1. Partitions
Define pk(n) as the number of solutions of n=x1+…+xk, x1 ≥ x2 ≥ …≥ xk ≥ 1, n is a positive integer. pk(n) = 0, for n < k.
Eg. 7 = = = =
So p3(7) = 4.
pk(n) is the number of solutions of n-k = y1+…+yk, with y1 ≥ … ≥ yk ≥ 0.
If s of the yi’s are positive, then there are ps(n-k) solutions (y1,…,yk).
Thus
pk(k) = 1, p1(n) = 1, p2(n) = ?

2. Eg. Prove p3(n) = {n2/12}.
Pf:
Let a3(n) denote the number of solutions of n = x1+x2+x3, x1 ≥ x2 ≥ x3 ≥ 0. Then a3(n) = p3(n+3).
Let y3=x3, y2=x2-x3, y1=x1-x2. Then a3(n) is the number of solutions of n = y1+2y2+3y3, yi ≥ 0. ▨

3. Thm 1. If k is fixed then pk(n)~nk-1/(k!(k-1)!) (n→∞).
Pf:
If n=x1+…+xk, x1 ≥…≥ xk ≥ 1, then the k! permutation of (x1,…,xk) yield solutions of positive integer solutions of n=x1+…+xk.
Thus k!pk(n) ≥ C(n-1, k-1).
If n=x1+…+xk, x1 ≥ x2 ≥ …≥ xk ≥ 1, if yi ≔ xi+(k-i), 1 ≤ i ≤ k, yi’s are distinct and y1+…+yk = n+k(k-1)/2. Therefore, ▨

4. The number of compositions of n into k parts is C(n-1, k-1)
The number of compositions of n into k parts is C(n-1, k-1). Thus the total number of compositions of n is
Let cnk denote the number of compositions of n into k parts. Define

5. Define p(n) ≔ number of unordered partition of n.
Eg. 5 = = = = = 4+1 = 3+2.
Thus p(5) = 7.
Clearly, p(n) is the number of solutions of n = x1+…+xn (x1 ≥ x2 ≥ …≥ xn ≥ 0)
It can be reformulated as the number of solutions of n = y1+2y2+…+nyn, where yi ≥ 0.
Thm 2. The generating function for the partition function is
Pf: ▨

6. Def: Ferrers diagram (or Young diagram)
The partition (5, 4, 2, 1) of 12 is represented by:
(4, 3, 2, 2, 1) is a conjugate of (5, 4, 2, 1).

7. Thm 3. The number of partitions of n into parts, the largest of which is k, is pk(n).
Pf: For each partition for which the largest part is k, the conjugate partition has k parts (and vice versa) ▨
Thm 4. The number of partitions of n into odd parts equals the number of partitions of n into unequal parts.
Pf:

8. Consider
In the expansion of this product; a partition of n into unequal parts contributes +1 to the coefficient of xn if the number of parts is even, and -1 if odd.
Let pe(n) : the number of partitions of n into an even number of unequal parts.
Similarly for po(n). Thus the coefficient of xn is pe(n) – po(n).
It was shown by Euler that pe(n) = po(n) unless n has the form
n = ω(m) ≔ (3m2-m)/2 or
n = ω(-m) ≔ (3m2+m)/2,
in which case pe(n) – po(n) = (-1)m.

9. base: the number of dots in the last row.
Thm 5.
Pf: Consider a Ferrers diagram of a partition of n into unequal parts. Eg. 23 =
base: the number of dots in the last row.
slope: the longest 45。 line segment joining the last point in the top row with other points of the diagram.
slope=3
base=2

10. Define two operations A and B:
A: If b ≤ s then remove the base and adjoin it to the diagram at the right to form a new slope parallel to the original one, unless b=s and the base and the slope share a common point.
B: If b > s, remove the slope and adjoin it at the bottom as a new base, except if b = s+1 and the base and the slope share a common point.

11. The exception for A occurs only if
n = b + (b+1) + … + (2b-1) = (3b2-b)/2 = ω(b).
The exception for B occurs if
n = (s+1) + (s+2) + … + 2s = (3s2+s)/2 = ω(-s).
In all the other cases pe(n) = po(n), since there is a 1-1 correspondence between partitions of n into an even number of unequal parts and partitions of n into an odd number of unequal parts.
In the exceptional cases the difference is +1 or ▨

12. Thm 6. Let p(n) = 0 for n < 0. Then for n ≥ 1,
Pf: By Thm 2 and Thm 5, ▨

13. Thm 7. For n > 2, we have
Pf: Let f(t) = log P(t).
By Thm 2,
Let 0 < t < 1.

14. ▨

15. Def: A Young tableau of shape (n1, n2,…,nm) is a Ferrers diagram of squares where 1,…,n, have been inserted in such a way that all rows and columns are increasing.
We want to determine the number of Young tableaux of a given shape. 1 3 4 7 11 2 5 10 12 6 9 8

16. Def: An involution is a permutation which is its own inverse.
Eg. There are 10 involutions of [4].
Def:
deg(△)=?

17. Lemma 9. Let g(x1,…,xm;y) ≔ x1△(x1+y,x2,…,xm) +
Lemma 9. Let g(x1,…,xm;y) ≔ x1△(x1+y,x2,…,xm) x2△(x1,x2+y,…,xm) + … + xm△(x1,x2,…,xm+y) Then g(x1,…,xm;y)=(x1+…+xm+C(m,2)y)△(x1,…,xm).
Pf:
G is a homogeneous polynomial of degree deg△(x1,…,xm) with the variables x1,…,xm,y.
If interchange xi and xj, then g changes sign.
If y=0, then the assertion is obvious, i.e., g=(x1+…+xm)△(x1,…,xm)
Thus g=h(x1,…,xm)△(x1,…,xm)
h nest be homogeneous in x1,…,xm, y of total degree 1, i.e., h=a(x1+…+xm)+by. (why?)
Clearly, a=1.

18. To find b, consider ▨

19. Given a shape (n1,…,nm), where n1+…+nm=n, denote the number of Young tableaux by f(n1,…,nm). It must satisfy the following relations:
f(n1,…,nm)=0 unless n1 ≥ n2 ≥… ≥ nm ≥0.
f(n1,…,nm,0)=f(n1,…,nm).
f(n1,…,nm) = f(n1-1,n2,…,nm) + f(n1,n2-1,…,nm) + … f(n1,…,nm-1), if n1 ≥ n2 ≥ … ≥ nm ≥ 1.
f(n)=1, if n ≥ 0.

20. Thm 10.
for n1+m-1 ≥ n2+m-2 ≥ … ≥ nm.
Pf:
If ni+m-i = ni+1+m-i-1 for some i, then the right hand side is 0, i.e. there is no such tableaux.
We need to show the right-hand side satisfies (1)-(4).
All but (3) are trivial.
For (3) substituting xi = ni+m-i and y=-1 in lemma 9.
(Homework!) ▨

21. The “hook-length“ is the number of cells in the hook.
Def: The “hook” in a Young tableaux corresponding to a cell in a tableaux is the union of the cell and all cells to the right in the same row and all cells below it in the same column.
The “hook-length“ is the number of cells in the hook. 12 11 9 7 5 2 1 8 6 4 3
miss 10, 8, 6, 4, 3

22. Thm 11. The number of Young tableaux of a given shape
Thm 11. The number of Young tableaux of a given shape and a total of n cells is n! divided by the product of all hook-lengths.
Pf:
For a tableaux of shape (n1,…,nm), then the hook-length of cell (1,1) is n1+m-1, and for cell (i,1) is ni+m-i.
In row 1 we find a hook-lengths the numbers from 1 to n1+m-1 with the exception of (n1+m-1)- (nj+m-j) for 2 ≤ j ≤ m.
Similarly, row i contains from 1 to ni+m-i except for (ni+m-i)-(nj+m-j), i+1 ≤ j ≤ m.
It follows that the product of the hook-lengths is ▨