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a(n) = floor(5*n^2/2).
1

%I #18 Sep 08 2022 08:44:51

%S 0,2,10,22,40,62,90,122,160,202,250,302,360,422,490,562,640,722,810,

%T 902,1000,1102,1210,1322,1440,1562,1690,1822,1960,2102,2250,2402,2560,

%U 2722,2890,3062,3240,3422,3610,3802,4000,4202,4410,4622,4840,5062

%N a(n) = floor(5*n^2/2).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).

%F a(n) = 2n^2 + floor(n^2/2). [_Wesley Ivan Hurt_, Jun 14 2013]

%F G.f.: 2*x*(1+3*x+x^2)/((1+x)*(1-x)^3). [_Bruno Berselli_, Jun 14 2013]

%F a(n) = 2*A032527(n). [_Bruno Berselli_, Jun 14 2013]

%p A032526:=n->floor(5*n^2/2): seq(A032526(n), n=0..100); # _Wesley Ivan Hurt_, Feb 03 2017

%t Table[Floor[5 n^2/2], {n, 0, 50}] (* _Bruno Berselli_, Jun 14 2013 *)

%t LinearRecurrence[{2,0,-2,1},{0,2,10,22},50] (* _Harvey P. Dale_, Dec 14 2016 *)

%o (Magma) [Floor(5*n^2/2): n in [0..50]]; // _Bruno Berselli_, Jun 14 2013

%Y Cf. A032527.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_