User:Titus III

Update, 5/7/2022

I. Rogers-Selberg Mod 7 Identities

$A(q)={\frac {(q^{3},q^{4},q^{7};q^{7})_{\infty }}{(q^{2};q^{2})_{\infty }}}={\frac {f(-q^{3},-q^{4})}{f(-q^{2},-q^{4})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{7n})(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{2n})}}$

$B(q)={\frac {(q^{2},q^{5},q^{7};q^{7})_{\infty }}{(q^{2};q^{2})_{\infty }}}={\frac {f(-q^{2},-q^{5})}{f(-q^{2},-q^{4})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{7n})(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{2n})}}$

$C(q)={\frac {(q,q^{6},q^{7};q^{7})_{\infty }}{(q^{2};q^{2})_{\infty }}}\,={\frac {f(-q,-q^{6})}{f(-q^{2},-q^{4})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{7n})(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{2n})}}$

II. Bailey Mod 9 Identities

$A(q)={\frac {(q^{4},q^{5},q^{9};q^{9})_{\infty }}{(q^{3};q^{3})_{\infty }}}={\frac {f(-q^{4},-q^{5})}{f(-q^{3},-q^{6})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{9n})(1-q^{9n-4})(1-q^{9n-5})}{(1-q^{3n})}}$

$B(q)={\frac {(q^{2},q^{7},q^{9};q^{9})_{\infty }}{(q^{3};q^{3})_{\infty }}}={\frac {f(-q^{2},-q^{7})}{f(-q^{3},-q^{6})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{9n})(1-q^{9n-2})(1-q^{9n-7})}{(1-q^{3n})}}$

$C(q)={\frac {(q,q^{8},q^{9};q^{9})_{\infty }}{(q^{3};q^{3})_{\infty }}}\,={\frac {f(-q,-q^{8})}{f(-q^{3},-q^{6})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{9n})(1-q^{9n-1})(1-q^{9n-8})}{(1-q^{3n})}}$

III. Rogers Mod 14 Identities

$A(q)={\frac {(q^{6},q^{8},q^{14};q^{14})_{\infty }}{(q;q)_{\infty }}}\,={\frac {f(-q^{6},-q^{8})}{f(-q,-q^{2})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{14n})(1-q^{14n-6})(1-q^{14n-8})}{(1-q^{n})}}$

$B(q)={\frac {(q^{4},q^{10},q^{14};q^{14})_{\infty }}{(q;q)_{\infty }}}={\frac {f(-q^{4},-q^{10})}{f(-q,-q^{2})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{14n})(1-q^{14n-4})(1-q^{14n-10})}{(1-q^{n})}}$

$C(q)={\frac {(q^{2},q^{12},q^{14};q^{14})_{\infty }}{(q;q)_{\infty }}}={\frac {f(-q^{2},-q^{12})}{f(-q,-q^{2})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{14n})(1-q^{14n-2})(1-q^{14n-12})}{(1-q^{n})}}$

IV. Dyson Mod 27 Identities

$A(q)={\frac {(q^{12},q^{15},q^{27};q^{27})_{\infty }}{(q;q)_{\infty }}}={\frac {f(-q^{12},-q^{15})}{f(-q,-q^{2})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{27n})(1-q^{27n-12})(1-q^{27n-15})}{(1-q^{n})}}$

$B(q)={\frac {(q^{9},q^{18},q^{27};q^{27})_{\infty }}{(q;q)_{\infty }}}={\frac {f(-q^{9},-q^{18})}{f(-q,-q^{2})}}=\prod _{n=1}^{\infty }{\frac {(1-q^{27n})(1-q^{27n-9})(1-q^{27n-18})}{(1-q^{n})}}$

$C(q)={\frac {(q^{6},q^{21},q^{27};q^{27})_{\infty }}{(q;q)_{\infty }}}\,={\frac {f(-q^{6},-q^{21})}{f(-q,-q^{2})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{27n})(1-q^{27n-6})(1-q^{27n-21})}{(1-q^{n})}}$

$D(q)={\frac {(q^{3},q^{24},q^{27};q^{27})_{\infty }}{(q;q)_{\infty }}}\,={\frac {f(-q^{3},-q^{24})}{f(-q,-q^{2})}}\,=\prod _{n=1}^{\infty }{\frac {(1-q^{27n})(1-q^{27n-3})(1-q^{27n-24})}{(1-q^{n})}}$

Update, 11/5/2012

This is similar to the 11/3/2012 update. Let,

$m={\frac {12}{n-1}}$

be a positive integer for some odd number n. Thus, there are only 4 possibilities: $n=3,5,7,13$. Given the standard Ramanujan theta functions $\phi (q),\;\psi (q),\;f(-q)$, then,

$\left({\frac {\phi (-q^{2})\phi (-q^{6n})}{\phi (-q^{6})\phi (-q^{2n})}}\right)^{m}\,+\,q^{3}\left({\frac {\psi (q)\psi (q^{3n})}{\psi (q^{3})\psi (q^{n})}}\right)^{m}-\,q^{3}\left({\frac {\psi (-q)\psi (-q^{3n})}{\psi (-q^{3})\psi (-q^{n})}}\right)^{m}-\,2mq^{2}\left({\frac {f(-q^{2})f(-q^{6n})}{f(-q^{6})f(-q^{2n})}}\right)^{m}=1$

For example, for n = 5, hence m = 3, we have,

$\left({\frac {\phi (-q^{2})\phi (-q^{30})}{\phi (-q^{6})\phi (-q^{10})}}\right)^{3}\,+\,q^{3}\left({\frac {\psi (q)\psi (q^{15})}{\psi (q^{3})\psi (q^{5})}}\right)^{3}-\,q^{3}\left({\frac {\psi (-q)\psi (-q^{15})}{\psi (-q^{3})\psi (-q^{5})}}\right)^{3}-\,6q^{2}\left({\frac {f(-q^{2})f(-q^{30})}{f(-q^{6})f(-q^{10})}}\right)^{3}=1$

and so on for the other three n.

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Update, 11/4/2012

Define,

$h_{k}=(-1)^{k-1}q^{k(3k-25)/50}\,{\frac {f(-q^{2k},-q^{25-2k})}{f(-q^{k},-q^{25-k})}},\;\;{\text{for}}\;k=1,...,12.$

(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 $j={\frac {\eta (\tau )}{\eta (25\tau )}}$, analogous to the case for n = 13 discovered by Ramanujan. Since one pair is a constant, $h_{5}h_{10}=-1$, then the remaining five are the roots of a quintic (naturally enough) given by,

$1+(10+10j+4j^{2}+j^{3})x+(5+15j+11j^{2}+5j^{3}+j^{4})x^{2}-2(5+5j+3j^{2}+j^{3})x^{3}+j^{2}x^{4}+x^{5}=0$

with j as the eta quotient given above. The roots are then,

$[x_{1},\,x_{2},\,x_{3},\,x_{4},\,x_{5}]=[h_{1}h_{7},\,h_{2}h_{11},\,h_{3}h_{4},\,h_{6}h_{8},\,h_{9}h_{12}]$

(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!

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Update, 11/3/2012

Here is a generalization of one identity described by Berndt as "...fascinating, but with no direct proof" (Ramanujan's Notebooks III, p.322). Let,

$k={\frac {24}{n-1}}$

be a positive integer for some odd number n. Of course, there are only 6 possibilities: $n=3,5,7,9,13,25$. Given the standard Ramanujan theta functions $\phi (q),\;\psi (q),\;f(-q)$, then it is proposed that,

${\frac {\phi ^{k}(-q^{2n})}{\phi ^{k}(-q^{2})}}+q^{3}\left({\frac {\psi ^{k}(q^{n})}{\psi ^{k}(q)}}-{\frac {\psi ^{k}(-q^{n})}{\psi ^{k}(-q)}}\right)-2kq^{2}\left({\frac {f^{k}(-q^{2n})}{f^{k}(-q^{2})}}\right)=1$

and, as suggested by Michael Somos, by using a Fricke involution,

${\frac {\phi ^{k}(-q^{2})}{\phi ^{k}(-q^{2n})}}+{\frac {1}{q^{3}}}\left({\frac {\psi ^{k}(q)}{\psi ^{k}(q^{n})}}-{\frac {\psi ^{k}(-q)}{\psi ^{k}(-q^{n})}}\right)-{\frac {2k}{q^{2}}}\left({\frac {f^{k}(-q^{2})}{f^{k}(-q^{2n})}}\right)=n^{k/2}$

Thus, for n = 3, we have,

${\frac {\phi ^{12}(-q^{6})}{\phi ^{12}(-q^{2})}}+q^{3}\left({\frac {\psi ^{12}(q^{3})}{\psi ^{12}(q)}}-{\frac {\psi ^{12}(-q^{3})}{\psi ^{12}(-q)}}\right)-24q^{2}\left({\frac {f^{12}(-q^{6})}{f^{12}(-q^{2})}}\right)=1$

and,

${\frac {\phi ^{12}(-q^{2})}{\phi ^{12}(-q^{6})}}+{\frac {1}{q^{3}}}\left({\frac {\psi ^{12}(q)}{\psi ^{12}(q^{3})}}-{\frac {\psi ^{12}(-q)}{\psi ^{12}(-q^{3})}}\right)-{\frac {24}{q^{2}}}\left({\frac {f^{12}(-q^{2})}{f^{12}(-q^{6})}}\right)=3^{6}$

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.

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Update, 11/2/2012

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,

$h_{k}=(-1)^{k-1}q^{k(3k-7)/14}\,{\frac {f(-q^{2k},-q^{7-2k})}{f(-q^{k},-q^{7-k})}}$

hence,

$h_{1}={\frac {f(-q^{2},-q^{5})}{q^{2/7}f(-q,-q^{6})}}={\frac {1}{q^{2/7}}}\prod _{n=1}^{\infty }{\frac {(1-q^{7n-2})(1-q^{7n-5})}{(1-q^{7n-1})(1-q^{7n-6})}}$

$h_{2}={\frac {-f(-q^{3},-q^{4})}{q^{1/7}f(-q^{2},-q^{5})}}={\frac {-1}{q^{1/7}}}\prod _{n=1}^{\infty }{\frac {(1-q^{7n-3})(1-q^{7n-4})}{(1-q^{7n-2})(1-q^{7n-5})}}$

$h_{3}={\frac {q^{3/7}f(-q,-q^{6})}{f(-q^{3},-q^{4})}}={\frac {1}{q^{-3/7}}}\prod _{n=1}^{\infty }{\frac {(1-q^{7n-1})(1-q^{7n-6})}{(1-q^{7n-3})(1-q^{7n-4})}}$

functions highly analogous to the Rogers–Ramanujan continued fraction. (Kindly note that the even-index $h_{k}$ is negative.) Given the Dedekind eta function $\eta (\tau )$, let,

$c=\left({\frac {\eta (\tau )}{\eta (7\tau )}}\right)^{4}$

then Ramanujan found that the 3 roots of the cubic,

$x^{3}-(57+14c+c^{2})x^{2}-(289+126c+19c^{2}+c^{3})x+1=0$

are,

$x_{1}={h_{1}}^{7},\;\;x_{2}={h_{2}}^{7},\;\;x_{3}={h_{3}}^{7}$

Note also another use for the eta quotient $c$ is,

${\begin{aligned}&j(\tau )={\frac {(y^{2}+5y+1)^{3}(y^{2}+13y+49)}{y}}\\&j(\tau )-1728={\frac {(y^{4}+14y^{3}+63y^{2}+70y-7)^{2}}{y}}\end{aligned}}$

where $y={\frac {7^{2}}{c}}.$

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,

$h_{k}=(-1)^{k-1}q^{k(-13+3k)/26}\,{\frac {f(-q^{2k},-q^{13-2k})}{f(-q^{k},-q^{13-k})}}$

hence,

$h_{1}={\frac {f(-q^{2},-q^{11})}{q^{5/13}f(-q,-q^{12})}},\;\;h_{2}={\frac {-f(-q^{4},-q^{9})}{q^{7/13}f(-q^{2},-q^{11})}},\;\;h_{3}={\frac {f(-q^{6},-q^{7})}{q^{6/13}f(-q^{3},-q^{10})}}$

$h_{4}={\frac {-f(-q^{5},-q^{8})}{q^{2/13}f(-q^{4},-q^{9})}},\;\;h_{5}={\frac {q^{5/13}f(-q^{3},-q^{10})}{f(-q^{5},-q^{8})}},\;\;h_{6}={\frac {-q^{15/13}f(-q,-q^{12})}{f(-q^{6},-q^{7})}}$

(As before, the even-index $h_{k}$ are negative.) Let,

$d=\left({\frac {\eta (\tau )}{\eta (13\tau )}}\right)^{2}$

Ramanujan discovered that the 3 roots of the cubic,

$d={\frac {y^{3}+y^{2}-4y+1}{y(1-y)}}$

are,

$y_{1}=h_{1}h_{5},\;\;y_{2}=h_{2}h_{3},\;\;y_{3}=h_{4}h_{6}$

Note: Incidentally, $y^{3}+y^{2}-4y+1=0$ is a well-known equation for $y=2\cos(2\pi /13)+2\cos(10\pi /13)$ and similar roots.

However, if a modular equation can be found between, say, $h_{1},h_{5}$, then we can eliminate $h_{5}$, and have an equation solely in $h_{1}$ and $d$. After some effort using Mathematica's integer relations algorithm, I found that, let,

$u=h_{1},\,v=h_{5},\,x=h_{1}h_{5}$

then,

${\begin{aligned}u^{2}v^{2}(uv-1)^{5}(u^{13}+v^{13})&=1-7x+19x^{2}-27x^{3}+35x^{4}-68x^{5}+100x^{6}-84x^{7}\\&+69x^{8}-86x^{9}+70x^{10}-29x^{11}+11x^{12}-4x^{13}+x^{14}\end{aligned}}$

(The same relation exists between the other pairs.) Eliminating $h_{5}$, one gets the sextic in $z=h_{1}^{13}$,

$z^{6}+P_{1}z^{5}+P_{2}z^{4}+P_{3}z^{3}+P_{4}z^{2}+P_{5}z-1=0$

where the $P_{i}$ are polynomials in the eta quotient $d$ of degrees 7, 13, 18, 20, 15, respectively. Explicitly, the first one is,

${\begin{aligned}P_{1}&=-(h_{1}^{13}+h_{2}^{13}+h_{3}^{13}+h_{4}^{13}+h_{5}^{13}+h_{6}^{13})\\&=d^{7}+13d^{6}+91d^{5}+377d^{4}+962d^{3}+1040d^{2}-845d-4083\end{aligned}}$

though the others are too tedious to write down. One can then ascertain that the 6 roots of the sextic are in fact $z_{i}=h_{i}^{13}$.

(P.S. I do not see this sextic, nor the modular relation between the $h_{i}$, 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.)

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Update, 9/23/2012

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 $\eta (\tau )=q^{1/24}f(-q)$, for convenience I'll use $\eta (\tau )$ instead.

I. p = 2

For comparison, Ramanujan found modular relations between $P'={\tfrac {f(-q)}{f(-q^{n})}}$ and $Q'={\tfrac {f(-q^{2})}{f(-q^{2n})}}$ for n = 3, 5, 7, 9, 13, 25. For example, he found,

1. Define $P={\tfrac {\eta (t)}{\eta (13t)}},\;Q={\tfrac {\eta (2t)}{\eta (26t)}},\;V_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}+{\big (}{\tfrac {P}{Q}}{\big )}^{k}$, then,

$PQ+{\tfrac {13}{PQ}}=V_{3}-4V_{1}$

II. p = 11

But there are also relations between $P'={\tfrac {f(-q)}{f(-q^{n})}}$ and $Q'={\tfrac {f(-q^{11})}{f(-q^{11n})}}$ for n = 2, 3, (and 5, 7, 13?) with the first two as,

1. Define $P={\tfrac {\eta (t)}{\eta (2t)}},\;Q={\tfrac {\eta (11t)}{\eta (22t)}},\;R_{k}=(PQ)^{k}+({\tfrac {2}{PQ}})^{k}$, then,

$R_{5}+11R_{3}+44R_{1}={\big (}{\tfrac {Q}{P}}{\big )}^{6}-{\big (}{\tfrac {P}{Q}}{\big )}^{6}$

2. Define $P={\tfrac {\eta (t)}{\eta (3t)}},\;Q={\tfrac {\eta (11t)}{\eta (33t)}},\;S_{k}=(PQ)^{k}+({\tfrac {3}{PQ}})^{k}$, then,

$S_{5}+11(S_{4}+6S_{3}+23S_{2}+63S_{1}+126)={\big (}{\tfrac {Q}{P}}{\big )}^{6}+{\big (}{\tfrac {P}{Q}}{\big )}^{6}$

3. For comparison, define $P={\tfrac {\eta (t)}{\eta (33t)}},\;Q={\tfrac {\eta (3t)}{\eta (11t)}},\;T_{k}=({\tfrac {P}{Q}})^{k}-({\tfrac {3Q}{P}})^{k}$, then,

$T_{5}+2T_{4}+3T_{3}-8T_{2}+9T_{1}=(PQ)^{3}-({\tfrac {11}{PQ}})^{3}$

No.3 is equivalent to Somos' level 33 identity.

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III. p = 13

Likewise, there are modular relations between $P'={\tfrac {f(-q)}{f(-q^{n})}}$ and $Q'={\tfrac {f(-q^{13})}{f(-q^{13n})}}$ for n = 2, 3, 5, 7, with the first two being,

1. Define $P={\tfrac {\eta (t)}{\eta (2t)}},\;Q={\tfrac {\eta (13t)}{\eta (26t)}},\;U_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}-{\big (}{\tfrac {P}{Q}}{\big )}^{k}$, then,

$U_{7}-13U_{5}+52U_{3}-78U_{1}=(PQ)^{6}+{\big (}{\tfrac {2}{PQ}})^{6}$

2. Define $P={\tfrac {\eta (t)}{\eta (3t)}},\;Q={\tfrac {\eta (13t)}{\eta (39t)}},\;V_{k}={\big (}{\tfrac {Q}{P}}{\big )}^{k}+{\big (}{\tfrac {P}{Q}}{\big )}^{k}$, then,

$V_{7}+13(-V_{6}+4V_{5}-2V_{4}-19V_{3}+28V_{2}+17V_{1}-50)=(PQ)^{6}+{\big (}{\tfrac {3}{PQ}})^{6}$

I do not know if there is a p = 17 identity similar to the ones above.

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Update, 9/16/2012

Given the Dedekind eta function $\eta (\tau )$. Let p be a prime and define $m=(p-1)/2$,

1. Let p be a prime of form $p=12v+5$. Then for $n=2,4,8,14$:

$\sum _{k=0}^{p-1}{\Big (}e^{\pi imk/12}\eta {\big (}{\tfrac {\tau +mk}{p}}{\big )}{\Big )}^{n}=-{\big (}{\sqrt {p}}\,\eta (p\tau ){\big )}^{n}$

2. Let p be a prime of form $p=12v+11$. Then for $n=2,6,10,14$:

$\sum _{k=0}^{p-1}{\Big (}e^{\pi imk/12}\eta {\big (}{\tfrac {\tau +mk}{p}}{\big )}{\Big )}^{n}={\big (}{\sqrt {p}}\,\eta (p\tau ){\big )}^{n}$

Are these two multi-grade identities true?

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Update, 9/11/2012

In "An Identity for the Dedekind eta function involving two independent complex variables", given two complex numbers $a,b$ with imaginary part > 0, Berndt and Hart gave the identity,

$\eta ^{3}{\big (}{\tfrac {a}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b}{3}}{\big )}+i\eta ^{3}{\big (}{\tfrac {a+1}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b+1}{3}}{\big )}-\eta ^{3}{\big (}{\tfrac {a+2}{3}}{\big )}\eta ^{3}{\big (}{\tfrac {b+2}{3}}{\big )}=3^{3}\eta ^{3}(3a)\eta ^{3}(3b)$

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,

$\sum _{k=0}^{p-1}e^{2\pi ik/4}\eta ^{3}{\big (}{\tfrac {a+k}{p}}{\big )}\eta ^{3}{\big (}{\tfrac {b+k}{p}}{\big )}=p^{3}\eta ^{3}(pa)\eta ^{3}(pb)$

where p is ANY PRIME of form $p=4n-1$, with the Hart-Berndt identity simply the case $p=3$. 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.

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Update, 4/10/2009

Conjecture 1. Based on Simon Plouffe's work on pi. (April 10, 2009)

Let q = e^π and k be of the form 4m+3. Then it is true that,

${\Big (}{\frac {a}{b}}{\Big )}\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{k-1}}{q^{n}-1}}-{\frac {2^{k-1}+1}{q^{2n}-1}}+{\frac {1}{q^{4n}-1}}{\Big )}$

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	$2^{2}*3^{2}*5$
7	13	$3^{4}*5^{2}*7$
11	32072	$3^{5}*5^{3}*7^{2}*11*13$
15	219824	$3^{8}*5^{3}*7^{2}*13*17$

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.

Let q = e^π and k = 4m-1 (note this minor change), then

$r\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{k-1}}{q^{n}-1}}-{\frac {2^{k-1}+1}{q^{2n}-1}}+{\frac {1}{q^{4n}-1}}{\Big )}$

Where r is a rational number defined by,

$r=2^{4m-5}\sum _{j=0}^{2m}{\frac {(-1)^{j}(16^{m}+4-4^{j+1})B[2j]B[4m-2j]}{(2j)!(4m-2j)!}}$

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 = e^π and k be of the form 4m+1. Then it is true that,

$r\pi ^{k}=\sum _{n=1}^{\infty }{\frac {1}{n^{k}}}{\Big (}{\frac {2^{6m+1}+(-16)^{m}}{q^{n}-1}}-{\frac {2^{6m+1}+(-16)^{m}+(-1)^{m}}{q^{2n}-1}}+{\frac {(-1)^{m}}{q^{4n}-1}}{\Big )}$

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: