There's something special about Ramanujan! -- part 2

We can see that,

$a_{1} = 1111$

$a_{2} = 1^3+1^3+1^3+1^3 = 4$

$a_{3} = 4^3=64$

$a_{4} = 6^3+4^3 =280$

$a_{5} = 2^3+8^3+0^3 = 520$

$a_{6} = 5^3+2^3+0^3 = 133$

$a_{7} = 1^3 + 3^3+3^3 = 55$

$a_{8} = 5^3 + 5^3 = 250$

$a_{9} = 2^3+5^3 + 0^3 = 133$

and the series continues so on. The $1728^{th} \, or \, a_{1728}$ term will be the same as $a_{6}$ so the $1729^{th} \, or \,a_{1729}$ term will be the same as $a_{7}$ $a_{1729} = a_{7} = \boxed {55}$

From the 6th term the series is repeating the order $133,55,250\dots\dots$

So the $1729th$ term will be equal to $\boxed{55}$

the series is like 1111,4,64,280,520,133,55,250,133,55,250....................................and so on