A110422 - OEIS

%I #25 Jun 03 2024 13:35:23

%S 1,2,-1,-2,6,8,-6,-8,15,18,-15,-18,28,32,-28,-32,45,50,-45,-50,66,72,

%T -66,-72,91,98,-91,-98,120,128,-120,-128,153,162,-153,-162,190,200,

%U -190,-200,231,242,-231,-242,276,288,-276,-288,325,338,-325,-338,378,392,-378,-392,435,450,-435,-450,496,512,-496,-512,561

%N a(n) = sum( (-1)^(r+1)*(n-r)*r, r = 1..floor(n/2) ).

%C a(4n)=-a(4n-2); a(4n+1)=-a(4n-1). If sum in definition is not alternating one obtains A023855. - _Emeric Deutsch_, Aug 08 2005

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,-4,6,-6,6,-4,2,-1).

%F a(2n) = (1/2)n-(-1)^n*(1/2)n^2; a(2n-1) = (1/2)n-(1/4)+(-1)^n*(1/4)(2n^2-2n+1). - _Emeric Deutsch_, Aug 08 2005

%F a(n) = (-1)^((2*n-5+(-1)^n)/4)*(2*n^2+1-(-1)^n+4*n*(-1)^((2*n-5+(-1)^n)/4))/16. - _Luce ETIENNE_, Oct 30 2014

%F G.f.: x^2*(2*x^3-x^2+1) / ((x-1)^2*(x^2+1)^3). - _Colin Barker_, Oct 30 2014

%e a(8) = -6 because 7*1-6*2+5*3-4*4 = -6.

%p a:=n->sum((-1)^(r+1)*(n-r)*r,r=1..floor(n/2)): seq(a(n),n=2..70); # _Emeric Deutsch_, Aug 08 2005

%t CoefficientList[Series[(2 x^3 - x^2 + 1)/((x - 1)^2 (x^2 + 1)^3), {x, 0, 70}], x] (* _Vincenzo Librandi_, Oct 30 2014 *)

%t LinearRecurrence[{2,-4,6,-6,6,-4,2,-1},{1,2,-1,-2,6,8,-6,-8},70] (* _Harvey P. Dale_, Apr 04 2020 *)

%o (PARI) Vec(x^2*(2*x^3-x^2+1)/((x-1)^2*(x^2+1)^3) + O(x^100)) \\ _Colin Barker_, Oct 30 2014

%Y Cf. A023855.

%K sign,easy

%O 2,2

%A _Amarnath Murthy_, Aug 01 2005

%E Corrected and extended by _Emeric Deutsch_, Aug 08 2005