Advanced Discrete Math Generating function Problem
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I am suppose to prove that the number of partitions of $n$ in which each part appears $2$, $3$, or $5$ times equals the number partitions of $n$ in to parts which are congruent to $2$, $3$, $6$, $9$, or $10$ modulo $12$.
I tried using Remmels Thm in order to prove this but I ran into a problem classifying the pairwise disjoint multisets. I now think the approach that should be used is find the generating function for each and show equality using algebra. I think the generating function for
$A$: partitions of $n$ in which each part appears $2$, $3$, or $5$ times
$$
G(A) = (1+x^2+x^4+...)(1+x^3+x^6+..)(1+x^5+x^10+...) = frac{1}{1-x^2}frac{1}{1-x^3}frac{1}{1-x^5}
$$
$B$: partitions of $n$ into parts which are congruent to $2$,$3$,$6$,$9$, or $10$ modulo $12$
$$
G(B) = frac{1}{(1-x^{12k-2})(1-x^{12k-3})(1-x^{12k-6})(1-x^{12k-9})(1-x^{12k-10})}
$$
Is this correct? Not sure how to manipulate the functions into each other
combinatorics discrete-mathematics generating-functions
edited 2 days ago
Joey Kilpatrick
67118
asked 2 days ago
fireshock
364
add a comment |
up vote
0
down vote
favorite
I am suppose to prove that the number of partitions of $n$ in which each part appears $2$, $3$, or $5$ times equals the number partitions of $n$ in to parts which are congruent to $2$, $3$, $6$, $9$, or $10$ modulo $12$.
I tried using Remmels Thm in order to prove this but I ran into a problem classifying the pairwise disjoint multisets. I now think the approach that should be used is find the generating function for each and show equality using algebra. I think the generating function for
$A$: partitions of $n$ in which each part appears $2$, $3$, or $5$ times
$$
G(A) = (1+x^2+x^4+...)(1+x^3+x^6+..)(1+x^5+x^10+...) = frac{1}{1-x^2}frac{1}{1-x^3}frac{1}{1-x^5}
$$
$B$: partitions of $n$ into parts which are congruent to $2$,$3$,$6$,$9$, or $10$ modulo $12$
$$
G(B) = frac{1}{(1-x^{12k-2})(1-x^{12k-3})(1-x^{12k-6})(1-x^{12k-9})(1-x^{12k-10})}
$$
Is this correct? Not sure how to manipulate the functions into each other
combinatorics discrete-mathematics generating-functions
edited 2 days ago
Joey Kilpatrick
67118
asked 2 days ago
fireshock
364
2
A: No. It should be $prod_{i geq 1} left(1 + x^{2i} + x^{3i} + x^{5i}right)$. What you computed is instead the number of partitions of $n$ in which each part equals $2$, $3$ or $5$.
– darij grinberg
2 days ago
1
And your second GF is $$prod_{k=1}^inftyfrac1{(1-x^{2k-2})(1-x^{2k-3})(1-x^{2k-6})(1-x^{2k-9})(1-x^{2k-10})}.$$
– Lord Shark the Unknown
2 days ago
Okay thanks! I have been working on it but I don't seem to get the algebra
– fireshock
2 days ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am suppose to prove that the number of partitions of $n$ in which each part appears $2$, $3$, or $5$ times equals the number partitions of $n$ in to parts which are congruent to $2$, $3$, $6$, $9$, or $10$ modulo $12$.
I tried using Remmels Thm in order to prove this but I ran into a problem classifying the pairwise disjoint multisets. I now think the approach that should be used is find the generating function for each and show equality using algebra. I think the generating function for
$A$: partitions of $n$ in which each part appears $2$, $3$, or $5$ times
$$
G(A) = (1+x^2+x^4+...)(1+x^3+x^6+..)(1+x^5+x^10+...) = frac{1}{1-x^2}frac{1}{1-x^3}frac{1}{1-x^5}
$$
$B$: partitions of $n$ into parts which are congruent to $2$,$3$,$6$,$9$, or $10$ modulo $12$
$$
G(B) = frac{1}{(1-x^{12k-2})(1-x^{12k-3})(1-x^{12k-6})(1-x^{12k-9})(1-x^{12k-10})}
$$
Is this correct? Not sure how to manipulate the functions into each other
combinatorics discrete-mathematics generating-functions
edited 2 days ago
Joey Kilpatrick
67118
asked 2 days ago
fireshock
364
I am suppose to prove that the number of partitions of $n$ in which each part appears $2$, $3$, or $5$ times equals the number partitions of $n$ in to parts which are congruent to $2$, $3$, $6$, $9$, or $10$ modulo $12$.
I tried using Remmels Thm in order to prove this but I ran into a problem classifying the pairwise disjoint multisets. I now think the approach that should be used is find the generating function for each and show equality using algebra. I think the generating function for
$A$: partitions of $n$ in which each part appears $2$, $3$, or $5$ times
$$
G(A) = (1+x^2+x^4+...)(1+x^3+x^6+..)(1+x^5+x^10+...) = frac{1}{1-x^2}frac{1}{1-x^3}frac{1}{1-x^5}
$$
$B$: partitions of $n$ into parts which are congruent to $2$,$3$,$6$,$9$, or $10$ modulo $12$
$$
G(B) = frac{1}{(1-x^{12k-2})(1-x^{12k-3})(1-x^{12k-6})(1-x^{12k-9})(1-x^{12k-10})}
$$
Is this correct? Not sure how to manipulate the functions into each other
combinatorics discrete-mathematics generating-functions
combinatorics discrete-mathematics generating-functions
edited 2 days ago
Joey Kilpatrick
67118
asked 2 days ago
fireshock
364
edited 2 days ago
Joey Kilpatrick
67118
asked 2 days ago
fireshock
364
edited 2 days ago
Joey Kilpatrick
67118
edited 2 days ago
Joey Kilpatrick
67118
edited 2 days ago
Joey Kilpatrick
67118
67118
asked 2 days ago
fireshock
364
asked 2 days ago
fireshock
364
asked 2 days ago
fireshock
364
364
2
A: No. It should be $prod_{i geq 1} left(1 + x^{2i} + x^{3i} + x^{5i}right)$. What you computed is instead the number of partitions of $n$ in which each part equals $2$, $3$ or $5$.
– darij grinberg
2 days ago
1
And your second GF is $$prod_{k=1}^inftyfrac1{(1-x^{2k-2})(1-x^{2k-3})(1-x^{2k-6})(1-x^{2k-9})(1-x^{2k-10})}.$$
– Lord Shark the Unknown
2 days ago
Okay thanks! I have been working on it but I don't seem to get the algebra
– fireshock
2 days ago
add a comment |
2
A: No. It should be $prod_{i geq 1} left(1 + x^{2i} + x^{3i} + x^{5i}right)$. What you computed is instead the number of partitions of $n$ in which each part equals $2$, $3$ or $5$.
– darij grinberg
2 days ago
1
And your second GF is $$prod_{k=1}^inftyfrac1{(1-x^{2k-2})(1-x^{2k-3})(1-x^{2k-6})(1-x^{2k-9})(1-x^{2k-10})}.$$
– Lord Shark the Unknown
2 days ago
Okay thanks! I have been working on it but I don't seem to get the algebra
– fireshock
2 days ago
2
2
A: No. It should be $prod_{i geq 1} left(1 + x^{2i} + x^{3i} + x^{5i}right)$. What you computed is instead the number of partitions of $n$ in which each part equals $2$, $3$ or $5$.
– darij grinberg
2 days ago
A: No. It should be $prod_{i geq 1} left(1 + x^{2i} + x^{3i} + x^{5i}right)$. What you computed is instead the number of partitions of $n$ in which each part equals $2$, $3$ or $5$.
– darij grinberg
2 days ago
1
1
And your second GF is $$prod_{k=1}^inftyfrac1{(1-x^{2k-2})(1-x^{2k-3})(1-x^{2k-6})(1-x^{2k-9})(1-x^{2k-10})}.$$
– Lord Shark the Unknown
2 days ago
And your second GF is $$prod_{k=1}^inftyfrac1{(1-x^{2k-2})(1-x^{2k-3})(1-x^{2k-6})(1-x^{2k-9})(1-x^{2k-10})}.$$
– Lord Shark the Unknown
2 days ago
Okay thanks! I have been working on it but I don't seem to get the algebra
– fireshock
2 days ago
Okay thanks! I have been working on it but I don't seem to get the algebra
– fireshock
2 days ago
add a comment |
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2
A: No. It should be $prod_{i geq 1} left(1 + x^{2i} + x^{3i} + x^{5i}right)$. What you computed is instead the number of partitions of $n$ in which each part equals $2$, $3$ or $5$.
– darij grinberg
2 days ago
1
And your second GF is $$prod_{k=1}^inftyfrac1{(1-x^{2k-2})(1-x^{2k-3})(1-x^{2k-6})(1-x^{2k-9})(1-x^{2k-10})}.$$
– Lord Shark the Unknown
2 days ago
Okay thanks! I have been working on it but I don't seem to get the algebra
– fireshock
2 days ago