%I #36 Sep 08 2022 08:44:28
%S 0,1,2,3,4,125,726,2527,6728,15129,30250,55451,95052,154453,240254,
%T 360375,524176,742577,1028178,1395379,1860500,2441901,3160102,4037903,
%U 5100504,6375625,7893626,9687627,11793628,14250629,17100750,20389351
%N a(n) = n + n*(n-1)*(n-2)*(n-3)*(n-4).
%H Vincenzo Librandi, <a href="/A001095/b001095.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F G.f.: x*(1 - 4*x + 6*x^2 - 4*x^3 + 121*x^4)/(1-x)^6. - _Colin Barker_, Jun 25 2012
%F From _G. C. Greubel_, Aug 26 2019: (Start)
%F a(n) = n + 5!*binomial(n,5).
%F E.g.f.: x*(1 + x^4)*exp(x). (End)
%p seq(n + 5!*binomial(n,5), n=0..35); # _G. C. Greubel_, Aug 26 2019
%t Table[n+Times@@(n-Range[0,4]),{n,0,40}] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,2,3,4,125},40] (* _Harvey P. Dale_, Oct 08 2017 *)
%o (Magma) [n + n*(n-1)*(n-2)*(n-3)*(n-4): n in [0..35]]; // _Vincenzo Librandi_, Apr 30 2011
%o (PARI) vector(35, n, (n-1) + 5!*binomial(n-1,5)) \\ _G. C. Greubel_, Aug 26 2019
%o (Sage) [n + 120*binomial(n,5) for n in (0..35)] # _G. C. Greubel_, Aug 26 2019
%o (GAP) List([0..35], n-> n + 120*Binomial(n,5)); # _G. C. Greubel_, Aug 26 2019
%Y Equals A052787(n) + n.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_, Ray Wills (rwills(AT)vmprofs.estec.esa.nl)
%E More terms from _James A. Sellers_, Sep 19 2000