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A293296 a(n) = 2*n^2 - floor(n/4).

%I A293296 #17 Jan 26 2023 19:50:21
%S A293296 0,2,8,18,31,49,71,97,126,160,198,240,285,335,389,447,508,574,644,718,
%T A293296 795,877,963,1053,1146,1244,1346,1452,1561,1675,1793,1915,2040,2170,
%U A293296 2304,2442,2583,2729,2879,3033,3190,3352,3518,3688,3861,4039,4221,4407,4596
%N A293296 a(n) = 2*n^2 - floor(n/4).
%H A293296 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,1,-2,1).
%F A293296 a(n) = [x^n] (-x*(2+4*x+4*x^2+3*x^3+3*x^4)/((x+1)*(x^2+1)*(x-1)^3)).
%F A293296 a(n) = n! [x^n] (3*exp(x)-exp(-x)+14*exp(x)*x+16*exp(x)*x^2-2*cos(x)-2*sin(x))/8.
%F A293296 a(n) = a(n-6) - 2*a(n-5) + a(n-4) - a(n-2) + 2*a(n-1) for n >= 6.
%F A293296 (-1)^n*(a(n+3) - 3*a(n+2) + 3*a(n+1) - a(n)) = sqrt(n^2 mod 8) = A007877(n).
%p A293296 a := n -> 2*n^2 - floor(n/4): seq(a(n), n=0..48);
%t A293296 LinearRecurrence[{2, -1, 0, 1, -2, 1}, {0, 2, 8, 18, 31, 49}, 49]
%t A293296 Table[2n^2-Floor[n/4],{n,0,60}] (* _Harvey P. Dale_, Jan 08 2022 *)
%o A293296 (PARI) a(n) = 2*n^2-n\4; \\ _Altug Alkan_, Oct 08 2017
%o A293296 (Python)
%o A293296 def A293296(n): return (n**2<<1)-(n>>2) # _Chai Wah Wu_, Jan 26 2023
%Y A293296 Cf. A001105, A007877.
%K A293296 nonn,easy
%O A293296 0,2
%A A293296 _Peter Luschny_, Oct 08 2017