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%I #23 Mar 07 2023 02:27:16
%S 600,2475,5600,9975,15600,22475,30600,39975,50600,62475,75600,89975,
%T 105600,122475,140600,159975,180600,202475,225600,249975,275600,
%U 302475,330600,359975,390600,422475,455600,489975,525600,562475,600600,639975,680600,722475,765600
%N a(n) = 625*n^2 - 25.
%C The identity (50*n^2 - 1)^2 - (625*n^2 - 25)*(2*n)^2 = 1 can be written as A157919(n)^2 - a(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Feb 10 2012
%H Vincenzo Librandi, <a href="/A157918/b157918.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F From _Vincenzo Librandi_, Feb 10 2012: (Start)
%F G.f.: -25*x*(24 + 27*x - x^2)/(x-1)^3.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
%F From _Amiram Eldar_, Mar 07 2023: (Start)
%F Sum_{n>=1} 1/a(n) = (1 - cot(Pi/5)*Pi/5)/50.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/5)*Pi/5 - 1)/50. (End)
%t LinearRecurrence[{3, -3, 1}, {600, 2475, 5600}, 50] (* _Vincenzo Librandi_, Feb 10 2012 *)
%o (Magma) I:=[600, 2475, 5600]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // _Vincenzo Librandi_, Feb 10 2012
%o (PARI) for(n=1, 40, print1(625*n^2 - 25", ")); \\ _Vincenzo Librandi_, Feb 10 2012
%Y Cf. A005843, A157919.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 09 2009
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