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
On
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:
$$71^2+72^2+73^2+74^2+75^2+76^2+77^2+78^2=44444$$
On
rare. Seems to be impossible for numeral bases $5, 11.$ I reported the numbers in base ten, you need to imagine how to write them in the specified base. Base 15 does not seem to work, except for 13; that is, as the base gets larger, we get sums of consecutive squares that are smaller than the base and thus require just one digit. Base 16 ("hexadecimal") does well, we get $1111111_{16}.$
base 17
54 2 5
90 2 6
126 4 7
2456 4 19
end base 17
base 16
85 6 7
204 1 8
255 5 9
221 10 11
819 1 13
2730 8 20
17895697 624 666
end base 16
base 14
30 1 4
90 2 6
135 3 7
20685 25 42
end base 14
base 13
14 1 3
140 1 7
126 4 7
366 8 11
4760 8 24
4760 23 29
11900 18 34
23800 18 42
11900 35 42
end base 13
base 12
13 2 3
91 1 6
7540 15 29
9425 5 30
end base 12
base 10
55 1 5
77 4 6
1111 11 16
44444 71 78
444444 51 113
end base 10
base 9
30 1 4
50 3 5
91 1 6
728 7 13
44286 16 51
332150 53 104
end base 9
base 8
54 2 5
365 10 12
365 13 14
1755 5 17
3510 10 22
end base 8
base 7
285 1 9
end base 7
base 6
14 1 3
86 3 6
1036 9 15
9331 19 32
end base 6
base 4
5 1 2
85 6 7
255 5 9
2730 8 20
end base 4
base 3
13 2 3
728 7 13
end base 3
base 2
255 5 9
end base 2
This is the meat of the program, C++ with GMP
mpz_class choose3(mpz_class n)
{
return n * (n-1) * (n-2) / 6;
}
mpz_class squarepyramid(mpz_class n)
{
return choose3(n+2 ) + choose3(n+1 ) ;
}
int main()
{
mpz_class base = 10;
cout << " base " << base << endl;
for(mpz_class x = 2; x <= 12200; ++x) {
for(mpz_class y = 0; y < x -1; ++y) {
// cout << x << " " << squarepyramid(x) << endl;
mpz_class m = squarepyramid(x) - squarepyramid(y) ;
mpz_class n = 1;
while (n < m) n *= base;
n -= 1;
n /= (base - 1);
if( m % n == 0 && m >= base) cout << m << " " << y+1 << " " << x << endl;
}}
cout << " end base " << base << endl;
return 0 ;
}
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!
On
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.
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.