本文给出第2类Stirling数,Bernoulli数与Euler数的解析表示式: s_2(m+1,n)=(-1)~n/n1 sum form j=1 to n(-1)~j(?)_j^(-m+1) B_n=sum form k=1 to n 1/(k+1) sum form j=1 to k (-1)~j(?)_j^(-n) E_(2n) =1/(2n+1)[sum from p=0 to n-1 sum from k=1 to 2(n-p) sum from j=1 to k (-1)^(j-1)/(k+1)·(?)(?)(4j)~2(n-p)+4n+1]因此解决了它们的计算问题。
定义了Norland Bernoulli多项式和Norland Eurler多项式,证明了恒等式: B_(m_1·m_2·…·m_p)^((k))(x_1,x_2,…x_p,y_1,y_2,…y_k)=(1/2^(sum from i to p(m_i)))((sum from s_1=0 to m_1)(sum from s_2=0 to m_2)…(sum from s_p=0 to m_p)(m_1 s_1)…(m_p s_p))E_(s_1·s_2·…·s_p)^((k))(x_1,x_2,…x_p,y_1,y_2,…y_k) B_(m_1-s_1,m_2-s_2,…,m_p-s_p)^((k))(x_1,x_2,…,x_p,y_1,y_2,…,y_k)
