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%I #26 Mar 07 2023 02:27:12
%S 650,2525,5650,10025,15650,22525,30650,40025,50650,62525,75650,90025,
%T 105650,122525,140650,160025,180650,202525,225650,250025,275650,
%U 302525,330650,360025,390650,422525,455650,490025,525650,562525,600650,640025,680650,722525,765650
%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 A157916(n)^2 - a(n)*A005843(n)^2 = 1. - _Vincenzo Librandi_, Feb 10 2012
%H Vincenzo Librandi, <a href="/A157915/b157915.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: x*(650 + 575*x + 25*x^2)/(1-x)^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) = (coth(Pi/5)*Pi/5 - 1)/50.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/5)*Pi/5)/50. (End)
%t 625Range[40]^2+25 (* _Harvey P. Dale_, Apr 05 2011 *)
%t LinearRecurrence[{3, -3, 1}, {650, 2525, 5650}, 40] (* _Vincenzo Librandi_, Feb 10 2012 *)
%o (Magma) I:=[650, 2525, 5650]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 10 2012
%o (PARI) for(n=1, 40, print1(625*n^2 + 25", ")); \\ _Vincenzo Librandi_, Feb 10 2012
%Y Cf. A005843, A157916.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 09 2009
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