OFFSET
0,1
COMMENTS
a(n)^2 = Sum_{j=0..10} (x(n)+j)^2 = 11*(x(n)+5)^2 + 110 and b(n) = x(n)+5 give the Pell equation a(n)^2 - 11*b(n)^2 = 110 with the 2 fundamental solutions (11; 1) and (77; 23) and the solution (10; 3) for the unit form. A198949(n+1) = b(n); A106521(n) = x(n) and x(0) = -4.
General: If the sum of the squares of c neighboring numbers is a square with c = 3*k^2-1 and 1 <= k, then a(n)^2 = Sum_{j=0..c-1} (x(n)+j)^2 and b(n) = 2*x(n)+c-1 give the Pell equation a(n)^2 - c*(b(n)/2)^2 = binomial(c+1,3)/2. a(n) = 2*e1*a(n-k) - a(n-2*k); b(n) = 2*e1*b(n-k) - b(n-2*k); a(n) = e1*a(n-k) + c*e2*b(n-k); b(n) = e2*a(n-k) + e1*b(n-k) with the solution (e1; e2) for the unit form.
LINKS
V. Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 1 p. 7.
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).
FORMULA
a(n) = 20*a(n-2) - a(n-4); b(n) = 20*b(n-2) - b(n-4);
a(n) = 10*a(n-2) + 33*b(n-2); b(n) = 3*a(n-2) + 10*b(n-2).
a(n) = a(n-1) + 20*a(n-2) - 20*a(n-3) - a(n-4) + a(n-5).
G.f.: 11 * (1-x)*(1+8*x+x^2) / (1 - 20*x^2 + x^4).
With r=sqrt(11); s=10+3*r; t=10-3*r:
a(2*n) = ((11+r)*s^n + (11-r)*t^n)/2.
a(2*n+1) = ((77+23*r) * s^n + (77-23*r)*t^n)/2.
a(n) = 11 * A198947(n+1). - Bill McEachen, Dec 01 2022
EXAMPLE
For n=6, Sum_{z=17132..17142} z^2 = 3230444569;
a(6) = sqrt(3230444569) = 56837;
b(6) = sqrt((a(6)^2-110)/11) = 17137; x(6) = b(6)-5 = 17132.
MAPLE
s:=0: n:=-1:
for j from -5 to 5 do s:=s+j^2: end do:
for z from -4 to 100000 do
s:=s-(z-1)^2+(z+10)^2: r:=sqrt(s):
if (r=floor(r)) then
n:=n+1: a(n):=r: x(n):=z:
b(n):=sqrt((s-110)/11):
print(n, a(n), b(n), x(n)):
end if:
end do:
MATHEMATICA
LinearRecurrence[{0, 20, 0, -1}, {11, 77, 143, 1529}, 30] (* Harvey P. Dale, Aug 15 2022 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Weisenhorn, Oct 28 2012
STATUS
approved