A218395 - OEIS

%I #70 Dec 12 2022 15:00:11

%S 11,77,143,1529,2849,30503,56837,608531,1133891,12140117,22620983,

%T 242193809,451285769,4831736063,9003094397,96392527451,179610602171,

%U 1923018812957,3583208949023,38363983731689,71484568378289,765356655820823,1426108158616757

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

%C 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.

%C 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.

%H V. Pletser, <a href="http://arxiv.org/abs/1409.7972">Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials</a>, arXiv preprint arXiv:1409.7972 [math.NT], 2014. See Table 1 p. 7.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,20,0,-1).

%F a(n) = 20*a(n-2) - a(n-4); b(n) = 20*b(n-2) - b(n-4);

%F a(n) = 10*a(n-2) + 33*b(n-2); b(n) = 3*a(n-2) + 10*b(n-2).

%F a(n) = a(n-1) + 20*a(n-2) - 20*a(n-3) - a(n-4) + a(n-5).

%F G.f.: 11 * (1-x)*(1+8*x+x^2) / (1 - 20*x^2 + x^4).

%F With r=sqrt(11); s=10+3*r; t=10-3*r:

%F a(2*n) = ((11+r)*s^n + (11-r)*t^n)/2.

%F a(2*n+1) = ((77+23*r) * s^n + (77-23*r)*t^n)/2.

%F a(n) = 11 * A198947(n+1). - _Bill McEachen_, Dec 01 2022

%e For n=6, Sum_{z=17132..17142} z^2 = 3230444569;

%e a(6) = sqrt(3230444569) = 56837;

%e b(6) = sqrt((a(6)^2-110)/11) = 17137; x(6) = b(6)-5 = 17132.

%p s:=0: n:=-1:

%p for j from -5 to 5 do s:=s+j^2: end do:

%p for z from -4 to 100000 do

%p   s:=s-(z-1)^2+(z+10)^2: r:=sqrt(s):

%p   if (r=floor(r)) then

%p     n:=n+1: a(n):=r: x(n):=z:

%p     b(n):=sqrt((s-110)/11):

%p     print(n,a(n),b(n),x(n)):

%p   end if:

%p end do:

%t LinearRecurrence[{0,20,0,-1},{11,77,143,1529},30] (* _Harvey P. Dale_, Aug 15 2022 *)

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

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

%K nonn,easy

%O 0,1

%A _Paul Weisenhorn_, Oct 28 2012