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Ramanujan's Sum Identity

If

\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 $n=0$ 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.




© 1996-9 Eric W. Weisstein
1999-05-25