Editing Generating function (section)

====The Stirling numbers modulo small integers====

The [[Stirling numbers of the first kind#Congruences|main article]] on the Stirling numbers generated by the finite products
<math display="block">S_n(x) := \sum_{k=0}^n \begin{bmatrix} n \\ k \end{bmatrix} x^k = x(x+1)(x+2) \cdots (x+n-1)\,,\quad n \geq 1\,, </math>

provides an overview of the congruences for these numbers derived strictly from properties of their generating function as in Section 4.6 of Wilf's stock reference ''Generatingfunctionology''.
We repeat the basic argument and notice that when reduces modulo 2, these finite product generating functions each satisfy

<math display="block">S_n(x) = [x(x+1)] \cdot [x(x+1)] \cdots = x^{\left\lceil \frac{n}{2} \right\rceil} (x+1)^{\left\lfloor \frac{n}{2} \right\rfloor}\,, </math>

which implies that the parity of these [[Stirling numbers]] matches that of the binomial coefficient

<math display="block">\begin{bmatrix} n \\ k \end{bmatrix} \equiv \binom{\left\lfloor \frac{n}{2} \right\rfloor}{k - \left\lceil \frac{n}{2} \right\rceil} \pmod{2}\,, </math>

and consequently shows that {{math|{{resize|150%|[}}{{su|p=''n''|b=''k''|a=c}}{{resize|150%|]}}}} is even whenever {{math|''k'' < ⌊ {{sfrac|''n''|2}} ⌋}}.

Similarly, we can reduce the right-hand-side products defining the Stirling number generating functions modulo 3 to obtain slightly more complicated expressions providing that
<math display="block">\begin{align}
\begin{bmatrix} n \\ m \end{bmatrix} & \equiv
 [x^m] \left(
 x^{\left\lceil \frac{n}{3} \right\rceil} (x+1)^{\left\lceil \frac{n-1}{3} \right\rceil}
 (x+2)^{\left\lfloor \frac{n}{3} \right\rfloor}
 \right) && \pmod{3} \\
 & \equiv
 \sum_{k=0}^{m} \begin{pmatrix} \left\lceil \frac{n-1}{3} \right\rceil \\ k \end{pmatrix}
 \begin{pmatrix} \left\lfloor \frac{n}{3} \right\rfloor \\ m-k - \left\lceil \frac{n}{3} \right\rceil \end{pmatrix} \times
 2^{\left\lceil \frac{n}{3} \right\rceil + \left\lfloor \frac{n}{3} \right\rfloor -(m-k)} && \pmod{3}\,.
\end{align}</math>