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Solutions k to the Diophantine equation k = 2n^2 = m^2+1.
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%I #10 Jul 02 2023 18:18:18

%S 2,50,1682,57122,1940450,65918162,2239277042,76069501250,

%T 2584123765442,87784138523762,2982076586042450,101302819786919522,

%U 3441313796169221282,116903366249966604050,3971273138702695316402

%N Solutions k to the Diophantine equation k = 2n^2 = m^2+1.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/NSWNumber.html">NSW Number</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35, -35, 1).

%F G.f.: (2x^2 - 20x + 2)/((1-x)(1 - 34x + x^2)).

%F a(n) = -(sinh((2n - 1) arctanh(sqrt(2))))^2 = 1 -(cosh((2n - 1) arctanh(sqrt(2))))^2. - _Artur Jasinski_, Oct 30 2008

%t Table[Round[N[ -(Sinh[(2 n - 1) ArcTanh[Sqrt[2]]])^2, 100]], {n, 1, 20}] (* _Artur Jasinski_, Oct 30 2008 *)

%Y Corresponding solutions of n are A001653 and m are A002315.

%Y A008843(n-1) + 1.

%K nonn

%O 1,1

%A _Eric W. Weisstein_, Oct 23 2003