Ramanujan's Sum Identity

$\begin{displaymath} {1+53x+9x^2\over 1-82x-82x^2+x^3}=\sum_{n=1}^\infty a_n x^n \end{displaymath}$

(1)

$\begin{displaymath} {2-26x-12x^2\over 1-82x-82x^2+x^3}=\sum_{n=0}^\infty b_n x^n \end{displaymath}$

(2)

$\begin{displaymath} {2+8x-10x^2\over 1-82x-82x^2+x^3}=\sum_{n=0}^\infty c_n x^n, \end{displaymath}$

(3)

then

$\begin{displaymath} {a_n}^3+{b_n}^3={c_n}^3+(-1)^n. \end{displaymath}$

(4)

Hirschhorn (1995) showed that

$\displaystyle a_n$	$\textstyle =$	$\displaystyle {\textstyle{1\over 85}} [(64+8\sqrt{85}\,)\alpha^n+(64-8\sqrt{85}\,)\beta^n-43(-1)^n]$
			(5)
$\displaystyle b_n$	$\textstyle =$	$\displaystyle {\textstyle{1\over 85}} [(77+7\sqrt{85}\,)\alpha^n+(77-7\sqrt{85}\,)\beta^n+16(-1)^n]$
			(6)
$\displaystyle c_n$	$\textstyle =$	$\displaystyle {\textstyle{1\over 85}} [(93+9\sqrt{85}\,)\alpha^n+(93-9\sqrt{85}\,)\beta^n-16(-1)^n],$
			(7)

where

$\displaystyle \alpha$	$\textstyle =$	$\displaystyle {\textstyle{1\over 2}}(83+9\sqrt{85}\,)$	(8)
$\displaystyle \beta$	$\textstyle =$	$\displaystyle {\textstyle{1\over 2}}(83-9\sqrt{85}\,).$	(9)

Hirschhorn (1996) showed that checking the first seven cases

to 6 is sufficient to prove the result.

References

Hirschhorn, M. D. ``An Amazing Identity of Ramanujan.'' Math. Mag. 68, 199-201, 1995.

Hirschhorn, M. D. ``A Proof in the Spirit of Zeilberger of an Amazing Identity of Ramanujan.'' Math. Mag. 69, 267-269, 1996.