Is $11^2+12^2+13^2+14^2+15^2+16^2=1111$ special? 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
 A: I've heard this kind of thing called a curio, or "curiosity."  The kind of thing that makes you go, "Huh ... neat!"
Piquito's answer is interesting.  I wouldn't be so bold as to say it's the answer, because there are lots of other ways to dissect and reassemble things.  For example:
$$1111 = 555 + 555 + 1 = 10 \cdot 111  + 1.$$
The formula for the sum of the squares of six consecutive integers, starting with $n$, is $S(n) = 6n^2 + 30n + 55.$  This will let you probe that space (I didn't see any other examples of repeated-digit numbers, and I looked up to around $n=90000$.)
Or, extending a bit, the sum of $m$ consecutive squares, starting with the square of $n$ is
$$S'(m,n) = mn^2 + m(m-1)n + \frac{(m-1)m(2m-1)}{6}.$$
For example
$$S'(6,11) = 1111$$
Canvassing this space would result in a subset of OEIS sequence A180436, "Palindromic numbers which are sum of consecutive squares."  The next one above $1111$ that has just one digit in its representation is $44444$.  I'm resisting the temptation to figure out which squares this one is a sum of.  (OK, wythagoras figured it out.)
Anyway, if there is a relationship or an underlying rule, it's well-hidden.  But it's fun to look!
A: 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.
A: The reason is that $$11^2+12^2+13^2+14^2+15^2+16^2=600+420+91$$ in which the summands are such that successively give $1,1,1,1$  when making the sum. In fact,
$$(10+1)^2+(10+2)^2+(10+3)^2+(10+4)^2+(10+5)^2+(10+6)^2=6\cdot10^2+20(1+2+3+4+5+6)+(1^2+2^2+3^2+4^2+5^2+6^2)=6\cdot10^2+20(\frac{6\cdot7}{2})+\frac{6(6+1)(2\cdot6+1)}{6}=600+420+91$$
I don't know if it is generalizable.
A: It is probably coincidence. When considering sums of 16 or less consecutive squares, where the smallest is at most $200^2$, I found only one other such sum:
$$71^2+72^2+73^2+74^2+75^2+76^2+77^2+78^2=44444$$
