TITLE: Is $11^2+12^2+13^2+14^2+15^2+16^2=1111$ special?
QUESTION [21 upvotes]: Is this pure coincidence or is this a special case of some well-known number-theoretic result?
If the latter is true, is there some notable generalization?
EDIT: Thanks to the interesting answers below, a follow-up question is now on MathOverflow: https://mathoverflow.net/questions/239033/repdigit-numbers-which-are-sum-of-consecutive-squares?noredirect=1#comment591447_239033

REPLY [5 votes]: It looks like the $n$'th positive integer representable as a sum of consecutive squares is asymptotically on the order of $\alpha n^\beta$ for some positive constants $\alpha$ and $\beta$ (on the basis of the solutions $\le 10^6$, I'd estimate $\alpha \approx  0.871$, $\beta \approx 1.436$, but all that really matters for this posting is $\beta > 1$).  If so, the probability that a random number $N$ is so representable is on the order of $\dfrac{N^{1/\beta - 1}}{\alpha^{1/\beta} \beta}$.  Now there are $9$ base-10 repunits with $d$ digits, so (if these repunits are "typical") the expected number of those that are representable is
on the order of a constant times $10^{d(1/\beta-1)}$.  Since $\sum_{d=1}^\infty 10^{d(1/\beta-1)} < \infty$, we should expect only finitely many base-10 repunits to be representable as sums of consecutive squares.
Of course this is just heuristic (we really have no reason to think the repunits are "typical"), so should not be taken too seriously.