Re: Question about integer partitions
- From: "The Qurqirish Dragon" <qurqirishd@xxxxxxx>
- Date: 30 Apr 2005 05:25:15 -0700
The number should be easy to find. Try this:
Let P(n) be the number of partitions of n, for n>=1, and let P(n)=0 for
n<=0 (this last bit is just to make the formula I will provide later
simpler)
Thus, there are P(n-k) partitions with at least one partition of k
there are (n-2k) paritions with at least two partitional of k
and so on.
Thus, the total patitions of size k in the list of paritions of n are:
P(n-k) + P(n-2k) + P(n-3k) + . . .
Since P(n-jk)=0 if j>= n/k, this sum will terminate.
Any partition of n with one k will be counted exactly once, those with
two ks will be counted twice, and so on, so the double counting in the
formula seems to be exactly what you wanted to have.
.
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