How this expression leads to the given sequence Here given is a sequence from OEIS.

The sequence is triangle of coefficients from fractional iteration of e^x - 1. Few terms are:
1, 1, 3, 1, 13, 18, 1, 50, 205, 180, 1, 201, 1865, 4245, 2700, 1, 875, 16674

The expression of finding the sequence is also given as:
A(n;x) for n-th row satisfies
 A(n;x) = Sum_{k=0..n-1} Stirling_2(n, k)*A(k;x)*x, A(1;x) = 1.

The tabular  view shows the entries row wise. Stirling_2(n,k) is most probably stirling numbers of the second type

I am not able to get how above expression is resulting in the given sequence.
In summation, k begins from 0, but nothing is mentioned about A(0;x). I assumed it to be 0, but still can't get the above values.
Please explain how the first few terms are resulting from the expression.
Stirling numbers of the second type satisfy
$$ \newcommand{\stirling}[2]{\genfrac\{\}{0pt}{}{#1}{#2}}\stirling{n}{0} = 0, \qquad (n \neq 0)$$
and so there is no need to define $A(0;x)$, and so the formula seems to work. You can also play with the Mathematica code which can be found in the entry corresponding to the sequence. I am puzzled, however, what is meant by "fractional iteration of $e^x-1$".

The recurrence given is for generating functions. Let's rewrite it:
$$
A_1(x) = 1, \qquad A_n(x) = \sum_{k=1}^{n-1} \stirling{n}{k} A_k(x) x.
$$
Using a table for the Stirling numbers, we get
$$
\begin{align*}
A_1(x) &amp;= 1 \\
A_2(x) &amp;= \stirling{2}{1} A_1(x) x = 1 \cdot 1 \cdot x = x \\
A_3(x) &amp;= \stirling{3}{1} A_1(x) x + \stirling{3}{2} A_2(x) x = 1 \cdot 1 \cdot x + 3 \cdot x \cdot x = x + 3x^2 \\
A_4(x) &amp;= \stirling{4}{1} A_1(x) x + \stirling{4}{2} A_2(x) x + \stirling{4}{3} A_3(x) x \\ &amp;= 1 \cdot 1 \cdot x + 7 \cdot x \cdot x + 6 (x + 3x^2) x  = x + 13x^2 + 18x^3
\end{align*}
$$
and so on. The numbers in the table are $a(n,k) = A_n(x)|_{x^k}$ :
$$
\begin{align*}
a(2,1) &amp;= 1, \\
a(3,1) &amp;= 1, \\
a(3,2) &amp;= 3, \\
a(4,1) &amp;= 1, \\
a(4,2) &amp;= 13, \\
a(4,3) &amp;= 18,
\end{align*}
$$

If you prefer an actual recurrence on the coefficients, here is what you get:
$$
\begin{align*}
a(n,l) = \begin{cases} 1 &amp; l = 1, \\ \sum_{k=l}^{n-1} \stirling{n}{k} a(k,l-1) &amp; l &gt; 1. \end{cases}
\end{align*}
$$