The power of q
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
The power of q
Michael D. Hirschhorn
A course of lectures presented at Wits, July 2014
Part IV
[email protected]
Even parts distinct
Partitions with even parts distinct
Let ped(n) be the number of partitions of n with even parts
distinct.
For example, if n = 5, the partitions are
5, 4 + 1, 3 + 2, 3 + 1 + 1, 2 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1,
and ped(5) = 6.
The generating function is
X
ped(n)q n =
n≥0
(−q 2 ; q 2 )∞
= (−q; q)∞ (−q 2 ; q 2 )∞ .
(q; q 2 )∞
It is also true that
X
ped(n)q n =
n≥0
=
(q 2 ; q 2 )∞ (q 4 ; q 4 )∞
(q 4 ; q 4 )∞
=
(q; q)∞ (q 2 ; q 2 )∞
(q; q)∞
1
(q, q 2 , q 3 ; q 4 )∞
The power of q
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
Even parts distinct
The power of q
so
ped(n) = p{1,2,3;4} (n)
the number of partitions of n into parts not multiples of 4.
(“4–regular partitions”)
In the case n = 5, the relevant partitions are
5, 3 + 2, 3 + 1 + 1, 2 + 2 + 1, 2 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1
and p{1,2,3;4} (5) = 6 = ped(5).
James Sellers and I discovered that ped(n) has some nice
arithmetic properties.
I shall show that ped(n) is divisible by 12 for (at least) one
in nine values of n.
Indeed, I will prove the Ramanujan–like identity,
X
n≥0
ped(9n + 7)q n = 12
Y (1 − q 2n )4 (1 − q 3n )6 (1 − q 4n )
.
(1 − q n )11
n≥1
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
Even parts distinct
The power of q
X
ped(n)q n = (−q; q)∞ (−q 2 ; q 2 )∞
n≥0
(−q 3 ; q 3 )∞
(−q, −q 2 , q 3 ; q 3 )∞
(q 3 ; q 3 )∞
(−q 6 ; q 6 )∞
× 6 6
(−q 2 , −q 4 , q 6 ; q 6 )∞
(q ; q )∞
∞
(q 6 ; q 6 )∞ (q 12 ; q 12 )∞ X
2
2
=
q (3m +m)/2+(3n +n) .
3
3
2
6
6
2
(q ; q )∞ (q ; q )∞ m,n=−∞
=
Split the sum according to the residue modulo 3 of m + 2n.
If m + 2n ≡ 0 let m = r − 2t, n = r + t,
if m + 2n ≡ 1, let m = r + 1 − 2t, n = r + t,
if m + 2n ≡ −1, let m = r − 1 − 2t, n = r + t
and we obtain
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
Even parts distinct
The power of q
X
n≥0
ped(n)q
n
=
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
(q 12 ; q 12 )∞
(q 3 ; q 3 )2∞ (q 6 ; q 6 )∞
∞
X
×
q (9r
r ,t=−∞
∞
X
2
2 +3r )/2+9t 2
q (9r
+q
Even parts distinct
2 +9r )/2+(9t 2 −6t)
r ,t=−∞
+q
∞
X
r ,t=−∞
!
q
(9r 2 −3r )/2+(9t 2 +6t)
The power of q
X
n≥0
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
(q 12 ; q 12 )∞
ped(n)q = 3 3 2 6 6
(q ; q )∞ (q ; q )∞
n
× (−q 3 , −q 6 , q 9 ; q 9 )∞ (−q 9 , −q 9 , q 18 ; q 18 )∞
Even parts distinct
+q(−q 3 , −q 6 , q 9 ; q 9 )∞ (−q 3 , −q 15 , q 18 ; q 18 )∞
+2q 2 (−q 9 , −q 9 , q 9 ; q 9 )∞ (−q 3 , −q 15 , q 18 ; q 18 )∞
f12
= 2
f3 f6
=
5
2 2
f6 f92 f18
f6 f92 f62 f9 f36
2 f18 f6 f9 f36
+
2q
+
q
2
f3 f18 f92 f36
f3 f18 f3 f12 f18
f9 f2 f12 f18
4
f12 f18
f18 f36
f62 f93 f36
+
q
+ 2q 2 3 .
3
2
4
2
f3 f36
f3 f18
f3
The power of q
It follows that
X
ped(3n)q n =
n≥0
X
ped(3n + 1)q n =
n≥0
X
f22 f33 f12
,
f14 f62
ped(3n + 2)q n = 2
n≥0
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
f4 f64
,
2
f13 f12
Even parts distinct
f2 f6 f12
.
f13
In particular,
X
f 3 f12 f 2
φ(−q 3 )ψ(−q 3 )
,
ped(3n + 1)q n = 3 2 · 24 =
φ(−q)2
f6
f1
n≥0
where
φ(−q) =
∞
X
2
(−1)n q n = (q, q, q 2 ; q 2 )∞ =
−∞
ψ(−q) =
∞
X
−∞
(−1)n q 2n
2 +n
f12
,
f2
= (q, q 3 , q 4 ; q 4 )∞ =
f1 f4
.
f2
The power of q
So
X
(−1)n ped(3n + 1)q n =
φ(q 3 )ψ(q 3 )
φ(q)2
n≥0
.
Now,
φ(ωq)φ(ω 2 q)
1
=
.
φ(q)
φ(q)φ(ωq)φ(ω 2 q)
The denominator,
φ(q)φ(ωq)φ(ω 2 q) =
φ(q 3 )4
.
φ(q 9 )
This relies on the facts that
φ(q) =
E (q 2 )5
E (q)2 E (q 4 )2
and
E (q)E (ωq)E (ω 2 q) =
E (q 3 )4
.
E (q 9 )
(We saw a similar calculation when we proved RMBI.)
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
Even parts distinct
The power of q
Also,
φ(q) = φ(q 9 ) + 2qX (q 3 ),
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
where
X (q) = (−q, −q 5 , q 6 ; q 6 )∞ .
Even parts distinct
So the numerator,
φ(ωq)φ(ω 2 q) = (φ(q 9 ) + 2ωqX (q 3 ))(φ(q 9 ) + 2ω 2 qX (q 3 ))
= φ(q 9 )2 − 2qφ(q 9 )X (q 3 ) + 4q 2 X (q 3 )2 .
It follows that
φ(q 9 )
(−1) ped(3n + 1)q = φ(q )ψ(q )
φ(q 3 )4
n≥0
2
× φ(q 9 )2 − 2qφ(q 9 )X (q 3 ) + 4q 2 X (q 3 )2
X
n
n
3
3
2
It follows that
X
φ(q 3 )2
(−1)n ped(9n + 7)q n = φ(q)ψ(q)
12φ(q 3 )2 X (q)2
φ(q)8
n≥0
= 12
φ(q 3 )4 ψ(q)X (q)2
φ(q)7
and
X
ped(9n + 7)q n
n≥0
φ(−q 3 )4 ψ(−q)X (−q)2
φ(−q)7
7 2 4 2 2
f2
f3
f1 f4
f1 f6
= 12
2
f6
f2
f2 f3
f1
= 12
= 12
as claimed earlier.
f23 f36 f4
,
f111
The power of q
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
Even parts distinct
We can also show that for α ≥ 1, modulo 3,
X
32α − 1
n
2α
(−1) ped 3 n +
q n ≡ ψ(q)φ(q 3 ),
8
n≥0
X
n≥0
n
2α+1
(−1) ped 3
32α+2 − 1
n+
8
q n ≡ φ(q)ψ(q 3 ).
and that for α ≥ 1,
17 × 32α − 1
2α+1
ped 3
n+
≡ 0 (mod 3),
8
19 × 32α+1 − 1
ped 32α+2 n +
≡ 0 (mod 3).
8
The power of q
Michael D.
Hirschhorn
A course of
lectures presented
at Wits, July 2014
Part IV
Even parts distinct