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Let A={2,6,10,14,...} 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
5+1,4+1+1,3+3,3+1+1+1,1+1+1+1+1+1,and there are 5 of each type.

 Jul 9, 2020
 #1
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For n >= 2, the number of such partition of n is n - 1.  You can prove this using induction.

 Jul 9, 2020
 #2
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IS there a way to solve this problem with generating functions?

Guest Jul 9, 2020

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