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The y-values in the solution to 19*x^2 - 18 = y^2.
2

%I #24 May 06 2024 06:55:05

%S 1,39,571,911,13299,194141,309739,4521621,66007369,105310349,

%T 1537337841,22442311319,35805208921,522690344319,7630319841091,

%U 12173665722791,177713179730619,2594286303659621,4139010540540019,60421958418066141,882049712924430049

%N The y-values in the solution to 19*x^2 - 18 = y^2.

%C When are both n+1 and 19*n+1 perfect squares? This gives the equation 19*x^2 - 18 = y^2.

%H Vincenzo Librandi, <a href="/A200409/b200409.txt">Table of n, a(n) for n = 1..300</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,340,0,0,-1).

%F a(n) = 340*a(n-3) - a(n-6), a(1)=1, a(2)=39, a(3)=571, a(4)=911, a(5)=13299, a(6)=194141.

%F G.f.: x*(x+1)*(x^4 + 38*x^3 + 533*x^2 + 38*x + 1) / (x^6 - 340*x^3 + 1). - _Colin Barker_, Sep 01 2013

%e a(7) = 340*911 - 1 = 309739.

%t LinearRecurrence[{0, 0, 340, 0, 0, -1}, {1, 39, 571, 911, 13299,194141}, 50]

%o (Magma) I:=[1, 39, 571, 911, 13299, 194141]; [n le 6 select I[n] else 340*Self(n-3)-Self(n-6): n in [1..30]]; // _Vincenzo Librandi_, Nov 18 2011

%o (PARI) Vec(x*(x+1)*(x^4+38*x^3+533*x^2+38*x+1)/(x^6-340*x^3+1) + O(x^100)) \\ _Colin Barker_, Sep 01 2013

%Y Cf. A200407, A199773, A199774, A199798.

%K nonn,easy

%O 1,2

%A _Sture Sjöstedt_, Nov 17 2011