Lah number

In mathematics, the (signed and unsigned) Lah numbers are coefficients expressing rising factorials in terms of falling factorials and vice versa. They were discovered by Ivo Lah in 1954. Explicitly, the unsigned Lah numbers $$L(n, k)$$ are given by the formula involving the binomial coefficient

$$ L(n,k) = {n-1 \choose k-1} \frac{n!}{k!}$$

for $$n \geq k \geq 1$$.

Unsigned Lah numbers have an interesting meaning in combinatorics: they count the number of ways a set of $n$  elements can be partitioned into $k$  nonempty linearly ordered subsets. Lah numbers are related to Stirling numbers.

For $n \geq 1$, the Lah number $L(n, 1)$ is equal to the factorial $n!$  in the interpretation above, the only partition of $\{1, 2, 3 \}$  into 1 set can have its set ordered in 6 ways:$$\{(1, 2, 3)\}, \{(1, 3, 2)\}, \{(2, 1, 3)\}, \{(2, 3, 1)\}, \{(3, 1, 2)\}, \{(3, 2, 1)\}$$$L(3, 2)$  is equal to 6, because there are six partitions of $\{1, 2, 3 \}$  into two ordered parts:$$\{1, (2, 3) \}, \{1, (3, 2) \}, \{2, (1, 3) \}, \{2, (3, 1) \}, \{3, (1, 2) \}, \{3, (2, 1) \}$$$L(n, n)$  is always 1 because the only way to partition $\{1, 2, \ldots, n\}$  into $$n$$ non-empty subsets results in subsets of size 1, that can only be permuted in one way. In the more recent literature, Karamata–Knuth style notation has taken over. Lah numbers are now often written as$$L(n,k) = \left\lfloor {n \atop k} \right\rfloor$$

Table of values
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,$$x^{(n)} = \sum_{k=0}^n L(n,k) (x)_k$$and$$(x)_n = \sum_{k=0}^n (-1)^{n-k} L(n,k)x^{(k)}.$$For example,$$x(x+1)(x+2) = {\color{red}6}x + {\color{red}6}x(x-1) + {\color{red}1}x(x-1)(x-2)$$and$$x(x-1)(x-2) = {\color{red}6}x - {\color{red}6}x(x+1) + {\color{red}1}x(x+1)(x+2),$$

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 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 $$\alpha = -1$$$$ n! L_n^{(-1)}(x) =\sum_{k=0}^n L(n,k) (-x)^k$$This 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&mdash;$$O(n \log n)$$&mdash;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.