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%I #26 Mar 19 2024 09:32:29
%S 5,5,10,10,15,15,20,20,25,25,30,30,35,35,40,40,45,45,50,50,55,55,60,
%T 60,65,65,70,70,75,75,80,80,85,85,90,90,95,95,100,100,105,105,110,110,
%U 115,115,120,120,125,125,130,130,135,135,140,140,145,145,150,150,155,155
%N a(n) = 5*n - a(n-1), with n>1, a(1)=5.
%H Vincenzo Librandi, <a href="/A166598/b166598.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F G.f.: 5*x/((1+x)*(x-1)^2). - _Vincenzo Librandi_, Sep 15 2013
%F From _G. C. Greubel_, May 18 2016: (Start)
%F a(n) = a(n-1) + a(n-2) - a(n-3).
%F E.g.f.: (5/2)*(x*cosh(x) + (1+x)*sinh(x)).
%F a(n) = 5*A004526(n+1) = 5*A008619(n-1) = 5*A110654(n). (End)
%t RecurrenceTable[{a[1]==5,a[n]==5n-a[n-1]},a,{n,70}] (* or *) Flatten[ {#,#}&/@(5Range[40])] (* _Harvey P. Dale_, Nov 29 2011 *)
%t CoefficientList[Series[5 / ((1 + x) (x - 1)^2), {x, 0, 60}], x] (* _Vincenzo Librandi_, Sep 15 2013 *)
%t LinearRecurrence[{1,1,-1}, {5, 5, 10}, 50] (* _G. C. Greubel_, May 18 2016 *)
%o (Magma) [n le 1 select (n+4) else 5*n-Self(n-1): n in [1..70] ]; // _Vincenzo Librandi_, Sep 14 2013
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Oct 18 2009
%E Corrected a(59) by _Harvey P. Dale_, Nov 29 2011
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