Proof number-partitions $P_n = sum_{k=1}^{n} P_{n,k}$ in positive summands












0














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?




  1. The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$


  2. The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$


  3. 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 $P_{n,k}$



enter image description here



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







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    0














    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?




    1. The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$


    2. The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$


    3. 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 $P_{n,k}$



    enter image description here



    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







    share|cite|improve this question

























      0












      0








      0







      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?




      1. The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$


      2. The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$


      3. 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 $P_{n,k}$



      enter image description here



      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







      share|cite|improve this question













      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?




      1. The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$


      2. The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$


      3. 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 $P_{n,k}$



      enter image description here



      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






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      asked Nov 29 '18 at 22:43









      Math DummyMath Dummy

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          1 Answer
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          We can use combinatorial arguments here.




          1. 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.

          2. 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, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.

          3. 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 $P_n - P_{n-1}$.






          share|cite|improve this answer





















          • 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











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          0














          We can use combinatorial arguments here.




          1. 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.

          2. 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, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.

          3. 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 $P_n - P_{n-1}$.






          share|cite|improve this answer





















          • 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
















          0














          We can use combinatorial arguments here.




          1. 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.

          2. 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, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.

          3. 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 $P_n - P_{n-1}$.






          share|cite|improve this answer





















          • 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














          0












          0








          0






          We can use combinatorial arguments here.




          1. 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.

          2. 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, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.

          3. 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 $P_n - P_{n-1}$.






          share|cite|improve this answer












          We can use combinatorial arguments here.




          1. 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.

          2. 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, $8 = 1 + 1 + 3 + 3 mapsto 1 + 3 = 4$.

          3. 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 $P_n - P_{n-1}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















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