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Need to get page number, ISBN for reference to enumerative combinatorics vol 2, and perhaps write up some more examples from there. Dmharvey 19:07, 31 May 2005 (UTC)
The first formula accompanying the quadruple factorial discussion might be made more clear with some parentheses- i.e., (2n)!/n! rather than 2n!/n!. I know that with just a little thought it is obvious, but since factorial has precedence over multiplication it caught me for a moment. By the way, this is a very nice article. :)
D. Andre, Note: Calcul des probabilites. Solution directe du probleme resolu par M. Bertrand, Comptes Rendus de l’Academie des Sciences, Paris, vol 105(1887), p. 436., where the reflection principle was first used (I think?)
Dmharvey 12:07, 1 Jun 2005 (UTC)
I have deleted the following from the "history" section.
It doesn't belong under History. Perhaps it belongs in the list of combinatorial applications. But the text quality needs to be improved, and I don't really know what it's about. Dmharvey Talk 17:11, 6 Jun 2005 (UTC)
Ming An-tu discovered them before but they were published in 1839. 24.203.251.69 03:27, 31 October 2005 (UTC)
I've created the 3x3 grid image in SVG for the article in the spanish wikipedia:
I don't know about licensing, but I want to release it on the public domain as a trivial work. I hope it will be useful. Thank you.
As an architect I try to understand things with graphics. See: http://home.versateladsl.be/vt649464/TrapVeelhNummering.PDF If someone wants to use it I can give a version without text.-- Bleuprint ( talk) 05:12, 8 February 2008 (UTC)
Dmharvey wrote in the TODO list above:
I've added Bertrand's ballot theorem to the See also section, of course, some more details are worth including. -- Kompik 11:28, 11 March 2006 (UTC)
Does anyone have any references for this hankel matrices stuff? Thanks. Dmharvey 11:15, 31 May 2006 (UTC)
1 1 2 5 14 42 132 1 2 5 14 42 132 429 2 5 14 42 132 429 1430 5 14 42 132 429 1430 4862 14 42 132 429 1430 4862 16796 42 132 429 1430 4862 16796 58786 132 429 1430 4862 16796 58786 208012
1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 2 2 0 0 0 0 0 0 0 1 3 5 5 0 0 0 0 0 0 1 4 9 14 14 0 0 0 0 0 1 5 14 28 42 42 0 0 0 0 1 6 20 48 90 132 132 0 0 0 1 7 27 75 165 297 429 429 0 0 1 8 35 110 275 572 1001 1430 1430 0 1 9 44 154 429 1001 2002 3432 4862 4862
1 0 0 0 0 0 0 1 1 0 0 0 0 0 2 3 1 0 0 0 0 5 9 5 1 0 0 0 14 28 20 7 1 0 0 42 90 75 35 9 1 0 132 297 275 154 54 11 1
1 2 5 14 42 2 5 14 42 132 5 14 42 132 429 14 42 132 429 1430 42 132 429 1430 4862
1 0 0 0 0 0 0 2 1 0 0 0 0 0 5 4 1 0 0 0 0 14 14 6 1 0 0 0 42 48 27 8 1 0 0 132 165 110 44 10 1 0 429 572 429 208 65 12 1
The Catalan Recurrence
C(x)=SIGMA C(i) C(n-1)
has not been solved properly. A Quadratic eqn. appears out of nowhere. Can someone explain where it came from?
Is it possible to explain why a particular root of the Quadratic equation is used instead of the another one? In the expanded presentation that is no justification on that. —Preceding unsigned comment added by 207.46.55.31 ( talk • contribs)
Acortis ( talk) 18:45, 30 May 2021 (UTC) The recurrence relation : does not appear ro be correct!!
# python C = {} C[0] = 1 for n in range(5): C[n+1] = 2*(2*n+1)//(n+2)*C[n] print(C)
{0: 1, 1: 1, 2: 2, 3: 4, 4: 8, 5: 24}
Darcourse ( talk) 06:45, 20 June 2021 (UTC)
Perhaps these two problems with Catalan solutions are already in the harder stuff in the article. If not, do they belong?
1. How many ways can 2n ordered objects be arranged in a 2 x n matrix so that the elements are strictly increasing left to right and top down? (sometimes posed as a photographer arranging 2n people in 2 rows so that the heights increase left to right and front to back)
2. How many ways can n non-intersecting diagonals be drawn in a 2n-gon? (How many ways can 2n people seated at a circular table shake hands with no one's arms crossing) Jd2718 18:00, 5 November 2006 (UTC)
The sum of the entries of every row is Twentythreethousand ( talk) 16:36, 31 May 2008 (UTC)
Where should we send the reader who typed "Catalan numbers" into the search box, looking for how to count to ten in Catalan? -- Damian Yerrick ( talk | stalk) 21:31, 23 August 2008 (UTC)
The reflection method is described both here and in Bertrand's ballot theorem. I added a more obvious link in the article; should there be an attempt to merge? Also, it's not entirely incorrect to call it André's reflection method, he had the basic idea but without the reflection part. The reflections are essentially only a way of giving a one-one proof that .-- RDBury ( talk) 17:11, 8 February 2009 (UTC)
The article says that "A Dyck word is a string .. such that no initial segment of the string has more Y's than X's" then it lists several examples including XXXYYY. But: X, XX, XXX, XXXY are all initial segments of XXXYYY and each of them has more X's than Y's.-- Jirka6 ( talk) 00:25, 10 April 2009 (UTC)
The nice picture with green trees is correct, the description was not. There is an infinite number of binary trees (not necessarily full) with n leaves. Also, "ordered" is already implied by "binary". The picture actually belongs to (what used to be) the next item, I fixed that. -- Ikska ( talk)
What do you think? Is it worthy of inclusion in the article that a Catalan number can be written in terms of a generalized binomial coefficient?:
using the Gamma function. Thanks -- Quantling ( talk) 17:01, 21 April 2010 (UTC)
I have added a new bijective proof aimed to directly derive the fascinating factor in the formula. I think it makes a nice complement to the existing bijective proofs as it is not based on a geometric interpretation but instead uses Dyke words and calculates the factor algebraically by comparing coefficients.
I tried several approaches and found this to be the most economical. I made this proof up myself as I could not find a similar one with the same features on the web. Still it should be simple and straightforward enough to be immediately obvious and verifiable by anyone with basic math skills in order to justify its inclusion into wikipedia. If it has been done before (which seems likely), please feel free to add a reference.
Bernhard Oemer ( talk) 16:02, 10 January 2011 (UTC)
What the heck is this section refering to? Where does the name "quadruple factorial" come from? Is it notable at all? Right now I'm inclined to delete it because I can't figure out what it's about, but if someone can tell me what it means, then I'd be happy to reverse my position. (Maybe it would be better as a subsection somewhere?) -- JBL ( talk) 13:32, 17 September 2012 (UTC)
I have just finished some copyediting of the second proof to improve the English and flow of the proof. I tried to preserve as much of the section as I could, but it did require some significant changes to increase clarity and deal with some of the harder parts of the proof. I did not, at this time, put back the historical sentence concerning Andre's reflection method, since there is an editor who objects to that, and so, it should be discussed here. My feeling is that it should be returned since it has encyclopedic value and we are writing an encyclopedia, not a textbook. The statement is of interest and I should point out a comment concerning it made above on this talk page. Bill Cherowitzo ( talk) 06:41, 8 January 2015 (UTC)
The last formula in the first equation of the page:
Is incorrect (or at least gives inconsistent answers with the other formulas), for example for
n = 0, 1, 2, 3, 4, 5
this gives:
1, 1, 2, 4, 14, 36
but should be:
1, 1, 2, 5, 14, 42
— Preceding unsigned comment added by Agold1982 ( talk • contribs)
"Mingantu" felt bogus to me, so I Googled it, and found the more plausible entry, "Minggatu," above. My feeling is that this "Mingantu" thingie, and the related renaming of his hometown as "Ming Antu," apparently, by some Chinese bureaucrat, are the work of perhaps well-intentioned but semi-literate functionaries. The Han government in Beijing is faced with all sorts of ethnic frictions, and they're scurrying around trying to solve them, but clearly some of the people involved are acting without really understanding what they're doing. (We have the same sort of thing in the West, with all the halfwits who pronounce the Chinese capital a Frenchified "bay-zhhing." It's furrin, but if I make it sound French that will show people how cosmopolitan and sophisticated I am, right? Feh!)
Imho, we ought to go with the way the guy pronounced it himself, and in Roman letters that would be "Minggatu," to the minimum possible distortion.
Fixing this will involve inserting the Wikipedia reference as #14, and renumbering the higher footnote numbers, which I'm afraid is beyond my Wiki skills. Could somebody else maybe do that?
David Lloyd-Jones ( talk) 12:31, 6 November 2017 (UTC)
Should there be a mention of inversion of power series? One way to define the Catalan numbers is to say that they are the coefficients in the inverse series of x(1-x), regarded as a power series in x. In other words, if you set y=x(1-x) and solve for x as a series in y, you get them as coefficients. Maybe this follows pretty readily from one of the other descriptions given here. I don't feel confident enough to wade in and make a change to the article. Ishboyfay ( talk) 23:22, 16 November 2017 (UTC)
Knowing that some Catalan number counts the number of triangulations of a polygon is part of the excitement of what mathematics is all about. Knowing which Catalan number it is, through the mnemonic that "the n-th Catalan number counts the case of n triangles" is intended to be the frosting on the cake. I'd like to put in the count of triangles without it getting reverted. Thank you, 64.132.59.226 ( talk) 13:32, 30 January 2018 (UTC)
The discussion has ground to a halt so let's try the BOLD, revert, discuss cycle. Because the edit I am making attempts to address the prior criticism, if you find yourself compelled to revert the edit then also comment here as to your reasons. Make your comments constructive; "it would be better if we ..." helps much more than "that is worse because ..." because the former helps to find the right direction to go, but the latter lists only one of many possible directions that should be avoided. 64.132.59.226 ( talk) 13:36, 2 February 2018 (UTC)
In the list of applications, applications 2, 3, and 13 are stating the same thing. Also, applications 4 and 12 are redundant. I am going to remove applications 3, 12, and 13 upon consensus.
Plaba123 ( talk) 16:23, 15 October 2018 (UTC)
The second recursive formula was shifted one to the right, it returned 1 for C_0, 2 for C_1, 5 for C_2 and so on. OEIS gives the same recursive definition ( https://oeis.org/A000108):
2*(2*n-1)*a(n-1) = (n+1)*a(n)
I see that my change was reverted, could it be incorporated? Pranomostro ( talk) 21:06, 18 February 2019 (UTC) comment added by Pranomostro ( talk • contribs) 20:51, 18 February 2019 (UTC)
{{
re}}
or {{
u}}
template, depending). Anyway, about the recurrence relation, you might want to double check that you're using the right value for n. The formula that's already here is correct and agrees with what's listed at the OEIS. –
Deacon Vorbis (
carbon •
videos)
02:11, 19 February 2019 (UTC)@ Deacon Vorbis: I'm sorry if I violated some wikiquette by notifying you on your talk page, thanks for the pointers :D
Now, about the recurrence relation, the following was my thought process:
When plugging in the indices [0;13]:
The given values for the catalan numbers were:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012
S1: gives the following values for C_0, C_1, C_2 etc.:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012
OEIS lists the catalan numbers the same way:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012
S2: gives the following numbers (and is used in the current article):
1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900
Note that one is missing.
S3: . This is equivalent to 2*(2*n-1)*a(n-1) = (n+1)*a(n) from OEIS and returns the following sequence:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012
Here is the code I used, in python3:
from math import factorial as fac
def binomial(x, y):
try:
binom = fac(x) // fac(y) // fac(x - y)
except ValueError:
binom = 0
return binom
def s1(x):
return binomial(2*x, x)/(1+x)
def s2(x):
if x==0:
return 1
else:
return (2*(2*x+1))/(x+2)*s2(x-1)
def s3(x):
if x==0:
return 1
else:
return ((2*(2*x-1))/(x+1))*s3(x-1)
arg=list(range(0,13))
print(list(map(s1,arg)))
print(list(map(s2,arg)))
print(list(map(s3,arg)))
There might still be good reasons to keep the current version. First of all, there is a proof for it in this article, and secondly, it is a lot more common. On the other hand, it may be misleading (it misled me at first as well). I hope I made my case clear here, and will stop pestering you now.
Have a nice day.
Pranomostro ( talk) 21:16, 19 February 2019 (UTC)
Using WA, one can use the following query: (5*4^(n-3) * (Pochhammer 7/2,n-3)) / (Pochhammer 5,n-3) for n=3,10
Thank you @JayBeeEll for your recent concern about my edit to this article. Would you please give more information about your comment "this is not an example of what is being discussed"? The paragraph is about the number of associative ways to multiply an ordered set of n+1 objects. For the matrix chain multiplication problem, the n+1 objects are matrices. To me, it is the premier example where people care about the association order for the multiplications. Thank you — Quantling ( talk | contribs) 17:38, 8 February 2021 (UTC)
That parenthetical works for me. Thank you — Quantling ( talk | contribs) 19:53, 8 February 2021 (UTC)
If any talk-page watcher has any interest in taking this to a GA or improve the overall quality, I will appreciate their collaboration. Will start drafting at User:TrangaBellam/Catalan Number. Thanks, TrangaBellam ( talk) 18:00, 21 April 2022 (UTC)
I think using "we" in the proofs is kinda weird and should be removed — Preceding unsigned comment added by 2604:3D09:1580:9600:C884:4058:E2C4:3F13 ( talk) 03:19, 17 May 2022 (UTC)
@ Darcourse et al. There is use of the method of images in the second proof of the Catalan numbers formula. While the method of images is used for continuous cases, such as energy potentials in the presence of boundary conditions, it is also used for discrete cases. A random walk that starts at the origin on the real line, that takes steps of ±1 , and that is prohibited from going to −1 — in our case the number of '(' minus the number of ')' cannot be −1 — can be modeled by adding the random walk to a "negative" of the random walk that starts at −2. For any given number of steps, both the random walk from the origin and the one from −2 have equal chance of ending up at −1, and thus adding that "negative" "image" that starts at −2 to the original random walk from the origin ends up zeroing all values at −1, which is exactly what we need when we are enumerating Catalan numbers.
I propose to mention this in the section on the second proof, via something like
but I am open to alternatives. Thanks — Quantling ( talk | contribs) 14:57, 16 October 2023 (UTC)
If you take the variable c and run it through the function for the Mandelbrot set, , an infinite number of times, you get an infinite series denoted by:
where is the nth Catalan number. And the curve in the complex plane for which this infinite series equals two is precisely the boundary of the Mandelbrot set. Denelson 83 02:17, 19 January 2024 (UTC)
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Need to get page number, ISBN for reference to enumerative combinatorics vol 2, and perhaps write up some more examples from there. Dmharvey 19:07, 31 May 2005 (UTC)
The first formula accompanying the quadruple factorial discussion might be made more clear with some parentheses- i.e., (2n)!/n! rather than 2n!/n!. I know that with just a little thought it is obvious, but since factorial has precedence over multiplication it caught me for a moment. By the way, this is a very nice article. :)
D. Andre, Note: Calcul des probabilites. Solution directe du probleme resolu par M. Bertrand, Comptes Rendus de l’Academie des Sciences, Paris, vol 105(1887), p. 436., where the reflection principle was first used (I think?)
Dmharvey 12:07, 1 Jun 2005 (UTC)
I have deleted the following from the "history" section.
It doesn't belong under History. Perhaps it belongs in the list of combinatorial applications. But the text quality needs to be improved, and I don't really know what it's about. Dmharvey Talk 17:11, 6 Jun 2005 (UTC)
Ming An-tu discovered them before but they were published in 1839. 24.203.251.69 03:27, 31 October 2005 (UTC)
I've created the 3x3 grid image in SVG for the article in the spanish wikipedia:
I don't know about licensing, but I want to release it on the public domain as a trivial work. I hope it will be useful. Thank you.
As an architect I try to understand things with graphics. See: http://home.versateladsl.be/vt649464/TrapVeelhNummering.PDF If someone wants to use it I can give a version without text.-- Bleuprint ( talk) 05:12, 8 February 2008 (UTC)
Dmharvey wrote in the TODO list above:
I've added Bertrand's ballot theorem to the See also section, of course, some more details are worth including. -- Kompik 11:28, 11 March 2006 (UTC)
Does anyone have any references for this hankel matrices stuff? Thanks. Dmharvey 11:15, 31 May 2006 (UTC)
1 1 2 5 14 42 132 1 2 5 14 42 132 429 2 5 14 42 132 429 1430 5 14 42 132 429 1430 4862 14 42 132 429 1430 4862 16796 42 132 429 1430 4862 16796 58786 132 429 1430 4862 16796 58786 208012
1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 2 2 0 0 0 0 0 0 0 1 3 5 5 0 0 0 0 0 0 1 4 9 14 14 0 0 0 0 0 1 5 14 28 42 42 0 0 0 0 1 6 20 48 90 132 132 0 0 0 1 7 27 75 165 297 429 429 0 0 1 8 35 110 275 572 1001 1430 1430 0 1 9 44 154 429 1001 2002 3432 4862 4862
1 0 0 0 0 0 0 1 1 0 0 0 0 0 2 3 1 0 0 0 0 5 9 5 1 0 0 0 14 28 20 7 1 0 0 42 90 75 35 9 1 0 132 297 275 154 54 11 1
1 2 5 14 42 2 5 14 42 132 5 14 42 132 429 14 42 132 429 1430 42 132 429 1430 4862
1 0 0 0 0 0 0 2 1 0 0 0 0 0 5 4 1 0 0 0 0 14 14 6 1 0 0 0 42 48 27 8 1 0 0 132 165 110 44 10 1 0 429 572 429 208 65 12 1
The Catalan Recurrence
C(x)=SIGMA C(i) C(n-1)
has not been solved properly. A Quadratic eqn. appears out of nowhere. Can someone explain where it came from?
Is it possible to explain why a particular root of the Quadratic equation is used instead of the another one? In the expanded presentation that is no justification on that. —Preceding unsigned comment added by 207.46.55.31 ( talk • contribs)
Acortis ( talk) 18:45, 30 May 2021 (UTC) The recurrence relation : does not appear ro be correct!!
# python C = {} C[0] = 1 for n in range(5): C[n+1] = 2*(2*n+1)//(n+2)*C[n] print(C)
{0: 1, 1: 1, 2: 2, 3: 4, 4: 8, 5: 24}
Darcourse ( talk) 06:45, 20 June 2021 (UTC)
Perhaps these two problems with Catalan solutions are already in the harder stuff in the article. If not, do they belong?
1. How many ways can 2n ordered objects be arranged in a 2 x n matrix so that the elements are strictly increasing left to right and top down? (sometimes posed as a photographer arranging 2n people in 2 rows so that the heights increase left to right and front to back)
2. How many ways can n non-intersecting diagonals be drawn in a 2n-gon? (How many ways can 2n people seated at a circular table shake hands with no one's arms crossing) Jd2718 18:00, 5 November 2006 (UTC)
The sum of the entries of every row is Twentythreethousand ( talk) 16:36, 31 May 2008 (UTC)
Where should we send the reader who typed "Catalan numbers" into the search box, looking for how to count to ten in Catalan? -- Damian Yerrick ( talk | stalk) 21:31, 23 August 2008 (UTC)
The reflection method is described both here and in Bertrand's ballot theorem. I added a more obvious link in the article; should there be an attempt to merge? Also, it's not entirely incorrect to call it André's reflection method, he had the basic idea but without the reflection part. The reflections are essentially only a way of giving a one-one proof that .-- RDBury ( talk) 17:11, 8 February 2009 (UTC)
The article says that "A Dyck word is a string .. such that no initial segment of the string has more Y's than X's" then it lists several examples including XXXYYY. But: X, XX, XXX, XXXY are all initial segments of XXXYYY and each of them has more X's than Y's.-- Jirka6 ( talk) 00:25, 10 April 2009 (UTC)
The nice picture with green trees is correct, the description was not. There is an infinite number of binary trees (not necessarily full) with n leaves. Also, "ordered" is already implied by "binary". The picture actually belongs to (what used to be) the next item, I fixed that. -- Ikska ( talk)
What do you think? Is it worthy of inclusion in the article that a Catalan number can be written in terms of a generalized binomial coefficient?:
using the Gamma function. Thanks -- Quantling ( talk) 17:01, 21 April 2010 (UTC)
I have added a new bijective proof aimed to directly derive the fascinating factor in the formula. I think it makes a nice complement to the existing bijective proofs as it is not based on a geometric interpretation but instead uses Dyke words and calculates the factor algebraically by comparing coefficients.
I tried several approaches and found this to be the most economical. I made this proof up myself as I could not find a similar one with the same features on the web. Still it should be simple and straightforward enough to be immediately obvious and verifiable by anyone with basic math skills in order to justify its inclusion into wikipedia. If it has been done before (which seems likely), please feel free to add a reference.
Bernhard Oemer ( talk) 16:02, 10 January 2011 (UTC)
What the heck is this section refering to? Where does the name "quadruple factorial" come from? Is it notable at all? Right now I'm inclined to delete it because I can't figure out what it's about, but if someone can tell me what it means, then I'd be happy to reverse my position. (Maybe it would be better as a subsection somewhere?) -- JBL ( talk) 13:32, 17 September 2012 (UTC)
I have just finished some copyediting of the second proof to improve the English and flow of the proof. I tried to preserve as much of the section as I could, but it did require some significant changes to increase clarity and deal with some of the harder parts of the proof. I did not, at this time, put back the historical sentence concerning Andre's reflection method, since there is an editor who objects to that, and so, it should be discussed here. My feeling is that it should be returned since it has encyclopedic value and we are writing an encyclopedia, not a textbook. The statement is of interest and I should point out a comment concerning it made above on this talk page. Bill Cherowitzo ( talk) 06:41, 8 January 2015 (UTC)
The last formula in the first equation of the page:
Is incorrect (or at least gives inconsistent answers with the other formulas), for example for
n = 0, 1, 2, 3, 4, 5
this gives:
1, 1, 2, 4, 14, 36
but should be:
1, 1, 2, 5, 14, 42
— Preceding unsigned comment added by Agold1982 ( talk • contribs)
"Mingantu" felt bogus to me, so I Googled it, and found the more plausible entry, "Minggatu," above. My feeling is that this "Mingantu" thingie, and the related renaming of his hometown as "Ming Antu," apparently, by some Chinese bureaucrat, are the work of perhaps well-intentioned but semi-literate functionaries. The Han government in Beijing is faced with all sorts of ethnic frictions, and they're scurrying around trying to solve them, but clearly some of the people involved are acting without really understanding what they're doing. (We have the same sort of thing in the West, with all the halfwits who pronounce the Chinese capital a Frenchified "bay-zhhing." It's furrin, but if I make it sound French that will show people how cosmopolitan and sophisticated I am, right? Feh!)
Imho, we ought to go with the way the guy pronounced it himself, and in Roman letters that would be "Minggatu," to the minimum possible distortion.
Fixing this will involve inserting the Wikipedia reference as #14, and renumbering the higher footnote numbers, which I'm afraid is beyond my Wiki skills. Could somebody else maybe do that?
David Lloyd-Jones ( talk) 12:31, 6 November 2017 (UTC)
Should there be a mention of inversion of power series? One way to define the Catalan numbers is to say that they are the coefficients in the inverse series of x(1-x), regarded as a power series in x. In other words, if you set y=x(1-x) and solve for x as a series in y, you get them as coefficients. Maybe this follows pretty readily from one of the other descriptions given here. I don't feel confident enough to wade in and make a change to the article. Ishboyfay ( talk) 23:22, 16 November 2017 (UTC)
Knowing that some Catalan number counts the number of triangulations of a polygon is part of the excitement of what mathematics is all about. Knowing which Catalan number it is, through the mnemonic that "the n-th Catalan number counts the case of n triangles" is intended to be the frosting on the cake. I'd like to put in the count of triangles without it getting reverted. Thank you, 64.132.59.226 ( talk) 13:32, 30 January 2018 (UTC)
The discussion has ground to a halt so let's try the BOLD, revert, discuss cycle. Because the edit I am making attempts to address the prior criticism, if you find yourself compelled to revert the edit then also comment here as to your reasons. Make your comments constructive; "it would be better if we ..." helps much more than "that is worse because ..." because the former helps to find the right direction to go, but the latter lists only one of many possible directions that should be avoided. 64.132.59.226 ( talk) 13:36, 2 February 2018 (UTC)
In the list of applications, applications 2, 3, and 13 are stating the same thing. Also, applications 4 and 12 are redundant. I am going to remove applications 3, 12, and 13 upon consensus.
Plaba123 ( talk) 16:23, 15 October 2018 (UTC)
The second recursive formula was shifted one to the right, it returned 1 for C_0, 2 for C_1, 5 for C_2 and so on. OEIS gives the same recursive definition ( https://oeis.org/A000108):
2*(2*n-1)*a(n-1) = (n+1)*a(n)
I see that my change was reverted, could it be incorporated? Pranomostro ( talk) 21:06, 18 February 2019 (UTC) comment added by Pranomostro ( talk • contribs) 20:51, 18 February 2019 (UTC)
{{
re}}
or {{
u}}
template, depending). Anyway, about the recurrence relation, you might want to double check that you're using the right value for n. The formula that's already here is correct and agrees with what's listed at the OEIS. –
Deacon Vorbis (
carbon •
videos)
02:11, 19 February 2019 (UTC)@ Deacon Vorbis: I'm sorry if I violated some wikiquette by notifying you on your talk page, thanks for the pointers :D
Now, about the recurrence relation, the following was my thought process:
When plugging in the indices [0;13]:
The given values for the catalan numbers were:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012
S1: gives the following values for C_0, C_1, C_2 etc.:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012
OEIS lists the catalan numbers the same way:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012
S2: gives the following numbers (and is used in the current article):
1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900
Note that one is missing.
S3: . This is equivalent to 2*(2*n-1)*a(n-1) = (n+1)*a(n) from OEIS and returns the following sequence:
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012
Here is the code I used, in python3:
from math import factorial as fac
def binomial(x, y):
try:
binom = fac(x) // fac(y) // fac(x - y)
except ValueError:
binom = 0
return binom
def s1(x):
return binomial(2*x, x)/(1+x)
def s2(x):
if x==0:
return 1
else:
return (2*(2*x+1))/(x+2)*s2(x-1)
def s3(x):
if x==0:
return 1
else:
return ((2*(2*x-1))/(x+1))*s3(x-1)
arg=list(range(0,13))
print(list(map(s1,arg)))
print(list(map(s2,arg)))
print(list(map(s3,arg)))
There might still be good reasons to keep the current version. First of all, there is a proof for it in this article, and secondly, it is a lot more common. On the other hand, it may be misleading (it misled me at first as well). I hope I made my case clear here, and will stop pestering you now.
Have a nice day.
Pranomostro ( talk) 21:16, 19 February 2019 (UTC)
Using WA, one can use the following query: (5*4^(n-3) * (Pochhammer 7/2,n-3)) / (Pochhammer 5,n-3) for n=3,10
Thank you @JayBeeEll for your recent concern about my edit to this article. Would you please give more information about your comment "this is not an example of what is being discussed"? The paragraph is about the number of associative ways to multiply an ordered set of n+1 objects. For the matrix chain multiplication problem, the n+1 objects are matrices. To me, it is the premier example where people care about the association order for the multiplications. Thank you — Quantling ( talk | contribs) 17:38, 8 February 2021 (UTC)
That parenthetical works for me. Thank you — Quantling ( talk | contribs) 19:53, 8 February 2021 (UTC)
If any talk-page watcher has any interest in taking this to a GA or improve the overall quality, I will appreciate their collaboration. Will start drafting at User:TrangaBellam/Catalan Number. Thanks, TrangaBellam ( talk) 18:00, 21 April 2022 (UTC)
I think using "we" in the proofs is kinda weird and should be removed — Preceding unsigned comment added by 2604:3D09:1580:9600:C884:4058:E2C4:3F13 ( talk) 03:19, 17 May 2022 (UTC)
@ Darcourse et al. There is use of the method of images in the second proof of the Catalan numbers formula. While the method of images is used for continuous cases, such as energy potentials in the presence of boundary conditions, it is also used for discrete cases. A random walk that starts at the origin on the real line, that takes steps of ±1 , and that is prohibited from going to −1 — in our case the number of '(' minus the number of ')' cannot be −1 — can be modeled by adding the random walk to a "negative" of the random walk that starts at −2. For any given number of steps, both the random walk from the origin and the one from −2 have equal chance of ending up at −1, and thus adding that "negative" "image" that starts at −2 to the original random walk from the origin ends up zeroing all values at −1, which is exactly what we need when we are enumerating Catalan numbers.
I propose to mention this in the section on the second proof, via something like
but I am open to alternatives. Thanks — Quantling ( talk | contribs) 14:57, 16 October 2023 (UTC)
If you take the variable c and run it through the function for the Mandelbrot set, , an infinite number of times, you get an infinite series denoted by:
where is the nth Catalan number. And the curve in the complex plane for which this infinite series equals two is precisely the boundary of the Mandelbrot set. Denelson 83 02:17, 19 January 2024 (UTC)