Lah number

Let ${\displaystyle x^{(n)}}$ represent the rising factorial ${\displaystyle x(x+1)(x+2)\cdots (x+n-1)}$ and let ${\displaystyle (x)_{n}}$ represent the falling factorial ${\displaystyle x(x-1)(x-2)\cdots (x-n+1)}$.

Then ${\displaystyle x^{(n)}=\sum _{k=1}^{n}L(n,k)(x)_{k}}$ and ${\displaystyle (x)_{n}=\sum _{k=1}^{n}(-1)^{n-k}L(n,k)x^{(k)}.}$

For example,

$${\displaystyle x(x+1)(x+2)={\color {red}6}x+{\color {red}6}x(x-1)+{\color {red}1}x(x-1)(x-2).}$$

Compare the third row of the table of values.

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

${\displaystyle L(n,k)={\frac {n!(n-1)!}{k!(k-1)!}}\cdot {\frac {1}{(n-k)!}}=\left({\frac {n!}{k!}}\right)^{2}{\frac {k}{n(n-k)!}}}$

${\displaystyle L(n,k+1)={\frac {n-k}{k(k+1)}}L(n,k).}$

${\displaystyle L(n+1,k)=(n+k)L(n,k)+L(n,k-1)}$ where ${\displaystyle L(n,0)=0}$, ${\displaystyle L(n,k)=0}$ for all ${\displaystyle k>n}$, and ${\displaystyle L(1,1)=1.}$

${\displaystyle L(n,k)=\sum _{j}\left[{n \atop j}\right]\left\{{j \atop k}\right\},}$ where ${\displaystyle \left[{n \atop j}\right]}$ are the Stirling numbers of the first kind and ${\displaystyle \left\{{j \atop k}\right\}}$ are the Stirling numbers of the second kind, ${\displaystyle L(0,0)=1}$, and ${\displaystyle L(n,k)=0}$ for all ${\displaystyle k>n}$.

${\displaystyle L(n,1)=n!}$

${\displaystyle L(n,2)=(n-1)n!/2}$

${\displaystyle L(n,3)=(n-2)(n-1)n!/12}$

${\displaystyle L(n,n-1)=n(n-1)}$

${\displaystyle L(n,n)=1}$

${\displaystyle \sum _{n\geq k}L(n,k){\frac {x^{n}}{n!}}={\frac {1}{k!}}\left({\frac {x}{1-x}}\right)^{k}}$

Below is a table of values for the Lah numbers:

• Stirling numbers
• Pascal matrix

In most recent years, Lah Number is used in Steganography, data hiding in image. So few researchers[6] [7] like Dr. Sudipta Kumar Ghosal have exploited it in this domain as alternative of DCT, DFT and DWT because of low complexity ${\displaystyle O(nlogn)}$ of calculation of integer coefficients of the said transform.