$) gives $G=P e^{g}$. $\square$

VERIFICATION

• Lemma 2 is a standard factorisation for entire functions with finitely many zeros; I verified the logarithm construction carefully: $H$ nonvanishing on simply connected $\mathbb{C}$ implies $H'/H$ entire and integrable path-independently.
• Neither lemma resolves whether such a factorisation can be compatible with lacunary Maclaurin support $\{n_k\}$ and the condition $n_k/k\to\infty$.

FINAL

UNRESOLVED

(i) Strongest proved partial result: The gap condition $n_k/k\to\infty$ forces $\limsup (n_{k+1}-n_k)=\infty$ (Lemma 1). If $f$ had a value $w$ that is assumed only finitely often, then $f-w$ must factor as $P e^{g}$ with $P$ a polynomial and $g$ entire (Lemma 2).

(ii) First gap (crisp): Show (or refute) that an entire lacunary series with $n_k/k\to\infty$ cannot admit a decomposition $f-w=P e^{g}$ for any $w\in\mathbb{C}$, where $P$ is a polynomial and $g$ is entire.

(iii) Top 3 next moves:

1. Coefficient-obstruction attempt: derive necessary algebraic/differential recurrences for Maclaurin coefficients of $P e^{g}$ and prove these recurrences force the support to have $n_k=O(k)$ (or at least violate $n_k/k\to\infty$).
2. Computational search for structured candidates: attempt to explicitly construct (via solving for coefficients) a nonvanishing entire function with sparse support $\{n_k\}$, starting from a desired lacunary series and taking a convergent logarithm $g=\log f$.
3. Use growth/order: combine Lemma 1 (unbounded gaps) with value-distribution theorems that rule out exceptional values for functions with sufficiently sparse exponents unless the order is finite (where Pólya's result applies).

(iv) Minimal counterexample structure (if the conjecture is false): would be an entire $f(z)=\sum a_k z^{n_k}$ with $n_k/k\to\infty$ for which some value $w$ has only finitely many preimages, hence $f-w=P e^{g}$; such a counterexample would amount to a nonvanishing entire function (up to a polynomial factor) whose Maclaurin series has Fabry gaps.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.