A218395 Numbers whose square is the sum of the squares of 11 consecutive integers.

11, 77, 143, 1529, 2849, 30503, 56837, 608531, 1133891, 12140117, 22620983, 242193809, 451285769, 4831736063, 9003094397, 96392527451, 179610602171, 1923018812957, 3583208949023, 38363983731689, 71484568378289, 765356655820823, 1426108158616757

Offset: 0

Author

Paul Weisenhorn, Oct 28 2012

Keywords

• nonn
• easy

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.

Examples

			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.
		

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).

Crossrefs

c=2: A001653(n+1) = a(n); A002315(n) = b(n); A001652(n) = x(n).

Cf. A001032 (11 is a term of that sequence), A198947.

Programs

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 *)

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