On the 2–, 3–, 4– and 6–dissections of
Ramanujan’s cubic continued fraction and its
reciprocal
Michael D. Hirschhorn and Roselin
July 2, 2010
Dedicated to the memory of K. Venkatachaliengar on the occasion of the Centenary of his birth.
2000 Mathematics Subject Classification: 30B70
Keywords: Ramanujan, cubic continued fraction, dissection.
Abstract
We give 2–, 3–, 4– and 6–dissections of Ramanujan’s cubic continued
fraction, with each component a single product or if not a single product
then a sum of two products.
1
Introduction
The Rogers–Ramanujan continued fraction is
R(q) :=
1
q
q2
q3
( q, q 4 ; q 5 )∞
= 2 3 5 ,
1 + 1 + 1 + 1 + ···
(q , q ; q )∞
|q| < 1.
Ramanujan [13, p. 50] gave the 2–dissections of this continued fraction and its
reciprocal,
R(q) =
(q 4 , q 6 , q 14 , q 16 ; q 20 )∞
(q 4 , q 4 , q 16 , q 16 ; q 20 )∞
−
q
(q 2 , q 10 , q 10 , q 18 ; q 20 )∞
(q 8 , q 10 , q 10 , q 12 ; q 20 )∞
and
1/R(q) =
(q 8 , q 8 , q 12 , q 12 ; q 20 )∞
(q 2 , q 8 , q 12 , q 18 ; q 20 )∞
+
q
(q 6 , q 10 , q 10 , q 14 ; q 20 )∞
(q 4 , q 10 , q 10 , q 16 ; q 20 )∞
1
and these were first proved by Andrews [2].
Here we use the standard notation
(a; q)∞ =
∞
Y
(1 − aq k ), (a1 , · · · , an ; q)∞ =
k=0
n
Y
(ak ; q)∞ .
k=1
Ramanujan [13, p. 50] also gave 5–dissections of R(q) and its reciprocal, and
these results were improved upon and proved by Hirschhorn [8]. In so doing, he
was able to demonstrate the periodic behaviour of the sign of the coefficients in
the series expansion of R(q) and its reciprocal, first observed by Szekeres [16].
(It would be surprising if Ramanujan did not also observe this periodicity, given
the tables in [13, p. 49].) In the same paper, Hirschhorn conjectured formulas
for the 4–dissection of R(q) and its reciprocal, and these were first proved by
Lewis and Liu [12].
Gordon’s continued fraction is
G(q) :=
1
q2
q4
( q, q 7 ; q 8 )∞
=
,
1 + q + 1 + q3 + 1 + q5 + · · ·
(q 3 , q 5 ; q 8 )∞
|q| < 1.
Hirschhorn [10] found the 8–dissection of G(q) and its reciprocal, thereby demonstrating the periodicity of the sign of the coefficients in the expansion of G(q) and
its reciprocal, and in particular that certain coefficients are zero, a phenomenon
first observed and proved by Richmond and Szekeres [14]. These themes have
been generalised by Andrews and Bressoud [3], Alladi and Gordon [1] and Chan
and Yesilyurt [5].
Ramanujan’s cubic continued fraction is
RC(q) :=
q + q2
q2 + q4
1
,
+ ···
1+ 1 +
1
|q| < 1.
Ramanujan [13, p. 44] states that
RC(q) =
( q, q 5 ; q 6 )∞
(q 3 , q 3 ; q 6 )∞
and, indeed, this is a special case of a general result of Ramanujan [13, p. 41],
proved by Hirschhorn [7], as is R(q), as is also
q + q2
q4
q3 + q6
1
( q, q 7 ; q 8 )∞
= 3 5 8
= G(q).
+ ···
1 + 1 + 1 +
1
(q , q ; q )∞
.
The object of this note is to establish the following results. The first two
are the 2– and 3–dissections of RC(q) and its reciprocal. In these dissections
2
each component is a single product. The third and fourth results are 4– and 6–
dissections of RC(q) and its reciprocal. In these dissections, not all components
are single products. Those components that are not single products are shown
to be expressible as sums of two products.
Note that the 2–and 4–dissections of the continued fraction 1/RC(q) were given
by B. Srivastava [15].
Theorem 1.1. The 2–dissections of RC(q) and its reciprocal are given by
RC(q)
=
1/RC(q) =
(q 4 ; q 4 )2∞ (q 12 ; q 12 )2∞
(q 2 ; q 2 )2∞ (q 12 ; q 12 )6∞
−
q
,
(q 6 ; q 6 )4∞
(q 4 ; q 4 )2∞ (q 6 ; q 6 )6∞
(q 4 ; q 4 )3∞
(q 2 ; q 2 )∞ (q 12 ; q 12 )3∞
+
q
.
(q 2 ; q 2 )∞ (q 6 ; q 6 )∞ (q 12 ; q 12 )∞
(q 4 ; q 4 )∞ (q 6 ; q 6 )3∞
Theorem 1.2. The 3–dissections of RC(q) and its reciprocal are given by
(q 6 ; q 6 )∞ (q 21 , q 33 , q 54 ; q 54 )∞
(q 3 ; q 3 )2∞
6 6
15 39 54 54
6 6
3 51 54 54
(q ; q )∞ (q , q , q ; q )∞
5 (q ; q )∞ (q , q , q ; q )∞
−q
−
q
,
(q 3 ; q 3 )2∞
(q 3 ; q 3 )2∞
RC(q) =
(q 3 ; q 3 )∞ (−q 12 , −q 15 , q 27 ; q 27 )∞
(q 6 ; q 6 )2∞
(q 3 ; q 3 )∞ (−q 3 , −q 24 , q 27 ; q 27 )∞
(q 3 ; q 3 )∞ (−q 6 , −q 21 , q 27 ; q 27 )∞
+ q2
.
+q
6
6
2
(q ; q )∞
(q 6 ; q 6 )2∞
1/RC(q) =
Theorem 1.3. The 4–dissections of RC(q) and its reciprocal are given by
RC(q) =
1/RC(q) =
(q 4 ; q 4 )2∞ (q 24 ; q 24 )14
∞
48 48 4
(q 12 ; q 12 )12
∞ (q ; q )∞
8 8 4 24 24 4
4 4 4
24 24 12
(q ; q )∞ (q ; q )∞
4 (q ; q )∞ (q ; q )∞
−q
+
q
(q 12 ; q 12 )8∞
(q 8 ; q 8 )4∞ (q 12 ; q 12 )12
∞
4 4 2
24 24 2
48 48 4
4 4 2
(q ; q )∞ (q ; q )∞ (q ; q )∞
(q ; q )∞ (q 24 ; q 24 )8∞
+4q 6
+ 2q 3
,
12
12
8
(q , q )∞
(q 12 ; q 12 )10
∞
(q 8 ; q 8 )2∞ (q 24 ; q 24 )2∞
(q 4 ; q 4 )∞ (q 16 ; q 16 )2∞ (q 24 ; q 24 )5∞
+q
8
8
12
12
5
48
48
2
(q ; q )∞ (q ; q )∞ (q ; q )∞
(q 12 ; q 12 )4∞
(q 8 ; q 8 )5∞ (q 48 ; q 48 )2∞
+q 2 4 4
(q ; q )∞ (q 12 ; q 12 )3∞ (q 16 ; q 16 )2∞ (q 24 ; q 24 )∞
(q 4 ; q 4 )2 (q 24 ; q 24 )6∞
−q 3 8 8 ∞
.
(q ; q )2∞ (q 12 ; q 12 )6∞
3
Theorem 1.4. The 6–dissections of RC(q) and its reciprocal are given by
RC(q)
12 12 3 18 18 2 24 30 54 54
(q ; q )∞ (q , q )∞ (q , q , q ; q )∞
=
(q 6 ; q 6 )5∞ (q 36 ; q 36 )∞
(q 12 ; q 12 )2∞ (q 36 ; q 36 )2∞ (q 6 , q 48 , q 54 ; q 54 )∞
−q 6
(q 6 ; q 6 )4∞ (q 18 ; q 18 )∞
12 12 2 42 66 108 108 2
(q ; q )∞ (q , q , q ; q )∞
−q
(q 6 ; q 6 )4∞
12 12 2
6 30 78 102 108 108 108
, q , q ; q )∞
12 (q ; q )∞ (q , q , q , q
+2q
(q 6 ; q 6 )4∞
12 12 3 18 18 2 12 42 54 54
(q ; q )∞ (q ; q )∞ (q , q , q ; q )∞
−q 2
(q 6 ; q 6 )5∞ (q 36 ; q 36 )∞
(q 12 ; q 12 )2∞ (q 36 ; q 36 )2∞ (q 24 , q 30 , q 54 ; q 54 )∞
−
(q 6 ; q 6 )4∞ (q 18 ; q 18 )∞
12 12 2 30 42 66 78 108 108 108
(q ; q )∞ (q , q , q , q , q , q ; q )∞
+q 3 2
(q 6 ; q 6 )4∞
12 12 2
6 102 108 108 2
, q ; q )∞
18 (q ; q )∞ (q , q
−q
(q 6 ; q 6 )4∞
12 12 3 18 18 2 6 48 54 54
(q ; q )∞ (q ; q )∞ (q , q .q ; q )∞
−q 4
(q 6 ; q 6 )5∞ (q 36 ; q 36 )∞
(q 12 ; q 12 )2∞ (q 36 ; q 36 )2∞ (q 12 , q 42 , q 54 ; q 54 )∞
+
(q 6 ; q 6 )4∞ (q 18 ; q 18 )∞
12 12 2 30 78 108 108 2
(q ; q )∞ (q , q , q ; q )∞
−q 5
(q 6 ; q 6 )4∞
(q 12 ; q 12 )2∞ (q 6 , q 42 , q 66 , q 102 , q 108 , q 108 ; q 108 )∞
,
−2q 6
(q 6 ; q 6 )4∞
1/RC(q)
18 18
36 36 2
24 84 108 108
; q )∞
(q ; q )∞ (q 48 , q 60 , q 108 ; q 108 )∞
6 (q ; q )∞ (q , q , q
−
q
=
(q 6 ; q 6 )2∞ (q 36 ; q 36 )∞
(q 6 ; q 6 )∞ (q 12 ; q 12 )∞ (q 18 ; q 18 )∞
(q 12 ; q 12 )∞ (q 42 , q 66 , q 108 ; q 108 )∞
+q
(q 6 ; q 6 )2
36 36 2 48 ∞60 108 108
18 18
12 96 108 108
(q ; q )∞ (q , q , q ; q )∞
; q )∞
6 (q ; q )∞ (q , q , q
+q 2
−
q
(q 6 ; q 6 )∞ (q 12 ; q 12 )∞ (q 18 ; q 18 )∞
(q 6 ; q 6 )2∞ (q 36 ; q 36 )∞
(q 12 ; q 12 )∞ (q 30 , q 78 , q 108 ; q 108 )∞
−q 3
(q 6 ; q 6 )2∞
18 18
36 36 2
12 96 108 108
; q )∞
(q ; q )∞ (q 24 , q 84 , q 108 ; q 108 )∞
6 (q ; q )∞ (q , q , q
+
q
−q 4
(q 6 ; q 6 )2∞ (q 36 ; q 36 )∞
(q 6 ; q 6 )∞ (q 12 ; q 12 )∞ (q 18 ; q 18 )∞
(q 12 ; q 12 )∞ (q 6 , q 102 , q 108 ; q 108 )∞
−q 11
.
(q 6 ; q 6 )2∞
4
Theorem 1.5. If we write
RC(q) =:
X
an q n ,
1/RC(q) =:
n≥0
X
bn q n ,
n≥0
then the sign of the an is periodic with period 3, and the sign of the bn is periodic
with period 6. Indeed,
b6n > 0, a3n = b6n+1 > 0, b6n+2 > 0, a3n+1 = b6n+3 < 0, b6n+4 < 0,
a3n+2 = b6n+5 < 0
except a2 = b5 = b8 = 0.
2
Preliminaries
Lemma 2.1. The following 2–dissections hold:
(q 3 ; q 3 )3∞
(q; q)∞
(q; q)∞
(q 3 ; q 3 )3∞
=
=
(q 4 ; q 4 )3∞ (q 6 ; q 6 )2∞
(q 12 ; q 12 )3∞
+
q
,
(q 2 ; q 2 )2∞ (q 12 ; q 12 )∞
(q 4 ; q 4 )∞
(q 2 ; q 2 )∞ (q 4 ; q 4 )2∞ (q 12 ; q 12 )2∞
(q 2 ; q 2 )3∞ (q 12 ; q 12 )6∞
−
q
.
(q 6 ; q 6 )7∞
(q 4 ; q 4 )2∞ (q 6 ; q 6 )9∞
Proof. Let
c∗ (q) =
∞
2
2
1 X
q m +mn+n +m+n .
3 m,n=−∞
It is shown in Hirschhorn, Garvan and Borwein [11], using nothing deeper than
Jacobi’s triple product identity, that [11, (1.7), (1.36)],
c∗ (q) =
(q 4 ; q 4 )3 (q 6 ; q 6 )2
(q 3 ; q 3 )3∞
= 2 2 2∞ 12 12 ∞ + q c∗ (q 4 ).
(q; q)∞
(q ; q )∞ (q ; q )∞
This is the first identity. Also,
(q; q)∞
(q 3 ; q 3 )3∞
=
=
=
1
c∗ (−q)
=
c∗ (q)
c∗ (q)c∗ (−q)
4 4 3 6 6 2
2 2 3
(q ; q )∞ (q ; q )∞
(q 12 ; q 12 )3∞
(q ; q )∞ (q 12 ; q 12 )3∞
−q 4 4
(q 4 ; q 4 )∞ (q 6 ; q 6 )9∞
(q 2 ; q 2 )2∞ (q 12 ; q 12 )∞
(q ; q )∞
(q 2 ; q 2 )∞ (q 4 ; q 4 )2∞ (q 12 ; q 12 )2∞
(q 2 ; q 2 )3∞ (q 12 ; q 12 )6∞
−q 4 4 2 6 6 9 ,
(q 6 ; q 6 )7∞
(q ; q )∞ (q ; q )∞
which is the second identity.
5
Lemma 2.2. The following 3–dissections hold:
(q 2 ; q 2 )∞
(q; q)∞
(q 6 ; q 6 )∞ (−q 12 , −q 15 , q 27 ; q 27 )∞ + q(−q 6 , −q 21 , q 27 ; q 27 )∞
(q 3 ; q 3 )2∞
=
+q 2 (−q 3 , −q 24 , q 27 ; q 27 )∞ ,
(q; q)∞
(q 2 ; q 2 )∞
(q 3 ; q 3 )∞ 21 33 54 54
(q , q , q ; q )∞ − q(q 15 , q 39 , q 54 ; q 54 )∞
(q 6 ; q 6 )2∞
=
−q 5 (q 3 , q 51 , q 54 ; q 54 )∞ .
Proof. The first identity follows from Entry 31 [4, p. 48] by setting a = q, b = q 2
and n = 3, the second by setting a = −q, b = −q 5 and n = 3.
Lemma 2.3. The following 2–dissection holds:
1
(q; q)4∞
=
(q 4 ; q 4 )14
∞
2
8 ; q 8 )4
(q ; q 2 )14
(q
∞
∞
+ 4q
(q 4 ; q 4 )2∞ (q 8 ; q 8 )4∞
.
(q 2 ; q 2 )10
∞
Proof. With
φ(q) =
∞
X
2
qn =
−∞
X 2
(q 2 ; q 2 )5∞
(q 2 ; q 2 )2∞
and ψ(q) =
q (n +n)/2 =
2
4
4
2
(q; q)∞ (q ; q )∞
(q; q)∞
n≥0
it is easy to show that (Berndt [4, p. 40], Hirschhorn [9])
φ(q)2 = φ(q 2 )2 + 4qψ(q 4 )2 .
That is,
(q 2 ; q 2 )10
(q 4 ; q 4 )10
(q 8 ; q 8 )4∞
∞
∞
=
+
4q
.
(q; q)4∞ (q 4 ; q 4 )4∞
(q 2 ; q 2 )4∞ (q 8 ; q 8 )4∞
(q 4 ; q 4 )2∞
The result follows.
Lemma 2.4. The following 2–dissection holds:
1
(q; q)∞ (q 3 ; q 3 )∞
=
(q 8 ; q 8 )2∞ (q 12 ; q 12 )5∞
2
2
2
(q ; q )∞ (q 4 ; q 4 )∞ (q 6 ; q 6 )4∞ (q 24 ; q 24 )2∞
(q 4 ; q 4 )5 (q 24 ; q 24 )2
+q 2 2 4 6 6 2∞ 8 8 2 ∞ 12 12 .
(q ; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )∞
6
Proof. It can be shown that (Berndt [4, (36.8)], Cooper and Hirschhorn [6,
lemma (xxxiii)])
ψ(q)ψ(q 3 ) = φ(q 6 )ψ(q 4 ) + qφ(q 2 )ψ(q 12 ).
That is,
(q 2 ; q 2 )2∞ (q 6 ; q 6 )2∞
(q 8 ; q 8 )2∞
(q 12 ; q 12 )5∞
=
(q; q)∞ (q 3 ; q 3 )∞
(q 6 ; q 6 )2∞ (q 24 ; q 24 )2∞ (q 4 ; q 4 )∞
(q 4 ; q 4 )5
(q 24 ; q 24 )2∞
+q 2 2 2 8∞ 8 2
.
(q ; q )∞ (q ; q )∞ (q 12 ; q 12 )∞
The result follows.
Lemma 2.5. The following 3–dissections hold:
(q 2 ; q 2 )2∞
(q; q)∞
=
(q 6 ; q 6 )∞ (q 9 ; q 9 )2∞
(q 18 ; q 18 )2
+q 9 9 ∞,
3
3
18
18
(q ; q )∞ (q ; q )∞
(q ; q )∞
(q, q)∞
=
(q 12 , q 15 , q 27 ; q 27 )∞ − q(q 6 , q 21 , q 27 ; q 27 )∞ − q 2 (q 3 , q 24 , q 27 ; q 27 )∞ .
Proof. The first identity follows from the first equality in Corollary (ii) [4, p.
49], the second from Entry 31 [4, p. 48] by setting a = −q, b = −q 2 and
n = 3.
3
Proofs of Theorems
Proof. (Theorem (1.1)) We have, by Lemma (2.1),
RC(q) =
=
=
(q; q)∞ (q 6 ; q 6 )3∞
(q; q 2 )∞
=
(q 3 ; q 6 )3∞
(q 2 ; q 2 )∞ (q 3 ; q 3 )3∞
2 2
6 6 3
(q ; q )∞ (q ; q )∞ (q 4 ; q 4 )2∞ (q 12 ; q 12 )2∞
(q 2 ; q 2 )3∞ (q 12 ; q 12 )6∞
−q 4 4 2 6 6 9
(q 2 ; q 2 )∞
(q 6 ; q 6 )7∞
(q ; q )∞ (q ; q )∞
4 4 2
12 12 2
2 2 2
12 12 6
(q ; q )∞ (q ; q )∞
(q ; q ) (q ; q )
− q 4 4 ∞2 6 6 6∞
(q 6 ; q 6 )4∞
(q ; q )∞ (q ; q )∞
7
and
1/RC(q) =
=
=
(q 2 ; q 2 )∞ (q 3 ; q 3 )3∞
(q 6 ; q 6 )3∞ (q; q)∞
4 4 3 6 6 2
(q ; q )∞ (q ; q )∞
(q 2 ; q 2 )∞
(q 12 ; q 12 )3∞
+q 4 4
(q 6 ; q 6 )3∞ (q 2 ; q 2 )2∞ (q 12 ; q 12 )∞
(q ; q )∞
4 4 3
2 2
(q ; q )∞
(q ; q )∞ (q 12 ; q 12 )3∞
+
q
.
(q 2 ; q 2 )∞ (q 6 ; q 6 )∞ (q 12 ; q 12 )∞
(q 4 ; q 4 )∞ (q 6 ; q 6 )3∞
Proof. (Theorem (1.2)).
We have, by Lemma (2.2),
(q 6 ; q 6 )3∞ (q; q)∞
(q 3 ; q 3 )3∞ (q 2 ; q 2 )∞
(q 6 ; q 6 )∞ 21 33 54 54
(q , q , q ; q )∞ − q(q 15 , q 39 , q 54 ; q 54 )∞
(q 3 ; q 3 )2∞
RC(q) =
=
−q 5 (q 3 , q 51 , q 54 ; q 54 )∞
(q 6 ; q 6 )∞ (q 21 , q 33 , q 54 ; q 54 )∞
(q 6 ; q 6 )∞ (q 15 , q 39 , q 54 ; q 54 )∞
−
q
(q 3 ; q 3 )2∞
(q 3 ; q 3 )2∞
6 6
(q ; q )∞ (q 3 , q 51 , q 54 ; q 54 )∞
−q 5
(q 3 ; q 3 )2∞
=
and
1/RC(q) =
=
(q 3 ; q 3 )3∞ (q 2 ; q 2 )∞
(q 6 ; q 6 )3∞ (q; q)∞
(q 3 ; q 3 )∞ (−q 12 , −q 15 , q 27 ; q 27 )∞ + q(−q 6 , −q 21 , q 27 ; q 27 )∞
(q 6 ; q 6 )2∞
+q 2 (−q 3 , −q 24 , q 27 ; q 27 )∞
=
(q 3 ; q 3 )∞ (−q 12 , −q 15 , q 27 ; q 27 )∞
(q 6 ; q 6 )2∞
(q 3 ; q 3 )∞ (−q 6 , −q 21 , q 27 ; q 27 )∞
+q
(q 6 ; q 6 )2∞
(q 3 ; q 3 )∞ (−q 3 , −q 24 , q 27 ; q 27 )∞
.
+q 2
(q 6 ; q 6 )2∞
8
Proof. (Theorem (1.3)) We have, by Theorem (1.1) and Lemma (2.3) with q 6
for q and Lemma (2.1) with q 2 for q,
RC(q) =
=
=
(q 4 ; q 4 )2∞ (q 12 ; q 12 )2∞
(q 2 ; q 2 )2 (q 12 ; q 12 )6
− q 4 4 ∞2 6 6 6∞
6
6
4
(q ; q )∞
(q ; q )∞ (q ; q )∞
(q 4 ; q 4 )2∞ (q 12 ; q 12 )2∞
24 24 2
48 48 4
(q 24 ; q 24 )14
∞
6 (q ; q )∞ (q ; q )∞
×
+
4q
48 48 4
(q 12 ; q 12 )14
(q 12 ; q 12 )10
∞ (q ; q )∞
∞
(q 12 ; q 12 )6∞
−q 4 4 2
(q ; q )∞
4 4
2
4 4 3
24 24 6
(q ; q )∞ (q 8 ; q 8 )2∞ (q 24 ; q 24 )2∞
2 (q ; q )∞ (q ; q )∞
×
−
q
(q 12 ; q 12 )7∞
(q 8 ; q 8 )2∞ (q 12 ; q 12 )9∞
(q 4 ; q 4 )2∞ (q 24 ; q 24 )14
∞
48 48 4
(q 12 ; q 12 )12
∞ (q ; q )∞
8 8 4 24 24 4
4 4 4
24 24 12
(q ; q )∞ (q ; q )∞
4 (q ; q )∞ (q ; q )∞
+
q
−q
(q 12 ; q 12 )8∞
(q 8 ; q 8 )4∞ (q 12 ; q 12 )12
∞
4 4 2
24 24 2
48 48 4
4 4 2
(q
;
q
)
(q
;
q
)
(q
;
q
)
(q
;
q
)∞ (q 24 ; q 24 )8∞
∞
∞
∞
+4q 6
+ 2q 3
12
12
8
(q , q )∞
(q 12 ; q 12 )10
∞
and, by Theorem (1.1) and Lemma (2.4) with q 2 for q and Lemma (2.1) with
q 2 for q,
=
=
1/RC(q) =
(q 4 ; q 4 )3∞
2
2
(q ; q )∞ (q 6 ; q 6 )∞ (q 12 ; q 12 )∞
(q 4 ; q 4 )3∞
(q 12 ; q 12 )∞
+q
(q 2 ; q 2 )∞ (q 12 ; q 12 )3∞
(q 4 ; q 4 )∞ (q 6 ; q 6 )3∞
(q 16 ; q 16 )2∞ (q 24 ; q 24 )5∞
4
4
2
(q ; q )∞ (q 8 ; q 8 )∞ (q 12 ; q 12 )4∞ (q 48 ; q 48 )2∞
(q 8 ; q 8 )5 (q 48 ; q 48 )2∞
+q 2 4 4 4 12 12 ∞
(q ; q )∞ (q ; q )2∞ (q 16 ; q 16 )2∞ (q 24 ; q 24 )∞
4 4 3
24 24 6
(q 12 ; q 12 )3
(q 4 ; q 4 )∞ (q 8 ; q 8 )2∞ (q 24 ; q 24 )2∞
2 (q ; q )∞ (q ; q )∞
+q 4 4 ∞
−
q
(q ; q )∞
(q 12 ; q 12 )7∞
(q 8 ; q 8 )2∞ (q 12 ; q 12 )9∞
(q 8 ; q 8 )2∞ (q 24 ; q 24 )2∞
(q 4 ; q 4 )∞ (q 16 ; q 16 )2∞ (q 24 ; q 24 )5∞
+q
8
8
12
12
5
48
48
2
(q ; q )∞ (q ; q )∞ (q ; q )∞
(q 12 ; q 12 )4∞
(q 8 ; q 8 )5∞ (q 48 ; q 48 )2∞
(q 4 ; q 4 )2 (q 24 ; q 24 )6
+q 2 4 4
− q 3 8 8 2∞ 12 12 6∞ .
12
12
3
16
16
2
24
24
(q ; q )∞ (q ; q )∞ (q ; q )∞ (q ; q )∞
(q ; q )∞ (q ; q )∞
9
Proof. (Theorem (1.4))
We have, by Theorem (1.1) and Lemmas (2.2) and (2.5),
36 36 2
(q 12 ; q 12 )2∞ (q 12 ; q 12 )∞ (q 18 ; q 18 )2∞
2 (q ; q )∞
RC(q) =
×
+ q 18 18
(q 6 ; q 6 )4∞
(q 6 ; q 6 )∞ (q 36 ; q 36 )∞
(q ; q )∞
(q 24 , q 30 , q 54 ; q 54 )∞ − q 2 (q 12 , q 42 , q 54 ; q 54 )∞ − q 4 (q 6 , q 48 , q 54 ; q 54 )∞
(q 12 ; q 12 )6∞
(q 6 ; q 6 )6∞
(q 6 ; q 6 )
∞ 42 66 108 108
×
(q , q , q ; q )∞ − q 2 (q 30 , q 78 , q 108 ; q 108 )∞
(q 12 ; q 12 )2∞
−q
−q 10 (q 6 , q 102 , q 108 ; q 108 )∞
=
2
(q 12 ; q 12 )3∞ (q 18 , q 18 )2∞ (q 24 , q 30 , q 54 ; q 54 )∞
(q 6 ; q 6 )5∞ (q 36 ; q 36 )∞
12 12 2
36 36 2
6 48 54 54
6 (q ; q )∞ (q ; q )∞ (q , q , q ; q )∞
−q
(q 6 ; q 6 )4∞ (q 18 ; q 18 )∞
12 12 2 42 66 108 108 2
(q ; q )∞ (q , q , q ; q )∞
−q
(q 6 ; q 6 )4∞
12 12 2
6 30 78 102 108 108 108
, q , q ; q )∞
12 (q ; q )∞ (q , q , q , q
+2q
(q 6 ; q 6 )4∞
12 12 3 18 18 2 12 42 54 54
(q ; q )∞ (q ; q )∞ (q , q , q ; q )∞
−q 2
(q 6 ; q 6 )5∞ (q 36 ; q 36 )∞
(q 12 ; q 12 )2∞ (q 36 ; q 36 )2∞ (q 24 , q 30 , q 54 ; q 54 )∞
−
(q 6 ; q 6 )4 (q 18 ; q 18 )∞
12 12 2 30 42 66 78 108 ∞
(q ; q )∞ (q , q , q , q , q , q 108 ; q 108 )∞
+q 3 2
(q 6 ; q 6 )4∞
12 12 2
6 102 108 108 2
, q ; q )∞
18 (q ; q )∞ (q , q
−q
(q 6 ; q 6 )4∞
12 12 3 18 18 2 6 48 54 54
(q ; q )∞ (q ; q )∞ (q , q .q ; q )∞
−q 4
(q 6 ; q 6 )5∞ (q 36 ; q 36 )∞
(q 12 ; q 12 )2∞ (q 36 ; q 36 )2∞ (q 12 , q 42 , q 54 ; q 54 )∞
+
(q 6 ; q 6 )4∞ (q 18 ; q 18 )∞
12 12 2 30 78 108 108 2
(q ; q )∞ (q , q , q ; q )∞
−q 5
(q 6 ; q 6 )4∞
(q 12 ; q 12 )2∞ (q 6 , q 42 , q 66 , q 102 , q 108 , q 108 ; q 108 )∞
,
−2q 6
(q 6 ; q 6 )4∞
10
and
1/RC(q) =
1
(q 6 ; q 6 )∞ (q 12 ; q 12 )∞
36 36 2
(q 12 ; q 12 )∞ (q 18 ; q 18 )2∞
2 (q ; q )∞
+
q
(q 6 ; q 6 )∞ (q 36 ; q 36 )∞
(q 18 ; q 18 )∞
× (q 48 , q 60 , q 108 ; q 108 )∞ − q 4 (q 24 , q 84 , q 108 ; q 108 )∞
−q 8 (q 12 , q 96 , q 108 ; q 108 )∞
(q 12 ; q 12 )3∞
(q 6 ; q 6 )3∞
(q 6 ; q 6 )
∞ 42 66 108 108
×
(q , q , q ; q )∞ − q 2 (q 30 , q 78 , q 108 ; q 108 )∞
12
12
(q ; q )2∞
−q 10 (q 6 , q 102 , q 108 ; q 108 )∞
+q
=
(q 18 ; q 18 )2∞ (q 48 , q 60 , q 108 ; q 108 )∞
(q 6 ; q 6 )2∞ (q 36 ; q 36 )∞
(q 36 ; q 36 )2 (q 24 , q 84 , q 108 ; q 108 )∞
−q 6 6 6 ∞ 12 12
(q ; q )∞ (q ; q )∞ (q 18 ; q 18 )∞
(q 12 ; q 12 )∞ (q 42 , q 66 , q 108 ; q 108 )∞
+q
(q 6 ; q 6 )2
36 36 2 48 ∞60 108 108
(q ; q )∞ (q , q , q ; q )∞
+q 2
(q 6 ; q 6 )∞ (q 12 ; q 12 )∞ (q 18 ; q 18 )∞
18 18 2
12 96 108 108
; q )∞
6 (q ; q )∞ (q , q , q
−q
(q 6 ; q 6 )2∞ (q 36 ; q 36 )∞
12 12
30 78 108 108
(q ; q )∞ (q , q , q ; q )∞
−q 3
(q 6 ; q 6 )2
18 18 2 24 ∞
(q ; q )∞ (q , q 84 , q 108 ; q 108 )∞
−q 4
(q 6 ; q 6 )2∞ (q 36 ; q 36 )∞
36 36 2
12 96 108 108
; q )∞
6 (q ; q )∞ (q , q , q
+q
(q 6 ; q 6 )∞ (q 12 ; q 12 )∞ (q 18 ; q 18 )∞
12 12
6 102 108 108
(q ; q )∞ (q , q , q ; q )∞
−q 11
.
(q 6 ; q 6 )2∞
Proof. (Theorem (1.5))
We have, from Theorems (1.2) and (1.4),
11
X
a3n q n =
n≥0
X
b6n+1 q n
n≥0
2
= 1/(q, q, q , q 3 , q 3 , q 4 , q 5 , q 5 , q 6 , q 7 , q 8 , q 9 , q 9 , q 10 ,
q 11 , q 12 , q 13 , q 13 , q 14 , q 15 , q 15 , q 16 , q 17 , q 17 ; q 18 )∞ ,
X
a3n+1 q n =
n≥0
X
b6n+3 q n
n≥0
2
= −1/(q, q, q , q 3 , q 3 , q 4 , q 5 , q 6 , q 7 , q 7 , q 8 , q 9 , q 9 , q 10 ,
q 11 , q 11 , q 12 , q 13 , q 14 , q 15 , q 15 , q 16 , q 17 , q 17 ; q 18 )∞ ,
X
a3n+2 q n =
n≥0
X
b6n+5 q n
n≥0
= −q/(q, q 2 , q 3 , q 3 , q 4 , q 5 , q 5 , q 6 , q 7 , q 7 , q 8 , q 9 , q 9 , q 10 ,
q 11 , q 11 , q 12 , q 13 , q 13 , q 14 , q 15 , q 15 , q 16 , q 17 ; q 18 )∞ .
Also, from Theorem (1.2),
X
b3n q n =
n≥0
=
(−q 4 , −q 5 , q 9 ; q 9 )∞
(−q 4 , −q 5 , q 9 ; q 9 )∞
=
ψ(q)
(−q, −q 2 , q 4 ; q 4 )∞
(−q 4 , −q 14 , q 18 ; q 18 )∞ (−q 5 , q 9 , −q 13 ; q 18 )∞
,
(q 4 ; q 4 )∞
(−q; q 2 )∞
so
X
(−q 4 , −q 14 , q 18 ; q 18 )∞ (q 5 , −q 9 , q 13 ; q 18 )∞
(q 4 ; q 4 )∞
(q; q 2 )∞
P∞
2
(−q 9 ; q 18 )∞ −∞ q 9n −5n
= 4 4
(q ; q )∞ (q, q 3 , q 7 , q 9 , q 11 , q 15 , q 17 ; q 18 )∞
(−1)n b3n q n =
n≥0
= f (q 4 , q 14 )f (q 9 , q 27 )/(q, q 3 , q 4 , q 7 , q 8 , q 9 , q 11 , q 12 , q 15 , q 16 , q 17 ,
q 19 , q 20 , q 21 , q 24 , q 25 , q 27 , q 28 , q 29 , q 32 , q 33 , q 35 , q 36 , q 36 ; q 36 )∞ ,
12
and similarly,
X
(−1)n b3n+1 q n
n≥0
= f (q 2 , q 16 )f (q 9 , q 27 )/(q, q 3 , q 4 , q 5 , q 8 , q 9 , q 12 , q 13 , q 15 , q 16 , q 17 ,
q 19 , q 20 , q 21 , q 23 , q 24 , q 27 , q 28 , q 31 , q 32 , q 33 , q 35 , q 36 , q 36 ; q 36 )∞ ,
X
(−1)n b3n+2 q n
n≥0
= f (q 8 , q 10 )f (q 9 , q 27 )/(q, q 3 , q 4 , q 5 , q 7 , q 8 , q 9 , q 11 , q 12 , q 13 , q 15 , q 16 ,
q 20 , q 21 , q 23 , q 24 , q 25 , q 27 , q 28 , q 29 , q 31 , q 32 , q 33 , q 36 , q 36 ; q 36 )∞ .
The results follow.
Acknowledgement. The latter part of the proof of Theorem (1.5) is essentially
the proof of Chan and Yesilyurt [5, §4].
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[email protected]
[email protected]
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