Number of partitions of 2N into N parts

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SUMMARY

The number of partitions of an even integer 2N into N parts is definitively equal to the number of partitions of N. This conclusion is supported by a bijective proof utilizing Ferrer's diagrams, which demonstrates that the number of partitions of 2N with the largest part N corresponds directly to the partitions of N. Additionally, a more general principle is established: for positive integers m and j, where m > j >= m/2, the number of partitions of m into j parts equals the number of partitions of m-j.

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FaustoMorales
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The number of partitions of an even number 2N into N parts appears to be equal to the number of partitions of N.

Is this known? If so: Can anyone provide a reference of the corresponding proof? Thanks in advance for any information on this.
 
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The number of partitions of any integer m into j parts equals the number of partitions of m with largest part j (easy bijective proof using Ferrer's diagrams). Thus the number of partitions of 2n into n parts equals the number of partitions of 2n with largest part n. The bijection between partitions of 2n with largest part n and partitions of n is straightforward.
 
That takes care of it. Thanks!
 
JCVD said:
The number of partitions of any integer m into j parts equals the number of partitions of m with largest part j (easy bijective proof using Ferrer's diagrams). Thus the number of partitions of 2n into n parts equals the number of partitions of 2n with largest part n. The bijection between partitions of 2n with largest part n and partitions of n is straightforward.

This clever idea can be used to prove something more general: Given positive integers m and j, with m > j >= m/2, the number of partitions of m into j parts is the number of partitions of m-j.
 

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