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a(n) = 2*floor(3*n*(n+1)/4).
1

%I #25 Sep 08 2022 08:46:15

%S 0,2,8,18,30,44,62,84,108,134,164,198,234,272,314,360,408,458,512,570,

%T 630,692,758,828,900,974,1052,1134,1218,1304,1394,1488,1584,1682,1784,

%U 1890,1998,2108,2222,2340,2460,2582,2708,2838,2970,3104,3242,3384,3528,3674,3824,3978,4134,4292

%N a(n) = 2*floor(3*n*(n+1)/4).

%H G. C. Greubel, <a href="/A268810/b268810.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,-4,4,-3,1).

%F a(n) = (3*n^2 - 3*n + cos(n*Pi/2) + sin(n*Pi/2) - 1)/2.

%F a(n) = A268428(n) - 149.

%F a(n) = (1/4+i/4)*((i-1)+(-i)^n-i*i^n-(-3*i+3)*n+(-3*i+3)*n^2), where i is the imaginary unit.

%F a(n) = (3*n^2 - 3*n + (-1)^binomial(n+1,2) - 1)/2.

%F G.f.: (2*(x^3 + x^2 + x)/((1 - x)^3*(x^2 + 1))).

%t Table[2 Floor[3 n (n + 1)/4], {n, 0, 60}]

%t LinearRecurrence[{3, -4, 4, -3, 1}, {0, 2, 8, 18, 30}, 60]

%o (Magma) [2*Floor(3*n*(n+1)/4): n in [0..60]]; // _Vincenzo Librandi_, Feb 15 2016

%o (PARI) vector(60, n, n--; 2*floor(3*n*(n+1)/4)) \\ _G. C. Greubel_, Nov 04 2018

%Y Cf. A087960, A268428.

%K nonn,easy

%O 0,2

%A _Mikk Heidemaa_, Feb 13 2016