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a(1)=0; for n>1, a(n) = 4*n^3 - 3*n^2 - 3*n + 4.
2

%I #13 Oct 18 2022 15:25:59

%S 0,18,76,200,414,742,1208,1836,2650,3674,4932,6448,8246,10350,12784,

%T 15572,18738,22306,26300,30744,35662,41078,47016,53500,60554,68202,

%U 76468,85376,94950,105214,116192,127908,140386,153650,167724,182632,198398,215046,232600

%N a(1)=0; for n>1, a(n) = 4*n^3 - 3*n^2 - 3*n + 4.

%C This was once thought (mistakenly) to be a formula for A262402.

%H Peter Kagey, <a href="/A296363/b296363.txt">Table of n, a(n) for n = 1..10000</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.: -2 * (x^3-2*x^2-2*x-9) * x^2 / (x-1)^4.

%t CoefficientList[Series[- 2 x (x^3 - 2 x^2 - 2 x - 9)/(x - 1)^4, {x, 0, 30}], x] (* _Vincenzo Librandi_, Sep 23 2015 *)

%t Join[{0},LinearRecurrence[{4, -6, 4, -1},{18, 76, 200, 414},38]] (* _Ray Chandler_, Sep 23 2015 *)

%t Join[{0},Table[4n^3-3n^2-3n+4,{n,2,40}]] (* _Harvey P. Dale_, Mar 28 2019 *)

%o (Magma) [0] cat [4*n^3-3*n^2-3*n+4: n in [2..40]]; // _Vincenzo Librandi_, Sep 23 2015

%o (PARI) a(n)=if(n>1, 4*n^3 - 3*n^2 - 3*n + 4, 0) \\ _Charles R Greathouse IV_, Oct 18 2022

%Y Cf. A262402.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_, Dec 16 2017