Young Tableaux generating function The number of young tableaux of $n$ cells is known to satisfy the recurrence $a_{n+1} = a_{n} + na_{n-1}$. I am trying to find the generating function but I keep getting something dependent on $n$. Here's what I did so far:

Denote by $f(x) = \sum_{n\geq 1}a_nx^n$. We have $\sum_{n \geq 1} a_{n+1}x^n = f(x) + nxf(x)$ (if we assume $a_1 = 1, a_2 = 2$). We can infer that $\sum_{n \geq 1} a_{n+1}x^n = \frac{f(x)}{x} - 1$. 


The first step is to look in the OEIS and see that this is sequence A000085. The sequence grows so fast that the o.g.f. has radius of convergence $0$. This strongly suggests looking at the e.g.f instead, which in the OEIS entry is given as $\exp(x+x^2/2).$ The question now is how to derive this e.g.f. from the recursion $\;a_{n+1}=a_n+na_{n-1}\;$ and initial values $\;a_0=a_1=1,\;a_2=2.$
There are several methods.  First, the OEIS mentions that the e.g.f. $A(x)$ for A000085 satisfies D.E.
$A'(x) = A(x)(1+x),\;$ and as mentioned in another answer to this question, and thus leading to $A(x)=\exp(x+x^2/2).\;$ 
A little different is $\;A''(x) = A'(x)(1+x) + A(x)\;$ which comes directly either from the recursion $\;a_{n+2}=a_{n+1}+(n+1)a_n\;$ or the derivative of the first D.E. for $A(x).$
