I understand the complete Bell polynomial $B_n$ satisfies the identity:
$$B_n(x_1+y_1,x_2+y_2,...,x_n+y_n) = \sum_{k=0}^n \left(\matrix{ n \\ k }\right) B_{n-k}(x_1,x_2,..,x_{n-k})\, B_k(y_1,y_2,...,y_k) $$
Does anybody know of a similar identity for Bell polynomials of the second kind?
$$B_{n,k}(x_1+y_1,x_2+y_2,...,x_{n-k+1}+y_{n-k+1}) = \;???$$
Any information is appreciated, thank you!
They satisfy (writing the argument as a vector to save notation)
$$B_{n, k}(x + y) = \sum_{i=0}^n \sum_{j=0}^k {n \choose i} B_{n-i, k-j}(x) B_{i, j}(y)$$
and this follows immediately from the fact that the generating function
$$F(x, t, s) = \sum_{n, k} B_{n, k}(x) \frac{t^n}{n!} s^k = \exp \left( s \sum_{i \ge 1} \frac{x_i t^i}{i!} \right)$$
satisfies $F(x + y, t, s) = F(x, t, s) F(y, t, s)$.