A while back I wrote that the number of ordered partitions ( called compositions in the Handbook of Discrete Mathematics ) of n of length k is C(n-1,k-1). It seems that the case where no 1's are used reduces that number to F(n-1) ( where F means the Fibonacci sequence ). I haven't been able to figure out a proof yet though. I just find it fascinating how often this number actually pops up.

Example.

The partitions of 6 are.

6

5-1

4-2

4-1-1

3-3

3-2-1

3-1-1-1

2-2-2

2-2-1-1

2-1-1-1-1

1-1-1-1-1-1

Or, a total of 11.

Ordered partitions with no 1s used are:

6

4-2

2-4

3-3

2-2-2

Or a total of 5, which is equal to, as expected F(5) = 5.

Partitions of length three are:

4-1-1

3-2-1

2-2-2

Or, a total of 3.

Ordered partitions of length three are:

4-1-1

1-4-1

1-1-4

3-2-1

3-1-2

2-1-3

2-3-1

1-2-3

1-3-2

2-2-2

Or a total of 10, which is equal to, as expected C(5,2) = 10.

13-2016 Open letter to Open Source for You (OSFY)

5 months ago

Don't know if you're still interested ... why the Fibonacci sequence?

ReplyDeleteWhat you call ordered partitions are often called compositions. They can be analysed recursively by ignoring say the last element of each set in the partition (since ordering is important this is well defined). Let c_n be the number of combinations of n. This consideration shows that

c_n = c_{n-1} + c_{n-2} + \dots + c_1 + c_0.

This shows recursively that c_n = 2^{n-1}.

(There are other proofs involving binary expressions, but I don't see them generalizing to your question).

If you want all compositions without a 1, this recursive relation becomes

d_n = d_{n-2} + d_{n-3} + \dots + d_1 + d_0.

(This gives you Fibonacci)

If you want all compositions without a 1 or a 2 you get

e_n = e_{n-3} + e_{n-4} + \dots + e_1 + e_0.

And you need to use "initial values", c_0=1, and d_0=0,d_1=0, d_2=1, and e_0=e_1=e_2=0, e_3=1.

Etc.

Hope that is s

Thank you very much for the reply. ( Just fyi, this website is MathJax ( = LaTeX ) enabled. ) If you mean "Let $c_n$ be the number of compositions of $n$." then I got it!

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