User talk:GUITARD Thomas

Stirling numbers of the first kind‎
Hi GUITARD Thomas,

I have reverted your addition to the article Stirling numbers of the first kind‎, for two reasons: first, the formulas you added were not correct (they suffered from mis-indexing; for example, $$\left[{n\atop n-1}\right] = {n \choose 2} = \sum_{x=0}^{n - 1}x$$, which is close to but not exactly what you wrote. Second, the formulas you gave are general formulas valid (with appropriate variation) for every n and k, whereas you added them to a section that is specifically about the fact that for certain cases, the formula for Stirling numbers is particularly simple.  It seems to me like the formulas you added (when corrected) belong more naturally alongside some of the formulas in the section Stirling_numbers_of_the_first_kind, for example.

Finally, it would be nice ideally if you could include a citation for your formulas to a reliable source, such as a relevant textbook.

All the best, JBL (talk) 03:56, 27 December 2018 (UTC)


 * Hi JBL,


 * Sorry for the inconvenience.
 * I am not a mathematician.
 * I do not really have a source, I researched the harmonic number.
 * http://www.les-mathematiques.net/phorum/read.php?5,1686208,1720280#msg-1720280
 * Effectively. it would be more correct to write.
 * $$\left[{n\atop n-1}\right] = {n \choose 2} = \sum_{x=0}^{n - 1}x$$ ,
 * $$\left[{n\atop n-2}\right] = \frac{1}{4} (3n-1) {n \choose 3}\quad = \sum_{x=0}^{n - 2}x\sum_{y=0}^{x}y$$ ,
 * $$\quad\left[{n\atop n-3}\right] = {n \choose 2} {n \choose 4} = \sum_{x=0}^{n - 3}x\sum_{y=0}^{x}y\sum_{z=0}^{y}z$$
 * But now I'm not sure.
 * Thanks, GUITARD Thomas 27 December 2018


 * Hi GUITARD Thomas,
 * Thanks for your response. No inconvenience, I'm sure -- Wikipedia is surprisingly complicated, it takes time to figure out the various rules, policies, and so on.  In Richard P. Stanley's book Enumerative Combinatorics, volume 1 (second edition; available online here) I found something very similar on page 34 -- I will think about the best way to add it to the article later today.  (In general, Wikipedia is not a good place to publish original research -- we have a policy about it called WP:OR -- but I would say that rule is often observed in the breach when it comes to mathematics.)
 * Regards,
 * JBL (talk) 16:12, 28 December 2018 (UTC)
 * (P.S. I have reformatted your comment slightly to correct a link and do indentation; I hope that's all right.)


 * Hi JBL,
 * In Factorial-related https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind#Factorial-related_sums sums, I found http://www.les-mathematiques.net/phorum/read.php?5,1626850.
 * I do not know if it has any interest.
 * Cordially.
 * Thanks, GUITARD Thomas 29 December 2018


 * Hi GUITARD Thomas,
 * As you can see, I have now added some version of the identities you mentioned (I chose the version that appears in Stanley's EC1, namely that the sums should be like $$\sum_{x = 0}^{n - 1} \sum_{y = 0}^{x - 1} \sum_{z = 0}^{y - 1} xyz$$) -- I have not carefully checked whether your version is also correct. (Probably it is, I would guess it's not too hard to prove one from the other.)  About your subsequent message, I would describe the formula you give as a straightforward consequence of the formula for Stirling numbers, so I am not particularly inclined to go searching for other sources to see if it is already written down somewhere (to avoid our prohibition on original research).  Of course, if you find such a source, you are welcome to add it.
 * (Frankly, in my opinion, the article would benefit more from trimming of material than adding -- it was not so long ago the focus of editing by a perhaps over-eager graduate student, who added a lot of material that is only of very specialized interest. Identities like the ones you mention seem more natural and pretty.  But they do require a source!)
 * All the best,
 * JBL (talk) 21:37, 29 December 2018 (UTC)