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a(n) = n^4/6 - 2*n^3/3 - n^2/6 + 5*n/3 + 1.
2

%I #30 Sep 08 2022 08:46:20

%S 2,1,0,5,26,77,176,345,610,1001,1552,2301,3290,4565,6176,8177,10626,

%T 13585,17120,21301,26202,31901,38480,46025,54626,64377,75376,87725,

%U 101530,116901,133952,152801,173570,196385,221376,248677,278426,310765,345840,383801,424802,469001

%N a(n) = n^4/6 - 2*n^3/3 - n^2/6 + 5*n/3 + 1.

%H Colin Barker, <a href="/A299198/b299198.txt">Table of n, a(n) for n = 1..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 a(n) = (n - 3)*(n + 1)*(n^2 - 2*n - 2)/6 = A299120(n-1) + A299120(1-n).

%F From _Colin Barker_, Feb 05 2018: (Start)

%F G.f.: x*(2 - 9*x + 15*x^2 - 5*x^3 + x^4) / (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>5.

%F (End)

%F E.g.f.: exp(x)*(6 + 6*x - 6*x^2 + 2*x^3 + x^4)/6. - _Iain Fox_, Feb 09 2018

%F 6*a(n) = A067998(n)^2 - 5*A067998(n) + 6. - _Bruno Berselli_, Apr 11 2018

%e For n=2, a(2) = 1^4/6 - 2*1^3/3 - 1^2/6 + 5*1/3 + 1 = 2.

%p seq(n^4/6-2*n^3/3-n^2/6+5*n/3+1,n=1..50); # _Muniru A Asiru_, Feb 04 2018

%t f[n_] := n^4/6 - 2 n^3/3 - n^2/6 + 5 n/3 + 1; Array[f, 50] (* or *)

%t CoefficientList[ Series[(-2 + 9 x - 15 x^2 + 5 x^3 - x^4)/(-1 + x)^5, {x, 0, 50}], x] (* or *)

%t LinearRecurrence[{5, -10, 10, -5, 1}, {2, 1, 0, 5, 26}, 50] (* _Robert G. Wilson v_, Feb 09 2018 *)

%o (Magma) [n^4/6-2*n^3/3-n^2/6+5*n/3+1: n in [1..50]];

%o (GAP) List([1..50], n -> n^4/6-2*n^3/3-n^2/6+5*n/3+1); # _Muniru A Asiru_, Feb 04 2018

%o (PARI) Vec(x*(2 - 9*x + 15*x^2 - 5*x^3 + x^4) / (1 - x)^5 + O(x^50)) \\ _Colin Barker_, Feb 05 2018

%o (Julia) [div((n-3)*(n+1)*(n^2-2*n-2),6) for n in 1:50] |> println # _Bruno Berselli_, Apr 11 2018

%Y Cf. A001263, A067998, A077415, A299120, A299146.

%K nonn,easy

%O 1,1

%A _Juri-Stepan Gerasimov_, Feb 04 2018