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The function in this section seems to be Euler's function mentioned a few lines earlier. - Herbmuell ( talk) 18:25, 7 August 2016 (UTC)
Hi, could someone please check the formula ½m(3m − 1) that is given for generalized pentagonal numbers in the section "Generating function" in this article? it seems to me that this is the formula for pentagonal numbers and leaves out the sequence 0, 1, -1, 2, -2, 3, -3, 4 that is needed for generalized pentagonal numbers.
Thanks, Joachim Neumann ( talk) 15:25, 8 May 2016 (UTC)
The title of this article should really be Partition (Combinatorics). While partitions may have applications in number theory, they are usually thought of as combinatorial objects, and this article certainly treats them this way. 128.138.150.145 ( talk) 18:36, 4 October 2013 (UTC)
I have developed a method which enables to find the possible number of partition of any given numerical number. To view my developed method then click on the below link. https://drive.google.com/folderview?id=0B4wrAuhEkQyvMXJuWEVQaHR5MXM&usp=sharing — Preceding unsigned comment added by 59.184.32.163 ( talk) 11:44, 29 April 2014 (UTC)
Let q(n,m) be the number of partitions of n with parts <= m and at least one part = m. There is a recurrence relation:
q(n,m) = q(n-m,m) + q(n-1,m-1)
Taking q(n,m) = 0 or q(n,m) = 1 in the following base cases:
(where each case assumes the preceding cases do not apply)
n = m then q(n,m)= 1
n = 0 then q(n,m)= 0
m > n then q(n,m)= 0
m = 1 then q(n,m)= 1
m <= 0 then q(n,m)= 0
Is this correct? If so, should it be included in the article?
Best wishes, Jos Koot Jos.koot ( talk) 12:52, 29 May 2014 (UTC)
Problem 15A. Show that pk(n) = pk−1(n − 1) + pk(n − k) and use this to prove (15.2).
Here we have a reference I think, a very good one, for it is shown as an exercise that a student should be able to solve. If the other sources mentioned by Deltahedron can be regarded as references too, the solution seems to be solved. Just include them. mho. Jos.koot ( talk) 21:40, 30 May 2014 (UTC)
I reverted [1] a good faith edit [2] that changed the sentence "the Young diagram uses boxes" to "the Young diagram uses polyominos", and a third editor has in turn reverted me [3]. My reason is that a Young diagram is not the same thing as a polyomino: a Young diagram satisfies much more stringent conditions; the theory of partitions is not closely connected with the theory of polyominoes; and the word "boxes" is quite apt when studying generalisations of Young diagrams in combinatorics such as Young tableaux, where we are told [a] Young tableau is obtained by filling in the boxes of the Young diagram with symbols taken from some alphabet [my emphasis]. Replacing "boxes" by "polyominoes" here would make nonsense of the description. The third editor's rationale "polyominoes (connected sets of squares) is more precise; they are not made of cardboard" seems irrelevant: as I have said, polynomino is not a precise description, it is less rather than more useful and I certainly did not maintain that Young diagrams are made of cardboard, and cannot see how anyone would imagine that I did. So -- any other opinions? Deltahedron ( talk) 07:58, 14 June 2014 (UTC) Additional: it would help the discussion to point to independent reliable sources that refer to Young diagrams as using polyominoes. Deltahedron ( talk) 15:35, 14 June 2014 (UTC)
The sum of Permutation of all Possible Partition for a given number (n) is equal to 2^(n-1). To have a look then click on the link below. https://drive.google.com/folderview?id=0B4wrAuhEkQyvY0g0QlhfQ3h0UU0&usp=sharing_eid
At the moment the subsection "Representation of partitions" starts of saying that Ferrer diagrams use a row of dots for each part, giving an example. However, all subsequent examples seem to use columns of dots instead. There should be consistency, but I don't know what the standard choice is. — Preceding unsigned comment added by KernelClass ( talk • contribs) 10:23, 16 September 2014 (UTC)
One possible generating function for such partitions, taking k fixed and n variable, is
More generally, if T is a set of positive integers then the number of partitions of n, all of whose parts belong to T, has generating function
Isn't there a contradiction between the given example and the general formula ? Why is there a leading in the example ?
LeCrayonVert ( talk) 09:44, 1 February 2015 (UTC)
There is something about Leibniz under the heading "See also", but there should be more, as he seems to have introduced partitions. — Preceding unsigned comment added by 79.180.28.37 ( talk • contribs) 12:13, 13 January 2016 (UTC)
In
Using the same conjugation trick as above, one may show that the number pk(n) of partitions of n into exactly k parts is equal to the number of partitions of n in which the largest part has size k.[23] The function pk(n) satisfies the recurrence
pk(n) = pk(n − k) + pk − 1(n − 1)
with initial values p0(0) = 1 and pk(n) = 0 if n ≤ 0 or k ≤ 0.
shouldn't the second addend be p_(k-1)(n)? — Preceding unsigned comment added by 69.12.245.117 ( talk • contribs) 00:54, 21 January 2017 (UTC)
Shouldn't the formula for be " if or "? At least that is the formula that Abramowitz and Stegun p. 826, 24.2.2 eq. II(A) give. Kdpw ( talk) 22:51, 10 April 2017 (UTC)
@Joel B. Lewis, that is wrong. It is not the partition that is distinct, it is the parts that are. ImTheIP ( talk) 16:20, 28 October 2018 (UTC)
In this section the recurrence
has been the object of several edits and reverts. It is correct; proof: given a partition of either all parts are ≥ 2, and one gets a partition of by removing one to each part, or one gets a partition of by removing one part equal to 1.
However, another recurrence is also true:
Proof: Substract one to the lowest part.
As this recurrence is simpler (and the initial conditions are much simpler, as there is no need to consider negative integers), this recurrence must appear in this section, and deserve probably to be more emphasized than the one which is presently given. I have no reference under hand, but this is so simple that it must exists, if there is no mistake in my sketched proof. D.Lazard ( talk) 14:06, 15 September 2019 (UTC)
The Asymptotics section begins like this:
The asymptotic expression for p(n) implies that
where .
If A is a set of natural numbers, we let pA(n) denote the number of partitions of n into elements of A. If A possesses positive natural density α then
No, no, no, no, no. Do not discuss the logarithm of the asymptotic formula when the asymptotic formula has not even been stated yet.
First things first.
This ought to be 100% obvious to anyone writing for an encyclopedia. 2600:1700:E1C0:F340:51D:10F4:9B05:6C06 ( talk) 05:48, 20 September 2019 (UTC)
The function: y=(0.00005)x^9 - (0.00005)x^8 +(0.00005)x^7 -(0.00005)x^6+(0.01)x^5-(0.07)x^4+(0.36)x^3-(0.75)x^2+x+0.85 is a better fit for values of P(n) from 0 to 31. I apologize for my previous typing mistakes. enoraB otreboR 79.42.168.169 ( talk) 11:26, 25 April 2023 (UTC)
This is the
talk page for discussing improvements to the
Integer partition article. This is not a forum for general discussion of the article's subject. |
Article policies
|
Find sources: Google ( books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL |
Archives: 1 |
This
level-5 vital article is rated C-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
The function in this section seems to be Euler's function mentioned a few lines earlier. - Herbmuell ( talk) 18:25, 7 August 2016 (UTC)
Hi, could someone please check the formula ½m(3m − 1) that is given for generalized pentagonal numbers in the section "Generating function" in this article? it seems to me that this is the formula for pentagonal numbers and leaves out the sequence 0, 1, -1, 2, -2, 3, -3, 4 that is needed for generalized pentagonal numbers.
Thanks, Joachim Neumann ( talk) 15:25, 8 May 2016 (UTC)
The title of this article should really be Partition (Combinatorics). While partitions may have applications in number theory, they are usually thought of as combinatorial objects, and this article certainly treats them this way. 128.138.150.145 ( talk) 18:36, 4 October 2013 (UTC)
I have developed a method which enables to find the possible number of partition of any given numerical number. To view my developed method then click on the below link. https://drive.google.com/folderview?id=0B4wrAuhEkQyvMXJuWEVQaHR5MXM&usp=sharing — Preceding unsigned comment added by 59.184.32.163 ( talk) 11:44, 29 April 2014 (UTC)
Let q(n,m) be the number of partitions of n with parts <= m and at least one part = m. There is a recurrence relation:
q(n,m) = q(n-m,m) + q(n-1,m-1)
Taking q(n,m) = 0 or q(n,m) = 1 in the following base cases:
(where each case assumes the preceding cases do not apply)
n = m then q(n,m)= 1
n = 0 then q(n,m)= 0
m > n then q(n,m)= 0
m = 1 then q(n,m)= 1
m <= 0 then q(n,m)= 0
Is this correct? If so, should it be included in the article?
Best wishes, Jos Koot Jos.koot ( talk) 12:52, 29 May 2014 (UTC)
Problem 15A. Show that pk(n) = pk−1(n − 1) + pk(n − k) and use this to prove (15.2).
Here we have a reference I think, a very good one, for it is shown as an exercise that a student should be able to solve. If the other sources mentioned by Deltahedron can be regarded as references too, the solution seems to be solved. Just include them. mho. Jos.koot ( talk) 21:40, 30 May 2014 (UTC)
I reverted [1] a good faith edit [2] that changed the sentence "the Young diagram uses boxes" to "the Young diagram uses polyominos", and a third editor has in turn reverted me [3]. My reason is that a Young diagram is not the same thing as a polyomino: a Young diagram satisfies much more stringent conditions; the theory of partitions is not closely connected with the theory of polyominoes; and the word "boxes" is quite apt when studying generalisations of Young diagrams in combinatorics such as Young tableaux, where we are told [a] Young tableau is obtained by filling in the boxes of the Young diagram with symbols taken from some alphabet [my emphasis]. Replacing "boxes" by "polyominoes" here would make nonsense of the description. The third editor's rationale "polyominoes (connected sets of squares) is more precise; they are not made of cardboard" seems irrelevant: as I have said, polynomino is not a precise description, it is less rather than more useful and I certainly did not maintain that Young diagrams are made of cardboard, and cannot see how anyone would imagine that I did. So -- any other opinions? Deltahedron ( talk) 07:58, 14 June 2014 (UTC) Additional: it would help the discussion to point to independent reliable sources that refer to Young diagrams as using polyominoes. Deltahedron ( talk) 15:35, 14 June 2014 (UTC)
The sum of Permutation of all Possible Partition for a given number (n) is equal to 2^(n-1). To have a look then click on the link below. https://drive.google.com/folderview?id=0B4wrAuhEkQyvY0g0QlhfQ3h0UU0&usp=sharing_eid
At the moment the subsection "Representation of partitions" starts of saying that Ferrer diagrams use a row of dots for each part, giving an example. However, all subsequent examples seem to use columns of dots instead. There should be consistency, but I don't know what the standard choice is. — Preceding unsigned comment added by KernelClass ( talk • contribs) 10:23, 16 September 2014 (UTC)
One possible generating function for such partitions, taking k fixed and n variable, is
More generally, if T is a set of positive integers then the number of partitions of n, all of whose parts belong to T, has generating function
Isn't there a contradiction between the given example and the general formula ? Why is there a leading in the example ?
LeCrayonVert ( talk) 09:44, 1 February 2015 (UTC)
There is something about Leibniz under the heading "See also", but there should be more, as he seems to have introduced partitions. — Preceding unsigned comment added by 79.180.28.37 ( talk • contribs) 12:13, 13 January 2016 (UTC)
In
Using the same conjugation trick as above, one may show that the number pk(n) of partitions of n into exactly k parts is equal to the number of partitions of n in which the largest part has size k.[23] The function pk(n) satisfies the recurrence
pk(n) = pk(n − k) + pk − 1(n − 1)
with initial values p0(0) = 1 and pk(n) = 0 if n ≤ 0 or k ≤ 0.
shouldn't the second addend be p_(k-1)(n)? — Preceding unsigned comment added by 69.12.245.117 ( talk • contribs) 00:54, 21 January 2017 (UTC)
Shouldn't the formula for be " if or "? At least that is the formula that Abramowitz and Stegun p. 826, 24.2.2 eq. II(A) give. Kdpw ( talk) 22:51, 10 April 2017 (UTC)
@Joel B. Lewis, that is wrong. It is not the partition that is distinct, it is the parts that are. ImTheIP ( talk) 16:20, 28 October 2018 (UTC)
In this section the recurrence
has been the object of several edits and reverts. It is correct; proof: given a partition of either all parts are ≥ 2, and one gets a partition of by removing one to each part, or one gets a partition of by removing one part equal to 1.
However, another recurrence is also true:
Proof: Substract one to the lowest part.
As this recurrence is simpler (and the initial conditions are much simpler, as there is no need to consider negative integers), this recurrence must appear in this section, and deserve probably to be more emphasized than the one which is presently given. I have no reference under hand, but this is so simple that it must exists, if there is no mistake in my sketched proof. D.Lazard ( talk) 14:06, 15 September 2019 (UTC)
The Asymptotics section begins like this:
The asymptotic expression for p(n) implies that
where .
If A is a set of natural numbers, we let pA(n) denote the number of partitions of n into elements of A. If A possesses positive natural density α then
No, no, no, no, no. Do not discuss the logarithm of the asymptotic formula when the asymptotic formula has not even been stated yet.
First things first.
This ought to be 100% obvious to anyone writing for an encyclopedia. 2600:1700:E1C0:F340:51D:10F4:9B05:6C06 ( talk) 05:48, 20 September 2019 (UTC)
The function: y=(0.00005)x^9 - (0.00005)x^8 +(0.00005)x^7 -(0.00005)x^6+(0.01)x^5-(0.07)x^4+(0.36)x^3-(0.75)x^2+x+0.85 is a better fit for values of P(n) from 0 to 31. I apologize for my previous typing mistakes. enoraB otreboR 79.42.168.169 ( talk) 11:26, 25 April 2023 (UTC)