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?

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 $\displaystyle p(2,n)$. But there is no easy way to define that function.

Re: Partition of a number with no 1s

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

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

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

That's why I don't understand why we should consider the function $\displaystyle p(2,n)$

Re: Partition of a number with no 1s

Quote:

Originally Posted by

**wilhelm** I'm sorry, I don't see that; $\displaystyle p(2,n)$ 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.