Download INTEGER PARTITIONS: EXERCISE SHEET 6 (APRIL 27 AND MAY 4)

INTEGER PARTITIONS: EXERCISE SHEET 6 (APRIL
27 AND MAY 4)
Exercise 1: Reverse bijection for Schur’s theorem
Question 1. Show that the transformation from P1 to P4 seen during
the lecture is equivalent to the following process:
As long as there exists some number that is not at least 3 greater
than the number below, subtract 3 from this number, add 3 to the
number below, and exchange these two numbers.
Example:
11
18 + 3 = 21
21
21
18
11 − 3 = 8
8
6+3=9
P1 =
7→
7→
7→
= P4 .
5
5
3+3=6
8−3=5
3
3
5−3=2
2
Question 2. Show that the following process is the reverse bijection
of the above:
Start by splitting parts of P4 that are multiple of 3 into pairs of parts
differing by at most 2.
Example :
21
11 + 10
9
5+4
P4 =
7→
= P40 .
5
5
2
2
0
We obtain a partition P4 with no multiples of 3. Now as long as the
smallest part of some pair is less than 3 greater than the part below,
subtract 3 from the largest part of the pair, add 3 to the part below,
and switch their positions. This process ends with a partition into
parts that are not multiples of 3, where parts differing by at most two
are paired up, starting from the smallest part.
Example :
11 + 10
11 + 10
11
11 + 10
5
+
4
8
8
10
+8
P40 =
7→
7→
7→
= P.
5
4+2
5
5
2
2
2+1
2+1
INTEGER PARTITIONS: EXERCISE SHEET 6 (APRIL 27 AND MAY 4)
Exercise 2: Refinement of Schur’s theorem
The goal of this exercise is to prove the following refinement of
Schur’s theorem due to Gleissberg.
Theorem 1 (Gleissberg). Let C(m, n) denote the number of partitions
of n into m distinct parts congruent to 1 or 2 modulo 3. Let D(m, n)
denote the number of partitions of n into m parts (where parts divisible
by 3 are counted twice), where parts differ by at least 3 and no two
consecutive multiples of three appear. Then for all m, n ≥ 0, C(m, n) =
D(m, n).
Question 1. Let π` (m, n) denote the number of partitions counted by
D(m, n) such that the largest part does not exceed `. Prove that for
all `, m, n positive integers,
π3`+1 (m, n) = π3` (m, n) + π3`−2 (m − 1, n − 3` − 1),
π3`+2 (m, n) = π3`+1 (m, n) + π3`−1 (m − 1, n − 3` − 2),
π3`+3 (m, n) = π3`+2 (m, n) + π3`−1 (m − 2, n − 3` − 3).
Question 2. Define, for |q| < 1, |t| < 1,
∞ X
∞
X
π` (m, n)tm q n .
a` (t, q) = 1 +
m=1 n=1
What is lim`→∞ a` (t, q)?
Question 3. Prove that
a3`−1 (tq 3 , q) = (1+tq 3`+1 +tq 3`+2 )a3`−4 (tq 3 , q)+t2 q 3`+3 (1−q 3`−3 )a3`−7 (tq 3 , q).
Question 4. Show that
a3`+3 (t, q) = (1 + tq 3`+1 + tq 3`+2 )a3` (t, q) + t2 q 3`+3 (1 − q 3`−3 )a3`−3 (t, q).
Question 5. What are the initial values a−1 (tq 3 , q), a2 (tq 3 , q), a3 (t, q), a6 (t, q)?
Verify that
(1)
a3 (t, q) = (1 + tq)(1 + tq 2 )a−1 (tq 3 , q),
and
(2)
a6 (t, q) = (1 + tq)(1 + tq 2 )a2 (tq 3 , q).
Question 6. Deduce that for all ` ≥ 0 :
a3`+3 (t, q) = (1 + tq)(1 + tq 2 )a3`−1 (tq 3 , q).
Question 7. Conclude by finding lim`→∞ a` (t, q).