Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Stirling number
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Lah numbers== {{main|Lah numbers}} The Lah numbers <math>L(n,k) = {n-1 \choose k-1} \frac{n!}{k!}</math> are sometimes called Stirling numbers of the third kind.<ref>{{cite book | last1=Sándor | first1=Jozsef | last2=Crstici | first2=Borislav | title=Handbook of Number Theory II | publisher=[[Kluwer Academic Publishers]] | year=2004 | url=https://books.google.com/books?id=B2WZkvmFKk8C&dq=%22Stirling+numbers+of+the+third+kind%22&pg=PA464 | isbn=9781402025464 | page=464}}</ref> By convention, <math>L(0,0)=1</math> and <math>L(n,k)=0</math> if <math>n<k</math> or <math>k = 0 < n</math>. These numbers are coefficients expressing falling factorials in terms of rising factorials and vice versa: :<math>x^{(n)} = \sum_{k=0}^n L(n,k) (x)_k\quad</math> and <math>\quad(x)_n = \sum_{k=0}^n (-1)^{n-k} L(n,k)x^{(k)}.</math> As above, this means they express the change of basis between the bases <math>(x)_0,(x)_1,(x)_2,\cdots</math> and <math>x^{(0)},x^{(1)},x^{(2)},\cdots</math>, completing the diagram. In particular, one formula is the inverse of the other, thus: : <math>\sum_{j=k}^n (-1)^{j-k} L(n,j) L(j,k) = \delta_{n,k}.</math> Similarly, composing the change of basis from <math>x^{(n)}</math> to <math>x^n</math> with the change of basis from <math>x^n</math> to <math>(x)_{n}</math> gives the change of basis directly from <math>x^{(n)}</math> to <math>(x)_{n}</math>: :<math> L(n,k) = \sum_{j=k}^n \biggl[{n \atop j}\biggr] \biggl\{{\!j\! \atop \!k\!}\biggr\} ,</math> and similarly for other compositions. In terms of matrices, if <math>L</math> denotes the matrix with entries <math>L_{nk}=L(n,k)</math> and <math>L^{-}</math> denotes the matrix with entries <math>L^{-}_{nk}=(-1)^{n-k}L(n,k)</math>, then one is the inverse of the other: <math> L^{-} = L^{-1}</math>. Composing the matrix of unsigned Stirling numbers of the first kind with the matrix of Stirling numbers of the second kind gives the Lah numbers: <math>L = |s| \cdot S</math>. [[Enumerative combinatorics|Enumeratively]], <math display="inline">\left\{{\!n\! \atop \!k\!}\right\}, \left[{n \atop k}\right] , L(n,k)</math> can be defined as the number of partitions of ''n'' elements into ''k'' non-empty unlabeled subsets, where each subset is endowed with no order, a [[cyclic order]], or a linear order, respectively. In particular, this implies the inequalities: : <math>\biggl\{{\!n\! \atop \!k\!}\biggr\} \leq \biggl[{n \atop k}\biggr] \leq L(n,k).</math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)