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A059859 Sum of squares of first n quarter-squares (A002620).

%I A059859 #28 Aug 17 2024 03:23:52
%S A059859 0,0,1,5,21,57,138,282,538,938,1563,2463,3759,5523,7924,11060,15156,
%T A059859 20340,26901,35001,45001,57101,71742,89166,109902,134238,162799,
%U A059859 195923,234339,278439,329064,386664,452200,526184,609705,703341
%N A059859 Sum of squares of first n quarter-squares (A002620).
%H A059859 Vincenzo Librandi, <a href="/A059859/b059859.txt">Table of n, a(n) for n = 0..1000</a>
%H A059859 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-8,6,6,-8,0,3,-1).
%F A059859 If n is even, a(n) = n*(n+2)*(2*n^3+n^2-2*n+4)/160; if n is odd, a(n) = (n^2-1)*(2*n^3+5*n^2+2*n-5)/160.
%F A059859 From _R. J. Mathar_, Feb 15 2010: (Start)
%F A059859 a(n) = 3*a(n-1) - 8*a(n-3) + 6*a(n-4) + 6*a(n-5) - 8*a(n-6) + 3*a(n-8) - a(n-9).
%F A059859 G.f.: x^2*(1+2*x+6*x^2+2*x^3+x^4) / ((1+x)^3*(x-1)^6). (End)
%F A059859 a(n) = Sum_{i=1..n} floor(i^2/4)^2. - _Enrique Pérez Herrero_, Mar 20 2012
%F A059859 a(n) = (2*n*(2*n^4+5*n^3-5*n+3) + 5*(2*n*(n+1)-1)*(-1)^n + 5)/320. - _Bruno Berselli_, Mar 21 2012
%p A059859 A059859 := n->add(A002620(i)^2,i=0..n);
%p A059859 f1 := n->1/160*(n-1)*(1+n)*(2*n^3+5*n^2+2*n-5); f2 := n->1/160*n*(n+2)*(2*n^3+n^2-2*n+4); A059859 := n-> if n mod 2 = 0 then f2(n) else f1(n); fi;
%t A059859 a[n_] := Sum[Floor[i^2/4]^2, {i,1,n}]; Table[a[n], {n, 0, 100}] (* _Enrique Pérez Herrero_, Mar 20 2012 *)
%Y A059859 Cf. A002620.
%K A059859 nonn,easy
%O A059859 0,4
%A A059859 _N. J. A. Sloane_, Feb 26 2001