|
%I #9 Mar 03 2024 15:42:35
%S 697,857,1117,3065,4285,6005,17693,24853,34913,103093,144833,203473,
%T 600865,844145,1185925,3502097,4920037,6912077,20411717,28676077,
%U 40286537,118968205,167136425,234807145,693397513,974142473,1368556333
%N Positive numbers y such that y^2 is of the form x^2+(x+857)^2 with integer x.
%C (-185, a(1)) and (A129857(n), a(n+1)) are solutions (x, y) to the Diophantine equation x^2+(x+857)^2 = y^2.
%C lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
%C lim_{n -> infinity} a(n)/a(n-1) = (907+210*sqrt(2))/857 for n mod 3 = {0, 2}.
%C lim_{n -> infinity} a(n)/a(n-1) = (1208787+678878*sqrt(2))/857^2 for n mod 3 = 1.
%H G. C. Greubel, <a href="/A160206/b160206.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 6, 0, 0, -1).
%F a(n) = 6*a(n-3) -a(n-6) for n > 6; a(1)=697, a(2)=857, a(3)=1117, a(4)=3065, a(5)=4285, a(6)=6005.
%F G.f.: (1-x)*(697+1554*x+2671*x^2+1554*x^3+697*x^4)/(1-6*x^3+x^6).
%F a(3*k-1) = 857*A001653(k) for k >= 1.
%e (-185, a(1)) = (-185, 697) is a solution: (-185)^2+(-185+857)^2 = 34225+451584 = 485809 = 697^2.
%e (A129857(1), a(2)) = (0, 857) is a solution: 0^2+(0+857)^2 = 734449 = 857^2.
%e (A129857(3), a(4)) = (1696, 3065) is a solution: 1696^2+(1696+857)^2 = 2876416+6517809 = 9394225 = 3065^2.
%t LinearRecurrence[{0,0,6,0,0,-1}, {697,857,1117,3065,4285,6005}, 50] (* _G. C. Greubel_, May 14 2018 *)
%o (PARI) {forstep(n=-188, 10000000, [3, 1], if(issquare(2*n^2 +1714*n +734449, &k), print1(k, ",")))}
%o (PARI) x='x+O('x^30); Vec((1-x)*(697+1554*x+2671*x^2+1554*x^3 +697*x^4 )/(1-6*x^3+x^6)) \\ _G. C. Greubel_, May 14 2018
%o (Magma) I:=[697,857,1117,3065,4285,6005]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..40]]; // _G. C. Greubel_, May 14 2018
%Y Cf. A129857, A001653, A156035 (decimal expansion of 3+2*sqrt(2)), A160207 (decimal expansion of (907+210*sqrt(2))/857), A160208 (decimal expansion of (1208787+678878*sqrt(2))/857^2).
%K nonn
%O 1,1
%A _Klaus Brockhaus_, May 18 2009
|
