I. Rogers-Selberg Mod 7 Identities
A
(
q
)
=
(
q
3
,
q
4
,
q
7
;
q
7
)
∞
(
q
2
;
q
2
)
∞
=
f
(
−
q
3
,
−
q
4
)
f
(
−
q
2
,
−
q
4
)
=
∏
n
=
1
∞
(
1
−
q
7
n
)
(
1
−
q
7
n
−
3
)
(
1
−
q
7
n
−
4
)
(
1
−
q
2
n
)
{\displaystyle A(q)={\frac {(q^{3},q^{4},q^{7};q^{7})_{\infty }}{(q^{2};q^{2})_{\infty }}}={\frac {f(-q^{3},-q^{4})}{f(-q^{2},-q^{4})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{7n})(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{2n})}}}
B
(
q
)
=
(
q
2
,
q
5
,
q
7
;
q
7
)
∞
(
q
2
;
q
2
)
∞
=
f
(
−
q
2
,
−
q
5
)
f
(
−
q
2
,
−
q
4
)
=
∏
n
=
1
∞
(
1
−
q
7
n
)
(
1
−
q
7
n
−
2
)
(
1
−
q
7
n
−
5
)
(
1
−
q
2
n
)
{\displaystyle B(q)={\frac {(q^{2},q^{5},q^{7};q^{7})_{\infty }}{(q^{2};q^{2})_{\infty }}}={\frac {f(-q^{2},-q^{5})}{f(-q^{2},-q^{4})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{7n})(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{2n})}}}
C
(
q
)
=
(
q
,
q
6
,
q
7
;
q
7
)
∞
(
q
2
;
q
2
)
∞
=
f
(
−
q
,
−
q
6
)
f
(
−
q
2
,
−
q
4
)
=
∏
n
=
1
∞
(
1
−
q
7
n
)
(
1
−
q
7
n
−
1
)
(
1
−
q
7
n
−
6
)
(
1
−
q
2
n
)
{\displaystyle C(q)={\frac {(q,q^{6},q^{7};q^{7})_{\infty }}{(q^{2};q^{2})_{\infty }}}\,={\frac {f(-q,-q^{6})}{f(-q^{2},-q^{4})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{7n})(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{2n})}}}
II. Bailey Mod 9 Identities
A
(
q
)
=
(
q
4
,
q
5
,
q
9
;
q
9
)
∞
(
q
3
;
q
3
)
∞
=
f
(
−
q
4
,
−
q
5
)
f
(
−
q
3
,
−
q
6
)
=
∏
n
=
1
∞
(
1
−
q
9
n
)
(
1
−
q
9
n
−
4
)
(
1
−
q
9
n
−
5
)
(
1
−
q
3
n
)
{\displaystyle A(q)={\frac {(q^{4},q^{5},q^{9};q^{9})_{\infty }}{(q^{3};q^{3})_{\infty }}}={\frac {f(-q^{4},-q^{5})}{f(-q^{3},-q^{6})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{9n})(1-q^{9n-4})(1-q^{9n-5})}{(1-q^{3n})}}}
B
(
q
)
=
(
q
2
,
q
7
,
q
9
;
q
9
)
∞
(
q
3
;
q
3
)
∞
=
f
(
−
q
2
,
−
q
7
)
f
(
−
q
3
,
−
q
6
)
=
∏
n
=
1
∞
(
1
−
q
9
n
)
(
1
−
q
9
n
−
2
)
(
1
−
q
9
n
−
7
)
(
1
−
q
3
n
)
{\displaystyle B(q)={\frac {(q^{2},q^{7},q^{9};q^{9})_{\infty }}{(q^{3};q^{3})_{\infty }}}={\frac {f(-q^{2},-q^{7})}{f(-q^{3},-q^{6})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{9n})(1-q^{9n-2})(1-q^{9n-7})}{(1-q^{3n})}}}
C
(
q
)
=
(
q
,
q
8
,
q
9
;
q
9
)
∞
(
q
3
;
q
3
)
∞
=
f
(
−
q
,
−
q
8
)
f
(
−
q
3
,
−
q
6
)
=
∏
n
=
1
∞
(
1
−
q
9
n
)
(
1
−
q
9
n
−
1
)
(
1
−
q
9
n
−
8
)
(
1
−
q
3
n
)
{\displaystyle C(q)={\frac {(q,q^{8},q^{9};q^{9})_{\infty }}{(q^{3};q^{3})_{\infty }}}\,={\frac {f(-q,-q^{8})}{f(-q^{3},-q^{6})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{9n})(1-q^{9n-1})(1-q^{9n-8})}{(1-q^{3n})}}}
III. Rogers Mod 14 Identities
A
(
q
)
=
(
q
6
,
q
8
,
q
14
;
q
14
)
∞
(
q
;
q
)
∞
=
f
(
−
q
6
,
−
q
8
)
f
(
−
q
,
−
q
2
)
=
∏
n
=
1
∞
(
1
−
q
14
n
)
(
1
−
q
14
n
−
6
)
(
1
−
q
14
n
−
8
)
(
1
−
q
n
)
{\displaystyle A(q)={\frac {(q^{6},q^{8},q^{14};q^{14})_{\infty }}{(q;q)_{\infty }}}\,={\frac {f(-q^{6},-q^{8})}{f(-q,-q^{2})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{14n})(1-q^{14n-6})(1-q^{14n-8})}{(1-q^{n})}}}
B
(
q
)
=
(
q
4
,
q
10
,
q
14
;
q
14
)
∞
(
q
;
q
)
∞
=
f
(
−
q
4
,
−
q
10
)
f
(
−
q
,
−
q
2
)
=
∏
n
=
1
∞
(
1
−
q
14
n
)
(
1
−
q
14
n
−
4
)
(
1
−
q
14
n
−
10
)
(
1
−
q
n
)
{\displaystyle B(q)={\frac {(q^{4},q^{10},q^{14};q^{14})_{\infty }}{(q;q)_{\infty }}}={\frac {f(-q^{4},-q^{10})}{f(-q,-q^{2})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{14n})(1-q^{14n-4})(1-q^{14n-10})}{(1-q^{n})}}}
C
(
q
)
=
(
q
2
,
q
12
,
q
14
;
q
14
)
∞
(
q
;
q
)
∞
=
f
(
−
q
2
,
−
q
12
)
f
(
−
q
,
−
q
2
)
=
∏
n
=
1
∞
(
1
−
q
14
n
)
(
1
−
q
14
n
−
2
)
(
1
−
q
14
n
−
12
)
(
1
−
q
n
)
{\displaystyle C(q)={\frac {(q^{2},q^{12},q^{14};q^{14})_{\infty }}{(q;q)_{\infty }}}={\frac {f(-q^{2},-q^{12})}{f(-q,-q^{2})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{14n})(1-q^{14n-2})(1-q^{14n-12})}{(1-q^{n})}}}
IV. Dyson Mod 27 Identities
A
(
q
)
=
(
q
12
,
q
15
,
q
27
;
q
27
)
∞
(
q
;
q
)
∞
=
f
(
−
q
12
,
−
q
15
)
f
(
−
q
,
−
q
2
)
=
∏
n
=
1
∞
(
1
−
q
27
n
)
(
1
−
q
27
n
−
12
)
(
1
−
q
27
n
−
15
)
(
1
−
q
n
)
{\displaystyle A(q)={\frac {(q^{12},q^{15},q^{27};q^{27})_{\infty }}{(q;q)_{\infty }}}={\frac {f(-q^{12},-q^{15})}{f(-q,-q^{2})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{27n})(1-q^{27n-12})(1-q^{27n-15})}{(1-q^{n})}}}
B
(
q
)
=
(
q
9
,
q
18
,
q
27
;
q
27
)
∞
(
q
;
q
)
∞
=
f
(
−
q
9
,
−
q
18
)
f
(
−
q
,
−
q
2
)
=
∏
n
=
1
∞
(
1
−
q
27
n
)
(
1
−
q
27
n
−
9
)
(
1
−
q
27
n
−
18
)
(
1
−
q
n
)
{\displaystyle B(q)={\frac {(q^{9},q^{18},q^{27};q^{27})_{\infty }}{(q;q)_{\infty }}}={\frac {f(-q^{9},-q^{18})}{f(-q,-q^{2})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{27n})(1-q^{27n-9})(1-q^{27n-18})}{(1-q^{n})}}}
C
(
q
)
=
(
q
6
,
q
21
,
q
27
;
q
27
)
∞
(
q
;
q
)
∞
=
f
(
−
q
6
,
−
q
21
)
f
(
−
q
,
−
q
2
)
=
∏
n
=
1
∞
(
1
−
q
27
n
)
(
1
−
q
27
n
−
6
)
(
1
−
q
27
n
−
21
)
(
1
−
q
n
)
{\displaystyle C(q)={\frac {(q^{6},q^{21},q^{27};q^{27})_{\infty }}{(q;q)_{\infty }}}\,={\frac {f(-q^{6},-q^{21})}{f(-q,-q^{2})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{27n})(1-q^{27n-6})(1-q^{27n-21})}{(1-q^{n})}}}
D
(
q
)
=
(
q
3
,
q
24
,
q
27
;
q
27
)
∞
(
q
;
q
)
∞
=
f
(
−
q
3
,
−
q
24
)
f
(
−
q
,
−
q
2
)
=
∏
n
=
1
∞
(
1
−
q
27
n
)
(
1
−
q
27
n
−
3
)
(
1
−
q
27
n
−
24
)
(
1
−
q
n
)
{\displaystyle D(q)={\frac {(q^{3},q^{24},q^{27};q^{27})_{\infty }}{(q;q)_{\infty }}}\,={\frac {f(-q^{3},-q^{24})}{f(-q,-q^{2})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{27n})(1-q^{27n-3})(1-q^{27n-24})}{(1-q^{n})}}}
This is similar to the 11/3/2012 update. Let,
m
=
12
n
−
1
{\displaystyle m={\frac {12}{n-1}}}
be a positive integer for some odd number n . Thus, there are only 4 possibilities:
n
=
3
,
5
,
7
,
13
{\displaystyle n=3,5,7,13}
. Given the standard Ramanujan theta functions
ϕ
(
q
)
,
ψ
(
q
)
,
f
(
−
q
)
{\displaystyle \phi (q),\;\psi (q),\;f(-q)}
, then,
(
ϕ
(
−
q
2
)
ϕ
(
−
q
6
n
)
ϕ
(
−
q
6
)
ϕ
(
−
q
2
n
)
)
m
+
q
3
(
ψ
(
q
)
ψ
(
q
3
n
)
ψ
(
q
3
)
ψ
(
q
n
)
)
m
−
q
3
(
ψ
(
−
q
)
ψ
(
−
q
3
n
)
ψ
(
−
q
3
)
ψ
(
−
q
n
)
)
m
−
2
m
q
2
(
f
(
−
q
2
)
f
(
−
q
6
n
)
f
(
−
q
6
)
f
(
−
q
2
n
)
)
m
=
1
{\displaystyle \left({\frac {\phi (-q^{2})\phi (-q^{6n})}{\phi (-q^{6})\phi (-q^{2n})}}\right)^{m}\,+\,q^{3}\left({\frac {\psi (q)\psi (q^{3n})}{\psi (q^{3})\psi (q^{n})}}\right)^{m}-\,q^{3}\left({\frac {\psi (-q)\psi (-q^{3n})}{\psi (-q^{3})\psi (-q^{n})}}\right)^{m}-\,2mq^{2}\left({\frac {f(-q^{2})f(-q^{6n})}{f(-q^{6})f(-q^{2n})}}\right)^{m}=1}
For example, for n = 5, hence m = 3, we have,
(
ϕ
(
−
q
2
)
ϕ
(
−
q
30
)
ϕ
(
−
q
6
)
ϕ
(
−
q
10
)
)
3
+
q
3
(
ψ
(
q
)
ψ
(
q
15
)
ψ
(
q
3
)
ψ
(
q
5
)
)
3
−
q
3
(
ψ
(
−
q
)
ψ
(
−
q
15
)
ψ
(
−
q
3
)
ψ
(
−
q
5
)
)
3
−
6
q
2
(
f
(
−
q
2
)
f
(
−
q
30
)
f
(
−
q
6
)
f
(
−
q
10
)
)
3
=
1
{\displaystyle \left({\frac {\phi (-q^{2})\phi (-q^{30})}{\phi (-q^{6})\phi (-q^{10})}}\right)^{3}\,+\,q^{3}\left({\frac {\psi (q)\psi (q^{15})}{\psi (q^{3})\psi (q^{5})}}\right)^{3}-\,q^{3}\left({\frac {\psi (-q)\psi (-q^{15})}{\psi (-q^{3})\psi (-q^{5})}}\right)^{3}-\,6q^{2}\left({\frac {f(-q^{2})f(-q^{30})}{f(-q^{6})f(-q^{10})}}\right)^{3}=1}
and so on for the other three n .
--------------END--------------
Define,
h
k
=
(
−
1
)
k
−
1
q
k
(
3
k
−
25
)
/
50
f
(
−
q
2
k
,
−
q
25
−
2
k
)
f
(
−
q
k
,
−
q
25
−
k
)
,
for
k
=
1
,
.
.
.
,
12.
{\displaystyle h_{k}=(-1)^{k-1}q^{k(3k-25)/50}\,{\frac {f(-q^{2k},-q^{25-2k})}{f(-q^{k},-q^{25-k})}},\;\;{\text{for}}\;k=1,...,12.}
(Note, as usual, that the even-index h_k have a negative sign.) It turns out that appropriate pairs of h_k are roots of a polynomial whose coefficients are in
j
=
η
(
τ
)
η
(
25
τ
)
{\displaystyle j={\frac {\eta (\tau )}{\eta (25\tau )}}}
, analogous to the case for n = 13 discovered by Ramanujan. Since one pair is a constant,
h
5
h
10
=
−
1
{\displaystyle h_{5}h_{10}=-1}
, then the remaining five are the roots of a quintic (naturally enough) given by,
1
+
(
10
+
10
j
+
4
j
2
+
j
3
)
x
+
(
5
+
15
j
+
11
j
2
+
5
j
3
+
j
4
)
x
2
−
2
(
5
+
5
j
+
3
j
2
+
j
3
)
x
3
+
j
2
x
4
+
x
5
=
0
{\displaystyle 1+(10+10j+4j^{2}+j^{3})x+(5+15j+11j^{2}+5j^{3}+j^{4})x^{2}-2(5+5j+3j^{2}+j^{3})x^{3}+j^{2}x^{4}+x^{5}=0}
with j as the eta quotient given above. The roots are then,
[
x
1
,
x
2
,
x
3
,
x
4
,
x
5
]
=
[
h
1
h
7
,
h
2
h
11
,
h
3
h
4
,
h
6
h
8
,
h
9
h
12
]
{\displaystyle [x_{1},\,x_{2},\,x_{3},\,x_{4},\,x_{5}]=[h_{1}h_{7},\,h_{2}h_{11},\,h_{3}h_{4},\,h_{6}h_{8},\,h_{9}h_{12}]}
(Note : The quintic looks vaguely familiar to me, it seems I've come across it before, or something similar in the course of my research into the Rogers-Ramanujan continued fraction, but I cannot recall precisely in what context.)
Update April 28, 2022 : After almost 10 years, I finally found the answer to my "note". This is just the EMMA LEHMER QUINTIC !
--------------END--------------
Here is a generalization of one identity described by Berndt as "...fascinating, but with no direct proof" (Ramanujan's Notebooks III , p.322). Let,
k
=
24
n
−
1
{\displaystyle k={\frac {24}{n-1}}}
be a positive integer for some odd number n . Of course, there are only 6 possibilities:
n
=
3
,
5
,
7
,
9
,
13
,
25
{\displaystyle n=3,5,7,9,13,25}
. Given the standard Ramanujan theta functions
ϕ
(
q
)
,
ψ
(
q
)
,
f
(
−
q
)
{\displaystyle \phi (q),\;\psi (q),\;f(-q)}
, then it is proposed that,
ϕ
k
(
−
q
2
n
)
ϕ
k
(
−
q
2
)
+
q
3
(
ψ
k
(
q
n
)
ψ
k
(
q
)
−
ψ
k
(
−
q
n
)
ψ
k
(
−
q
)
)
−
2
k
q
2
(
f
k
(
−
q
2
n
)
f
k
(
−
q
2
)
)
=
1
{\displaystyle {\frac {\phi ^{k}(-q^{2n})}{\phi ^{k}(-q^{2})}}+q^{3}\left({\frac {\psi ^{k}(q^{n})}{\psi ^{k}(q)}}-{\frac {\psi ^{k}(-q^{n})}{\psi ^{k}(-q)}}\right)-2kq^{2}\left({\frac {f^{k}(-q^{2n})}{f^{k}(-q^{2})}}\right)=1}
and, as suggested by Michael Somos, by using a Fricke involution ,
ϕ
k
(
−
q
2
)
ϕ
k
(
−
q
2
n
)
+
1
q
3
(
ψ
k
(
q
)
ψ
k
(
q
n
)
−
ψ
k
(
−
q
)
ψ
k
(
−
q
n
)
)
−
2
k
q
2
(
f
k
(
−
q
2
)
f
k
(
−
q
2
n
)
)
=
n
k
/
2
{\displaystyle {\frac {\phi ^{k}(-q^{2})}{\phi ^{k}(-q^{2n})}}+{\frac {1}{q^{3}}}\left({\frac {\psi ^{k}(q)}{\psi ^{k}(q^{n})}}-{\frac {\psi ^{k}(-q)}{\psi ^{k}(-q^{n})}}\right)-{\frac {2k}{q^{2}}}\left({\frac {f^{k}(-q^{2})}{f^{k}(-q^{2n})}}\right)=n^{k/2}}
Thus, for n = 3, we have,
ϕ
12
(
−
q
6
)
ϕ
12
(
−
q
2
)
+
q
3
(
ψ
12
(
q
3
)
ψ
12
(
q
)
−
ψ
12
(
−
q
3
)
ψ
12
(
−
q
)
)
−
24
q
2
(
f
12
(
−
q
6
)
f
12
(
−
q
2
)
)
=
1
{\displaystyle {\frac {\phi ^{12}(-q^{6})}{\phi ^{12}(-q^{2})}}+q^{3}\left({\frac {\psi ^{12}(q^{3})}{\psi ^{12}(q)}}-{\frac {\psi ^{12}(-q^{3})}{\psi ^{12}(-q)}}\right)-24q^{2}\left({\frac {f^{12}(-q^{6})}{f^{12}(-q^{2})}}\right)=1}
and,
ϕ
12
(
−
q
2
)
ϕ
12
(
−
q
6
)
+
1
q
3
(
ψ
12
(
q
)
ψ
12
(
q
3
)
−
ψ
12
(
−
q
)
ψ
12
(
−
q
3
)
)
−
24
q
2
(
f
12
(
−
q
2
)
f
12
(
−
q
6
)
)
=
3
6
{\displaystyle {\frac {\phi ^{12}(-q^{2})}{\phi ^{12}(-q^{6})}}+{\frac {1}{q^{3}}}\left({\frac {\psi ^{12}(q)}{\psi ^{12}(q^{3})}}-{\frac {\psi ^{12}(-q)}{\psi ^{12}(-q^{3})}}\right)-{\frac {24}{q^{2}}}\left({\frac {f^{12}(-q^{2})}{f^{12}(-q^{6})}}\right)=3^{6}}
and so on for the other five n , with the case n = 7 given in page 322 which was the inspiration for this generalization. I have no rigorous proof for this "family", but one can easily see via Mathematica that the proposed equality indeed holds true for hundreds of decimal places.
--------------END--------------
It seems Ramanujan missed certain aspects of theta quotients at p = 13.
I. Case p = 7
To illustrate, define the quotients for p = 7 as,
h
k
=
(
−
1
)
k
−
1
q
k
(
3
k
−
7
)
/
14
f
(
−
q
2
k
,
−
q
7
−
2
k
)
f
(
−
q
k
,
−
q
7
−
k
)
{\displaystyle h_{k}=(-1)^{k-1}q^{k(3k-7)/14}\,{\frac {f(-q^{2k},-q^{7-2k})}{f(-q^{k},-q^{7-k})}}}
hence,
h
1
=
f
(
−
q
2
,
−
q
5
)
q
2
/
7
f
(
−
q
,
−
q
6
)
=
1
q
2
/
7
∏
n
=
1
∞
(
1
−
q
7
n
−
2
)
(
1
−
q
7
n
−
5
)
(
1
−
q
7
n
−
1
)
(
1
−
q
7
n
−
6
)
{\displaystyle h_{1}={\frac {f(-q^{2},-q^{5})}{q^{2/7}f(-q,-q^{6})}}={\frac {1}{q^{2/7}}}\prod _{n=1}^{\infty }{\frac {(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{7n-1})(1-q^{7n-6})}}}
h
2
=
−
f
(
−
q
3
,
−
q
4
)
q
1
/
7
f
(
−
q
2
,
−
q
5
)
=
−
1
q
1
/
7
∏
n
=
1
∞
(
1
−
q
7
n
−
3
)
(
1
−
q
7
n
−
4
)
(
1
−
q
7
n
−
2
)
(
1
−
q
7
n
−
5
)
{\displaystyle h_{2}={\frac {-f(-q^{3},-q^{4})}{q^{1/7}f(-q^{2},-q^{5})}}={\frac {-1}{q^{1/7}}}\prod _{n=1}^{\infty }{\frac {(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-2})(1-q^{7n-5})}}}
h
3
=
q
3
/
7
f
(
−
q
,
−
q
6
)
f
(
−
q
3
,
−
q
4
)
=
1
q
−
3
/
7
∏
n
=
1
∞
(
1
−
q
7
n
−
1
)
(
1
−
q
7
n
−
6
)
(
1
−
q
7
n
−
3
)
(
1
−
q
7
n
−
4
)
{\displaystyle h_{3}={\frac {q^{3/7}f(-q,-q^{6})}{f(-q^{3},-q^{4})}}={\frac {1}{q^{-3/7}}}\prod _{n=1}^{\infty }{\frac {(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{7n-3})(1-q^{7n-4})}}}
functions highly analogous to the Rogers–Ramanujan continued fraction . (Kindly note that the even-index
h
k
{\displaystyle h_{k}}
is negative .) Given the Dedekind eta function
η
(
τ
)
{\displaystyle \eta (\tau )}
, let,
c
=
(
η
(
τ
)
η
(
7
τ
)
)
4
{\displaystyle c=\left({\frac {\eta (\tau )}{\eta (7\tau )}}\right)^{4}}
then Ramanujan found that the 3 roots of the cubic,
x
3
−
(
57
+
14
c
+
c
2
)
x
2
−
(
289
+
126
c
+
19
c
2
+
c
3
)
x
+
1
=
0
{\displaystyle x^{3}-(57+14c+c^{2})x^{2}-(289+126c+19c^{2}+c^{3})x+1=0}
are,
x
1
=
h
1
7
,
x
2
=
h
2
7
,
x
3
=
h
3
7
{\displaystyle x_{1}={h_{1}}^{7},\;\;x_{2}={h_{2}}^{7},\;\;x_{3}={h_{3}}^{7}}
Note also another use for the eta quotient
c
{\displaystyle c}
is,
j
(
τ
)
=
(
y
2
+
5
y
+
1
)
3
(
y
2
+
13
y
+
49
)
y
j
(
τ
)
−
1728
=
(
y
4
+
14
y
3
+
63
y
2
+
70
y
−
7
)
2
y
{\displaystyle {\begin{aligned}&j(\tau )={\frac {(y^{2}+5y+1)^{3}(y^{2}+13y+49)}{y}}\\&j(\tau )-1728={\frac {(y^{4}+14y^{3}+63y^{2}+70y-7)^{2}}{y}}\end{aligned}}}
where
y
=
7
2
c
.
{\displaystyle y={\frac {7^{2}}{c}}.}
II. Case p = 13
It turns out that for p = 13, then the 13th power of the analogous theta quotients are the 6 roots of a sextic .
Define,
h
k
=
(
−
1
)
k
−
1
q
k
(
−
13
+
3
k
)
/
26
f
(
−
q
2
k
,
−
q
13
−
2
k
)
f
(
−
q
k
,
−
q
13
−
k
)
{\displaystyle h_{k}=(-1)^{k-1}q^{k(-13+3k)/26}\,{\frac {f(-q^{2k},-q^{13-2k})}{f(-q^{k},-q^{13-k})}}}
hence,
h
1
=
f
(
−
q
2
,
−
q
11
)
q
5
/
13
f
(
−
q
,
−
q
12
)
,
h
2
=
−
f
(
−
q
4
,
−
q
9
)
q
7
/
13
f
(
−
q
2
,
−
q
11
)
,
h
3
=
f
(
−
q
6
,
−
q
7
)
q
6
/
13
f
(
−
q
3
,
−
q
10
)
{\displaystyle h_{1}={\frac {f(-q^{2},-q^{11})}{q^{5/13}f(-q,-q^{12})}},\;\;h_{2}={\frac {-f(-q^{4},-q^{9})}{q^{7/13}f(-q^{2},-q^{11})}},\;\;h_{3}={\frac {f(-q^{6},-q^{7})}{q^{6/13}f(-q^{3},-q^{10})}}}
h
4
=
−
f
(
−
q
5
,
−
q
8
)
q
2
/
13
f
(
−
q
4
,
−
q
9
)
,
h
5
=
q
5
/
13
f
(
−
q
3
,
−
q
10
)
f
(
−
q
5
,
−
q
8
)
,
h
6
=
−
q
15
/
13
f
(
−
q
,
−
q
12
)
f
(
−
q
6
,
−
q
7
)
{\displaystyle h_{4}={\frac {-f(-q^{5},-q^{8})}{q^{2/13}f(-q^{4},-q^{9})}},\;\;h_{5}={\frac {q^{5/13}f(-q^{3},-q^{10})}{f(-q^{5},-q^{8})}},\;\;h_{6}={\frac {-q^{15/13}f(-q,-q^{12})}{f(-q^{6},-q^{7})}}}
(As before, the even-index
h
k
{\displaystyle h_{k}}
are negative .) Let,
d
=
(
η
(
τ
)
η
(
13
τ
)
)
2
{\displaystyle d=\left({\frac {\eta (\tau )}{\eta (13\tau )}}\right)^{2}}
Ramanujan discovered that the 3 roots of the cubic,
d
=
y
3
+
y
2
−
4
y
+
1
y
(
1
−
y
)
{\displaystyle d={\frac {y^{3}+y^{2}-4y+1}{y(1-y)}}}
are,
y
1
=
h
1
h
5
,
y
2
=
h
2
h
3
,
y
3
=
h
4
h
6
{\displaystyle y_{1}=h_{1}h_{5},\;\;y_{2}=h_{2}h_{3},\;\;y_{3}=h_{4}h_{6}}
Note: Incidentally,
y
3
+
y
2
−
4
y
+
1
=
0
{\displaystyle y^{3}+y^{2}-4y+1=0}
is a well-known equation for
y
=
2
cos
(
2
π
/
13
)
+
2
cos
(
10
π
/
13
)
{\displaystyle y=2\cos(2\pi /13)+2\cos(10\pi /13)}
and similar roots.
However, if a modular equation can be found between, say,
h
1
,
h
5
{\displaystyle h_{1},h_{5}}
, then we can eliminate
h
5
{\displaystyle h_{5}}
, and have an equation solely in
h
1
{\displaystyle h_{1}}
and
d
{\displaystyle d}
. After some effort using Mathematica's integer relations algorithm, I found that, let,
u
=
h
1
,
v
=
h
5
,
x
=
h
1
h
5
{\displaystyle u=h_{1},\,v=h_{5},\,x=h_{1}h_{5}}
then,
u
2
v
2
(
u
v
−
1
)
5
(
u
13
+
v
13
)
=
1
−
7
x
+
19
x
2
−
27
x
3
+
35
x
4
−
68
x
5
+
100
x
6
−
84
x
7
+
69
x
8
−
86
x
9
+
70
x
10
−
29
x
11
+
11
x
12
−
4
x
13
+
x
14
{\displaystyle {\begin{aligned}u^{2}v^{2}(uv-1)^{5}(u^{13}+v^{13})&=1-7x+19x^{2}-27x^{3}+35x^{4}-68x^{5}+100x^{6}-84x^{7}\\&+69x^{8}-86x^{9}+70x^{10}-29x^{11}+11x^{12}-4x^{13}+x^{14}\end{aligned}}}
(The same relation exists between the other pairs.) Eliminating
h
5
{\displaystyle h_{5}}
, one gets the sextic in
z
=
h
1
13
{\displaystyle z=h_{1}^{13}}
,
z
6
+
P
1
z
5
+
P
2
z
4
+
P
3
z
3
+
P
4
z
2
+
P
5
z
−
1
=
0
{\displaystyle z^{6}+P_{1}z^{5}+P_{2}z^{4}+P_{3}z^{3}+P_{4}z^{2}+P_{5}z-1=0}
where the
P
i
{\displaystyle P_{i}}
are polynomials in the eta quotient
d
{\displaystyle d}
of degrees 7, 13, 18, 20, 15, respectively. Explicitly, the first one is,
P
1
=
−
(
h
1
13
+
h
2
13
+
h
3
13
+
h
4
13
+
h
5
13
+
h
6
13
)
=
d
7
+
13
d
6
+
91
d
5
+
377
d
4
+
962
d
3
+
1040
d
2
−
845
d
−
4083
{\displaystyle {\begin{aligned}P_{1}&=-(h_{1}^{13}+h_{2}^{13}+h_{3}^{13}+h_{4}^{13}+h_{5}^{13}+h_{6}^{13})\\&=d^{7}+13d^{6}+91d^{5}+377d^{4}+962d^{3}+1040d^{2}-845d-4083\end{aligned}}}
though the others are too tedious to write down. One can then ascertain that the 6 roots of the sextic are in fact
z
i
=
h
i
13
{\displaystyle z_{i}=h_{i}^{13}}
.
(P.S. I do not see this sextic, nor the modular relation between the
h
i
{\displaystyle h_{i}}
, in Ramanujan's Notebooks III , Entry 8, page 372, which discusses the theta quotients for p = 13. But I find it satisfying that the results for p = 7 can be extended to the next "lacunary" prime p = 13.)
----------- End -----------
In Ramanujan's Notebook IV , (entries 51-72, p.207-237), there are 23 of Ramanujan's P-Q modular equations . However, it seems he missed the prime orders p = 11,13. Since the Dedekind eta function is
η
(
τ
)
=
q
1
/
24
f
(
−
q
)
{\displaystyle \eta (\tau )=q^{1/24}f(-q)}
, for convenience I'll use
η
(
τ
)
{\displaystyle \eta (\tau )}
instead.
I. p = 2
For comparison, Ramanujan found modular relations between
P
′
=
f
(
−
q
)
f
(
−
q
n
)
{\displaystyle P'={\tfrac {f(-q)}{f(-q^{n})}}}
and
Q
′
=
f
(
−
q
2
)
f
(
−
q
2
n
)
{\displaystyle Q'={\tfrac {f(-q^{2})}{f(-q^{2n})}}}
for n = 3, 5, 7, 9, 13, 25. For example, he found,
1. Define
P
=
η
(
t
)
η
(
13
t
)
,
Q
=
η
(
2
t
)
η
(
26
t
)
,
V
k
=
(
Q
P
)
k
+
(
P
Q
)
k
{\displaystyle P={\tfrac {\eta (t)}{\eta (13t)}},\;Q={\tfrac {\eta (2t)}{\eta (26t)}},\;V_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}+{\big (}{\tfrac {P}{Q}}{\big )}^{k}}
, then,
P
Q
+
13
P
Q
=
V
3
−
4
V
1
{\displaystyle PQ+{\tfrac {13}{PQ}}=V_{3}-4V_{1}}
II. p = 11
But there are also relations between
P
′
=
f
(
−
q
)
f
(
−
q
n
)
{\displaystyle P'={\tfrac {f(-q)}{f(-q^{n})}}}
and
Q
′
=
f
(
−
q
11
)
f
(
−
q
11
n
)
{\displaystyle Q'={\tfrac {f(-q^{11})}{f(-q^{11n})}}}
for n = 2, 3, (and 5, 7, 13?) with the first two as,
1. Define
P
=
η
(
t
)
η
(
2
t
)
,
Q
=
η
(
11
t
)
η
(
22
t
)
,
R
k
=
(
P
Q
)
k
+
(
2
P
Q
)
k
{\displaystyle P={\tfrac {\eta (t)}{\eta (2t)}},\;Q={\tfrac {\eta (11t)}{\eta (22t)}},\;R_{k}=(PQ)^{k}+({\tfrac {2}{PQ}})^{k}}
, then,
R
5
+
11
R
3
+
44
R
1
=
(
Q
P
)
6
−
(
P
Q
)
6
{\displaystyle R_{5}+11R_{3}+44R_{1}={\big (}{\tfrac {Q}{P}}{\big )}^{6}-{\big (}{\tfrac {P}{Q}}{\big )}^{6}}
2. Define
P
=
η
(
t
)
η
(
3
t
)
,
Q
=
η
(
11
t
)
η
(
33
t
)
,
S
k
=
(
P
Q
)
k
+
(
3
P
Q
)
k
{\displaystyle P={\tfrac {\eta (t)}{\eta (3t)}},\;Q={\tfrac {\eta (11t)}{\eta (33t)}},\;S_{k}=(PQ)^{k}+({\tfrac {3}{PQ}})^{k}}
, then,
S
5
+
11
(
S
4
+
6
S
3
+
23
S
2
+
63
S
1
+
126
)
=
(
Q
P
)
6
+
(
P
Q
)
6
{\displaystyle S_{5}+11(S_{4}+6S_{3}+23S_{2}+63S_{1}+126)={\big (}{\tfrac {Q}{P}}{\big )}^{6}+{\big (}{\tfrac {P}{Q}}{\big )}^{6}}
3. For comparison, define
P
=
η
(
t
)
η
(
33
t
)
,
Q
=
η
(
3
t
)
η
(
11
t
)
,
T
k
=
(
P
Q
)
k
−
(
3
Q
P
)
k
{\displaystyle P={\tfrac {\eta (t)}{\eta (33t)}},\;Q={\tfrac {\eta (3t)}{\eta (11t)}},\;T_{k}=({\tfrac {P}{Q}})^{k}-({\tfrac {3Q}{P}})^{k}}
, then,
T
5
+
2
T
4
+
3
T
3
−
8
T
2
+
9
T
1
=
(
P
Q
)
3
−
(
11
P
Q
)
3
{\displaystyle T_{5}+2T_{4}+3T_{3}-8T_{2}+9T_{1}=(PQ)^{3}-({\tfrac {11}{PQ}})^{3}}
No.3 is equivalent to Somos' level 33 identity.
- - - - - - - - - - - - - - - - - - - - -
III. p = 13
Likewise, there are modular relations between
P
′
=
f
(
−
q
)
f
(
−
q
n
)
{\displaystyle P'={\tfrac {f(-q)}{f(-q^{n})}}}
and
Q
′
=
f
(
−
q
13
)
f
(
−
q
13
n
)
{\displaystyle Q'={\tfrac {f(-q^{13})}{f(-q^{13n})}}}
for n = 2, 3, 5, 7, with the first two being,
1. Define
P
=
η
(
t
)
η
(
2
t
)
,
Q
=
η
(
13
t
)
η
(
26
t
)
,
U
k
=
(
Q
P
)
k
−
(
P
Q
)
k
{\displaystyle P={\tfrac {\eta (t)}{\eta (2t)}},\;Q={\tfrac {\eta (13t)}{\eta (26t)}},\;U_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}-{\big (}{\tfrac {P}{Q}}{\big )}^{k}}
, then,
U
7
−
13
U
5
+
52
U
3
−
78
U
1
=
(
P
Q
)
6
+
(
2
P
Q
)
6
{\displaystyle U_{7}-13U_{5}+52U_{3}-78U_{1}=(PQ)^{6}+{\big (}{\tfrac {2}{PQ}})^{6}}
2. Define
P
=
η
(
t
)
η
(
3
t
)
,
Q
=
η
(
13
t
)
η
(
39
t
)
,
V
k
=
(
Q
P
)
k
+
(
P
Q
)
k
{\displaystyle P={\tfrac {\eta (t)}{\eta (3t)}},\;Q={\tfrac {\eta (13t)}{\eta (39t)}},\;V_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}+{\big (}{\tfrac {P}{Q}}{\big )}^{k}}
, then,
V
7
+
13
(
−
V
6
+
4
V
5
−
2
V
4
−
19
V
3
+
28
V
2
+
17
V
1
−
50
)
=
(
P
Q
)
6
+
(
3
P
Q
)
6
{\displaystyle V_{7}+13(-V_{6}+4V_{5}-2V_{4}-19V_{3}+28V_{2}+17V_{1}-50)=(PQ)^{6}+{\big (}{\tfrac {3}{PQ}})^{6}}
I do not know if there is a p = 17 identity similar to the ones above.
--------------END--------------
Given the Dedekind eta function
η
(
τ
)
{\displaystyle \eta (\tau )}
. Let p be a prime and define
m
=
(
p
−
1
)
/
2
{\displaystyle m=(p-1)/2}
,
1. Let p be a prime of form
p
=
12
v
+
5
{\displaystyle p=12v+5}
. Then for
n
=
2
,
4
,
8
,
14
{\displaystyle n=2,4,8,14}
:
∑
k
=
0
p
−
1
(
e
π
i
m
k
/
12
η
(
τ
+
m
k
p
)
)
n
=
−
(
p
η
(
p
τ
)
)
n
{\displaystyle \sum _{k=0}^{p-1}{\Big (}e^{\pi imk/12}\eta {\big (}{\tfrac {\tau +mk}{p}}{\big )}{\Big )}^{n}=-{\big (}{\sqrt {p}}\,\eta (p\tau ){\big )}^{n}}
2. Let p be a prime of form
p
=
12
v
+
11
{\displaystyle p=12v+11}
. Then for
n
=
2
,
6
,
10
,
14
{\displaystyle n=2,6,10,14}
:
∑
k
=
0
p
−
1
(
e
π
i
m
k
/
12
η
(
τ
+
m
k
p
)
)
n
=
(
p
η
(
p
τ
)
)
n
{\displaystyle \sum _{k=0}^{p-1}{\Big (}e^{\pi imk/12}\eta {\big (}{\tfrac {\tau +mk}{p}}{\big )}{\Big )}^{n}={\big (}{\sqrt {p}}\,\eta (p\tau ){\big )}^{n}}
Are these two multi-grade identities true?
--------------END--------------
In "An Identity for the Dedekind eta function involving two independent complex variables" , given two complex numbers
a
,
b
{\displaystyle a,b}
with imaginary part > 0, Berndt and Hart gave the identity,
η
3
(
a
3
)
η
3
(
b
3
)
+
i
η
3
(
a
+
1
3
)
η
3
(
b
+
1
3
)
−
η
3
(
a
+
2
3
)
η
3
(
b
+
2
3
)
=
3
3
η
3
(
3
a
)
η
3
(
3
b
)
{\displaystyle \eta ^{3}{\big (}{\tfrac {a}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b}{3}}{\big )}+i\eta ^{3}{\big (}{\tfrac {a+1}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b+1}{3}}{\big )}-\eta ^{3}{\big (}{\tfrac {a+2}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b+2}{3}}{\big )}=3^{3}\eta ^{3}(3a)\eta ^{3}(3b)}
and remarked that they, "...know of no other examples of a similar type." However, it seems the above is just the smallest member of an infinite family of cubes of the Dedekind eta function,
∑
k
=
0
p
−
1
e
2
π
i
k
/
4
η
3
(
a
+
k
p
)
η
3
(
b
+
k
p
)
=
p
3
η
3
(
p
a
)
η
3
(
p
b
)
{\displaystyle \sum _{k=0}^{p-1}e^{2\pi ik/4}\eta ^{3}{\big (}{\tfrac {a+k}{p}}{\big )}\eta ^{3}{\big (}{\tfrac {b+k}{p}}{\big )}=p^{3}\eta ^{3}(pa)\eta ^{3}(pb)}
where p is ANY PRIME of form
p
=
4
n
−
1
{\displaystyle p=4n-1}
, with the Hart-Berndt identity simply the case
p
=
3
{\displaystyle p=3}
. It is easy to test the family using Mathematica and see that it holds for hundreds of decimal digits, but I have no proof that it is generally true.
--------------END--------------
Conjecture 1 . Based on Simon Plouffe 's work on pi .[ 1] (April 10, 2009)
Let q = eπ and k be of the form 4m+3. Then it is true that,
(
a
b
)
π
k
=
∑
n
=
1
∞
1
n
k
(
2
k
−
1
q
n
−
1
−
2
k
−
1
+
1
q
2
n
−
1
+
1
q
4
n
−
1
)
{\displaystyle {\Big (}{\frac {a}{b}}{\Big )}\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{k-1}}{q^{n}-1}}-{\frac {2^{k-1}+1}{q^{2n}-1}}+{\frac {1}{q^{4n}-1}}{\Big )}}
where a,b are integral. (The denominator b turns out to be a highly factorable number.)
For the first few k , we have:
k
a
b
3
1
2
2
∗
3
2
∗
5
{\displaystyle 2^{2}*3^{2}*5}
7
13
3
4
∗
5
2
∗
7
{\displaystyle 3^{4}*5^{2}*7}
11
32072
3
5
∗
5
3
∗
7
2
∗
11
∗
13
{\displaystyle 3^{5}*5^{3}*7^{2}*11*13}
15
219824
3
8
∗
5
3
∗
7
2
∗
13
∗
17
{\displaystyle 3^{8}*5^{3}*7^{2}*13*17}
and so on. Anyone knows how to prove this conjecture?
UPDATE (May 18, 2009)
Turns out there is a closed-form formula for (a/b). This is based on Theorem 6.7 (page 11) of Linas Vepstas' On Plouffe's Ramanujan Identities .[ 2]
Let q = eπ and k = 4m-1 (note this minor change), then
r
π
k
=
∑
n
=
1
∞
1
n
k
(
2
k
−
1
q
n
−
1
−
2
k
−
1
+
1
q
2
n
−
1
+
1
q
4
n
−
1
)
{\displaystyle r\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{k-1}}{q^{n}-1}}-{\frac {2^{k-1}+1}{q^{2n}-1}}+{\frac {1}{q^{4n}-1}}{\Big )}}
Where r is a rational number defined by,
r
=
2
4
m
−
5
∑
j
=
0
2
m
(
−
1
)
j
(
16
m
+
4
−
4
j
+
1
)
B
[
2
j
]
B
[
4
m
−
2
j
]
(
2
j
)
!
(
4
m
−
2
j
)
!
{\displaystyle r=2^{4m-5}\sum _{j=0}^{2m}{\frac {(-1)^{j}(16^{m}+4-4^{j+1})B[2j]B[4m-2j]}{(2j)!(4m-2j)!}}}
and B[w] is a Bernoulli number .
Conjecture 2 . Still based on Plouffe's work on pi but now involves powers k = 4m+1 . (May 19, 2009)
Let q = eπ and k be of the form 4m+1 . Then it is true that,
r
π
k
=
∑
n
=
1
∞
1
n
k
(
2
6
m
+
1
+
(
−
16
)
m
q
n
−
1
−
2
6
m
+
1
+
(
−
16
)
m
+
(
−
1
)
m
q
2
n
−
1
+
(
−
1
)
m
q
4
n
−
1
)
{\displaystyle r\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{6m+1}+(-16)^{m}}{q^{n}-1}}-{\frac {2^{6m+1}+(-16)^{m}+(-1)^{m}}{q^{2n}-1}}+{\frac {(-1)^{m}}{q^{4n}-1}}{\Big )}}
where r is a rational number.
For the first few k = {1,5,9,13,...} we have:
r = {1/24, 1/63, 164/13365, 76192/9823275,...}
and so on. Is there a closed-form formula for r when k = 4m+1?
References :