The exponential generating function of this recurrence relation, $T_n=T_{n-1}\;+(n-1)\cdot T_{n-2}\;$, is
$$f(x)=e^{x + \frac{x^2}{2}}$$
Multiplying the exponential generating functions for each term, we get, $$=\ \sum_{j\geq0}\;\frac{1}{j!}x^j\cdot\sum_{k\geq0}\;\frac{1}{2^kk!}x^{2k}$$ $$=\ \sum_{j,\ \ k\geq0}\;\frac{1}{j!}\cdot\frac{1}{2^kk!}x^{j+2k}$$
Taking $n\ =\ j\ +\ 2k$, $$=\ \sum_{n-2k,\ \ k\geq0}\;\frac{1}{(n-2k)!}\cdot\frac{1}{2^kk!}x^n$$
What steps can I take from here to get to the 'closed' form of the exponential generating function, i.e.,
$$\;\sum_{n\ge0}\Big(\sum_{k=0}^{[n/2]}\frac{n!}{(n-2k)!2^kk!}\Big)\frac{x^n}{n!}.$$