On the Eulerian numbers \(\displaystyle{ M_ n = \max{}_{1{\leq{}}k{\leq{}}n}A(n,k)}\)
The Eulerian numbers \(A(n,k)\) have studied extensively. They can be defined and computed by \(A(n+1,k)=(n-k+2)A(n,k-1)+kA(n,k)\), \(A(n,1)=A(n,n)=1\) and are expressed by \(A(n,k)=\sum_{0\leq j\leq k}(- 1)^ j{{n+1} \choose j}(k-j)^ n\). The maximum \(M_ n\) on the line \(n=\text{Max}_{k=1,2,\dots,n} A(n,k)=A(n,[{n \over 2}]+1)\). Let \[ s(n)=\sum_{j=0}^ n {{n+2} \over j}(-1)^{j+1} \sum _{q=\max(0,[n/2]+2-j)} ^{n+1-j} q^ n. \] It is proved that \(M_ n=s(n)\) and that \(M_ n/n!\) form a decreasing sequence. Using Cauchy's integral formula and Laplace method, it is proved that \({{M_{2p-1} \over {(2p-1)!}}}\leq \sqrt {{3 \over {\pi p}}}(1-{3\over {40p}})\), \(p\geq 3\); \({M_{2p-1} \over {(2p-1)!}} \geq \sqrt{{3 \over {\pi p}}} (1-{3\over {40p}} - {{13} \over {4480p^ 2}})\), \(p\geq 1\); \({M_{2p+5} \over {(2p+5)!}} \leq {M_{2p} \over {(2p)!}} \leq {M_{2p+3} \over {(2p+3)!}}\).
scientific article; zbMATH DE number 94941
scientific article