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

[Ryan Propper]

Hello seqfans,

Sequence A073851 is defined as:

"Group the squares so that the sum of each group is a prime: (1, 4),
(9, 16,  25, 36, 49, 64), (81, 100), (121, 144, ..., 1521, 1600), ...
This is the sequence of the number of terms in each group."

Unfortunately, the fourth term of this sequence is wrong, as

sum_{k=11..40}(k^2) = 21755

is not prime.  I was trying to correct and extend this sequence but
was thwarted by a(4).  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?