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%I A115135 #15 Sep 08 2022 08:45:23
%S A115135 0,108,1407,1851,2407,9768,12340,15568,58435,73423,92235,342076,
%T A115135 429432,539076,1995255,2504403,3143455,11630688,14598220,18322888,
%U A115135 67790107,85086151,106795107,395111188,495919920,622448988,2302878255,2890434603
%N A115135 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2+(x+617)^2 = y^2.
%C A115135 Also values x of Pythagorean triples (x, x+617, y).
%C A115135 Corresponding values y of solutions (x, y) are in A160176.
%C A115135 lim_{n -> infinity} a(n)/a(n-3) = 3+2*sqrt(2).
%C A115135 lim_{n -> infinity} a(n)/a(n-1) = (633+100*sqrt(2))/617 for n mod 3 = {1, 2}.
%C A115135 lim_{n -> infinity} a(n)/a(n-1) = (755667+461578*sqrt(2))/617^2 for n mod 3 = 0.
%H A115135 G. C. Greubel, Table of n, a(n) for n = 1..1000
%H A115135 Index entries for linear recurrences with constant coefficients, signature (1,0,6,-6,0,-1,1).
%F A115135 a(n) = 6*a(n-3) -a(n-6) +1234 for n > 6; a(1)=0, a(2)=108, a(3)=1407, a(4)=1851, a(5)=2407, a(6)=9768.
%F A115135 G.f.: x*(108 +1299*x +444*x^2 -92*x^3 -433*x^4 -92*x^5)/((1-x)*(1 -6*x^3 +x^6)).
%F A115135 a(3*k+1) = 617*A001652(k) for k >= 0.
%t A115135 LinearRecurrence[{1,0,6,-6,0,-1,1}, {0,108,1407,1851,2407,9768,12340}, 50] (* _G. C. Greubel_, May 04 2018 *)
%o A115135 (PARI) {forstep(n=0, 10000000, [3, 1], if(issquare(2*n^2+1234*n+380689), print1(n, ",")))}
%o A115135 (PARI) x='x+O('x^30); Vec(x*(108 +1299*x +444*x^2 -92*x^3 -433*x^4 -92*x^5)/((1-x)*(1 -6*x^3 +x^6))) \\ _G. C. Greubel_, May 04 2018
%o A115135 (Magma) I:=[0,108,1407,1851,2407,9768,12340]; [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..30]]; // _G. C. Greubel_, May 04 2018
%Y A115135 Cf. A160176, A001652, A111258, A156035 (decimal expansion of 3+2*sqrt(2)), A160177 (decimal expansion of (633+100*sqrt(2))/617), A160178 (decimal expansion of (755667+461578*sqrt(2))/617^2).
%K A115135 nonn,easy
%O A115135 1,2
%A A115135 _Mohamed Bouhamida_, Jun 03 2007
%E A115135 Edited and two terms added by _Klaus Brockhaus_, May 18 2009
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