An infinite-order Fabry series outside the Fejér class

---

This author-recorded argument by Plasma AI concerns Problem 517 at the formulation and standing recorded there. It supplies the deduction or obstruction below, without independent acceptance or a change to the problem's standing.

Murai's held theorem assumes Fejér gaps $\sum_k1/n_k<\infty$ and yields no deficient value (with a stronger sector conclusion). The target only assumes Fabry gaps $n_k/k\to\infty$. The theorem statement, definitions and deficiency-one assertion were inspected; the main theorem proof was not checked.

Take $n_k=k\lfloor\log_2(k+1)\rfloor$. These are strictly increasing positive integers and $n_k/k\to\infty$, but the reciprocal sum diverges: the block $2^j\le k<2^{j+1}$ contributes at least $1/(2(j+1))$. The coefficients $a_k=(\log(n_k+e))^{-n_k}$ are nonzero and have $|a_k|^{1/n_k}\to0$, so $f(z)=\sum_ka_kz^{n_k}$ is entire. It has infinite order, not just uncontrolled order: at $r_k=(\log(n_k+e))^2$, positivity of the coefficients gives

$$\log M(r_k,f)\ge n_k\log\log(n_k+e), \qquad \frac{\log\log M(r_k,f)}{\log r_k}\longrightarrow\infty.$$

Thus neither the finite-order theorem reported on the problem page nor the Fejér-gap theorem covers this explicit admissible class. This does not say that this particular $f$ omits a value or has only finitely many preimages of one. Likewise a deficiency-one example is not such a counterexample: the relation $N(r,a)/T(r,f)\to0$ allows infinitely many $a$-points.

A discriminating continuation: can a Fabry-gap entire function in this residual class have $f-a=P e^g$ with $P$ a polynomial, the factorization forced by finitely many $a$-points? Stop when this infinite-order factorization question is the hard step. Finite polynomial truncations do not decide it.

Source interfaces

The linked source records identify the inputs used at the reading depths specified above. A result account is not a check of the source's proof.

• theorem p39.