+198 Let \(A=\{2,6,10,14,\ldots\}\) be the set of integers that are twice an odd number.
Prove that, for every positive integer n, the number of partitions of n in which no odd number appears more than once is equal to the number of partitions of n containing no element of A.
For example, for n=6, the partitions of the first type are
\(6,~~~5+1,~~~4+2,~~~3+2+1,~~~2+2+2,\)
and the partitions of the second type are
\(6,~~~5+1,~~~4+2,~~~3+2+1,~~~2+2+2,\)
and there are 5 of each type.
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I believe that I could show the generating functions of the two are the same, but I do not know where to start. So I'm thinking to first set some variables and functions, so I decided to let \(\operatorname{P}_0(n)\) be the set of partitions of n in which no odd number appears more than once, and let \(\operatorname{P}_1(n)\) be the set of partitions of n containing no element of A. How do I move out from there? If there is anyone who would guide me to the correct proof, or give a proof, their help would be appreciated. Thanks!
Try to find a bijection between the two sets- You are given a partition where no odd number appears more than once, try to find a way to turn that partition (maybe by splitting up some of the members of the partition, or by adding up some of them) into a partition that contains no element from the set A. Try to find a way to do that for ever partition, such that every partition that contains no element from the set A has EXACTLY one partition where no odd number appears more than once, that when applying your method/algorithm/function to will produce that partition.
Maybe melody/Cphill can explain that better, they are very smart and surely can solve your question
+198 Okay. I have found more on the problem. So if β∈P_0, let \(\overline{\beta}\) be the partition of n obtained by splitting each member of β that is in A into two halves and leaving all other numbers of the partition where they are. This clearly appears to map \(P_0(n)\) into \(P_1(n)\). However, I am having trouble showing that the map is actually a bijection. It seems clearly injective, but I just can't solve it from what I have now. Can I have another hint? Melody? CPhill??
