A198947 -id:A198947 - cp's OEIS frontend

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

Paul Weisenhorn, Oct 28 2012

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.

			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.

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).

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.

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:

LinearRecurrence[{0,20,0,-1},{11,77,143,1529},30] (* Harvey P. Dale, Aug 15 2022 *)

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

A221762 Numbers m such that 11*m^2 + 5 is a square.

Original entry on oeis.org

1, 2, 22, 41, 439, 818, 8758, 16319, 174721, 325562, 3485662, 6494921, 69538519, 129572858, 1387284718, 2584962239, 27676155841, 51569671922, 552135832102, 1028808476201, 11015040486199, 20524599852098, 219748673891878, 409463188565759

Offset: 1

Bruno Berselli, Jan 24 2013

Corresponding squares are: 16, 49, 5329, 18496, 2119936, 7360369, 843728209, 2929407376, ... (subsequence of A016778).
The Diophantine equation 11*x^2+k = y^2, for |k|<11, has integer solutions with the following k values:
k = -10, the nonnegative x values are in A198947;
k =  -8,            "                    2*A075839;
k =  -7,            "                    A221763;
k =  -2,            "                    A075839;
k =   1,            "                    A001084;
k =   4,            "                    A075844;
k =   5,            "                    this sequence;
k =   9,            "                    3*A001084.
Also, the Diophantine equation h*x^2+5 = y^2 has infinitely many integer solutions for h = 5, 11, 19, 20, 29, 31, 41, 44, 55, 59, ...
a(n+1)/a(n) tends alternately to (1+sqrt(11))^2/10 and (4+sqrt(11))^2/5.
a(n+2)/a(n) tends to A176395^2/2.

Bruno Berselli, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (0,20,0,-1).

Cf. A001084, A075839, A075844, A198947, A198949, A221763.

Cf. A049629 (numbers m such that 20*m^2 + 5 is a square), A075796 (numbers m such that 5*m^2 + 5 is a square).

m:=24; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+2*x^2+x^3)/(1-20*x^2+x^4)));

I:=[1,2,22,41]; [n le 4 select I[n] else 20*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Aug 18 2013

A221762:=proc(q)
local n;
for n from 1 to q do if type(sqrt(11*n^2+5), integer) then print(n);
fi; od; end:
A221762(1000); # Paolo P. Lava, Feb 19 2013

LinearRecurrence[{0, 20, 0, -1}, {1, 2, 22, 41}, 24]
CoefficientList[Series[(1 + 2 x + 2 x^2 + x^3)/(1 - 20 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 18 2013 *)

makelist(expand(((-11*(-1)^n+4*sqrt(11))*(10+3*sqrt(11))^floor(n/2)-(11*(-1)^n+4*sqrt(11))*(10-3*sqrt(11))^floor(n/2))/22), n, 1, 24);

G.f.: x*(1+2*x+2*x^2+x^3)/(1-20*x^2+x^4).
a(n) = -a(1-n) = ((-11*(-1)^n+4*t)*(10+3*t)^floor(n/2)-(11*(-1)^n+4*t)*(10-3*t)^floor(n/2))/22, where t=sqrt(11).
a(n) = 20*a(n-2) - a(n-4) for n>4, a(1)=1, a(2)=2, a(3)=22, a(4)=41.
a(n)*a(n-3)-a(n-1)*a(n-2) = -(3/2)*(9-7*(-1)^n).
a(n+1) + a(n-1) =  A198949(n), with a(0)=-1.
2*a(n-1) - a(n) =  A001084(n/2-1) for even n.

A198949 y-values in the solution to 11*x^2-10 = y^2.

Original entry on oeis.org

1, 23, 43, 461, 859, 9197, 17137, 183479, 341881, 3660383, 6820483, 73024181, 136067779, 1456823237, 2714535097, 29063440559, 54154634161, 579811987943, 1080378148123, 11567176318301, 21553408328299, 230763714378077, 429987788417857, 4603707111243239

Offset: 1

Sture Sjöstedt, Oct 31 2011

When are both n+1 and 11*n+1 perfect squares? This problem gives the equation 11*x^2-10 = y^2.

Vincenzo Librandi, Table of n, a(n) for n = 1..250
Index entries for linear recurrences with constant coefficients, signature (0, 20, 0, -1).

Cf. A198947.

LinearRecurrence[{0, 20, 0, -1}, {1,23,43,461}, 24]  (* Bruno Berselli, Nov 11 2011 *)

makelist(expand(((-(-1)^n-sqrt(11))*(10-3*sqrt(11))^floor(n/2)+(-(-1)^n+sqrt(11))*(10+3*sqrt(11))^floor(n/2))/2), n, 1, 24);  /* Bruno Berselli, Nov 14 2011 */

a(n+4) = 20*a(n+2)-a(n) with a(1)=1, a(2)=23, a(3)=43, a(4)=461.
G.f.: x*(1+x)*(1+22*x+x^2)/(1-20*x^2+x^4). - Bruno Berselli, Nov 04 2011
a(n) = ((-(-1)^n-t)*(10-3*t)^floor(n/2)+(-(-1)^n+t)*(10+3*t)^floor(n/2))/2 where t=sqrt(11). - Bruno Berselli, Nov 14 2011

More terms from Bruno Berselli, Nov 04 2011

A199336 x-values in the solution to 15*x^2 - 14 = y^2.

Original entry on oeis.org

1, 3, 5, 23, 39, 181, 307, 1425, 2417, 11219, 19029, 88327, 149815, 695397, 1179491, 5474849, 9286113, 43103395, 73109413, 339352311, 575589191, 2671715093, 4531604115, 21034368433, 35677243729, 165603232371, 280886345717, 1303791490535, 2211413522007

Offset: 1

Sture Sjöstedt, Nov 05 2011

When are both n+1 and 15*n+1 perfect squares? This problem gives the equation 15*x^2-14 = y^2.

Values of x (or y) in the solutions to x^2 - 8xy + y^2 + 14 = 0. - Colin Barker, Feb 05 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..250
Index entries for linear recurrences with constant coefficients, signature (0,8,0,-1).

Cf. A198947, A199338.

Essentially the second differences of A237262. Cf. also A322780.

m:=29; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-x)*(1+4*x+x^2)/(1-8*x^2+x^4))); // Bruno Berselli, Nov 08 2011

LinearRecurrence[{0, 8, 0, -1}, {1, 3, 5, 23}, 50] (* T. D. Noe, Nov 07 2011 *)

a(n+4) = 8*a(n+2) - a(n), a(1)=1, a(2)=3, a(3)=5, a(4)=23.

G.f.: x*(1-x)*(1+4*x+x^2)/(1-8*x^2+x^4). - Bruno Berselli, Nov 08 2011

More terms from T. D. Noe, Nov 07 2011

A238240 Positive integers n such that x^2 - 20xy + y^2 + n = 0 has integer solutions.

Original entry on oeis.org

18, 35, 50, 63, 72, 74, 83, 90, 95, 98, 99, 107, 140, 162, 171, 200, 215, 227, 252, 266, 275, 288, 296, 315, 332, 347, 359, 360, 362, 371, 380, 387, 392, 395, 396, 407, 428, 450, 491, 495, 530, 539, 560, 567, 602, 623, 626, 635, 648, 666, 684, 695, 711, 722, 743, 747, 755, 770, 791, 794, 800, 810

Offset: 1

Colin Barker, Feb 20 2014

nonn

Positive integers n such that x^2 - 99 y^2 + n = 0 has integer solutions. - Robert Israel, Oct 22 2024

			63 is in the sequence because x^2 - 20xy + y^2 + 63 = 0 has integer solutions, for example (x, y) = (1, 16).

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A075839 (n = 18), A221763 (n = 63), A198947 (n = 90), A001085 (n = 99).

Cf. A031363, A084917, A237351, A237599, A237606, A237609, A237610, A236330, A236331.

filter:= t -> [isolve(99*y^2 - z^2 = t)] <> []:
select(filter, [$1..1000]); # Robert Israel, Oct 22 2024

Corrected by Robert Israel, Oct 22 2024

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

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A221762 Numbers m such that 11*m^2 + 5 is a square.

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A198949 y-values in the solution to 11*x^2-10 = y^2.

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A199336 x-values in the solution to 15*x^2 - 14 = y^2.

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A238240 Positive integers n such that x^2 - 20xy + y^2 + n = 0 has integer solutions.

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