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%I #16 Mar 17 2023 07:14:37
%S 245,265,287,343,427,493,637,833,1097,1295,1715,2275,2693,3577,4753,
%T 6317,7483,9947,13223,15665,20825,27685,36805,43603,57967,77063,91297,
%U 121373,161357,214513,254135,337855,449155,532117,707413,940457,1250273
%N Positive numbers y such that y^2 is of the form x^2+(x+343)^2 with integer x.
%C (-147, a(1)), (-96, a(2)), (-63, a(3)) and (A118611(n), a(n+3)) are solutions (x, y) to the Diophantine equation x^2+(x+343)^2 = y^2.
%C lim_{n -> infinity} a(n)/a(n-7) = 3+2*sqrt(2).
%C lim_{n -> infinity} a(n)/a(n-1) = (3+2*sqrt(2)) / ((9+4*sqrt(2))/7)^2 for n mod 7 = {0, 1, 2, 4, 5}.
%C lim_{n -> infinity} a(n)/a(n-1) = ((9+4*sqrt(2))/7)^5 / (3+2*sqrt(2))^2 for n mod 7 = {3, 6}.
%H <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,6,0,0,0,0,0,0,-1).
%F a(n) = 6*a(n-7)-a(n-14) for n > 14; a(1)=245, a(2)=265, a(3)=287, a(4)=343, a(5)=427, a(6)=493, a(7)=637, a(8)=833, a(9)=1097, a(10)=1295, a(11)=1715, a(12)=2275, a(13)=2693, a(14)=3577.
%F G.f.: x * (1-x) * (245 +510*x +797*x^2 +1140*x^3 +1567*x^4 +2060*x^5 +2697*x^6 +2060*x^7 +1567*x^8 +1140*x^9 +797*x^10 +510*x^11 +245*x^12) / (1 -6*x^7 +x^14).
%F a(7*k-3) = 343*A001653(k) for k >= 1.
%e (-147, a(1)) = (-147, 245) is a solution: (-147)^2+(-147+343)^2 = 21609+38416 = 60025 = 245^2.
%e (A118611(1), a(4)) = (0, 343) is a solution: 0^2+(0+343)^2 = 117649 = 343^2.
%e (A118611(3), a(6)) = (132, 493) is a solution: 132^2+(132+343)^2 = 17424+225625 = 243049 = 493^2.
%t Sqrt[#]&/@Select[Table[x^2+(x+343)^2,{x,-150,885000}], IntegerQ[ Sqrt[#]]&] (* or *) LinearRecurrence[{0,0,0,0,0,0,6,0,0,0,0,0,0,-1},{245,265,287,343,427,493,637,833,1097,1295,1715,2275,2693,3577},40](* _Harvey P. Dale_, Dec 30 2011 *)
%o (PARI) {forstep(n=-148, 1000000, [1, 3], if(issquare(n^2+(n+343)^2, &k), print1(k, ",")))}
%Y Cf. A118611, A001653, A156035 (decimal expansion of 3+2*sqrt(2)), A156649 (decimal expansion of (9+4*sqrt(2))/7).
%K nonn
%O 1,1
%A _Klaus Brockhaus_, Feb 25 2009
%E G.f. adapted to the offset by _Bruno Berselli_, Apr 01 2011
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