Tamsui Oxford Journal of Information and Mathematical Sciences 31(1) (2017) 81-99
Aletheia University
On the expansion of Ramanujan's continued
fraction of order sixteen
y
A. Vanitha
Department of Mathematics, University of Mysore,
Manasagangotri, Mysuru 570 006, INDIA.
Received June 22, 2013, Accepted June 26, 2013.
Abstract
We give 2-, 4-, 8- and 16-dissections of a continued fraction of order
sixteen. We show that the sign of the coecients in the power series
expansion of the continued fraction of order sixteen is periodic with
period 16. We also give combinatorial interpretations for the coecients
in the power series expansion of the continued fraction of order sixteen.
Keywords and Phrases: Continued fractions, dissections, periodicity of
sign of coecients.
1. Introduction
Throughout this paper, we assume that jq j < 1 and use the standard notation
1
(a; q )1 = (1 aq n );
Y
n=0
(a1 ; a2 ; a3 ; : : : ; an ; q )1 = (a1 ; q )1 (a2 ; q )1 (a3 ; q )1 : : : ; (an ; q )1 ;
[z ; q ]1 := (z ; q )1 (z
and
1
; )1
q q
[z1 ; z2 ; z3 ; : : : ; zn ; q ]1 = [z1 ; q ]1 [z2 ; q ]1 [z3 ; q ]1 : : : ; [zn ; q ]1 :
2000 Mathematics Subject Classication. Primary 11A55,30B70.
y Corresponding author.
E-mail: [email protected]
A. Vanitha
82
The celebrated Rogers{Ramanujan continued fraction and its product representation is
1 q q2 q3
1 + 1 + 1 + 1 + ...
( ) :=
R q
X
where
1
=
(
f(
f
q;
2
q ;
)
;
q )
q
4
(1.1)
3
;
jabj < 1;
1
is Ramanujan's general theta function. The function f (a; b) satisfy the well
known Jacobi triple product identity
(
)=
f a; b
a
n(n+1)=2
n(n
b
1)=2
n=
(
)=(
f a; b
; )1 (
a ab
; )1 (ab; ab)1 :
b ab
On page 36 of his \lost" notebook, Ramanujan recorded four q -series representations of the famous Rogers-Ramanujan continued fraction. Recently, C.
Adiga, N. A. S. Bulkhali, Y. Simsek and H. M. Srivastava [3] have established
two q -series representations of Ramanujan's continued fraction found in his
\lost" notebook. They have also established three equivalent integral representations and modular equations for a special case of this continued fraction.
On employing q -identities of Ramanujan found in his lost notebook, Srivastava
et al. [19] established a number of results involving continued fractions of the
form involved in (1.1).
G. E. Andrews et al. [7] investigated combinatorial partition identities associated with the following general family:
(
R s; t; l; u; v; w
) :=
X
0 1
1 8@nA+tn
n=0
q
2
X
[n=u]
j =0
( 1)j
01
j
uv @ A+(w
q
(q ; q )n
2
uj
ul)j
(q uv ; q uv )j
:
(1.2)
In [18], Srivastava and M. P. Chaudhary, proved several results associated
with the family R(s; t; l; u; v; w) which depict the inter relationship between
q -product identities, continued fraction identities and combinatorial partition
identities.
S. Ramanujan [16, p. 50] gave the 2-dissections and 5-dissections of R(q )
and its reciprocal, and these were rst proved by Andrews [5] and M. D.
Hirschhorn [9] respectively. The periodic behavior of the sign of the coefcients in the series expansion of R(q ) and its reciprocal were observed by
On the expansion of Ramanujan's continued fraction of order sixteen
83
B. Richmond and G. Szekers [17]. Hirschhorn [9], conjectured formulas for
the 4-dissections of R(q ) and its reciprocal, and these were rst proved by
R. Lewis and Z. -G. Liu [13]. In [21], O. X. M. Yao and E. X. W. Xia gave
generalizations of Hirschhorn's formulas on the 4-dissections of R(q ) and its
reciprocal. Recently, Hirschhorn [11] gave a simple proof of the 2-dissections
and 4-dissections of R(q ) and its reciprocal by using Jacobi's triple product
identity.
The beautiful Ramanujan{Gollnitz{Gordon continued fraction and its product
representation is
( ) :=
G q
2
4
q
q
1
1 + q + 1 + q3 + 1 + q5 + . . .
=
(
f(
f
q;
3
q ;
)
:
q )
q
7
5
In [10], Hirschhorn found the 8-dissections of G(q ) and its reciprocal and also
proved that the sign of the coecients in the power series expansion of G(q )
and its reciprocal are periodic with period 8 and in particular that certain coecients are zero, a phenomenon rst observed and proved by Richmond and
Szekers [17]. Recently, S. H. Chan and H. Yesilyurt [8] show the periodicity of
signs of a large number of quotients of certain innite products. Their results
include as special cases, results of Ramanujan, Andrews, Bressoud, Richmond,
Szekers and Hirschhorn.
The famous Ramanujan's cubic continued fraction and its product representation is
5
1 q + q2 q2 + q4
f ( q;
q )
=
:
H (q ) :=
3
3
1 + 1 + 1 + ...
f( q ;
q )
In [20], B. Srivastava has studied the 2- and 4-dissections of the continued
fraction of H (q ) 1 . In [12], Hirschhorn and Roselin have studied the 2-, 3-, 4and 6-dissections of H (q ) and its reciprocal and also shows the sign of the coecients in the power series expansion of H (q ) and its reciprocal, are periodic
with period 3 and 6 respectively.
The fascinating continued fraction of order twelve and its product representation is
3
2
4
11
q (1
q )(1
q )
f ( q;
q )
(1 q )
U (q ) :=
=
:
5
7
(1 q 3 ) + (1 q 3 )(1 + q 6 ) + . . .
f( q ;
q )
In [14], B. L. S. Lin has studied the 2-, 3-, 4-, 6- and 12-dissections of U (q )
and its reciprocal and also shows the sign of the coecients in the power series
A. Vanitha
84
expansion of U (q ) and its reciprocal, are periodic with period 12.
The Ramanujan's continued fraction of order six and its product representation
is
5
f ( q;
q )
(q 1=2 q 3=2 ) (q 1=4 q 1=4 )(q 7=4 q 7=4 )
=
:
X (q ) :=
2
4
(1 q 3=2 ) +
(1 q 3=2 )(1 + q 3 )
+ ...
f( q ;
q )
In [4], Adiga et al. have studied the 2- and 4-dissections of X (q ) and its
reciprocal and also shows that the sign of the coecients in the power series expansion of X (q ) and its reciprocal, are periodic with period 2 and 6
respectively. In [2], Adiga, Bulkhali, D. Ranganatha and Srivastava have established several modular relations for the Rogers-Ramanujan type functions
of order eleven which are analogous to Ramanujan's forty identities for RogersRamanujan functions. Furthermore, they gave interesting partition theoretic
interpretation of some of the modular relations. Motivated by the above works
on continued fractions, we shall consider the power series expansion of the following continued fraction of order sixteen:
) q 4 (1
4
q ) + (1
)(1 q 5 ) q 4 (1
4
8
q )(1 + q ) +
(1
)(1 q 13 )
:
7
4
16
q ;
q )(1 + q )
+. . .
(1.3)
Ramanujan has recorded several continued fraction in his notebooks. One of
the fascinating continued fraction identities recorded by Ramanujan as Entry
12 in his second notebook [15] is
1
(a
(a2 q 3 ; q 4 )1 (b2 q 3 ; q 4 )1
=
2
4
2
4
(a q ; q )1 (b q ; q )1
(1 ab) + (1
)(b aq ) (a
2
ab)(1 + q ) + (1
( ) :=
A q
(
f(
f
q;
(1
)
=
9
q )
(1
q
15
q
q
3
(
A(q ) =
f(
f
( )=
B q
(
f(
f
q;
7
q ;
7
q ;
q;
a
15
bq
3
= q 3=2 and b = q 5=2 , we obtain
X
X
1
)
n
=
a(n)q ;
9
q )
n=0
1
9
q )
n
=
b(n)q :
15
q )
n=0
q
11
)(b aq 3 )
;
4
ab)(1 + q )
+ ...
jqj < 1; jabj < 1:
(1.4)
bq
For a proof of (1.4), see Adiga et al. [1].
In (1.4), replacing q by q 4 and then setting
(1.3).
For n 0, we dene a(n) and b(n) by
q
(1.5)
(1.6)
On the expansion of Ramanujan's continued fraction of order sixteen
85
The main object of this paper is to study dierent dissections of A(q ) and
B (q ). In Section 2, we prove the 2- and 4-dissections of A(q ) and B (q ). In
Section 3, we prove the 8-dissections of A(q ) and B (q ). Furthermore, we prove
that the signs of the coecients in the power series expansion of A(q ) and
B (q ) are periodic with period 16. In Section 4, we prove the 16-dissections of
A(q ) and B (q ). In the last section, we give combinatorial interpretations of
the coecients in the power series expansion of A(q ) and B (q ).
A(q) and B (q)
2. 2- and 4-Dissections of
In this section, we present 2- and 4-dissections of A(q ) and
our results, we need the following Lemmas.
( ). To prove
B q
Lemma 2.1. If ab = cd, then
(
) (
) = f (ac; bd)f (ad; bc) + af (b=c; ac2 d)f (b=d; acd2 ):
(2.1)
f a; b f c; d
Proof. Adding Entries 29(i) and 29(ii) in [1], we obtain the result.
Lemma 2.2. [1, p. 46, Entry 30 (iv)] One has
(
) (
f a; b f
) = f(
a;
b
2
a ;
) (
2
b f
ab;
)
(2.2)
ab :
Theorem 2.3. Let a(n) be as dened in (1.5). Then
X
1
X
1
a
(2n)q n =
n=0
a
[q 4 ; q 5 ; q 16 ]1
;
[q 7 ; q 8 ; q 16 ]1
[q 3 ; q 4 ; q 16 ]1
:
[q 7 ; q 8 ; q 16 ]1
(2n + 1)q n =
n=0
Proof. We have
1
X
( )
a n q
n
=
n=0
(
f(
f
By employing (2.1) with a =
f
(
q;
q
15
)f (q 7 ; q 9 ) = f (
8
q ;
q;
7
q ;
q
) f(
=
9
q )
f(
q
15
,b=
q
24
)f (
(2.3)
q
15
q
q;
7
q ;
(2.4)
)f (q 7 ; q 9 )
:
9
7
9
q )f (q ; q )
q
15
(2.5)
, c = q 7 and d = q 9 , we obtain
10
;
q
22
)
qf
(
8
q ;
q
24
)f (
6
q ;
q
26
(2.6)
):
A. Vanitha
86
Setting a = q 7 and b = q 9 in (2.2), we nd that
(
) (
7
9
f q ;q f
7
q ;
q
9
) = f(
q
14
;
q
18
)f (
q
16
;
q
16
Combining (2.5), (2.6) and (2.7), we obtain
1
8
24
10
22
8
24
f( q ;
q )f ( q ;
q )
qf ( q ;
q )f (
n
a(n)q =
14
18
16
16
f( q ;
q )f ( q ;
q )
n=0
X
):
(2.7)
6
q ;
q
26
)
:
It follows immediately that
1
8
24
10
22
f( q ;
q )f ( q ;
q )
2n
a(2n)q
=
;
14
18
16
16
f( q ;
q )f ( q ;
q )
n=0
1
8
24
6
26
f( q ;
q )f ( q ;
q )
2n+1
:
a(2n + 1)q
= q
14
18
16
16
f( q ;
q )f ( q ;
q )
n=0
X
X
Changing q to q 1=2 in the above equations, we obtain (2.3) and (2.4).
Theorem 2.4. Let b(n) be as dened in (1.6). Then
X
1
X
1
(2n)q
b
n=0
n
[q 4 ; q 5 ; q 16 ]1
=
;
[q; q 8 ; q 16 ]1
(2.8)
[q 3 ; q 4 ; q 16 ]1
:
[q; q 8 ; q 16 ]1
(2.9)
(2n + 1)q n =
b
n=0
Proof. The proof of Theorem 2.4 is similar to that of Theorem 2.3.
X
Theorem 2.5. We have
1
X
X
X
1
a
(4n)q n =
n=0
[q 2 ; q 6 ; q 6 ; q 16 ]1
;
[q 4 ; q 4 ; q 8 ; q 16 ]1
a
(4n + 1)q n =
[q 2 ; q 5 ; q 6 ; q 6 ; q 16 ]1
;
[q 4 ; q 4 ; q 7 ; q 8 ; q 16 ]1
a
(4n + 2)q n =
q
a
(4n + 3)q n =q
n=0
1
n=0
1
n=0
2
[q; q 2 ; q 2 ; q 6 ; q 16 ]1
;
[q 4 ; q 4 ; q 7 ; q 8 ; q 16 ]1
[q 2 ; q 2 ; q 3 ; q 6 ; q 16 ]1
:
[q 4 ; q 4 ; q 7 ; q 8 ; q 16 ]1
(2.10)
(2.11)
(2.12)
(2.13)
On the expansion of Ramanujan's continued fraction of order sixteen
87
Proof. From (2.3), we see that
X
1
a
(2n)q n =
n=0
[q 4 ; q 16 ]1 f (
[q 8 ; q 16 ]1 f (
By employing (2.1) with a =
f
(
5
q ;
q
11
)f (q 7 ; q 9 ) = f (
q
q
12
;
5
,b=
q
20
q
)f (
q
7
q ;
11
q
)f (q 7 ; q 9 )
:
9
7
9
q )f (q ; q )
5
q ;
11
(2.14)
, c = q 7 and d = q 9 , we obtain
14
;
q
18
)
5
q f
(
4
q ;
q
28
)f (
Combining above identity, (2.7) and (2.14), we obtain
f ( q12 ; q20 )f ( q14 ; q18 ) q5 f ( q4 ; q28 )f (
1
4
16
X
[
q ;q
]
1
n
a(2n)q =
14 ;
18 )f ( q 16 ;
16 )
[q8; q16]1
f( q
q
q
n=0
2
q ;
2
q ;
q
q
30
X
Proofs of (2.11) and (2.13) are similar.
Theorem 2.6. We have
X
X
X
1
X
1
n=0
(4n)q
b
n
[q 2 ; q 3 ; q 6 ; q 6 ; q 16 ]1
=
;
[q; q 4 ; q 4 ; q 8 ; q 16 ]1
(2.15)
b
(4n + 1)q n =
[q 2 ; q 2 ; q 6 ; q 7 ; q 16 ]1
;
[q; q 4 ; q 4 ; q 8 ; q 16 ]1
(2.16)
b
(4n + 2)q n =
[q 2 ; q 2 ; q 5 ; q 6 ; q 16 ]1
;
[q; q 4 ; q 4 ; q 8 ; q 16 ]1
(2.17)
b
(4n + 3)q n =
[q 2 ; q 6 ; q 6 ; q 16 ]1
:
[q 4 ; q 4 ; q 8 ; q 16 ]1
(2.18)
n=0
1
n=0
1
n=0
Proof. The proof of Theorem 2.6 is similar to that of Theorem 2.5.
By Theorems 2.3, 2.4, 2.5 and 2.6, we have the following corollary:
):
)
It follows immediately that
1
[q 2 ; q 6 ; q 6 ; q 16 ]1
n
a(4n)q =
;
[q 4 ; q 4 ; q 8 ; q 16 ]1
n=0
1
2
2
6
16
n
2 [q; q ; q ; q ; q ]1
a(4n + 2)q =
q
:
4
4
7
8
16
[
q ; q ; q ; q ; q ]1
n=0
X
30
:
A. Vanitha
88
Corollary 2.7. For n 0,
(4n) = b(4n + 3)
a
and
X
X
1
a
(2n)q
n=0
1
a
X
X
1
n
(2n)q
b
1
n=0
=
1
(2n + 1)q n
X
X
n
(2n + 1)q n
a
n=0
X
X
and
(4n)q
n=0
1
;
b
n=0
a
=
(4n + 1)q n
n=0
1
a
(4n + 2)q
n=0
a
(4n + 3)q
n=0
n
(4n + 1)q n
b
n=0
1
(4n + 2)q n
b
n=0
(4n + 3)q n
b
=
1
X
X
1
n
X
X
1
n
q
n=0
1
:
(4n)q
b
n
n=0
Remark 2.8. Theorem 2.5 and Theorem 2.6 can also be proved by applying
an identity proved by Andrews and D. Bressoud [6, Theorem 1].
3. 8-Dissections of
A(q) and B (q)
In this section, we present 8-dissections of A(q ) and B (q ) and also present the
signs of the coecients in the power series expansion of A(q ) and B (q ) are
periodic with period 16.
Theorem 3.1. We have
X
1
X
1
n=0
a
(8n)q n =
n=0
a
(8n + 1)q n =
[q; q 3 ; q 3 ; q 8 ]1
;
[q 2 ; q 2 ; q 4 ; q 8 ]1
[q; q 3 ; q 3 ; q 5 ; q 5 ; q 7 ; q 16 ]1
;
[q 2 ; q 2 ; q 4 ; q 4 ; q 6 ; q 8 ; q 16 ]1
(3.1)
(3.2)
On the expansion of Ramanujan's continued fraction of order sixteen
X
X
X
X
X
X
1
a
(8n + 2)q n =
q
n=0
1
n
a
(8n + 3)q =
a
(8n + 4)q n =0;
a
(8n + 5)q n =q 2
a
(8n + 6)q n =q
a
(8n + 7)q n =
q
n=0
1
[q; q; q 3 ; q 5 ; q 5 ; q 7 ; q 16 ]1
;
[q 2 ; q 2 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
2
[q; q; q 3 ; q 3 ; q 5 ; q 7 ; q 16 ]1
;
[q 2 ; q 4 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
89
(3.3)
(3.4)
(3.5)
n=0
1
n=0
1
n=0
1
n=0
[q; q; q 3 ; q 3 ; q 5 ; q 5 ; q 16 ]1
;
[q 2 ; q 4 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
[q; q; q 3 ; q 3 ; q 5 ; q 7 ; q 16 ]1
;
[q 2 ; q 2 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
[q; q; q 3 ; q 5 ; q 5 ; q 7 ; q 16 ]1
:
[q 2 ; q 2 ; q 4 ; q 4 ; q 6 ; q 8 ; q 16 ]1
(3.6)
(3.7)
(3.8)
Proof. Using Theorem 2.5, it is easy to prove the above eight identities as we
do in obtaining the 2-dissections of A(q ), so we omit the details.
Using Theorem 2.6, we establish the following eight identities.
Theorem 3.2. We have
X
1
X
X
X
X
1
(8n)q n =
[q 3 ; q 3 ; q 5 ; q 5 ; q 7 ; q 7 ; q 16 ]1
;
[q 2 ; q 4 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
(3.9)
(8n + 1)q n =
[q; q 3 ; q 5 ; q 5 ; q 7 ; q 7 ; q 16 ]1
;
[q 2 ; q 2 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
(3.10)
(8n + 2)q n =
[q; q 3 ; q 3 ; q 5 ; q 7 ; q 7 ; q 16 ]1
;
[q 2 ; q 2 ; q 4 ; q 4 ; q 6 ; q 8 ; q 16 ]1
(3.11)
(8n + 3)q n =
[q; q 3 ; q 3 ; q 5 ; q 5 ; q 7 ; q 16 ]1
;
[q 2 ; q 2 ; q 4 ; q 4 ; q 6 ; q 6 ; q 16 ]1
(3.12)
(8n + 4)q n =
[q; q 3 ; q 3 ; q 5 ; q 5 ; q 7 ; q 16 ]1
;
[q 2 ; q 2 ; q 4 ; q 4 ; q 6 ; q 8 ; q 16 ]1
(3.13)
b
n=0
b
n=0
1
b
n=0
1
b
n=0
1
n=0
b
A. Vanitha
90
X
X
X
1
(8n + 5)q n =
[q; q 3 ; q 3 ; q 5 ; q 7 ; q 7 ; q 16 ]1
;
[q 2 ; q 2 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
(3.14)
(8n + 6)q n =
[q; q 3 ; q 5 ; q 5 ; q 7 ; q 7 ; q 16 ]1
;
[q 2 ; q 4 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
(3.15)
b
n=0
1
b
n=0
1
(8n + 7)q n =0:
(3.16)
b
n=0
Corollary 3.3. We have
X
X
1
a
(8n)q
n=0
1
a
=
(8n + 1)q
n
n=0
X
X
X
X
1
n
(8n + 3)q
b
n=0
1
X
X
1
n
(8n + 4)q
a
(8n + 2)q
n=0
=
1
a
(8n + 6)q n
n=0
X
X
1
n
n
(8n + 1)q
b
X
X
X
X
1
a
(8n + 5)q n
(8n + 7)q
n=0
a
(8n + 2)q n
X
X
1
n
=
a
q
(8n + 3)q
a
(8n + 5)q n
n=0
X
X
(8n + 2)q n
n=0
1
(8n + 4)q n
b
n=0
(8n + 4)q n
:
(8n + 5)q n
b
n=0
Proof. Proof follows from Theorem 3.1 and Theorem 3.2.
Theorem 3.4. We have a(2) = a(3) = a(5) = a(6) = a(8n + 4) = 0: The
remaining coecients a(n) satisfy the inequalities
a(16n); a(16n
a(16n
n
b
b
n=0
1
;
(8n + 6)q
1
=
1
b
n=0
1
b
n
n=0
;
n=0
1
(8n + 7)q
(8n)q n
b
n=0
1
n
n=0
and
a
n
n=0
n=0
1
=
1
n=0
1
(8n + 1)q
n=0
;
b
a
X
X
1
n
+ 2); a(16n + 5); a(16n + 7); a(16n + 9); a(16n + 11); a(16n + 14)
>
0;
+ 1); a(16n + 3); a(16n + 6); a(16n + 8); a(16n + 10); a(16n + 13); a(16n + 15)
<
0:
On the expansion of Ramanujan's continued fraction of order sixteen
91
Proof. From (3.1), we have
X
1
[
; q 8 ]1
a(8n)( q ) =
[q 2 ; q 2 ; q 4 ; q 8 ]1
n=0
( q ; q 2 )1 ( q 3 ; q 5 ; q 8 )1
=
:
(q 2 ; q 4 ; q 6 ; q 8 )21
n
q;
3
q ;
q
3
From the above equality, it follows that a(16n) > 0 and a(16n + 8) < 0:
Similarly, we can determine the signs of the remaining subsequences for a(n):
Theorem 3.5. We have b(8) = b(8n + 7) = 0: The remaining coecients b(n)
satisfy the inequalities
b(16n); b(16n
b(16n
+ 1); b(16n + 2); b(16n + 3); b(16n + 4); b(16n + 5); b(16n + 6)
>
0;
+ 8); b(16n + 9); b(16n + 10); b(16n + 11); b(16n + 12); b(16n + 13); b(16n + 14)
<
0:
Proof. From (3.9), we have
X
1
(8n)(
b
q
)n =
[
3
q ;
n=0
; q 16 ]1
[q 2 ; q 4 ; q 4 ; q 6 ; q 6 ; q 8 ; q 16 ]1
3
q ;
5
q ;
5
q ;
7
q ;
q
7
From the above equality, it follows that b(16n) > 0 and b(16n + 8) < 0:
Similarly, we can determine the signs of the remaining subsequences for b(n):
4. 16-Dissections of
A(q) and B (q)
In this section, we present 16-dissections of A(q ) and B (q ). In these dissections
components are not single products. One can establish the following two
theorems on using 8-dissections of A(q ) and B (q ). The proof is similar to that
of 8-dissections of A(q ) and B (q ), so we omit the details.
A. Vanitha
92
Theorem 4.1. We have
1
X
1
X
n=0
n
q
n
q f
n
q
n
n
n
q
n
q
(
qP f
f
q;
f(
(
R f
f
f
f
(16 + 10) n =
n
q
a
n
q
a
n=0
1
X
a
n=0
1
X
n=0
a
(16 + 13) n = (
n
q
f
q
q
)(
q;
q f
6
f
q ;
q;
q ;
Q f
f
R f
f
10
q
q;
)(
4
q;
6
q;
6
q ;
q
q f
f
2
5
q ;
2
q ;
8
q ;
2
14
f
q;
8 f
q
3
13
q ;
q
4
q ;
4 f
q
8
q ;
8 f
q
qf
q ;
8
q ;
8 f
q
X
4 f
q
11
M
q ;
4 f
q
3
q
2
qf
q ;
4 f
q
q
14
Y
q
7
q ;
8 f
q
q
M
13
N
9
q
15
7
q ;
14
;
9
q ;
q;
;
9
q
3
f
f
q
7
q ;
q
N
9
q
q ;
8 f
q
14
8
q ;
13
15
2
f
7
q ;
q
8
q ;
2
q f
4
q ;
3
q
7
q ;
q f
11
4
9
q
)(
q
7
q
M
9
q
;
4
q
2 f
13
q
N
q ;
q
3
q ;
4 f
X
4 f
4 f
q
M
q f
4
q ;
2
q
4 f
q
5
q
2
q
8
q ;
2
q ;
q ;
q;
q
Y
7
q
9
3
f
q ;
8 f
14
q
7
11
f
8 f
7
q ;
q;
7
15
q
q
q ;
q ;
14
q
2
q ;
8 f
q
q
7
q ;
f
8 f
q ;
8
f
13
q ;
8
q ;
q
8
q ;
2
14
8 f
q
q ;
f
q f
4
q
8
q ;
qf
q ;
)
)(
)
)(
2
4 f
q
4
9
(
)(
) g
)(
)(
)
+ (
)(
) g
)(
)(
)
) + (
) g
)(
)(
)
+ (
)(
) g
)(
)(
)
+ (
)(
) g
)(
)(
)
3
q f
4
q ;
q ;
q
;
M
3
11
4
N
)
q ;
2 f
q ;
q
10
)(
q
q
)
) +
q ;
2 f
q ;
q
N
N
2 f
f
q
f
10
q
)f (
)(
)(
)(
)(
)(
q f
q ;
12
9
2 f
q
2
q
10
7
q ;
q ;
q ;
q
)(
f
2
q f
6
2 f
)(
q ;
q ;
q
2
q ;
11
10
q ;
;
2
f
q f
(
)(
)
q
)(
4 f
q
4
q ;
2 f
q
RY
2
q ;
)(
5
qP f
q;
2
q
2
4
q ;
2 f
q
7
2
q ;
q
2
q f
q
5
f
;
4
M
4
q ;
q ;
q f
q ;
q ;
2 f
q
f
q
5
f
q ;
)(
f(
(16 + 15) n = f ( (
n
10
q
q
2 f
q
2
q ;
q f
q f
(16 + 14) n = f ((
n
q
q ;
12
10
9
6
f
2
q ;
q f
q;
f
=0
1
X
q
q
2 f
q
Y
f
qP
7
2
q ;
6
4
q ;
q ;
11
q
q ;
q;
Q f
=0
q
4
6
)(
2
q ;
q ;
q ;
;
2 f
f
q f
q;
(
n
1
X
(16 + 11) n = f ((
n
a
)
qQ f
f
10
5
q f
)
)
)(
) + (
)(
) g
)(
)(
)(
)(
)
)(
) + (
)(
) g
)(
)(
)(
)(
)
)f (
) + (
) g
(
)(
)(
)(
)
)(
) + (
)(
) g
)(
)(
)(
)(
)
)(
) + (
)(
) g
)(
)(
)(
)(
)
q
q f
q;
2
RX
q ;
(
q
2
q ;
q
q;
f
f
q
q
6
q ;
f(
qP
q ;
)(
q;
f
q
2
q f
qQ f
(16 + 9) n = (
a
n=0
1
X
f
(16 + 8) n = (
a
n=0
1
X
)(
q;
(16 + 7) n = f ((
a
n=0
1
X
q
(16 + 6) n =
a
n=0
1
X
n
(16 + 5) n = (
a
n=0
1
X
f
(16 + 3) n =
a
n=0
1
X
X
(16 + 2) n = f ((
a
n=0
1
X
n q
(16 + 1) n = (
a
n=0
1
X
(16 ) n = (
a
q ;
M
9
14
q
N
9
;
;
;
q
13
q
q
15
q
M
9
9
N
;
;
;
;
On the expansion of Ramanujan's continued fraction of order sixteen
93
where,
P
X
Y
M
N
:=
:=
:=
:=
:=
f
(
(
(
(
(
q;
3
f q ;q
f q; q
7
3
f q ;q
3
f q ;q
)
)(
)(
)(
)(
q
5
7
;
7
f q ;q
7
f q ;q
5
9
3
f q ;q
5
:= (
)
)+ ( ) (
)+ ( ) (
)+ ( ) (
)+ ( ) (
Q
5
f q ;q
2
f
9
q ;
q
2
q f q; q
3
qf q ; q
13
f q; q
11
7
5
;
f q; q
5
f q ;q
q ;
q
5
)
;
;
;
11
3
3
f
15
15
f q ;q
7
:= (
)
)
)
)
R
f q; q
7
qf q; q
6
;
13
:
Theorem 4.2. We have
1
X
)f (
)
(
)(
)(
)(
n
1
X
)(
)
(
(16 + 1) n = f ((
)(
)(
)(
n
1
X
)(
)
(
(16 + 2) n = f ((
)(
)(
)(
n
=0
(16 ) n = (
b
n q
b
n
q
b
n
q
f
n=0
1
X
n=0
1
X
n
q
n
q
n
q
f
2
q;
b
n
q
f
1
X
n=0
(16 + 8) n = (
b
n
q
f
f
q
3
2
q ;
f
2
q ;
q;
2
q;
q
14
)(
)
5
q
X
9
N
4
q ;
2
2
q
2 f
)(
q f
6
q ;
q
q;
10
4
q ;
4
q f
q ;
2
q ;
q
12
2
q ;
q
9
2 f
q
3
f
q f
q ;
7
q ;
2
q ;
13
6
q
) g
Y
)(
)(
)(
)(
)(
14
q;
f
q
5
q ;
8 f
q
10
8
q ;
f
15
q
8 f
q
;
11
) g
)
M
15
q
q;
q;
)
q
15
) g
)
M
15
q
q;
;
;
;
q
f
q
8
q ;
q ;
4 f
q
q
8 f
q
q ;
4 f
q
qf
q ;
8
q ;
qf
4
q ;
3
qf
4 f
q
N
2 f
q
RX
q ;
13
2 f
q
q ;
2
q
q
)
q
4
)
;
(
)(
)(
)(
)(
)
(
)(
)(
)(
(
)f (
)
(
)(
)(
)(
2
q ;
Pf
q
7
f
;
4
q ;
q ;
q ;
)(
11
2 f
q
f
q f
q f
f
=0
14
q f
q;
Q f
P
q
)(
q;
f
=0
10
q f
q ;
(16 + 5) n = f ((
n
1
X
(16 + 6) n = f ((
n
b
q
q;
12
X
(16 + 4) n = (
b
6
f
q
2
q ;
q ;
R f
(16 + 3) n = (
4
q ;
q f
f
=0
b
(
q;
Q f
=0
1
X
Pf
13
N
4
q ;
N
2 f
q
f
2 f
q
qf
3
q ;
4
q ;
4 f
q
13
q ;
4 f
q
X
4 f
q
q
f
8
q ;
)(
) g
)(
)
)(
) g
)(
)
) g
)(
)
10
8
q ;
2
qf
4
q ;
q
6
q ;
f
q;
8 f
q
q
8
q ;
5
q ;
14
q;
5
f
q ;
8 f
q
q
8 f
q
11
q
15
q
Y
q
11
15
q
q;
q;
M
15
q
15
;
;
M
;
A. Vanitha
94
1
X
(16 + 9) n = f ((
n
1
X
(16 + 10) n = f( (
n
b
n
q
n
q
f
n=0
1
X
n=0
1
X
n=0
1
X
n=0
(16 + 11) n = (
b
n
q
n
q
n
q
n
q
6
f
10
)(
)(
5
f
q ;
2
q ;
)(
f
q;
q ;
q;
q f
f
q
6
q ;
q;
q
2
q ;
10
)(
q f
2
q ;
q;
q
q
RY
)(
f
f
2
q ;
14
)(
q f
q
)(
f
2
q ;
)(
f
2
q ;
q
)
11
)(
)
)(
2 f
q
2
q f
)(
Q f
P
q
q f
(16 + 14) n = f ((
b
)(
q;
(16 + 13) n = f( (
b
14
q f
q ;
q;
q
Y
f
(16 + 12) n = (
b
q;
R f
=0
1
X
q ;
f
=0
b
2
Q f
2
15
q
2 f
)
4 f
q
N
f
4
q
q ;
(
)(
q
2
q ;
4 f
f
3
q ;
8 f
q
14
q
)(
)(
)(
)(
10
8
q ;
8
q ;
7
f
q ;
8 f
q
q;
q
9
q
q;
) g
)
13
q
M
15
) g
)
M
15
q
q;
4
q ;
q
15
)(
2 f
q ;
q
)
11
4
)
N
f
4
q
q ;
)(
2 f
q
q
;
)
;
(
)(
2
q ;
4 f
f
4
q ;
4 f
q
14
8
q ;
(
)(
N
q
6
q ;
)(
)(
f
q
q
8
q ;
10
7
q ;
8 f
f
5.
Combinatorial Interpretations of
3
q ;
8 f
q
P; Q; R; X; Y ; M; N
q
q;
)(
)(
where
q;
9
) g
)
M
q
q
15
13
;
) g
)
M
15
q
are as dened in Theorem 4.1.
b(n)
;
;
)(
5
6
f
4
q ;
q ;
2 f
q
(
)(
N
a(n)
and
In this section, we present combinatorial interpretations of a(n) and b(n). For
simplicity, we dene
(q r ; q s )1 := (q r ; q s r ; q s )1 ;
where r and s are positive integers and r < s:
Denition 5.1. A positive integer
n has k colors if there are k copies of n
available and all of them are viewed as distinct objects. Partitions of positive
integer into parts with colors are called \colored partitions".
For example, if 2 is allowed to have two colors, say r (red) and g (green) and
odd parts being distinct, then all colored partitions of 3 are 3; 2r + 1; 2g + 1:
An important fact is that
1
;
u
(q ; q v )k1
;
On the expansion of Ramanujan's continued fraction of order sixteen
95
is the generating function for the number of partitions of n, where all the parts
are congruent to u (mod v ) and have k colors.
Theorem 5.2. Let P0 (n) denote the number of partitions of n with parts not
congruent to 8 (mod 16) and all having two colors except for parts congruent
to 1; 7 (mod 16) and odd parts being distinct.
Let P1 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 1; 6; 7
(mod 16) and odd parts being distinct.
Let P2 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 3; 4; 7
(mod 16) and odd parts being distinct.
Let P3 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 2; 5; 7
(mod 16) and odd parts being distinct.
Let P4 (n) denote the number of partitions of n with parts not congruent to 7
(mod 16) and all having two colors except for parts congruent to 2 (mod 16)
and odd parts being distinct.
Let P5 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 4; 5; 7
(mod 16) and odd parts being distinct.
Let P6 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 3; 6; 7
(mod 16) and odd parts being distinct.
Then, we have
( 1)n (8 ) = ( ) ( 1)n (8 + 1) = ( ) ( 1)n (8 + 2) = (
( 1)n (8 + 3) = ( 2) ( 1)n (8 + 5) = ( 2)
( 1)n (8 + 6) = ( 1) ( 1)n (8 + 7) = ( )
a
n
P0 n ;
1
n=0
n
P1 n ;
a
n
a
n
P3 n
;
a
n
P4 n
a
n
P5 n
;
a
n
P6 n :
Proof. Replacing q to
X
a
q
( 1)n a(8n)q n =
P2 n
1)
;
;
in (3.1), we obtain
(
q
1 ;
3 ; q 3 ; q 5 ; q 5 ; q 7 ; q 16 )1
:
(q 2 ; q 2 ; q 4 ; q 4 ; q 6 ; q 6 ; q 16 )1
q
Observe that the product on the right is the generating function for P0 (n) and
so
( 1)n a(8n) = P0 (n). Similar arguments can be used to derive the remaining equations.
A. Vanitha
96
Example 5.3. By Maple, we have been able to nd the following series expansion for A(q ):
( )=1
+2
A q
q
q
23
+
+
+
4 +4
2 +2
q
q
7
q
24
8
q
q
25
9
q
q
10
q
26
14
q
(3) = 4 =
P1 (3) =
4=
P2 (2) = 2 =
P3 (1) =
2=
P4 (1) = 2 =
P5 (2) =
2=
P6 (3) = 5 =
P0
q
27
The following table veries the case
2 +2
2 +2
n
15
q
29
q
16
q
30
2 +2
5 +7
q
17
q
31
q
18
q
32
+
2
7 +5
q
19
q
33
q
21
q
q
34
= 3 in Theorem 5.2.
(24) 3r ; 3g ; 2r + 1; 2g + 1
a(25)
3r ; 3g ; 2r + 1 ; 2g + 1
a(26)
2r ; 2g
a(27)
1r ; 1g
a(29)
1r ; 1g
a(30)
2r ; 2g
a(31)
3; 2r + 1 r ; 2r + 1 g ;
2g + 1 r ; 2g + 1 g
a
Theorem 5.4. Let P0 (n) denote the number of partitions of n with parts not
congruent to 1 (mod 16) and all having two colors except for parts congruent
to 2 (mod 16) and odd parts being distinct.
Let P1 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 1; 3; 4
(mod 16) and odd parts being distinct.
Let P2 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 1; 5; 6
(mod 16) and odd parts being distinct.
Let P3 (n) denote the number of partitions of n with parts not congruent to
8 (mod 16) and all having two colors except for parts congruent to 1; 7
(mod 16) and odd parts being distinct.
Let P4 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 1; 6; 7
(mod 16) and odd parts being distinct.
Let P5 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 1; 4; 5
(mod 16) and odd parts being distinct.
Let P6 (n) denote the number of partitions of n with parts congruent to all parts
(mod 16) and all having two colors except for parts congruent to 1; 2; 3
22
:
On the expansion of Ramanujan's continued fraction of order sixteen
(mod 16) and odd parts being distinct.
Then, we have
( 1)n b(8n + k) = Pk (n)
f or
97
0 k 6:
X
Proof. Replacing q to q in (3.9), we obtain
1
( q 3 ; q 3 ; q 5 ; q 5 ; q 7 ; q 7 ; q 16 )1
:
( 1)n b(8n)q n =
2 ; q 4 ; q 4 ; q 6 ; q 6 ; q 8 ; q 8 ; q 16 )1
(
q
n=0
Observe that the product on the right is the generating function for P0 (n) and
so
( 1)n b(8n) = P0 (n). Similar arguments can be used to derive the remaining equations.
Example 5.5. By Maple, we have been able to nd the following series expansion for B (q ):
B (q )
2
3
4
5
6
=1+q+q +q +q +q +q
+ 2q
19
+ 2q
20
+ 2q
21
+q
22
2q
24
q
9
q
3q
The following table veries the case
25
n
10
q
4q
11
26
q
4q
12
27
q
13
4q
q
28
14
4q
+q
29
16
+ 2q
2q
30
17
+
+ 2q
18
:
= 3 in Theorem 5.4.
(3) = 2 = b(24)
3r ; 3g
P1 (3) = 3 =
b(25)
3; 2r + 1 ; 2g + 1
P2 (3) = 4 =
b(26)
3r ; 3g ; 2r + 1 ; 2g + 1
P3 (3) = 4 =
b(27)
3r ; 3g ; 2r + 1 ; 2g + 1
P4 (3) = 4 =
b(28)
3r ; 3g ; 2r + 1 ; 2g + 1
P5 (3) = 4 =
b(29)
3r ; 3g ; 2r + 1 ; 2g + 1
P6 (3) = 2 =
b(30)
3; 2 + 1
Remark 5.6. Theorem 3.4 and Theorem 3.5 also follows from Theorem 5.2
and Theorem 5.4 respectively.
Acknowledgement: The author is thankful to Professor H.M. Srivastava
for his helpful comments and suggestions which improved the quality of the
paper. Author is also thankful to Professor Chadrashekar Adiga and Dr. M.
S. Surekha for the careful reading of the manuscript and helpful remarks. The
author is thankful to DST, New Delhi for awarding INSPIRE Fellowship [No.
DST/ INSPIRE Fellowship/2012/122], under which this work has been done.
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98
A. Vanitha
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