Proof number-partitions $P_n = sum_{k=1}^{n} P_{n,k}$ in positive summands
Let $n in mathbb{N}$. Let $P_n$ be the number of number-partitions of $n$ in positive summands, i.e. $$P_n = sum_{k=1}^{n} P_{n,k}$$
How can one prove the following?
The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$
The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$
The amount of number-partitions of $n$ in positive summands, which are all at least $2$, is $P_n - P_{n-1}$
I know that $$P_{n,k} = P(n-k,k)+P(n-1, k-1)$$ and the following picture shows example values of $P_{n,k}$
For example $P_{8,4}=5$, because
$$2+2+2+2 = 8 \ 1+1+3+3 = 8 \ 1+2+2+3 = 8 \ 1+1+2+4 = 8 \ 1+1+1+5 = 8$$
And I also know that $P_{n,n} = 1$, because $1+...+1 = n$ is the only number partition. As well as $P_{n,n-1} = 1$, because $1+...+1+2 = n$ is the only number partition.
And for all natural numbers $n$ and $k$ with $1 < k < n$ we have $$P_{n,k} = binom{n-1}{k-1}$$
I tried to prove it, but even with the given information I don't really know how to do it.
combinatorics binomial-coefficients
asked Nov 29 '18 at 22:43
Math DummyMath Dummy
276
add a comment |
Let $n in mathbb{N}$. Let $P_n$ be the number of number-partitions of $n$ in positive summands, i.e. $$P_n = sum_{k=1}^{n} P_{n,k}$$
How can one prove the following?
The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$
The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$
The amount of number-partitions of $n$ in positive summands, which are all at least $2$, is $P_n - P_{n-1}$
I know that $$P_{n,k} = P(n-k,k)+P(n-1, k-1)$$ and the following picture shows example values of $P_{n,k}$
For example $P_{8,4}=5$, because
$$2+2+2+2 = 8 \ 1+1+3+3 = 8 \ 1+2+2+3 = 8 \ 1+1+2+4 = 8 \ 1+1+1+5 = 8$$
And I also know that $P_{n,n} = 1$, because $1+...+1 = n$ is the only number partition. As well as $P_{n,n-1} = 1$, because $1+...+1+2 = n$ is the only number partition.
And for all natural numbers $n$ and $k$ with $1 < k < n$ we have $$P_{n,k} = binom{n-1}{k-1}$$
I tried to prove it, but even with the given information I don't really know how to do it.
combinatorics binomial-coefficients
asked Nov 29 '18 at 22:43
Math DummyMath Dummy
276
add a comment |
Let $n in mathbb{N}$. Let $P_n$ be the number of number-partitions of $n$ in positive summands, i.e. $$P_n = sum_{k=1}^{n} P_{n,k}$$
How can one prove the following?
The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$
The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$
The amount of number-partitions of $n$ in positive summands, which are all at least $2$, is $P_n - P_{n-1}$
I know that $$P_{n,k} = P(n-k,k)+P(n-1, k-1)$$ and the following picture shows example values of $P_{n,k}$
For example $P_{8,4}=5$, because
$$2+2+2+2 = 8 \ 1+1+3+3 = 8 \ 1+2+2+3 = 8 \ 1+1+2+4 = 8 \ 1+1+1+5 = 8$$
And I also know that $P_{n,n} = 1$, because $1+...+1 = n$ is the only number partition. As well as $P_{n,n-1} = 1$, because $1+...+1+2 = n$ is the only number partition.
And for all natural numbers $n$ and $k$ with $1 < k < n$ we have $$P_{n,k} = binom{n-1}{k-1}$$
I tried to prove it, but even with the given information I don't really know how to do it.
combinatorics binomial-coefficients
asked Nov 29 '18 at 22:43
Math DummyMath Dummy
276
Let $n in mathbb{N}$. Let $P_n$ be the number of number-partitions of $n$ in positive summands, i.e. $$P_n = sum_{k=1}^{n} P_{n,k}$$
How can one prove the following?
The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$
The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$
The amount of number-partitions of $n$ in positive summands, which are all at least $2$, is $P_n - P_{n-1}$
I know that $$P_{n,k} = P(n-k,k)+P(n-1, k-1)$$ and the following picture shows example values of $P_{n,k}$
For example $P_{8,4}=5$, because
$$2+2+2+2 = 8 \ 1+1+3+3 = 8 \ 1+2+2+3 = 8 \ 1+1+2+4 = 8 \ 1+1+1+5 = 8$$
And I also know that $P_{n,n} = 1$, because $1+...+1 = n$ is the only number partition. As well as $P_{n,n-1} = 1$, because $1+...+1+2 = n$ is the only number partition.
And for all natural numbers $n$ and $k$ with $1 < k < n$ we have $$P_{n,k} = binom{n-1}{k-1}$$
I tried to prove it, but even with the given information I don't really know how to do it.
combinatorics binomial-coefficients
combinatorics binomial-coefficients
asked Nov 29 '18 at 22:43
Math DummyMath Dummy
276
asked Nov 29 '18 at 22:43
Math DummyMath Dummy
276
asked Nov 29 '18 at 22:43
Math DummyMath Dummy
276
asked Nov 29 '18 at 22:43
Math DummyMath Dummy
276
asked Nov 29 '18 at 22:43
Math DummyMath Dummy
276
276
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1 Answer
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We can use combinatorial arguments here.
- Given a partition of $2n$ into even summands, you can recover a partition of $n$ into any summands by just taking half of each summand. For example, $8 = 2 + 2 + 4 mapsto 1 + 1 + 2 = 4$. It shouldn't be too hard to show that this actually defines a bijection between the sets.
- Similarly, given a partition of $2n$ where each summand appears an even number of times, just take one copy of each to get a partition of $n$. For example, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.
- Consider an arbitrary partition of $n$. Either all of its summands are at least 2, or at least one of them is equal to 1. The total count over both of these cases is $P_n$. But the latter case can be counted by noticing that removing one of the summands equal to 1 defines a partition of $n-1$. Again, it should be clear that this is a bijection, so there are $P_{n-1}$ partitions of the latter form. This gives that the number of partitions of the first variety is simply $P_n - P_{n-1}$.
answered Nov 29 '18 at 22:53
plattyplatty
3,370320
Hi, thanks for your answer. It helped me a lot. Can you maybe evaluate on 1.) how one can formally show that it defines a bijection between the sets? Thank you.
– Math Dummy
Nov 29 '18 at 23:00
Sure; one way to do so is to take an arbitrary $n$, explicitly describe the function, and then show that it is injective (no two different partitions give you the same output) and surjective (every output in the codomain is attained by some input in the domain). These can also be done by proving that an inverse function exists; here, the function is to either double the size of each summand, or to copy each summand; showing that it is indeed a function should not be too hard.
– platty
Nov 29 '18 at 23:05
add a comment |
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We can use combinatorial arguments here.
- Given a partition of $2n$ into even summands, you can recover a partition of $n$ into any summands by just taking half of each summand. For example, $8 = 2 + 2 + 4 mapsto 1 + 1 + 2 = 4$. It shouldn't be too hard to show that this actually defines a bijection between the sets.
- Similarly, given a partition of $2n$ where each summand appears an even number of times, just take one copy of each to get a partition of $n$. For example, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.
- Consider an arbitrary partition of $n$. Either all of its summands are at least 2, or at least one of them is equal to 1. The total count over both of these cases is $P_n$. But the latter case can be counted by noticing that removing one of the summands equal to 1 defines a partition of $n-1$. Again, it should be clear that this is a bijection, so there are $P_{n-1}$ partitions of the latter form. This gives that the number of partitions of the first variety is simply $P_n - P_{n-1}$.
answered Nov 29 '18 at 22:53
plattyplatty
3,370320
Hi, thanks for your answer. It helped me a lot. Can you maybe evaluate on 1.) how one can formally show that it defines a bijection between the sets? Thank you.
– Math Dummy
Nov 29 '18 at 23:00
Sure; one way to do so is to take an arbitrary $n$, explicitly describe the function, and then show that it is injective (no two different partitions give you the same output) and surjective (every output in the codomain is attained by some input in the domain). These can also be done by proving that an inverse function exists; here, the function is to either double the size of each summand, or to copy each summand; showing that it is indeed a function should not be too hard.
– platty
Nov 29 '18 at 23:05
add a comment |
We can use combinatorial arguments here.
- Given a partition of $2n$ into even summands, you can recover a partition of $n$ into any summands by just taking half of each summand. For example, $8 = 2 + 2 + 4 mapsto 1 + 1 + 2 = 4$. It shouldn't be too hard to show that this actually defines a bijection between the sets.
- Similarly, given a partition of $2n$ where each summand appears an even number of times, just take one copy of each to get a partition of $n$. For example, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.
- Consider an arbitrary partition of $n$. Either all of its summands are at least 2, or at least one of them is equal to 1. The total count over both of these cases is $P_n$. But the latter case can be counted by noticing that removing one of the summands equal to 1 defines a partition of $n-1$. Again, it should be clear that this is a bijection, so there are $P_{n-1}$ partitions of the latter form. This gives that the number of partitions of the first variety is simply $P_n - P_{n-1}$.
answered Nov 29 '18 at 22:53
plattyplatty
3,370320
Hi, thanks for your answer. It helped me a lot. Can you maybe evaluate on 1.) how one can formally show that it defines a bijection between the sets? Thank you.
– Math Dummy
Nov 29 '18 at 23:00
Sure; one way to do so is to take an arbitrary $n$, explicitly describe the function, and then show that it is injective (no two different partitions give you the same output) and surjective (every output in the codomain is attained by some input in the domain). These can also be done by proving that an inverse function exists; here, the function is to either double the size of each summand, or to copy each summand; showing that it is indeed a function should not be too hard.
– platty
Nov 29 '18 at 23:05
add a comment |
We can use combinatorial arguments here.
- Given a partition of $2n$ into even summands, you can recover a partition of $n$ into any summands by just taking half of each summand. For example, $8 = 2 + 2 + 4 mapsto 1 + 1 + 2 = 4$. It shouldn't be too hard to show that this actually defines a bijection between the sets.
- Similarly, given a partition of $2n$ where each summand appears an even number of times, just take one copy of each to get a partition of $n$. For example, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.
- Consider an arbitrary partition of $n$. Either all of its summands are at least 2, or at least one of them is equal to 1. The total count over both of these cases is $P_n$. But the latter case can be counted by noticing that removing one of the summands equal to 1 defines a partition of $n-1$. Again, it should be clear that this is a bijection, so there are $P_{n-1}$ partitions of the latter form. This gives that the number of partitions of the first variety is simply $P_n - P_{n-1}$.
answered Nov 29 '18 at 22:53
plattyplatty
3,370320
We can use combinatorial arguments here.
- Given a partition of $2n$ into even summands, you can recover a partition of $n$ into any summands by just taking half of each summand. For example, $8 = 2 + 2 + 4 mapsto 1 + 1 + 2 = 4$. It shouldn't be too hard to show that this actually defines a bijection between the sets.
- Similarly, given a partition of $2n$ where each summand appears an even number of times, just take one copy of each to get a partition of $n$. For example, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.
- Consider an arbitrary partition of $n$. Either all of its summands are at least 2, or at least one of them is equal to 1. The total count over both of these cases is $P_n$. But the latter case can be counted by noticing that removing one of the summands equal to 1 defines a partition of $n-1$. Again, it should be clear that this is a bijection, so there are $P_{n-1}$ partitions of the latter form. This gives that the number of partitions of the first variety is simply $P_n - P_{n-1}$.
answered Nov 29 '18 at 22:53
plattyplatty
3,370320
answered Nov 29 '18 at 22:53
plattyplatty
3,370320
answered Nov 29 '18 at 22:53
plattyplatty
3,370320
answered Nov 29 '18 at 22:53
plattyplatty
3,370320
3,370320
Hi, thanks for your answer. It helped me a lot. Can you maybe evaluate on 1.) how one can formally show that it defines a bijection between the sets? Thank you.
– Math Dummy
Nov 29 '18 at 23:00
Sure; one way to do so is to take an arbitrary $n$, explicitly describe the function, and then show that it is injective (no two different partitions give you the same output) and surjective (every output in the codomain is attained by some input in the domain). These can also be done by proving that an inverse function exists; here, the function is to either double the size of each summand, or to copy each summand; showing that it is indeed a function should not be too hard.
– platty
Nov 29 '18 at 23:05
add a comment |
Hi, thanks for your answer. It helped me a lot. Can you maybe evaluate on 1.) how one can formally show that it defines a bijection between the sets? Thank you.
– Math Dummy
Nov 29 '18 at 23:00
Sure; one way to do so is to take an arbitrary $n$, explicitly describe the function, and then show that it is injective (no two different partitions give you the same output) and surjective (every output in the codomain is attained by some input in the domain). These can also be done by proving that an inverse function exists; here, the function is to either double the size of each summand, or to copy each summand; showing that it is indeed a function should not be too hard.
– platty
Nov 29 '18 at 23:05
Hi, thanks for your answer. It helped me a lot. Can you maybe evaluate on 1.) how one can formally show that it defines a bijection between the sets? Thank you.
– Math Dummy
Nov 29 '18 at 23:00
Hi, thanks for your answer. It helped me a lot. Can you maybe evaluate on 1.) how one can formally show that it defines a bijection between the sets? Thank you.
– Math Dummy
Nov 29 '18 at 23:00
Sure; one way to do so is to take an arbitrary $n$, explicitly describe the function, and then show that it is injective (no two different partitions give you the same output) and surjective (every output in the codomain is attained by some input in the domain). These can also be done by proving that an inverse function exists; here, the function is to either double the size of each summand, or to copy each summand; showing that it is indeed a function should not be too hard.
– platty
Nov 29 '18 at 23:05
Sure; one way to do so is to take an arbitrary $n$, explicitly describe the function, and then show that it is injective (no two different partitions give you the same output) and surjective (every output in the codomain is attained by some input in the domain). These can also be done by proving that an inverse function exists; here, the function is to either double the size of each summand, or to copy each summand; showing that it is indeed a function should not be too hard.
– platty
Nov 29 '18 at 23:05
add a comment |
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