Editing Stirling number (section)

==Stirling numbers with negative integral values==
The Stirling numbers can be extended to negative integral values, but not all authors do so in the same way.<ref name="Loeb">{{cite journal |last1=Loeb |first1=Daniel E.|orig-year=Received 3 Nov 1989|title=A generalization of the Stirling numbers |journal=Discrete Mathematics |volume=103 |issue= 3|pages=259–269 |doi= 10.1016/0012-365X(92)90318-A|year=1992 |doi-access=free }}</ref><ref name=":0">{{cite web |url=http://www.fq.math.ca/Scanned/34-3/branson.pdf |archive-url=https://web.archive.org/web/20110827132927/http://www.fq.math.ca/Scanned/34-3/branson.pdf |archive-date=2011-08-27 |url-status=live |title=An extension of Stirling numbers |last=Branson |first=David |date=August 1994 |website=The Fibonacci Quarterly |access-date=Dec 6, 2017 }}</ref><ref name=":1">D.E. Knuth, 1992.</ref> Regardless of the approach taken, it is worth noting that Stirling numbers of first and second kind are connected by the relations:

: <math>\biggl[{n \atop k}\biggr] = \biggl\{{\!-k\! \atop \!-n\!}\biggr\} \quad \text{and} \quad
\biggl\{{\!n\! \atop \!k\!}\biggr\} = \biggl[{-k \atop -n}\biggr]</math>

when ''n'' and ''k'' are nonnegative integers.  So we have the following table for <math>\left[{-n \atop -k}\right]</math>:
{| cellspacing="0" cellpadding="5" style="text-align:right;" class="wikitable"
|-
| {{diagonal split header|''n''|''k''}}
! {{rh|align=right}} | −1
! {{rh|align=right}} | −2
! {{rh|align=right}} | −3
! {{rh|align=right}} | −4
! {{rh|align=right}} | −5
|-
! {{rh|align=right}} | −1
| 1
| 1
| 1
| 1
| 1
|-
! {{rh|align=right}} | −2
| 0
| 1
| 3
| 7
| 15
|-
! {{rh|align=right}} | −3
| 0
| 0
| 1
| 6
| 25
|-
! {{rh|align=right}} | −4
| 0
| 0
| 0
| 1
| 10
|-
! {{rh|align=right}} | −5
| 0
| 0
| 0
| 0
| 1
|}

Donald Knuth<ref name=":1" /> defined the more general Stirling numbers by extending a [[Stirling numbers of the second kind#Recurrence relation|recurrence relation]] to all integers.  In this approach, <math display=inline> \left[{n \atop k}\right]</math> and <math display=inline>\left\{{\!n\! \atop \!k\!}\right\}</math> are zero if ''n'' is negative and ''k'' is nonnegative, or if ''n'' is nonnegative and ''k'' is negative, and so we have, for ''any'' integers ''n'' and ''k'',
: <math>\biggl[{n \atop k}\biggr] = \biggl\{{\!-k\! \atop \!-n\!}\biggr\} \quad \text{and} \quad
\biggl\{{\!n\! \atop \!k\!}\biggr\} = \biggl[{-k \atop -n}\biggr].</math>

On the other hand, for positive integers ''n'' and ''k'', David Branson<ref name=":0" /> defined <math display=inline> \left[{-n \atop -k}\right]\!,</math> <math display=inline>\left\{{\!-n\! \atop \!-k\!}\right\}\!,</math> <math display=inline> \left[{-n \atop k}\right]\!,</math> and <math display=inline>\left\{{\!-n\! \atop \!k\!}\right\}</math> (but not <math display=inline> \left[{n \atop -k}\right]</math> or <math display=inline>\left\{{\!n\! \atop \!-k\!}\right\}</math>). In this approach, one has the following extension of the [[Stirling numbers of the second kind#Recurrence relation|recurrence relation]] of the Stirling numbers of the first kind:

:<math> \biggl[{-n \atop k}\biggr]
= \frac{(-1)^{n+1}}{n!}\sum_{i=1}^{n}\frac{(-1)^{i+1}}{i^k} \binom ni </math>,

For example, <math display=inline>\left[{-5 \atop k}\right] = \frac1{120}\Bigl(5-\frac{10}{2^k}+\frac{10}{3^k}-\frac 5{4^k}+\frac 1{5^k}\Bigr).</math>  This leads to the following table of values of <math display=inline>\left[{n \atop k}\right]</math> for negative integral ''n''.
{| cellspacing="0" cellpadding="5" style="text-align:center;" class="wikitable"
|-
| {{diagonal split header|''n''|''k''}}
! 0
! 1
! 2
! 3
! 4
|-
! −1
| 1
| 1
| 1
| 1
| 1
|-
! −2
| {{sfrac|{{val|-1}}|{{val|2}}}}
| {{sfrac|{{val|-3}}|{{val|4}}}}
| {{sfrac|{{val|-7}}|{{val|8}}}}
| {{sfrac|{{val|-15}}|{{val|16}}}}
| {{sfrac|{{val|-31}}|{{val|32}}}}
|-
! −3
| {{sfrac|{{val|1}}|{{val|6}}}}
| {{sfrac|{{val|11}}|{{val|36}}}}
| {{sfrac|{{val|85}}|{{val|216}}}}
| {{sfrac|{{val|575}}|{{val|1296}}}}
| {{sfrac|{{val|3661}}|{{val|7776}}}}
|-
! −4
| {{sfrac|{{val|-1}}|{{val|24}}}}
| {{sfrac|{{val|-25}}|{{val|288}}}}
| {{sfrac|{{val|-415}}|{{val|3456}}}}
| {{sfrac|{{val|-5845}}|{{val|41472}}}}
| {{sfrac|{{val|-76111}}|{{val|497664}}}}
|-
! −5
| {{sfrac|{{val|1}}|{{val|120}}}}
| {{sfrac|{{val|137}}|{{val|7200}}}}
| {{sfrac|{{val|12019}}|{{val|432000}}}}
| {{sfrac|{{val|874853}}|{{val|25920000}}}}
| {{sfrac|{{val|58067611}}|{{val|1555200000}}}}
|}

In this case <math display=inline>\sum_{n=1}^{\infty}\left[{-n \atop -k}\right]=B_{k} </math> where <math>B_{k}</math> is a [[Bell number]], and so one may define the negative Bell numbers by <math display=inline>\sum_{n=1}^{\infty}\left[{-n \atop k}\right]=:B_{-k}</math>.

For example, this produces <math display=inline>\sum_{n=1}^{\infty}\left[{-n \atop 1}\right]=B_{-1}=\frac 1e\sum_{j=1}^\infty\frac1{j\cdot j!}=\frac 1e\int_0^1\frac{e^t-1}{t}dt=0.4848291\dots</math>, generally <math display=inline>B_{-k}=\frac 1e\sum_{j=1}^\infty\frac1{j^kj!} </math>.