A221763 -id:A221763 - cp's OEIS frontend

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

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.

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

cp's OEIS Frontend

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

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