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 = 5+1+1 = 4+2+1 = 3+3+1 = 3+2+2. 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) = ?

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. ▨

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, ▨

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

Define p(n) ≔ number of unordered partition of n. Eg. 5 = 1+1+1+1+1 = 2+1+1+1 = 3+1+1 = 2+2+1 = 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: ▨

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).

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:

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.

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 = 7+6+5+3+2 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

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.

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 -1. ▨

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

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

▨

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

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

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 1+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.

To find b, consider ▨

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.

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!) ▨

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

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 ▨