Balanced Partition and Triangular Number
Shishuo Fu
Department of Mathematics
College of Mathematics and Statistics
Chongqing University
[email protected]
The Combinatorics of q-Series and Partitions
In Honor of Professor George Andrews’ 75th Birthday
Aug 4, 2013
Introduction
Balanced partition
The General Case
Outline
Introduction
Euler’s Theorem
Characteristic of a Partition
Balanced partition
The General Case
Further Thoughts
Further Thoughts
Introduction
Balanced partition
The General Case
Outline
Introduction
Euler’s Theorem
Characteristic of a Partition
Balanced partition
The General Case
Further Thoughts
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
D(n) := {partitions of n into distinct parts}
O(n) := {partitions of n into purely odd parts}
(Euler ) : |D(n)| = |O(n)|
"Distinct" versus "Even"?
TE(n) := {partitions of n into one triangular part and even parts}
(S. Vandervelde) : |D(n)| = |TE(n)|
∞
Y
(1−q 2m )(1+q m ) = T (q), where T (q) := 1+q+q 3 +q 6 +· · ·
m=1
Sam Vandervelde Balanced Partitions, THE RAMANUJAN
JOURNAL, 2010.
Introduction
Balanced partition
The General Case
Further Thoughts
D(n) := {partitions of n into distinct parts}
O(n) := {partitions of n into purely odd parts}
(Euler ) : |D(n)| = |O(n)|
"Distinct" versus "Even"?
TE(n) := {partitions of n into one triangular part and even parts}
(S. Vandervelde) : |D(n)| = |TE(n)|
∞
Y
(1−q 2m )(1+q m ) = T (q), where T (q) := 1+q+q 3 +q 6 +· · ·
m=1
• Jacobi triple product
Sam Vandervelde Balanced Partitions, THE RAMANUJAN
JOURNAL, 2010.
Introduction
Balanced partition
The General Case
Further Thoughts
D(n) := {partitions of n into distinct parts}
O(n) := {partitions of n into purely odd parts}
(Euler ) : |D(n)| = |O(n)|
"Distinct" versus "Even"?
TE(n) := {partitions of n into one triangular part and even parts}
(S. Vandervelde) : |D(n)| = |TE(n)|
∞
Y
(1−q 2m )(1+q m ) = T (q), where T (q) := 1+q+q 3 +q 6 +· · ·
m=1
• Jacobi triple product
• Gauss’ Triangular Number Theorem
Sam Vandervelde Balanced Partitions, THE RAMANUJAN
JOURNAL, 2010.
Introduction
Balanced partition
The General Case
Further Thoughts
Corollary
pd (n) =
∞
X
k =0
distinct
8
7-1
6-2
5-3
5-2-1
4-3-1
χ(π)
0
0
0
0
-2
0
1
1
p( (n − Tk )), where Tk = k (k + 1)
2
2
4+ evens
6, 2
0, 8
0, 6-2
0, 4-4
0, 4-2-2
0, 2-2-2-2
distinct
9
8-1
7-2
6-3
6-2-1
5-4
5-3-1
4-3-2
A refinement basing on the triangular parts?
χ(π)
-1
1
-1
1
-1
-1
-1
1
4+ evens
3, 6
3, 4-2
3, 2-2-2
1, 8
1, 6-2
1, 4-4
1, 4-2-2
1, 2-2-2-2
Introduction
Balanced partition
The General Case
Further Thoughts
Two Definitions
• Definition 1:
χ(π) := # odd parts in even position − # odd parts in odd
position. When χ(π) = 0, π is called balanced.
Introduction
Balanced partition
The General Case
Further Thoughts
Two Definitions
• Definition 1:
χ(π) := # odd parts in even position − # odd parts in odd
position. When χ(π) = 0, π is called balanced.
• Definition 2:
χ(π) :=
Pn
k
k =1 (−1) dk
d1 d2 d3 d4 d5 d6
1
2
3
4
3
1
Introduction
Balanced partition
The General Case
Further Thoughts
Two Definitions
• Definition 1:
χ(π) := # odd parts in even position − # odd parts in odd
position. When χ(π) = 0, π is called balanced.
• Definition 2:
χ(π) :=
Pn
k
k =1 (−1) dk
d1 d2 d3 d4 d5 d6
1
2
3
4
3
1
Introduction
Balanced partition
The General Case
Sequence of diagonal lengths
A sequence d1 , d2 , . . . , dn of nonnegative integers are the
diagonal lengths for some partition of n if and only if the
following three conditions are met:
1. d1 + d2 + · · · + dn = n,
Further Thoughts
Introduction
Balanced partition
The General Case
Sequence of diagonal lengths
A sequence d1 , d2 , . . . , dn of nonnegative integers are the
diagonal lengths for some partition of n if and only if the
following three conditions are met:
1. d1 + d2 + · · · + dn = n,
2. d1 = 1, d2 = 2, . . . , dm = m for some 1 ≤ m ≤ n,
Further Thoughts
Introduction
Balanced partition
The General Case
Sequence of diagonal lengths
A sequence d1 , d2 , . . . , dn of nonnegative integers are the
diagonal lengths for some partition of n if and only if the
following three conditions are met:
1. d1 + d2 + · · · + dn = n,
2. d1 = 1, d2 = 2, . . . , dm = m for some 1 ≤ m ≤ n,
3. dm ≥ dm+1 ≥ · · · ≥ dn .
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
Sequence of diagonal lengths
A sequence d1 , d2 , . . . , dn of nonnegative integers are the
diagonal lengths for some partition of n if and only if the
following three conditions are met:
1. d1 + d2 + · · · + dn = n,
2. d1 = 1, d2 = 2, . . . , dm = m for some 1 ≤ m ≤ n,
3. dm ≥ dm+1 ≥ · · · ≥ dn .
Rmk: In general this is not 1-1 correspondence, but for partition
into distinct parts, it is.
Introduction
Balanced partition
The General Case
Outline
Introduction
Euler’s Theorem
Characteristic of a Partition
Balanced partition
The General Case
Further Thoughts
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
Theorem (Conjectured by S. Vandervelde)
With n being a fixed nonnegative integer and k any integer, let
Dk (n) := {partitions of n into distinct parts having characteristic
1
k }. Then |Dk (n)| = p( (n − T2k )).
2
Introduction
Balanced partition
The General Case
Further Thoughts
Theorem (Conjectured by S. Vandervelde)
With n being a fixed nonnegative integer and k any integer, let
Dk (n) := {partitions of n into distinct parts having characteristic
1
k }. Then |Dk (n)| = p( (n − T2k )).
2
Corollary (S. Vandervelde)
The number of balanced partitions of 2n into distinct parts is
equal to the number of partitions of n.
Introduction
Balanced partition
The General Case
Further Thoughts
Theorem (Conjectured by S. Vandervelde)
With n being a fixed nonnegative integer and k any integer, let
Dk (n) := {partitions of n into distinct parts having characteristic
1
k }. Then |Dk (n)| = p( (n − T2k )).
2
Corollary (S. Vandervelde)
The number of balanced partitions of 2n into distinct parts is
equal to the number of partitions of n.
Proof
Construct a bijective map between the sequences of diagonal
lengths of 2n, and partitions of n. Forward map will pair light
and dark squares in the process of creating a partition of n,
while we use "configuration board" to get back to the sequence
of diagonal lengths.
Introduction
Balanced partition
The General Case
Configuration Board
..
.
8
6
4
2
1
3
5
7
9
···
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
Configuration Board
..
.
8
6
4
2
1
3
5
7
9
···
• First m (or m − 1) diagonals will lay out the Durfee square
for the partition of n.
Introduction
Balanced partition
The General Case
Further Thoughts
Configuration Board
..
.
8
6
4
2
1
3
5
7
9
···
• First m (or m − 1) diagonals will lay out the Durfee square
for the partition of n.
• For some 2k > m, we need
0 ≤ (d2k + d2k −2 + · · · + d2 ) − (d2k −1 + d2k −3 + · · · + d1 ) < k .
Introduction
Balanced partition
The General Case
Further Thoughts
Configuration Board
..
.
8
6
4
2
1
3
5
7
9
···
• First m (or m − 1) diagonals will lay out the Durfee square
for the partition of n.
• For some 2k > m, we need
0 ≤ (d2k + d2k −2 + · · · + d2 ) − (d2k −1 + d2k −3 + · · · + d1 ) < k .
χ(π) = 0 : (d2n + · · · + d2k +2 ) − (d2n−1 + · · · + d2k +1 ) ≤ 0.
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
Further Thoughts
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
Further Thoughts
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
Further Thoughts
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
Further Thoughts
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
Further Thoughts
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
Further Thoughts
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
Further Thoughts
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
• Backward map: dj = bj + bj−1 for j > 1, and d1 = b1 .
(6, 4, 3, 2, 2, 1) ⇒ (1, 2, 3, 4, 5, 5, 4, 4, 3, 2, 2, 1).
Further Thoughts
Introduction
Balanced partition
The General Case
• Example: (6, 4, 3, 1) ∼ (1, 2, 3, 4, 3, 1) ⇒ (3, 2, 2).
• Backward map: dj = bj + bj−1 for j > 1, and d1 = b1 .
10
8
6
4
2
1 3 5 7 9 11
(6, 4, 3, 2, 2, 1) ⇒ (1, 2, 3, 4, 5, 5, 4, 4, 3, 2, 2, 1).
Further Thoughts
Introduction
Balanced partition
The General Case
Outline
Introduction
Euler’s Theorem
Characteristic of a Partition
Balanced partition
The General Case
Further Thoughts
Further Thoughts
Introduction
Balanced partition
The General Case
Concerns
• Since k 6= 0, the "dark" and "light" won’t pair off.
Further Thoughts
Introduction
Balanced partition
The General Case
Concerns
• Since k 6= 0, the "dark" and "light" won’t pair off.
• We need to creat a triangular part as well.
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
Concerns
• Since k 6= 0, the "dark" and "light" won’t pair off.
• We need to creat a triangular part as well.
• We probably should peel off some diagonals to make the
triangular part and at the same time leaving the remaining
partition balanced.
Introduction
Balanced partition
The General Case
Further Thoughts
Concerns
• Since k 6= 0, the "dark" and "light" won’t pair off.
• We need to creat a triangular part as well.
• We probably should peel off some diagonals to make the
triangular part and at the same time leaving the remaining
partition balanced.
• The "tail" is somewhat irregular, so we probably should
consider the leading diagonals.
Introduction
Balanced partition
The General Case
Further Thoughts
Concerns
• Since k 6= 0, the "dark" and "light" won’t pair off.
• We need to creat a triangular part as well.
• We probably should peel off some diagonals to make the
triangular part and at the same time leaving the remaining
partition balanced.
• The "tail" is somewhat irregular, so we probably should
consider the leading diagonals.
• The diagonals won’t fit in the original configuration board.
Introduction
Balanced partition
The General Case
A Key Observation
Suppose χ(π) = k > 0, then we have:
Further Thoughts
Introduction
Balanced partition
The General Case
A Key Observation
Suppose χ(π) = k > 0, then we have:
i m ≥ 2k ;
Further Thoughts
Introduction
Balanced partition
The General Case
A Key Observation
Suppose χ(π) = k > 0, then we have:
i m ≥ 2k ;
ii d1 = 1, d2 = 2, · · · , d2k = 2k ;
Further Thoughts
Introduction
Balanced partition
The General Case
A Key Observation
Suppose χ(π) = k > 0, then we have:
i m ≥ 2k ;
ii d1 = 1, d2 = 2, · · · , d2k = 2k ;
Pn
j
iii
j=2k +1 (−1) dj = 0.
Further Thoughts
Introduction
Balanced partition
The General Case
A Key Observation
Suppose χ(π) = k > 0, then we have:
i m ≥ 2k ;
ii d1 = 1, d2 = 2, · · · , d2k = 2k ;
Pn
j
iii
j=2k +1 (−1) dj = 0.
Proof
Consider partial alternating sums ofP
dj ’s:
(−d1 , −d1 + d2 , −d1 + d2 − d3 , . . . , nj=1 (−1)j dj )
= (−1, 1, −2, 2, . . . , &, k ).
Further Thoughts
Introduction
Balanced partition
The General Case
New Configuration Board
..
.
8
6
4
2
1
3
5
7
9
···
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
New Configuration Board
..
.
..
.
8
8
6
6
4
2
1
3
5
7
9
···
⇒
4
2
1
3
5
7
9
···
Introduction
Balanced partition
The General Case
Further Thoughts
New Configuration Board
..
.
..
.
8
8
6
6
4
2
1
3
5
7
9
···
⇒
4
2
1
2k
3
5
7
9
···
Introduction
Balanced partition
The General Case
Proof of the General Case
I {d1 , d2 , . . . , d2k } → T2k
Further Thoughts
Introduction
Balanced partition
The General Case
Proof of the General Case
I {d1 , d2 , . . . , d2k } → T2k
II {d2k +1 , . . . , dn } → a partition of
1
(n − T2k )
2
Further Thoughts
Introduction
Balanced partition
The General Case
Proof of the General Case
I {d1 , d2 , . . . , d2k } → T2k
1
(n − T2k )
2
Rmk1 {d2k +1 , . . . , dm } will make out the a × (a + 2k ) Durfee
rectangle of the resulting partition.
II {d2k +1 , . . . , dn } → a partition of
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
Proof of the General Case
I {d1 , d2 , . . . , d2k } → T2k
1
(n − T2k )
2
Rmk1 {d2k +1 , . . . , dm } will make out the a × (a + 2k ) Durfee
rectangle of the resulting partition.
II {d2k +1 , . . . , dn } → a partition of
Rmk2 When 2k = m, the ending partition does not have
a × (a + 2k ) Durfee rectangle, which is never the case for
k = 0.
Introduction
Balanced partition
The General Case
Outline
Introduction
Euler’s Theorem
Characteristic of a Partition
Balanced partition
The General Case
Further Thoughts
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
Sliding UP?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
Introduction
Balanced partition
The General Case
Further Thoughts
Sliding UP?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
2) λi − λi+1 ≤ 1 for every i ≤ l − 1;
Introduction
Balanced partition
The General Case
Further Thoughts
Sliding UP?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
2) λi − λi+1 ≤ 1 for every i ≤ l − 1;
3) Balanced. (χ(λ) = 0)
Introduction
Balanced partition
The General Case
Further Thoughts
Sliding UP?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
2) λi − λi+1 ≤ 1 for every i ≤ l − 1;
3) Balanced. (χ(λ) = 0)
Theorem
The multiplicity of the weight in the basic representation of
certain affine algebra is exactly equal to the number of
partitions described above.
Introduction
Balanced partition
The General Case
Further Thoughts
Sliding UP?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
2) λi − λi+1 ≤ 1 for every i ≤ l − 1;
3) Balanced. (χ(λ) = 0)
Theorem
The multiplicity of the weight in the basic representation of
certain affine algebra is exactly equal to the number of
partitions described above.
Jiaoyang Huang, Peter Wear and Tianqi Wu, REU project
in University of Minnesota, 2013.
Introduction
Balanced partition
The General Case
Further Thoughts
Sliding UP?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
2) λi − λi+1 ≤ 1 for every i ≤ l − 1;
3) Balanced. (χ(λ) = 0)
Theorem
The multiplicity of the weight in the basic representation of
certain affine algebra is exactly equal to the number of
partitions described above.
Jiaoyang Huang, Peter Wear and Tianqi Wu, REU project
in University of Minnesota, 2013.
V. Kreiman, V. Lakshmibai, P. Magyar and J. Weyman,
Standard Bases For Affine SL(N)-Modules,
INTERNATIONAL MATHEMATICS RESEARCH NOTICES,
2005.
Introduction
Balanced partition
The General Case
Further Thoughts
Refinement?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
Introduction
Balanced partition
The General Case
Further Thoughts
Refinement?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
2) λi − λi+1 ≤ 1 for every i ≤ l − 1;
3) Balanced. (χ(λ) = 0)
Introduction
Balanced partition
The General Case
Further Thoughts
Refinement?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
2) λi − λi+1 ≤ 1 for every i ≤ l − 1;
3) Balanced. (χ(λ) = 0)
4) l is even.
Introduction
Balanced partition
The General Case
Further Thoughts
Refinement?
Suppose λ = (λ1 , λ2 , . . . , λl ) is a partition of 2n satisfying the
following conditions:
1) λl = 1;
2) λi − λi+1 ≤ 1 for every i ≤ l − 1;
3) Balanced. (χ(λ) = 0)
4) l is even.
Conjecture (OEIS.ORG A064174 )
The number of partitions of n with nonnegative rank is equal to
the number of partitions of 2n described above.
Introduction
Balanced partition
The General Case
Further Refinement?
Theorem
Let Dk (n) := {partitions of n into distinct parts having
1
characteristic k}. Then |Dk (n)| = p( (n − T2k )).
2
Further Thoughts
Introduction
Balanced partition
The General Case
Further Thoughts
Further Refinement?
Theorem
Let Dk (n) := {partitions of n into distinct parts having
1
characteristic k}. Then |Dk (n)| = p( (n − T2k )).
2
Theorem (Bessenrodt)
The number of partitions of n into distinct parts with alternating
sum l is equal to the number of partitions of n with l odd parts.
Introduction
Balanced partition
The General Case
Further Thoughts
Further Refinement?
Theorem
Let Dk (n) := {partitions of n into distinct parts having
1
characteristic k}. Then |Dk (n)| = p( (n − T2k )).
2
Theorem (Bessenrodt)
The number of partitions of n into distinct parts with alternating
sum l is equal to the number of partitions of n with l odd parts.
Theorem (Chen, Gao, Ji and Li)
The number of partitions of n into distinct parts with l odd parts
and alternating sum m is equal to the number of partitions of n
into exactly m odd parts and l parts repeated odd times.
Introduction
Balanced partition
The General Case
Further Thoughts
Further Refinement?
Theorem
Let Dk (n) := {partitions of n into distinct parts having
1
characteristic k}. Then |Dk (n)| = p( (n − T2k )).
2
Theorem (Bessenrodt)
The number of partitions of n into distinct parts with alternating
sum l is equal to the number of partitions of n with l odd parts.
Theorem (Chen, Gao, Ji and Li)
The number of partitions of n into distinct parts with l odd parts
and alternating sum m is equal to the number of partitions of n
into exactly m odd parts and l parts repeated odd times.
William Y. C. Chen, Henry Y. Gao, Kathy Q. Ji and Martin
Y. X. Li A Unification of Two Refinements of Euler’s Partition
Theorem, THE RAMANUJAN JOURNAL, 2010.
Introduction
Balanced partition
The General Case
Happy Birthday, Professor Andrews!
Further Thoughts