Lah number

Below is a table of values for the Lah numbers: The row sums are 1, 1, 3, 13, 73, 501, 4051, 37633, \dots . ==Rising and falling factorials==

Let x^{(n)} represent the rising factorial x(x+1)(x+2) \cdots (x+n-1) and let (x)_n represent the falling factorial x(x-1)(x-2) \cdots (x-n+1). The Lah numbers are the coefficients that express each of these families of polynomials in terms of the other. Explicitly,x^{(n)} = \sum_{k=0}^n L(n,k) (x)_kand(x)_n = \sum_{k=0}^n (-1)^{n-k} L(n,k)x^{(k)}.For example,x(x+1)(x+2) = {\color{red}6}x + {\color{red}6}x(x-1) + {\color{red}1}x(x-1)(x-2)andx(x-1)(x-2) = {\color{red}6}x - {\color{red}6}x(x+1) + {\color{red}1}x(x+1)(x+2), where the coefficients 6, 6, and 1 are exactly the Lah numbers L(3, 1), L(3, 2), and L(3, 3). ==Identities and relations==

The Lah numbers satisfy a variety of identities and relations. In Karamata–Knuth notation for Stirling numbers L(n,k) = \sum_{j=k}^n \left[{n\atop j}\right] \left\{{j\atop k}\right\}where \left[{n\atop j}\right] are the unsigned Stirling numbers of the first kind and \left\{{j\atop k}\right\} are the Stirling numbers of the second kind. : 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)! : 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)!} : k(k+1) L(n,k+1) = (n-k) L(n,k), for k>0. Recurrence relations The Lah numbers satisfy the recurrence relations \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} where L(n,0)=\delta_n, the Kronecker delta, and L(n,k)=0 for all k > n. Exponential generating function :\sum_{n\geq k} L(n,k)\frac{x^n}{n!} = \frac{1}{k!}\left( \frac{x}{1-x} \right)^k Derivative of exp(1/x) The n-th derivative of the function e^\frac1{x} can be expressed with the Lah numbers, as follows \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.For example, \frac{\textrm d}{\textrm dx} e^\frac1x = - \frac{1}{x^2} \cdot e^{\frac1x} \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} \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}} == Link to Laguerre polynomials ==

Generalized Laguerre polynomials L^{(\alpha)}_n(x) are linked to Lah numbers upon setting \alpha = -1 n! L_n^{(-1)}(x) =\sum_{k=0}^n L(n,k) (-x)^kThis formula is the default Laguerre polynomial in Umbral calculus convention. == Practical application ==

In recent years, Lah numbers have been used in steganography for hiding data in images. Compared to alternatives such as DCT, DFT and DWT, it has lower complexity of calculation—O(n \log n)—of their integer coefficients. The Lah and Laguerre transforms naturally arise in the perturbative description of the chromatic dispersion. In Lah-Laguerre optics, such an approach tremendously speeds up optimization problems. == See also ==