$(M_1)-(M_3)$ imply that the generating function $f_b(x)$ possesses the Hayman admissibility properties \[11\] (see also \[7; Sect. VIII.5\]) in the unit disc. For $0<r<1$, we introduce the functions: $$\label{capitalf}
 F_b(r) =\log{f_b(r)} =-\sum_{k=1}^\infty b_k\log{(1-r^k)},$$ $$\label{ar}
 \mathcal{A}_b(r) =rF_b^\prime(r) =r\frac{f_b^\prime(r)}{f_b(r)},$$ $$\label{br}
 \mathcal{B}_b(r) =r^2 F_b^{\prime\prime}(r)+rF_b^{\prime}(r) =r\frac{f_b^\prime(r)}{f_b(r)}
 +r^2\frac{f_b^{\prime\prime}(r)}{f_b(r)}
 -r^2\left(\frac{f_b^\prime(r)}{f_b(r)}\right)^2.$$ Furthermore, setting in (\[capitalf\])-(\[br\]) $r=e^{-\alpha}$, we shall obtain their asymptotic expansions as $\alpha\to 0^+$. For the sake of convenience, we also set $$\label{consth}
 h=h(\rho,A)=A\Gamma(\rho+1)\zeta(\rho+1).$$ The proof of the next lemma is contained in \[10; Lemma 2\].

Conditions $(M_1)$ and $(M_2)$ imply the following asymptotic expansions: $$\label{aalpha}
 \mathcal{A}_b(e^{-\alpha}) =h\alpha^{-\rho-1} +D(0)\alpha^{-1}
 +D^\prime(0)+O(\alpha^{C_0-1}),$$ $$\label{balpha}
 \mathcal{B}_b(e^{-\alpha})
 =\frac{d}{d\alpha}(-\mathcal{A}_b(e^{-\alpha}))
 =h(\rho+1)\alpha^{-\rho-2} +D(0)\alpha^{-2} +O(\alpha^{C_0-2}),$$ $$\label{terz}
 F_b^{\prime\prime\prime}(e^{-\alpha}) =O(\alpha^{-\rho-3}),$$ as $\alpha\to 0^+$, where $F_b , \mathcal{A}_b$ and $\mathcal{B}_b$ are defined by (\[capitalf\])-(\[br\]), respectively. Moreover, from (\[aalpha\]) it follows that the equation $$\label{eqhay}
 \mathcal{A}_b(e^{-\alpha})=n, \quad n\ge 1,$$ has a unique solution $\alpha=\alpha_n$, such that $\alpha_n\to 0$ as $n\to\infty$. An asymptotic expansion of this solution, as $n\to\infty$, is given by $$\label{alphan}
 \alpha_n=a(n) +\frac{D(0)}{(\rho+1)n} +O(n^{-1-\beta}),$$ where $\beta=\min{\left(\frac{C_0}{\rho+1},\frac{\rho}{\rho+1}\right)}$ and $a(n)$ are the normalizing constants given by (\[norm\]). $\rule{2mm}{2mm}$

We notice that (\[aalpha\]), (\[balpha\]) and (\[alphan\]) imply that $$\label{capture}
 \mathcal{A}_b(e^{-\alpha_n})\to\infty, \quad
 \mathcal{B}_b(e^{-\alpha_n})\to\infty, \quad n\to\infty,$$ that is, Hayman “capture” condition \[7; p. 565\] is satisfied with $r=r_n=e^{-\alpha_n}$. Our next step is to 