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

Author

Bruno Berselli, Jan 24 2013

Keywords

• nonn
• easy

Comments

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.

Links

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

Crossrefs

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).

Programs

Magma

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

Magma

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

Maple

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

Mathematica

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 *)

Maxima

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);

Formula

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.