Prove that there exists a sequence $A_{0}, A_{1}$, . . . of rational polynomials $A_{i}(x)\in \mathrm{Q}[x]$ with $A_{i}$ of degree $i$ such that $$ s(n)=\frac{n^{n-1}}{(1-\log 2)^{n-1/2}e^{n}}(\sum_{k=0}^{m}\frac{A_{k}(1-\log 2)}{n^{k}}+o(n^{-m})) $$ for all $m\in \mathbb{N}$ and $s(n)$ is given by $$ s(n)=\displaystyle \sum_{i=n+1}^{2n}t_{i,0}(n) $$ where $t_{i,j}(n)$ is recursively given by $$ t_{i,j}(n) = (i-2)t_{i-1,j}(n-1) + t_{i-1,j-1}(n-1) + (i-3)t_{i-2,j}(n-1) $$ and $i \in \{n+1,...,2n\}$ , $j \in \{0, 1, ...,2n-i\}$ where $n=1,2,3...$
P.S. We know that $$ 2\ \sum_{i=n+1}^{2n}t_{i,0}(n)=\sum_{i=n+1}^{2n}\sum_{j=0}^{2n-i}t_{i,j}(n)+\sum_{i=n}^{2n-2}\sum_{j=0}^{2n-2-i}t_{i,j}(n-1) $$ for all $n\geq 2$ and first few polynomials $A_{0}, A_{1}$, . . . are $$ A_{0}\ =\ 1 $$ $$ A_{1}\ =\ \frac{11}{24}-\frac{x}{12} $$ $$ A_{2}\ =\ \frac{265}{1152}-\frac{47x}{288}+\frac{x^{2}}{288} $$ $$ A_{3}\ =\ \frac{48703}{414720}-\frac{3649x}{13824}+\frac{107x^{2}}{6912}+\frac{139x^{3}}{51840} $$ $$ A_{4}\ =\ \frac{2333717}{39813120}-\frac{2019163x}{4976640}+\frac{16489x^{2}}{331776}+\frac{26549x^{3}}{1244160}-\frac{571x^{4}}{2488320} $$ $$ A_{5}\ =\ \frac{38180761}{1337720832}-\frac{293093189x}{477757440}+\frac{16859263x^{2}}{119439360}+ $$ $$ +\frac{6752203x^{3}}{59719680}-\frac{170729x^{4}}{59719680}-\frac{163879x^{5}}{209018880} $$
