Partitions in Combinatorics
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Let $2leq kleq n$. Prove that $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$ where $p_k(n)$ is the number of partitions of $n$ into $k$ pieces. Here's my proof:
Proof: Let $2leq kleq n$. Let $p_k(n)$ be the number of partitions of $n$ into $k$ parts. We can divide the partitions into two classes. First, consider all partitions that contain a part of size 1. By deleting this part, we are left with a partition of $n-1$ into $k-1$ parts. Thus there are $p_{k-1}(n-1)$ partitions in this class. Next, consider all partitions in which every part has size 2. Then by deleting 1 from every part, we are left with a partition of $n-k$ into $k$ parts. Thus there are $p_k(n-k)$ partitions in this class. Therefore, $p_k(n)=p_{k-1}(n-1+p_k(n-k)$ with the initial conditions that $p_1(n)=1$ and $p_k(n)=0$ for $n<k$.
combinatorics integer-partitions
asked Nov 30 '18 at 22:39
TNTTNT
596
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add a comment |
$begingroup$
Let $2leq kleq n$. Prove that $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$ where $p_k(n)$ is the number of partitions of $n$ into $k$ pieces. Here's my proof:
Proof: Let $2leq kleq n$. Let $p_k(n)$ be the number of partitions of $n$ into $k$ parts. We can divide the partitions into two classes. First, consider all partitions that contain a part of size 1. By deleting this part, we are left with a partition of $n-1$ into $k-1$ parts. Thus there are $p_{k-1}(n-1)$ partitions in this class. Next, consider all partitions in which every part has size 2. Then by deleting 1 from every part, we are left with a partition of $n-k$ into $k$ parts. Thus there are $p_k(n-k)$ partitions in this class. Therefore, $p_k(n)=p_{k-1}(n-1+p_k(n-k)$ with the initial conditions that $p_1(n)=1$ and $p_k(n)=0$ for $n<k$.
combinatorics integer-partitions
asked Nov 30 '18 at 22:39
TNTTNT
596
$endgroup$
$begingroup$
"deleting 1 from every part" -> "subtracting $1$ from every part". The proof is correct but maybe worth formalizing more. You're setting up a bijection between the class-1 partitions of $n$ into $k$ parts and the class-1 partitions of $n-1$ into $k-1$ parts, and likewise for class 2. And it is worth saying what the inverse maps are (note you're using the fact that if a partition has a part $1$, then its last part must be $1$).
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– darij grinberg
Dec 2 '18 at 10:10
add a comment |
$begingroup$
Let $2leq kleq n$. Prove that $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$ where $p_k(n)$ is the number of partitions of $n$ into $k$ pieces. Here's my proof:
Proof: Let $2leq kleq n$. Let $p_k(n)$ be the number of partitions of $n$ into $k$ parts. We can divide the partitions into two classes. First, consider all partitions that contain a part of size 1. By deleting this part, we are left with a partition of $n-1$ into $k-1$ parts. Thus there are $p_{k-1}(n-1)$ partitions in this class. Next, consider all partitions in which every part has size 2. Then by deleting 1 from every part, we are left with a partition of $n-k$ into $k$ parts. Thus there are $p_k(n-k)$ partitions in this class. Therefore, $p_k(n)=p_{k-1}(n-1+p_k(n-k)$ with the initial conditions that $p_1(n)=1$ and $p_k(n)=0$ for $n<k$.
combinatorics integer-partitions
asked Nov 30 '18 at 22:39
TNTTNT
596
$endgroup$
Let $2leq kleq n$. Prove that $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$ where $p_k(n)$ is the number of partitions of $n$ into $k$ pieces. Here's my proof:
Proof: Let $2leq kleq n$. Let $p_k(n)$ be the number of partitions of $n$ into $k$ parts. We can divide the partitions into two classes. First, consider all partitions that contain a part of size 1. By deleting this part, we are left with a partition of $n-1$ into $k-1$ parts. Thus there are $p_{k-1}(n-1)$ partitions in this class. Next, consider all partitions in which every part has size 2. Then by deleting 1 from every part, we are left with a partition of $n-k$ into $k$ parts. Thus there are $p_k(n-k)$ partitions in this class. Therefore, $p_k(n)=p_{k-1}(n-1+p_k(n-k)$ with the initial conditions that $p_1(n)=1$ and $p_k(n)=0$ for $n<k$.
combinatorics integer-partitions
combinatorics integer-partitions
asked Nov 30 '18 at 22:39
TNTTNT
596
asked Nov 30 '18 at 22:39
TNTTNT
596
edited Nov 30 '18 at 23:50
TNT
asked Nov 30 '18 at 22:39
TNTTNT
596
asked Nov 30 '18 at 22:39
TNTTNT
596
asked Nov 30 '18 at 22:39
TNTTNT
596
596
$begingroup$
"deleting 1 from every part" -> "subtracting $1$ from every part". The proof is correct but maybe worth formalizing more. You're setting up a bijection between the class-1 partitions of $n$ into $k$ parts and the class-1 partitions of $n-1$ into $k-1$ parts, and likewise for class 2. And it is worth saying what the inverse maps are (note you're using the fact that if a partition has a part $1$, then its last part must be $1$).
$endgroup$
– darij grinberg
Dec 2 '18 at 10:10
add a comment |
$begingroup$
"deleting 1 from every part" -> "subtracting $1$ from every part". The proof is correct but maybe worth formalizing more. You're setting up a bijection between the class-1 partitions of $n$ into $k$ parts and the class-1 partitions of $n-1$ into $k-1$ parts, and likewise for class 2. And it is worth saying what the inverse maps are (note you're using the fact that if a partition has a part $1$, then its last part must be $1$).
$endgroup$
– darij grinberg
Dec 2 '18 at 10:10
$begingroup$
"deleting 1 from every part" -> "subtracting $1$ from every part". The proof is correct but maybe worth formalizing more. You're setting up a bijection between the class-1 partitions of $n$ into $k$ parts and the class-1 partitions of $n-1$ into $k-1$ parts, and likewise for class 2. And it is worth saying what the inverse maps are (note you're using the fact that if a partition has a part $1$, then its last part must be $1$).
$endgroup$
– darij grinberg
Dec 2 '18 at 10:10
$begingroup$
"deleting 1 from every part" -> "subtracting $1$ from every part". The proof is correct but maybe worth formalizing more. You're setting up a bijection between the class-1 partitions of $n$ into $k$ parts and the class-1 partitions of $n-1$ into $k-1$ parts, and likewise for class 2. And it is worth saying what the inverse maps are (note you're using the fact that if a partition has a part $1$, then its last part must be $1$).
$endgroup$
– darij grinberg
Dec 2 '18 at 10:10
add a comment |
1 Answer
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Hint
Let $P_k(n)$ be the set of partitions of $n$ with exactly $k$ parts i.e. $p_{k}(n)=|P_k(n)|$. Classify $lambda=(lambda_1,dots,lambda_k)in P_k(n)$ (where the $lambda_i$ are weakly decreasing and positive) based on whether $lambda_k=1$ or $lambda_k>1$.
answered Nov 30 '18 at 23:28
Foobaz JohnFoobaz John
21.6k41352
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$begingroup$
Hint
Let $P_k(n)$ be the set of partitions of $n$ with exactly $k$ parts i.e. $p_{k}(n)=|P_k(n)|$. Classify $lambda=(lambda_1,dots,lambda_k)in P_k(n)$ (where the $lambda_i$ are weakly decreasing and positive) based on whether $lambda_k=1$ or $lambda_k>1$.
answered Nov 30 '18 at 23:28
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
Hint
Let $P_k(n)$ be the set of partitions of $n$ with exactly $k$ parts i.e. $p_{k}(n)=|P_k(n)|$. Classify $lambda=(lambda_1,dots,lambda_k)in P_k(n)$ (where the $lambda_i$ are weakly decreasing and positive) based on whether $lambda_k=1$ or $lambda_k>1$.
answered Nov 30 '18 at 23:28
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
add a comment |
$begingroup$
Hint
Let $P_k(n)$ be the set of partitions of $n$ with exactly $k$ parts i.e. $p_{k}(n)=|P_k(n)|$. Classify $lambda=(lambda_1,dots,lambda_k)in P_k(n)$ (where the $lambda_i$ are weakly decreasing and positive) based on whether $lambda_k=1$ or $lambda_k>1$.
answered Nov 30 '18 at 23:28
Foobaz JohnFoobaz John
21.6k41352
$endgroup$
Hint
Let $P_k(n)$ be the set of partitions of $n$ with exactly $k$ parts i.e. $p_{k}(n)=|P_k(n)|$. Classify $lambda=(lambda_1,dots,lambda_k)in P_k(n)$ (where the $lambda_i$ are weakly decreasing and positive) based on whether $lambda_k=1$ or $lambda_k>1$.
answered Nov 30 '18 at 23:28
Foobaz JohnFoobaz John
21.6k41352
edited Nov 30 '18 at 23:48
answered Nov 30 '18 at 23:28
Foobaz JohnFoobaz John
21.6k41352
answered Nov 30 '18 at 23:28
Foobaz JohnFoobaz John
21.6k41352
answered Nov 30 '18 at 23:28
Foobaz JohnFoobaz John
21.6k41352
21.6k41352
add a comment |
add a comment |
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$begingroup$
"deleting 1 from every part" -> "subtracting $1$ from every part". The proof is correct but maybe worth formalizing more. You're setting up a bijection between the class-1 partitions of $n$ into $k$ parts and the class-1 partitions of $n-1$ into $k-1$ parts, and likewise for class 2. And it is worth saying what the inverse maps are (note you're using the fact that if a partition has a part $1$, then its last part must be $1$).
$endgroup$
– darij grinberg
Dec 2 '18 at 10:10