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?
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The amount of number-partitions of an even number $n$ in positive summands (which are all even) is $P_{n/2}$
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The amount of number-partitions of an even number $n$ in positive summands, in which every summand has an even multiplicity, is $P_{n/2}$
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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.
Best Answer
We can use combinatorial arguments here.