Sequence A073851 and the primality of sum_{k=11..n}(k^2)

[Max]

Ryan Propper wrote:
> It seems that if
> 
> sum_{k=11..n}(k^2)
> 
> ever assumes any prime values, it does so only for extremely large n.
> Specifically, I conducted a brute force search with Mathematica and
> this sum is composite for all n <= 4*10^10.
> 
> This led me to wonder if sum_{k=11..n}(k^2) is composite for all n >=
> 11.  However, I've been unable to prove this result (which would then
> imply that A073851 is both finite and full, with a(4) omitted). 
> Mathematica gives the closed form formula
> 
> sum_{k=11..n}(k^2) = (2n^2 + 23n + 231)(n - 10)/6
> 
> Can anyone prove this is composite for all n or find a counterexample?

There are four cases possible:

sum_{k=11..n}(k^2) = [(2n^2 + 23n + 231)] * [(n - 10)/6]

sum_{k=11..n}(k^2) = [(2n^2 + 23n + 231)/2] * [(n - 10)/3]

sum_{k=11..n}(k^2) = [(2n^2 + 23n + 231)/3] * [(n - 10)/2]

sum_{k=11..n}(k^2) = [(2n^2 + 23n + 231)/6] * [(n - 10)]

where the numbers in brackets are integer.

But it is clear that for n>16, both factors are greater than 1 in whatever case. That proves sum_{k=11..n}(k^2) is composite for n>16.
And you proved the same it for n=11..15. Therefore, sum_{k=11..n}(k^2) is composite for n>11.

Max