%I #47 Dec 10 2022 17:12:36
%S 0,0,1,3,12,40,105,231,448,792,1305,2035,3036,4368,6097,8295,11040,
%T 14416,18513,23427,29260,36120,44121,53383,64032,76200,90025,105651,
%U 123228,142912,164865,189255,216256,246048,278817,314755,354060,396936
%N a(n) = binomial(n, 2) + 6*binomial(n, 4).
%H Colin Barker, <a href="/A327319/b327319.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F From _Colin Barker_, Sep 21 2019: (Start)
%F G.f.: x^2*(1 - 2*x + 7*x^2) / (1 - x)^5.
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4.
%F a(n) = (n*(-8 + 13*n - 6*n^2 + n^3)) / 4.
%F (End)
%F E.g.f.: (1/4)*exp(x)*x^2*(2 + x^2). - _Stefano Spezia_, Sep 21 2019
%e a(5) = binomial(5, 2) + 6*binomial(5, 4) = 10 + 6*5 = 40.
%t Table[Binomial[n, 2] + 6Binomial[n, 4], {n, 0, 39}] (* _Alonso del Arte_, Sep 18 2019 *)
%t LinearRecurrence[{5,-10,10,-5,1},{0,0,1,3,12},40] (* _Harvey P. Dale_, Dec 10 2022 *)
%o (PARI) a(n) = {binomial(n, 2) + 6 * binomial(n, 4)} \\ _Andrew Howroyd_, Sep 20 2019
%o (PARI) concat([0,0], Vec(x^2*(1 - 2*x + 7*x^2) / (1 - x)^5 + O(x^40))) \\ _Colin Barker_, Sep 25 2019
%K nonn,easy
%O 0,4
%A _Aaron Kemats_, Sep 17 2019