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a(n) = (n-1)*(n-2)^3 - A003878(n-3), with a(1) = a(2) = 0 and a(3) = 2.
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%I #28 Jan 03 2024 17:58:06

%S 0,0,2,21,60,121,207,321,466,645,861,1117,1416,1761,2155,2601,3102,

%T 3661,4281,4965,5716,6537,7431,8401,9450,10581,11797,13101,14496,

%U 15985,17571,19257,21046,22941,24945,27061,29292,31641,34111,36705

%N a(n) = (n-1)*(n-2)^3 - A003878(n-3), with a(1) = a(2) = 0 and a(3) = 2.

%H G. C. Greubel, <a href="/A075681/b075681.txt">Table of n, a(n) for n = 1..1000</a>

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

%F From _Ralf Stephan_, Mar 13 2003: (Start)

%F a(n) = (1/2)*(n^3 + 7*n^2 - 46*n + 50), for n>3.

%F G.f.: x^3*(2 + 13*x - 12*x^2 - x^3 + x^4)/(1-x)^4. (End)

%F From _G. C. Greubel_, Jan 01 2024: (Start)

%F a(n) = (1/2)*(n^3 + 7*n^2 - 46*n + 50) + (-1)^floor((n+2)/2)*binomial(5 -n,2)*[n<4].

%F E.g.f.: (1/2)*(50 - 38*x + 10*x^2 + x^3)*exp(x) - 25 - 6*x + 3*x^2/2! + x^3/3!. (End)

%p A075681:=n->1/2*n^3+7/2*n^2-23*n+25: (0,0,2,seq(A075681(n), n=4..50)); # _Wesley Ivan Hurt_, Sep 06 2015

%t CoefficientList[Series[x^2 (x^4 -x^3 -12 x^2 +13 x +2)/(1-x)^4, {x, 0, 40}], x] (* _Vincenzo Librandi_, Sep 07 2015 *)

%t LinearRecurrence[{4,-6,4,-1}, {0,0,2,21,60,121,207}, 50] (* _G. C. Greubel_, Jan 03 2024 *)

%o (Magma) [0,0,2] cat [1/2*n^3+7/2*n^2-23*n+25: n in [4..50]]; // _Vincenzo Librandi_, Sep 07 2015

%o (SageMath) [(1/2)*(n^3+7*n^2-46*n+50) +(-1)^((n+2)//2)*binomial(5-n,2)*int(n<4) for n in range(1,51)] # _G. C. Greubel_, Jan 01 2024

%Y Cf. A003878.

%K nonn,easy

%O 1,3

%A _Jon Perry_, Oct 12 2002

%E More terms from _Ralf Stephan_, Mar 13 2003