Changing $integer to 6 for example gives: 
6 
5 1 
4 2 
4 1 1 
3 3 1   # oops 
3 2 1
3 1 1 1 
2 2 2 1 # oops
2 2 1 1 
2 1 1 1 1   
1 1 1 1 1 1

[download]
You can check how many partitions of N exist for the first many values of N at this site. For N=10, for instance, there should be 77 and your script returns less than 50 (many of them even adding up to more than 10). Update: Other than an extra 1 on a few partitions here and there, it seems to get all of them though.

blokhead

In reply to Re^2: Generator of integer partitionts of n by blokhead
in thread Generator of integer partitionts of n by chiburashka

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