Number Partitioning.

[Patrick D. Rockwell]

Here is a question regarding combinatorics. I'm cross posting it
to sci.math.probability because any combinatorics probablem speaks
to the issue of probability which I'm interested in.

How many ways are there to partition a number N into a number of
smaller nonzero numbers arranged in decending order of value. For
example,

N=6, the answer would seem to be 11.

Partitions are as follows.

1 1 1 1 1 1
2 1 1 1 1
2 2 1 1
2 2 2
3 1 1 1
3 2 1
3 3
4 1 1
4 2
5 1
6

Actually, it doesn't matter if the order is ascending or
descending, as long as you use the same order throughout
the list. Thanks in advance.

[David W. Cantrell]

See <http://en.wikipedia.org/wiki/Partition_%28number_theory%29>,
for example.

David

[se...@btinternet.com]

On 23 May, 20:17, David W. Cantrell <DWCantr...@sigmaxi.net> wrote:

> David- Hide quoted text -
>
> - Show quoted text -

http://www.btinternet.com/~se16/js/partitions.htm
will help work them out, and allow some constraints

[Frederick Williams]

se...@btinternet.com wrote:

>
> http://www.btinternet.com/~se16/js/partitions.htm
> will help work them out, and allow some constraints

What does "If it seems that a surprising number of these end in 5 or 0,
that is because it is true." mean?

--
Remove "antispam" and ".invalid" for e-mail address.

[se...@btinternet.com]

On 28 May, 13:43, Frederick Williams <"Frederick
Williams"@antispamhotmail.co.uk.invalid> wrote:

It means that more than a fifth of partition numbers are a multiple of
five.

Ramanujan proved that the number of partitions of 5k+4 is always a
multiple of five; in addition about a fifth of the rest also are, so
about 36% of partition numbers are a multiple of five. For example
3610 of the first 10000 partition numbers end with 0 or 5.