Editing Lah number (section)

==Identities and relations==
The Lah numbers satisfy a variety of identities and relations.

In [[Jovan Karamata|Karamata]]–[[Donald Knuth|Knuth]] notation for [[Stirling numbers]]<math display="block"> L(n,k) = \sum_{j=k}^n \left[{n\atop j}\right] \left\{{j\atop k}\right\}</math>where <math display="inline">\left[{n\atop j}\right]</math> are the [[Stirling numbers of the first kind|unsigned Stirling numbers of the first kind]] and <math display="inline">\left\{{j\atop k}\right\}</math> are the [[Stirling numbers of the second kind]].
:<math> L(n,k) = {n-1 \choose k-1} \frac{n!}{k!} = {n \choose k} \frac{(n-1)!}{(k-1)!} = {n \choose k} {n-1 \choose k-1} (n-k)!</math>

:<math> L(n,k) = \frac{n!(n-1)!}{k!(k-1)!}\cdot\frac{1}{(n-k)!} = \left (\frac{n!}{k!} \right )^2\frac{k}{n(n-k)!}</math>

:<math> k(k+1) L(n,k+1) = (n-k) L(n,k)</math>, for <math>k>0</math>.
=== Recurrence relations ===
The Lah numbers satisfy the recurrence relations<math display="block">
\begin{align} L(n+1,k) &= (n+k) L(n,k) + L(n,k-1) \\
&= k(k+1) L(n, k+1) + 2k L(n, k) + L(n, k-1)
\end{align}
</math>where <math>L(n,0)=\delta_n</math>, the [[Kronecker delta]], and <math>L(n,k)=0</math> for all <math>k > n</math>.

=== Exponential generating function ===

:<math>\sum_{n\geq k} L(n,k)\frac{x^n}{n!} = \frac{1}{k!}\left( \frac{x}{1-x} \right)^k</math>

=== Derivative of exp(1/''x'') ===

The ''n''-th [[derivative]] of the function <math>e^\frac1{x}</math> can be expressed with the Lah numbers, as follows<ref>{{cite journal |last1=Daboul |first1=Siad |last2=Mangaldan |first2=Jan |last3=Spivey |first3=Michael Z. |last4=Taylor |first4=Peter J. |year=2013 |title=The Lah Numbers and the nth Derivative of <math>e^{1\over x}</math> |journal=Mathematics Magazine |volume=86 |pages=39–47 |doi=10.4169/math.mag.86.1.039 |jstor=10.4169/math.mag.86.1.039 |s2cid=123113404 |number=1}}</ref><math display="block"> \frac{\textrm d^n}{\textrm dx^n} e^\frac1x = (-1)^n \sum_{k=1}^n \frac{L(n,k)}{x^{n+k}} \cdot e^\frac1x.</math>For example,

<math> \frac{\textrm d}{\textrm dx} e^\frac1x = - \frac{1}{x^2} \cdot e^{\frac1x}</math>

<math> \frac{\textrm d^2}{\textrm dx^2}e^\frac1{x} = \frac{\textrm d}{\textrm dx} \left(-\frac1{x^2} e^{\frac1x} \right)= -\frac{-2}{x^3} \cdot e^{\frac1x} - \frac1{x^2} \cdot \frac{-1}{x^2} \cdot e^{\frac1x}= \left(\frac2{x^3} + \frac1{x^4}\right) \cdot e^{\frac1x}</math>

<math> \frac{\textrm d^3}{\textrm dx^3}  e^\frac1{x} = \frac{\textrm d}{\textrm dx} \left( \left(\frac2{x^3} + \frac1{x^4}\right) \cdot e^{\frac1x} \right) = \left(\frac{-6}{x^4} + \frac{-4}{x^5}\right) \cdot e^{\frac1x} + \left(\frac2{x^3} + \frac1{x^4}\right) \cdot \frac{-1}{x^2} \cdot e^{\frac1x} =-\left(\frac6{x^4} + \frac6{x^5} + \frac1{x^6}\right) \cdot e^{\frac{1}{x}}</math>