# Partition of a number with no 1s

• Nov 26th 2012, 09:36 AM
wilhelm
Partition of a number with no 1s
Hi.

I have problems solving this problem:

How many partitions of a number n are there, if none of the components of the partition equals 1?

What I've got so far is that for
n=1 we have 0,
n=2 and n=3 we have 1
n=4 and n=5 we have 2
n=6 and n=7 we have 3

but then for n=8 we get 7 and for n=9 we get 8.

Unless I'm making mistakes while counting. Anyway, I do not see any relations here.

Could someone help me?
• Nov 26th 2012, 10:36 AM
Plato
Re: Partition of a number with no 1s
Quote:

Originally Posted by wilhelm
How many partitions of a number n are there, if none of the components of the partition equals 1?
What I've got so far is that for
n=1 we have 0,
n=2 and n=3 we have 1
n=4 and n=5 we have 2
n=6 and n=7 we have 3
but then for n=8 we get 7 and for n=9 we get 8.
Unless I'm making mistakes while counting. Anyway, I do not see any relations here.

This is a rather deep question.
If you look at this webpage, you will see how deep.

The function you want is $p(2,n)$. But there is no easy way to define that function.
• Nov 26th 2012, 12:15 PM
wilhelm
Re: Partition of a number with no 1s
I'm sorry, I don't see that; $p(2,n)$ describes the number of n using 2 numbers, right? But in the problem above

$7=7+0=5+2=4+3=2+2+3$ but not $6+1,5+1+1,4+1+1+1,....$

$8=8+0=6+2=5+3=4+4=2+3+3=2+2+4=2+2+2+2$ but not for example $7+1, 6+1+1, 5+1+1+1, ...$

That's why I don't understand why we should consider the function $p(2,n)$
• Nov 26th 2012, 12:22 PM
Plato
Re: Partition of a number with no 1s
Quote:

Originally Posted by wilhelm
I'm sorry, I don't see that; $p(2,n)$ describes the number of n using 2 numbers, right? But in the problem above

Here is the quote:
For any given value of k, partitions counted by p(k, n) fit into exactly one of the following categories:
smallest addend is k
smallest addend is strictly greater than k.