Refinement of Some Partition Inequalities Partitions, The Partition Counting Function Restricted...
Transcript of Refinement of Some Partition Inequalities Partitions, The Partition Counting Function Restricted...
Refinement of Some Partition Inequalities
James Mc Laughlin
West Chester [email protected]
http://math.wcupa.edu/∼mclaughlin/
West Coast Number TheoryPacific Grove
December 18, 2015
Outline
q-series NotationInteger Partitions, The Partition Counting FunctionRestricted Partition FunctionsFerrers Diagram, Durfee SquarePartition Generating FunctionsPartition InequalitiesPartition Generating Functions that Track the Number ofPartsSome ExperimentationResultsConcluding Remarks
q-series Notation
q-products: (a;q)0 := 1 and for n ≥ 1,
(a;q)n := (1− a)(1− aq) · · · (1− aqn−1)
(q;q)n := (1− q)(1− q2) · · · (1− qn) (∗)(a1, . . . ,aj ;q)n := (a1;q)n · · · (aj ;q)n
(a;q)∞ := (1− a)(1− aq)(1− aq2) · · ·(a1, . . . ,aj ;q)∞ := (a1;q)∞ · · · (aj ;q)∞
The q-binomial theorem: if |z|, |q| < 1, then
∞∑n=0
(a;q)n
(q;q)nzn =
(az;q)∞(z;q)∞
. (1)
Special Cases of the q-binomial theorem
Special Cases of the q-binomial theorem:
∞∑n=0
zn
(q;q)n=
1(z;q)∞
, |z| < 1, |q| < 1. (2)
∞∑n=0
(−a)nqn(n−1)/2
(q;q)n= (a;q)∞, |q| < 1. (3)
Integer Partitions
Definition: A partition of a positive integer n is a way of writingn as a sum of positive integers, where order does not matter.
Example. The partitions of 5 are
54 + 13 + 23 + 1 + 12 + 2 + 12 + 1 + 1 + 11 + 1 + 1 + 1 + 1
The summands of a partition are called parts of the partition.
The number of partitions of n is given by the partition functionp(n).For example, p(5) = 7.
Restricted Partition Functions, I
Some well known examples of restricted partition functions arepO(n), the number of partitions of n into odd parts, and pD(n),the number of partitions of n into distinct parts.
pO(5) = 3 (5, 3 + 1 + 1, 1 + 1 + 1 + 1 + 1),pD(5) = 3 (5, 4 + 1, 3 + 2).
(PO(n) = PD(n), ∀n ∈ N)
Restricted Partition Functions, IILet p2,3,5(n) denote the number of partitions of n into parts≡ 2,3( mod 5), andP∗(n) denote the number of partitions of n where each partfrom 1 to the largest part occurs at least twice.
p2,3,5(10) = 4 2 + 2 + 2 + 2 + 23 + 3 + 2 + 27 + 38 + 2
P∗(10) = 4 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 12 + 2 + 1 + 1 + 1 + 1 + 1 + 12 + 2 + 2 + 1 + 1 + 1 + 12 + 2 + 2 + 2 + 1 + 1
(P2,3,5(n) = P∗(n), ∀n ∈ N)
Partition Generating Functions, ILet S be any set of positive integers, finite or infinite. Then thegenerating function for pS(n), the number of partitions of thepositive integer n with parts from S is
∞∑n=0
pS(n)qn =1∏
ai∈S 1− qai
= (1 + qa1 + q2a1 + q3a1 + . . . )
× (1 + qa2 + q2a2 + q3a2 + . . . )
× (1 + qa3 + q2a3 + q3a3 + . . . ) . . .
The generating function for p∗S(n), the number of partitions ofthe positive integer n with distinct parts from S is
∞∑n=0
p∗S(n)qn =
∏ai∈S
1 + qai
= (1 + qa1)(1 + qa2)(1 + qa3) . . .
Partition Generating Functions, II
Recall that p(n) is the number of (unrestricted) partitions of n.
∞∑n=0
p(n)qn =1∏∞
k=1 1− qk =1
(q;q)∞
=∞∑
k=0
qk
(q;q)k
=∞∑
k=0
qk2
(q;q)2k
= 1 +∞∑
k=1
qk
(qk ;q)∞= 1 +
1(q;q)∞
∞∑k=1
(q;q)k−1qk
Partition Generating Functions, III
Recall that pD(n) is the number of partitions of n into distinctpositive integers.
∞∑n=0
pD(n)qn =∞∏
k=1
(1 + qk ) = (−q;q)∞
= (1 + q)(1 + q2)(1 + q3) . . .
Partition Generating Functions, IV
Recall that p2,3,5(n) denote the number of partitions of n intoparts ≡ 2,3( mod 5).
∞∑n=0
p2,3,5(n)qn
=1
(1− q2)(1− q3)(1− q7)(1− q8)(1− q12)(1− q13) . . .
=1
(q2;q5)∞(q3;q5)∞=
1(q2,q3;q5)∞
Partition Inequalities, I
Fact: For each positive integer n,
p1,4,5(n)− p2,3,5(n) ≥ 0.
Alternatively, if the sequence {cn} is defined by
∞∑n=0
cnqn =1
(q,q4;q5)∞− 1
(q2,q3;q5)∞,
then cn ≥ 0,∀n ≥ 0.
Partition Inequalities, II
Proof.By the Rogers-Ramanujan identities,
1(q,q4;q5)∞
− 1(q2,q3;q5)∞
=∞∑
k=0
qk2
(q;q)k−∞∑
k=0
qk2+k
(q;q)k
=∞∑
k=1
qk2(1− qk )
(q;q)k
=∞∑
k=1
qk2
(q;q)k−1
Partition Inequalities, Variations and Extensions, I
Theorem (Berkovich and Garvan, 2005)
Suppose L > 0, and 1 < r < m − 1. If the sequence {en} isdefined by
∞∑n=0
enqn =1
(q,qm−1;qm)L− 1
(qr ,qm−r ;qm)L,
thenen ≥ 0, ∀n ≥ 0⇐⇒ r - m − r and m − r - r .
Partition Inequalities, Variations and Extensions, II
Theorem (Andrews, 2011)
If L > 0, and the sequence {fn} is defined by
∞∑n=0
fnqn =1
(q,q5,q6;q8)L− 1
(q2,q3,q7;q8)L,
thenfn ≥ 0,∀n ≥ 0.
Partition Inequalities, Variations and Extensions, III
Theorem (Berkovich and Grizzell, 2012)For any L > 0, and any odd y > 1, the q-series expansion of
1(q,qy+2,q2y ;q2y+2)L
− 1(q2,qy ,q2y+1;q2y+2)L
=∞∑
n=0
a(L, y ,n)qn
has only non-negative coefficients. Furthermore, the coefficienta(L, y ,n) is 0 if and only if either
n ∈ {2,4,6, . . . , y + 1} ∪ {y} or (L, y ,n) = (1,3,9).
Partition Inequalities, Variations and Extensions, IV
Theorem (Berkovich and Grizzell, 2012)For any L > 0, and any odd y > 1, and any x with1 < x ≤ y + 2, the q-series expansion of
1(q,qx ,q2y ;q2y+2)L
− 1(q2,qy ,q2y+1;q2y+2)L
=∞∑
n=0
a(L, x , y ,n)qn
has only non-negative coefficients. Furthermore, the coefficienta(L, y ,n) is 0 if and only if either . . . .
Partition Inequalities, Variations and Extensions, V
Theorem (Berkovich and Grizzell, 2013)
For any octuple of positive integers (L,m, x , y , z, r ,R, ρ), theq-series expansion of
1(qx ,qy ,qz ,qrx+Ry+ρz ;qm)L
− 1(qrx ,qRy ,qρz ,qx+y+z ;qm)L
=∞∑
n=0
a(L, x , y , z, r ,R, ρ,n)qn
has only non-negative coefficients.
Partition Inequalities, Variations and Extensions, VI
Theorem (Berkovich and Grizzell, 2013)
For any positive integers m,n, y, and z, with gcd(n, y) = 1, andintegers K and L, with K ≥ L ≥ 0,
1(qz ;qm)K (qnyz ;qnm)L
− 1(qyz ;qm)K (qnz ;qnm)L
=∞∑
k=0
a(K ,L, x , y , z,n,m, k)qk
has only non-negative coefficients.
Partition Generating Functions that Track the Numberof Parts
Let S be any set of positive integers, finite or infinite. Then thegenerating function for pS(m,n), the number of partitions of thepositive integer n with exactly m parts from S is
∞∑n=0
pS(m,n)smqn =1∏
ai∈S 1− sqai
= (1 + sqa1 + s2q2a1 + s3q3a1 + . . . )
× (1 + sqa2 + s2q2a2 + s3q3a2 + . . . )
× (1 + sqa3 + s2q2a3 + s3q3a3 + . . . ) . . .
The generating function for p∗S(n), the number of partitions ofthe positive integer n with distinct parts from S is
∞∑n=0
p∗S(n)qn =
∏ai∈S
1 + sqai
= (1 + sqa1)(1 + sqa2)(1 + sqa3) . . .
Partition Inequalities that Track the Number of Parts
Q. If the polynomials {fn(s)} are defined by
∞∑n=0
fn(s)qn =1
(sq, sq4;q5)∞− 1
(sq2, sq3;q5)∞,
are there situations where the coefficients in fn(s) are allnon-negative?
Experimental Output
n fn(s)1 s
2 − s + s2
3 − s + s3
4 s − s2 + s4
5 s5
6 s − s2 + s6
7 − s + s2 − s3 + s4 + s7
8 − s + s2 − s4 + s5 + s8
9 s − s2 + s6 + s9
10 s7 + s10
Experimental Output
11 s − s2 + s3 − s4 + s6 + s8 + s11
12 − s + 2s2 − s3 + s4 − s5 + s7 + s9 + s12
13 − s + s2 + s5 − s6 + s7 + s8 + s10 + s13
14 s − 2s2 + s3 + s6 − s7 + s8 + s9 + s11 + s14
15 s7 + s9 + s10 + s12 + s15
16 s − 2s2 + 2s3 − s4 + s6 + s8 + s10 + s11 + s13 + s16
17 − s + 2s2 − 3s3 + 3s4 − s5 + s7 + 2s9 + s11 + s12
+ s14 + s17
18 − s + 2s2 − s3 + 2s5 − 2s6 + s7 + s8 + 2s10 + s12
+ s13 + s15 + s18
19 s − 2s2 + 2s3 − s4 + 2s6 − s7 + s8 + s9 + s10 + 2s11
+ s13 + s14 + s16 + s19
20 s5 + s7 + s9 + s10 + s11 + 2s12 + s14 + s15 + s17 + s20
Experimentation, II
5 s5
10 s10 + s7
15 s15 + s12 + s10 + s9 + s7
20 s20 + s17 + s15 + s14 + 2s12 + s11 + s10 + s9 + s7 + s5
25 s25 + s22 + s20 + s19 + 2s17 + s16 + 2s15 + 2s14 + s13
+ 3s12 + s11 + 2s10 + 2s9 + 2s7 + s5
30 s30 + s27 + s25 + s24 + 2s22 + s21 + 2s20 + 2s19 + s18
+ 4s17 + 2s16 + 3s15 + 4s14 + s13 + 5s12 + 2s11 + 2s10
+ 3s9 + 3s7 + s5
35 s35 + s32 + s30 + s29 + 2s27 + s26 + 2s25 + 2s24 + s23
+ 4s22 + 2s21 + 4s20 + 5s19 + 2s18 + 7s17 + 4s16 + 5s15
+ 7s14 + 2s13 + 7s12 + 4s11 + 3s10 + 5s9 + 4s7 + s5
First Theorem
Theorem (Mc L. 2015)
Let M ≥ 5 be a positive integer, and let a and b be integerssuch that 1 ≤ a < b < M/2 and gcd(a,M) = gcd(b,M) = 1.Define the integers c(m,n) by
1(sqa, sqM−a;qM)∞
− 1(sqb, sqM−b;qM)∞
:=∑
m,n≥0
c(m,n)smqn. (4)
(i) Then c(m,Mn) ≥ 0 for all integers m ≥ 0,n ≥ 0.(ii) If, in addition, M is even, then c(m,Mn + M/2) ≥ 0 for allintegers m ≥ 0,n ≥ 0.
Partitions Interpretation
Corollary
Let M, a and b be as in Theorem 7. Letpa,M,m(n) = # partitions of n into exactly m parts, each≡ ±a( mod M),and letpb,M,m(n) = # partitions of n into exactly m parts, each≡ ±b( mod M).
Then
(i) pa,M,m(nM) ≥ pb,M,m(nM) for all integers n ≥ 1, and allintegers m, 1 ≤ m ≤ Mn.
(ii) If M is even, then pa,M,m(nM + M/2) ≥ pb,M,m(nM + M/2)for all integers n ≥ 0, and integers m with 1 ≤ m ≤ Mn + M/2.
Proof of First Theorem
Proof.We recall a special case of the q-binomial theorem:
1(z;q)∞
=∞∑
n=0
zn
(q;q)n. (5)
Hence
1(sqa, sqM−a;qM)∞
− 1(sqb, sqM−b;qM)∞
=∑
j,k≥0
sj+kqa(j−k)+kM
(qM ;qM)j(qM ;qM)k−
∑j,k≥0
sj+kqb(j−k)+kM
(qM ;qM)j(qM ;qM)k(6)
Set j + k =: m, so that j = m − k and the right side of (6)becomes
Proof, Continued
Proof Continued.∑m≥0
smm∑
k=0
qa(m−2k)+kM − qb(m−2k)+kM
(qM ;qM)m−k (qM ;qM)k(7)
Next, we restrict the values of k so that when the inner sum isexpanded as a power series, it contains only those powers of qwhose exponents are multiples of M(so that the series multiplying sm is
∑∞n=0 c(m,Mn)qMn).
Since gcd(a,M) = gcd(b,M) = 1, this means restricting k sothat M|(m − 2k).
If m is even, then k = m/2 is such a value, andqa(m−2k)+km − qb(m−2k)+km = 0 in this case.
Hence we need only consider those k in the intervals0 ≤ k < m/2 and m/2 < k ≤ m satisfyingm − 2k ≡ 0( mod M).
Proof, Continued
Proof Continued.
sm
∑0≤k<m/2
+∑
m/2<k≤m
qa(m−2k)+kM − qb(m−2k)+kM
(qM ;qM)m−k (qM ;qM)k
Note that(1) every k ′ in the upper interval may be expressed ask ′ = m − k , for some k in the lower interval;(2) every k in the lower interval can be similarly matched with ak ′ in the upper interval;
(3) m − 2k ≡ 0( mod M)⇐⇒ m − 2(m − k) ≡ 0( mod M);
(4) the denominators of the summands remain invariant underthe transformation k ↔ m − k .
Proof, Continued
Proof Continued.
∑m,n≥0
c(m,Mn)smqMn =∑m≥0
sm∑
0≤k<m/2M|m−2k
qa(m−2k)+kM − qb(m−2k)+kM
+ q−a(m−2k)+(m−k)M − q−b(m−2k)+(m−k)M
(qM ;qM)m−k (qM ;qM)k
=∑m≥0
sm∑
0≤k<m/2M|m−2k
qa(m−2k)+kM(1− q(m−2k)(b−a))(1− q(m−2k)(M−b−a))
(qM ;qM)m−k (qM ;qM)k
Proof, Continued
Proof Continued.Finally,- M|(m − 2k)(b − a) and M|(m − 2k)(M − b − a);
- the conditions on a and b give that they are different multiplesof M, each less than (m − k)M;
- the factors (1− q(m−2k)(b−a)) and (1− q(m−2k)(M−b−a)) arecancelled by two different factors in the q-product (qM ;qM)m−k ;
- the remaining factors in the denominators may be expandedas geometric series with only non-negative coefficients, and theclaim at (i) above follows;
- The claim at (ii) follows similarly, upon noting that
m − 2k ≡ M/2( mod M)
⇐⇒ m − 2(m − k) ≡ −M/2 ≡ M/2( mod M).
Second Theorem
Theorem (Mc L. 2015)
Let M ≥ 5 be a positive integer, and let a and b be integerssuch that 1 ≤ a < b < M/2 and gcd(a,M) = gcd(b,M) = 1.Define the integers d(m,n) by
(−sqa,−sqM−a;qM)∞ − (−sqb,−sqM−b;qM)∞
:=∑
m,n≥0
d(m,n)smqn. (8)
(i) Then d(m,Mn) ≥ 0 for all integers m ≥ 0,n ≥ 0.(ii) If, in addition, M is even, then d(m,Mn + M/2) ≥ 0 for allintegers m ≥ 0,n ≥ 0.
Partitions Interpretation
Corollary
Let M, a and b be as in Theorem 9. Let
p∗a,M,m(n) denote the number of partitions of n into exactly mdistinct parts ≡ ±a( mod M), and let
p∗b,M,m(n) denote the number of partitions of n into exactly mdistinct parts ≡ ±b( mod M).
Then
(i) p∗a,M,m(nM) ≥ p∗b,M,m(nM) for all integers n ≥ 1, and allintegers m, 1 ≤ m ≤ Mn.
(ii) If M is even, then p∗a,M,m(nM + M/2) ≥ p∗b,M,m(nM + M/2)for all integers n ≥ 0, and integers m with 1 ≤ m ≤ Mn + M/2.
Sketch of Proof
Sketch of Proof.Recall
(−a;q)∞ =∞∑
n=0
anqn(n−1)/2
(q;q)n. (9)
The application of this to the infinite products leads to
∑m,n≥0
d(m,n)smqn =∑m≥0
smm∑
k=0
(qa(m−2k)+kM − qb(m−2k)+kM)qM[(m−k)(m−k−1)/2+k(k−1)/2]
(qM ;qM)m−k (qM ;qM)k
The proof now follows similarly.
No Finite Analogues
It does not appear that replacing “∞” with a positive integer L inTheorems 7 and 9 “works”. In other words, if L is a positiveinteger, and
1(sqa, sqM−a;qM)L
− 1(sqb, sqM−b;qM)L
:=∑
m,n≥0
c(m,n)smqn, (10)
it does not appear to be the case that c(m,Mn) ≥ 0 for all mand n, and likewise for the other theorem.Perhaps there is some restricted version of such a theorem thatis valid?
Injective Proofs
Let M ≥ 5 and 1 ≤ a < b < M/2 be integers. For integers1 ≤ m ≤ n let pa,M,m(n) = # partitions of n into exactly m parts,each≡ ±a( mod M),and letpb,M,m(n) = # partitions of n into exactly m parts, each≡ ±b( mod M).Can you find an injection from the partitions counted bypb,M,m(Mn) to those counted by pa,M,m(Mn)?Mind Floss: (1) Consider the partition of kMb consisting kMparts of size b.Find a partition of kMb into kM parts, where each part is ≡ ±a(mod M).
(2) Consider the partition of kM(M − b) consisting kM parts ofsize M − b.Find a partition of kM(M − b) into kM parts, where each part is≡ ±a( mod M).