Entry 14

Ramanujan's sum of cubes identity is defined by the generating functions, $$\begin{aligned}
\sum_{n=0}^\infty a_n x^n &= \frac{1+53x+9x^2}{R_1}\\
\sum_{n=0}^\infty b_n x^n &= \frac{2-26x-12x^2}{R_1}\\
\sum_{n=0}^\infty c_n x^n &= \frac{2+8x-10x^2}{R_1}\\
\end{aligned}\tag{1}$$ where \(R_1 = 1-82x-82x^2+x^3\). Then $$a_n^3+b_n^3 = c_n^3 + (-1)^n\tag{2}$$ It turns out the \(a_n, b_n, c_n\) can also be expressed as $$\begin{aligned}
a_n &=-9p^2+176pq-851q^2 = 1,135,11151,\dots\\
b_n &=4(3p^2-56pq+263q^2) = 2,138,11468,\dots\\
c_n &=2(5p^2-90pq+409q^2) = 2,172,14258,\dots\\
d_n &= -(p^2-85q^2) = 1, -1,1,-1\dots
\end{aligned}\tag{3}$$ with \(p,q\) chosen to satisfy the Pell equation \(p^2-85q^2 =\mp 1\). (Actually, \(p,q\) are half-integers since one can use \(p^2-85q^2 =\mp 4\).) But Entry 13 shows there are infinitely many quadratic parametrizations to \((2)\). Thus we can find similar generating functions such as $$\begin{aligned}
\sum_{n=0}^\infty a_n x^n &= \frac{-9(417-5602x+x^2)}{R_2}\\
\sum_{n=0}^\infty b_n x^n &= \frac{8(-566-11315x+x^2)}{R_2}\\
\sum_{n=0}^\infty c_n x^n &= \frac{-6(877+6898x+x^2)}{R_2}\\
\end{aligned}\tag{4}$$ where \(R_2 = -1+184899x-184899x^2+x^3\). Then $$a_n^3+b_n^3 = c_n^3 + 1\tag{5}$$ and its quadratic parameterization $$\begin{aligned}
a_n &=3(3p^2-104pq+909q^2) = 3753, 693875529,\dots\\
b_n &=-2(4p^2-135pq+1119q^2) = 4528, 837313192,\dots\\
c_n &=6(p^2-37pq+348q^2) = 5262, 972979926,\dots\\
d_n &=p^2-321q^2  = 1,1,1,\dots
\end{aligned}\tag{6}$$ with \(p,q\) chosen to satisfy \(p^2-321q^2 = 1\). So there are infinitely many sum of cubes identity analogous to Ramanujan's.