Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By Plato
  • 1 Post By Plato

Math Help - Partition of a number with no 1s

  1. #1
    Junior Member
    Joined
    Nov 2012
    From
    Ukraine
    Posts
    43

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,607
    Thanks
    1574
    Awards
    1

    Re: Partition of a number with no 1s

    Quote Originally Posted by wilhelm View Post
    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.
    Thanks from wilhelm
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2012
    From
    Ukraine
    Posts
    43

    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)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,607
    Thanks
    1574
    Awards
    1

    Re: Partition of a number with no 1s

    Quote Originally Posted by wilhelm View Post
    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.
    Thanks from wilhelm
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partition of a set
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 25th 2012, 07:35 PM
  2. partition help
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: June 28th 2010, 03:35 AM
  3. Partition Construction
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2008, 09:57 AM
  4. Partition
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 1st 2008, 02:39 AM
  5. Partition help!!!
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: March 15th 2008, 08:57 PM

Search Tags


/mathhelpforum @mathhelpforum