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

%I #26 Feb 08 2023 18:36:57

%S 0,24,84,198,384,660,1044,1554,2208,3024,4020,5214,6624,8268,10164,

%T 12330,14784,17544,20628,24054,27840,32004,36564,41538,46944,52800,

%U 59124,65934,73248,81084,89460,98394,107904,118008,128724,140070,152064,164724,178068,192114,206880,222384,238644,255678,273504,292140,311604,331914,353088,375144,398100

%N a(n) = 3*n*(n^2 + 3*n + 4).

%C Numbers of unit triangles in a certain structure obtained from A006003.

%H Colin Barker, <a href="/A280304/b280304.txt">Table of n, a(n) for n = 0..1000</a>

%H Luce ETIENNE, <a href="/A280304/a280304.pdf">Illustration of initial terms</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F G.f.: 6*x*(x^2-2*x+4) / (1-x)^4.

%F a(n) = 6*(A006003(n+1)-1).

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - _Colin Barker_, Dec 31 2016

%e a(0) = 6*(1-1) = 0, a(1) = 6*(5-1) = 24, a(2) = 6*(15-1) = 84, a(3) = 6*(34-1) = 198, a(4) = 6*(65-1) = 384.

%p A280304:=n->3*n*(n^2 + 3*n + 4): seq(A280304(n), n=0..60); # _Wesley Ivan Hurt_, Dec 31 2016

%t Table[3 n (n^2 + 3 n + 4), {n, 0, 50}] (* or *)

%t CoefficientList[Series[6 x (x^2 - 2 x + 4)/(1 - x)^4, {x, 0, 50}], x] (* _Michael De Vlieger_, Dec 31 2016 *)

%t LinearRecurrence[{4,-6,4,-1},{0,24,84,198},60] (* _Harvey P. Dale_, Feb 08 2023 *)

%o (PARI) concat(0, Vec(6*x*(x^2-2*x+4) / (1-x)^4 + O(x^30))) \\ _Colin Barker_, Dec 31 2016

%o (Magma) [3*n*(n^2 + 3*n + 4) : n in [0..60]]; // _Wesley Ivan Hurt_, Dec 31 2016

%Y Cf. A003215, A005448, A006003, A033428, A213389, A269064.

%K nonn,easy

%O 0,2

%A _Luce ETIENNE_, Dec 31 2016