%I #17 Sep 08 2022 08:45:30
%S 0,21,504,597,704,3441,3980,4601,20540,23681,27300,120197,138504,
%T 159597,701040,807741,930680,4086441,4708340,5424881,23818004,
%U 27442697,31619004,138821981,159948240,184289541,809114280,932247141,1074118640
%N Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x+199)^2 = y^2.
%C Also values x of Pythagorean triples (x, x+199, y).
%C Corresponding values y of solutions (x, y) are in A159548.
%C For the generic case x^2+(x+p)^2 = y^2 with p = 2*m^2-1 a (prime) number in A066436 see A118673 or A129836.
%C Lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
%C Lim_{n -> infinity} a(n)/a(n-1) = (201+20*sqrt(2))/199 for n mod 3 = {1, 2}.
%C Lim_{n -> infinity} a(n)/a(n-1) = (91443+58282*sqrt(2))/199^2 for n mod 3 = 0.
%H G. C. Greubel, <a href="/A129993/b129993.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,6,-6,0,-1,1).
%F a(n) = 6*a(n-3) - a(n-6) + 398 for n > 6; a(1)=0, a(2)=21, a(3)=504, a(4)=597, a(5)=704, a(6)=3441.
%F G.f.: x*(21+483*x+93*x^2-19*x^3-161*x^4-19*x^5) / ((1-x)*(1-6*x^3+x^6)).
%F a(3*k+1) = 199*A001652(k) for k >= 0.
%F a(1)=0, a(2)=21, a(3)=504, a(4)=597, a(5)=704, a(6)=3441, a(7)=3980, a(n)=a(n-1)+6*a(n-3)-6*a(n-4)-a(n-6)+a(n-7). - _Harvey P. Dale_, Jun 03 2012
%t LinearRecurrence[{1,0,6,-6,0,-1,1},{0,21,504,597,704,3441,3980},30] (* _Harvey P. Dale_, Jun 03 2012 *)
%o (PARI) {forstep(n=0, 500000000, [1, 3], if(issquare(2*n^2+398*n+39601), print1(n, ",")))};
%o (Magma) I:=[0,21,504,597,704,3441,3980]; [n le 7 select I[n] else Self(n-1) + 6*Self(n-3) - 6*Self(n-4) - Self(n-6) + Self(n-7): n in [1..50]]; // _G. C. Greubel_, Mar 31 2018
%Y Cf. A159548, A066436, A118673, A118674, A129836, A001652, A156035 (decimal expansion of 3+2*sqrt(2)), A159549 (decimal expansion of (201+20*sqrt(2))/199), A159550 (decimal expansion of (91443+58282*sqrt(2))/199^2).
%K nonn,easy
%O 1,2
%A _Mohamed Bouhamida_, Jun 14 2007
%E Edited and two terms added by _Klaus Brockhaus_, Apr 14 2009