Talk:Stirling number

Confused

It seems to me that the first and second kind are confused.

The way it's done here is what I am accustomed to. Whether it ought to be done that way is another question, since the "second" kind seems to me more basic". Michael Hardy 00:40, 20 Jun 2005 (UTC)

Introduction

This article needs an introduction, but I'm not knowledgeable enough about maths to write one. It should explain what sort of thing Stirling numbers are, and should also mention that they are named after James_Stirling_(mathematician) and why. --Stemonitis 13:41, 26 Jun 2005 (UTC)

Thanks, Linas. Excellent work. --Stemonitis 11:46, 19 August 2005 (UTC)

Utility

Overall, I think the math topics would achieve greater utility, if there were more context included. By this I mean that most of the math topics go straight into the math itself without discussing why the topic is important or what influence it had on the field of mathematics or other fields. I would contribute, but I know little about math theory and history; that's why I've been reading the calculus history.

thanks, morgan (10 October 2005, User:216.231.36.160)

Bell numbers

They are P sub n, not B sub n. B sub n are Bernoulli numbers. -Iopq 11:38, 10 November 2005 (UTC)

Yes, well .. there are many notational clashes. Who uses P_n ? linas 00:56, 11 November 2005 (UTC)

B is common for Bell numbers. Duplicated notation is inevitable since there aren't enough letters in the known alphabets for all the needs of mathematics. Zaslav (talk) 22:19, 25 December 2011 (UTC)

Table formating

Hi,

I reverted the table formting effort. I admit the new table looked nicer, but I reverted for several reasons. First, we should use the same table format for all math articles, and this article introduces an incompatible layout. Second, editing raw html is absolutely miserable, especially if we want to change things later. It would be much better to create a new table tamplate, and use that. If you do not agree, then we should probably continue the conversaion at Wikipedia talk:WikiProject Mathematics, since there are other articles and other authors who are affected by such a change. Thanks. linas 00:39, 3 February 2006 (UTC)

I wasn't aware of such a standard. My apologies. Thanks for assuming good faith! Faya 04:25, 3 February 2006 (UTC)

Actually, I don't want to discourage you; I think it would be great if you could figure out how to create table templates that improve the look and style; that way, they could be deployed on a number of articles. If you do do this, let the wikiproject math know; they might like it. Actually, you might also ask before you start, see what people think about it. linas 05:15, 3 February 2006 (UTC)

Notation problems

This article states

$s(n,k)=\left[{\begin{matrix}n\\k\end{matrix}}\right].$

but Adamchick (quoted in the references section) uses

$s(n,k)=(-1)^{n-k}\left[{\begin{matrix}n\\k\end{matrix}}\right].$

Does anyone have a copy of "art of computing", and what notation does it use? There's some sign errors later in this article as a result of this confusion. linas 16:03, 9 April 2006 (UTC)

Provided the article makes clear the difference between signed and unsigned articles of the first kind, I don't think the notation matters. Generally this should be clear from context. --Rhebus 10:47, 8 April 2007 (UTC)

The notations present in this article should not contradict to existing ones. In particular, both The art of computer programming and Concrete Mathematics define $\left[{n \atop k}\right]$ as unsigned Stirling numbers. I've just corrected the article and made it use this notation consistently. Maxal (talk) 04:28, 21 April 2008 (UTC)

That's only two references. The notation S(n,k), s(n,k), c(n,k) is much more common and should be used in the article - according to me and what I've read. Our opinions are derived from what we've read. That doesn't make them authoritative. Zaslav (talk) 22:21, 25 December 2011 (UTC)

explicit formula

Someone added this:

Another explicit formula for S(n, k) (found in June 2004 by André Labossière) is the following:

$\left\{{\begin{matrix}n\\k\end{matrix}}\right\}=S(n,k)=1/(k-1)!\sum _{i=1..[2k+1+(-1)^{(}k-1)]/4}[C(k-1,2i-2)*(k-2i+2)^{n-1}-C(k-1,2i-1)*(k-2i+1)^{n-1}].$

After the first time something like this was added, I cleaned up the TeX usage, and now it's a mess again. Also, could others look closely to see whether this is correct? (I'll look at it more closely within a couple of days.) Michael Hardy 01:23, 31 May 2006 (UTC)

Um...I just attempted to wikify a formula, but would be nice if someone took another look. I don't quite see the pattern to the terms. --HappyCamper 06:19, 16 August 2006 (UTC)

Concerning this formula, someone should ask Mr. Labossière how it is to be interpreted, where it is proved, and what the general term of the inner series is. Take $k=n-3$, for example, i.e. $S(n,n-3)$ This gives a range of 1 to 3 for the inner variable i. Setting e.g. $i=2$, and evaluating the denominators of the inner series, we run into $(i-1)!=1!,(i-3)!=(-1)!,(i-5)!=(-3)!,(i-7)!=(-5)!$ etc. How are these factorials of negative numbers to be interpreted? Should we take them to be one? -Zahlentheorie 17:17, 16 August 2006 (UTC)

I think the formula might be from someone else...I just Wikified it...is there a reference for it somewhere? If not, I think we should remove it. It's not entirely clear how it's to be used, as you said. --HappyCamper 17:48, 16 August 2006 (UTC)

Going by the list of references it seems that it is by Mr. Labossière. Would you ask him to explain his formula? -Zahlentheorie 11:49, 17 August 2006 (UTC)

Okay, I will do that. Actually, I was referring to these edits...this formula seems a bit weird... [1] --HappyCamper 13:49, 17 August 2006 (UTC)

I don't see how this formula is supposed to evaluate to 1 for S(0,0) (listed in the triangle above). The one listed at http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html does give S(0,0) = 1 (and seems to work correctly for the rest). 71.57.124.41 21:36, 4 December 2006 (UTC)

Since S(n, k) is the number of (unordered) partitions of a set of size n into k parts, S(0, 0) must be 1. Remember: A partition of a set A is a set B of non-empty subsets of A such that no two intersect and their union is A. It is vacuously true that every member of the empty set is a non-empty set. So the empty set is the only partition of the empty set into 0 parts. Hence S(0, 0) = 1. Michael Hardy 21:40, 4 December 2006 (UTC)

Yes, S(0,0) should be 1. But is it according to this formula? When k=0, the sum is taken for j ranging from 1 to 0. I'm not sure what to make of that. Notice also that in this case we have 0 over 1, which means 0!/(1!(-1)!) - what is (-1)!? The formula I referred to, equation (9) on the above website, doesn't have these characteristics and is, I think, at least easier to understand. 71.57.124.41 22:07, 4 December 2006 (UTC)

A sum from 1 to 0 is normally taken to equal 0. Thus, the formula is incorrect. Of course, if a different convention is intended, it has to be explained. Zaslav (talk) 04:31, 22 November 2012 (UTC)

More identities

Hm, I've some more identities heuristically found by some matrix-operations with kind-1 and kind-2 numbers, such as a relation bernoulli-numbers and some more. But I think these would explode the article. On the other hand, I would like to see them here. Third - I've no proofs (though they may be easy to obtain) since I found them primarily by some heuristics.

So - what to do best?

--Gotti 10:57, 21 December 2006 (UTC)

You could post an example here, where people could evaluate whether it is appropriate, how to prove it and whether a proof would belong on the page. You could also post to a mathematics newsgroup and ask if anyone recognizes your identities or might have seen proofs somewhere. Good luck. -Zahlentheorie 17:21, 21 December 2006 (UTC)

Wikipedia is not the appropriate place to publish original research. If you think your identities are important and interesting, then arrange to have them published in a math journal. There are several journals on combinatorics that may be appropriate; the "Ramanujan journal" may be appealing as well. linas 02:45, 29 December 2006 (UTC)

Thanks for your answers. Well, the more I find resources about Stirling numbers and get through them I find most of these identities known - so at one hand, this seems to be no "original work" and at the other hand it might not be very interesting to repeat known facts simply in new clothes. So I think, it may be better to deal with this as "watering" my own "tiny encyclopedia of related matrices"... Talking about journals - there should be something like "mathematical minatures" whose profile would be to promote explorative math even if it is not at the front of the research. Well, I don't know of such a journal/magazine - perhaps someone of you? --Gotti 07:43, 26 January 2007 (UTC)

Please look at 05_1_stirling.pdf to see, whether this article is appropriate to be linked in "references". I tried to keep it as encyclopaedic as possible.

--Gotti 09:16, 14 February 2007 (UTC)

expansion n=3?

Does the expansion

$x^{n}=\sum _{k=1}^{n}s(n,k)(x)^{k}$ [2]

really work with with n=3 using the factors from the table of values? 2x-3x(x+1)+1x(x+1)(x+2)=x³+x? DirkEn 15:53, 21 December 2006 (UTC)

Yeah, that formula is just plain wrong. That formula is a mangled definition for Stirling numbers of the second kind, not the first kind. This rather serious error sat there un-noticed since the Feb 2004 edits by Micheal Hardy. Bummer. I just fixed it up. linas 00:52, 29 December 2006 (UTC)

crazy formula

I'm removing the following formula, added by User:132.203.216.20 in June 2006. [3] I'm removing it because I can't figure out how to generat the next term of the series. As a side note, it appears to be original research, and thus not appropriate for this article anyway. linas 01:02, 29 December 2006 (UTC)

$s(n,k)=(-1)^{n-k}\times {\frac {n!}{2^{n-k}(k-1)!}}$

$\times \left[\sum _{i=1}^{n-k}(-1)^{i-1}\left\{\left({\frac {1}{6^{i-1}(i-1)!(n-k-2i+2)!}}\right)-\left({\frac {(i-3)}{5\times 6^{i-2}(i-3)!(n-k-2i+4)!}}\right)+\left({\frac {(i-5)(21i-46)}{1050\times 6^{i-3}(i-5)!(n-k-2i+6)!}}\right)-\left({\frac {(i-7)(i-4)(7i-11)}{5250\times 6^{i-4}(i-7)!(n-k-2i+8)!}}\right)+\left({\frac {(i-9)\times 6\left(5040+2959C(i-10,1)+\cdots \right)}{1819125\times 6^{i-5}(i-9)!(n-k-2i+10)!}}\right)-\cdots \right\}\times n^{n-k-i}\right]$

s(n,k) = (-1)^(n-k) × ( n!/{(k-1)!×2^(n-k)} ) 
       × [ Sum_{i=1..n-k} (-1)^(i-1) × {

         ( 1/{6^(i-1)×(i-1)!×(n-k-2×i+2)!} )
       - ( {(i-3)}/{5×6^(i-2)×(i-3)!×(n-k-2×i+4)!} )
       + ( {(i-5)×(21×i-46)}/{1050×6^(i-3)×(i-5)!×(n-k-2×i+6)!} )
       - ( {(i-7)×(i-4)×(7×i-11)}/{5250×6^(i-4)×(i-7)!×(n-k-2×i+8)!} )
       + (   {(i-9)×6×(5040+2959×C(i-10,1)+...)}/{1819125×6^(i-5)×(i-9)!×(n-k-2×i+10)!} )
       - .....
      }×n^(n-k-i) ].

Article split

Since this article seemed to have gotten excessively long, I split it into three pieces: Stirling numbers of the first kind, Stirling numbers of the second kind, and this article, to be a catch-all for whatever doesn't fit there.

After performing this split, I did a quick review of the content, and removed anything that seemed strange or incorrect. I wasn't thrilled by the fact that the opening paragraphs had a major error that sat undetected for nearly three years (see above). I wasn't thrilled by the addition of the original research above, which sat there for six months. There's a bunch of other strange/suspect material that I moved off to the talk pages of the two articles. linas 02:53, 29 December 2006 (UTC)

Linas knows best  ;-)

It's not because Mister Linas doesn't understand how to find the next member of the first explicit 1st-echelon equation for s(n,k) - his above "crazy formula" - that he must deprive others of knowing it by vandalizing all the wikipedia definition of Stirling numbers, and doing so pretending also for more clarity !

Show me!

It would be nice of there were actually some examples of the first few Stirling numbers!

Link not working

Hsien-Kuei Hwang, Asymptotic Expansions for the Stirling Numbers of the First Kind (1994). —Preceding unsigned comment added by 81.110.35.40 (talk) 21:39, 30 June 2008 (UTC)

Efficient Computation

It would be nice if the article, or perhaps the specific articles for the two kinds of Stirling numbers, contained references to the most efficient known agorithms for computing Stirling numbers of both kinds, with the asymptotic running times. If these algorithms are short, they could be included on the page. Even just the result of analysis of the algorithms based on the basic recursions (essentially, maintaining one row at a time) would be helpful. It would also be helpful to include analysis of algorithms based on the identity for the second Stirling numbers, S2(n,k) = (1/n!) sum(j = 0 to n) (-1)^(k-j) (k choose j) j^n. I suspect that there are faster algorithms. 71.112.25.123 (talk) 22:15, 30 October 2009 (UTC)ATBS (minor edit 4Nov09)

Don't know whether this is useful:

\\ Pari/Gp: Stirlingnumbers 2'nd and 1'st kind
\\ lower triangular matrices
\\ references to matrix-cells begin at r=1,c=1 !

{makemat_S2(dim=n) = local(f1,f2,M);
 M = matid(dim);
 for(c=2,dim, 
      for(r=c,dim, 
            M[r,c] = 1*M[r-1,c-1] + (c-1)*M[r-1,c]) ); 
 return(M);}

{makemat_S1(dim=n) = local(f1,f2,M);
  M = matid(dim);
  for(r=3,dim, 
       for(c=2,r, M[r,c] = 1*M[r-1,c-1] + (r-2)* M[r-1,c]); 
      );
  \\ finally alternating signs are required
  f1=1;for(r=2,dim,f1*=-1;f2=f1;for(c=2,r-1,f2*=-1;M[r,c]*=f2));
 return(M);}

--Gotti 13:51, 1 November 2009 (UTC)

They're not very useful for the purposes I mentioned. The Pari code above is based on the usual Stirling recurrences that I mentioned. I don't know Pari, so I don't know for sure, but the code above uses matrices rather than vectors, and thus appears to be very inefficient with storage consumption. (It's conceivable that Pari recognizes unused data in matrices and purges it, but I doubt that it does that.) As I said, it would be helpful to have something that goes beyond obvious implementations that use the usual recurrences. This would not be the relatively straightforward vector implementations based on these recurrences. I don't even know if such an algorithm is known. But, even an analysis of the running time and storage for the usual recurrence-based algorithms would be helpful. This analysis for large values of n and k (think multiple precision) might not be trivial. 71.112.25.123 (talk) 18:47, 4 November 2009 (UTC)ATBS

Stirling no.s and descending powers

There seems to be a glitch in the formula linking Stirling no.s of the first kind to falling powers, or otherwise the triangle schema is misleading. From the triangle I would think that

$s(0,0)=1,s(1,0)=-1,s(1,1)=1,s(2,0)=2,s(2,1)=-3,s(2,2)=1,\ldots$.

If we define

$(x)_{k}:=(x-1)(x-2)\cdots (x-k)$

then everything works out well with

$(x)_{n}=\sum _{k=0}^{n}s(n,k)x^{k},$

but if we take the usual definition of

$(x)_{k}:=x(x-1)(x-2)\cdots (x-k+1),$

then the formula would read

$(x)_{n}=\sum _{k=1}^{n}s(n-1,k-1)x^{k}$

for $n\geq 1$. (Where) am I mistaken? — Preceding unsigned comment added by 130.83.2.27 (talk) 10:29, 17 February 2014 (UTC)

You've got the Stirling numbers wrong after s(0,0): s(1,0)=0, s(1,1)=1, s(2,0)=0, s(2,1)=-1, s(2,2)=1, etc. — Preceding unsigned comment added by 76.88.102.108 (talk) 02:21, 23 November 2015 (UTC)

q-analog?

This page is linked from q-analog. I'm not sure of the relation, but could someone who does add a description? Chris2crawford (talk) 13:50, 3 December 2020 (UTC)