1. Cool Results Involving Compositions
James Sellers
Director, Undergraduate Mathematics
Penn State University

2. Goals for the Talk
Share some important definitions and examples related to compositions
Discover (and prove) some cool results involving compositions
Attack a recent conjecture related to a function which counts certain compositions

3. Basic Definitions
A partition of an integer n is a non-increasing sequence of positive integers which sum to n.
Each integer used in a partition is known as a part.
A composition of an integer n is a partition of n where the order of the parts matters.

4. Example
The compositions of n = 3 are: 3
2+1
1+2
1+1+1

5. Notation Let c(n) be the number of compositions of n. c(3) = 4

6. First Cool Result Theorem: For all positive integers n,
c(n) = ??
Let’s discover the formula together by performing some calculations!

7. First Cool Result Theorem: For all positive integers n,
c(n) = 2n-1.
Proof: This can be proven in a number of ways. One of the easiest is to note that c(1) = 1 and that c(n) = 2c(n-1) (via a clean recurrence). Then just iterate.
The recurrence from n-1 to n goes something like this. Take all the compositions counted by c(n-1). First, append ‘+1’ to each of the compositions. This gives you compositions counted by c(n). Then, go back to the compositions of n-1 and actually add 1 to each of the right-most parts. These *must* be different from the other ones you built, and in fact are all of them! Thus, you have c(n) = c(n-1) + c(n-1) = 2c(n-1).

8. Quick Sidenote
No such simple formula exists for p(n), the function that counts the partitions of n. But once one allows the different orderings of the parts to be counted, so that one is determining c(n), this nice formula arises.

9. Second Cool Result
Let c’(n) be the number of compositions of n where 1’s are not allowed as parts.
Theorem: For all positive integers n,
c’(n) = F(n-1)
where F(n) is the nth Fibonacci number!

10. Second Cool Result Proof: We have c’(n) = c’(n-1) + c’(n-2)
and c’(2) = 1.
The result follows.
Add 1 to the right-most part of those counted by c’(n-1) and append a ‘+2’ to the end of each composition counted by c’(n-2).

11. Third Cool Result
Define c(m,n) as the number of compositions of n into exactly m parts.

12. Third Cool Result
Theorem: For positive integers n and m,

13. Fourth Cool Result
The number of compositions of n into exactly m parts with exactly j designated parts is given by

14. Fourth Cool Result
In particular, the number of compositions of 2m into exactly m parts with exactly m - j designated parts >1 is given by

15. Fourth Cool Result
This product of binomial coefficients was the focus of a recent article by G.E. Andrews and S.T. Soh. In particular, they proved that, for positive values of j,
is odd if and only if the value j is the largest power of 2 dividing m.

16. An Interesting Conjecture
In the same paper, Andrews and Soh also conjectured that, for positive values of j, the only time the values of
are not unique is in the cases m=5, j=3 and m=5, j=4.

17. Quick Digression
Believe it or not, I can’t end this talk without talking about Pythagorean triples!
We will say (x,y,z) is a Pythagorean triple if
x2 + y2 = z2.

18. Quick Digression
For our purposes, we will focus attention on the triples of the form (x, x+1, z). Such triples (where the legs are consecutive integers) were studied as far back as the 17th century by Pierre Fermat ( ).

19. Quick Digression
It is now well-known that there are infinitely many such “consecutive leg” triples. The first few are:
(3, 4, 5)
(20, 21, 29)
(119, 120, 169)

20. Quick Digression
These values satisfy nice linear recurrences, so they can be computed rather efficiently.

21. Andrews-Soh Conjecture (Again)
Recall that Andrews and Soh conjectured that, for positive values of j, the only time the values of
are not unique is in the cases m=5, j=3 and m=5, j=4.

22. Andrews-Soh Conjecture (Again)
We can prove that this is not the case and use the “consecutive leg” Pythagorean triples to do it!
Allow me do this on the chalkboard.

23. Closing Thoughts
It always pays to stay active, to read mathematical works often.
I hope I have encouraged some of you to consider studying compositions more closely.