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a(n) = n + n*(n-1)*(n-2)*(n-3).
3

%I #31 Oct 21 2022 21:09:26

%S 0,1,2,3,28,125,366,847,1688,3033,5050,7931,11892,17173,24038,32775,

%T 43696,57137,73458,93043,116300,143661,175582,212543,255048,303625,

%U 358826,421227,491428,570053,657750,755191,863072,982113,1113058

%N a(n) = n + n*(n-1)*(n-2)*(n-3).

%H Vincenzo Librandi, <a href="/A001094/b001094.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 G.f.: x*(1 -3*x +3*x^2 +23*x^3)/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009

%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=1, a(2)=2, a(3)=3, a(4)=28. - _Harvey P. Dale_, Feb 02 2012

%F From _G. C. Greubel_, Aug 26 2019: (Start)

%F a(n) = n + 4!*binomial(n,4).

%F E.g.f.: x*(1+x^3)*exp(x). (End)

%p seq(n + 4!*binomial(n,4), n=0..35); # _G. C. Greubel_, Aug 26 2019

%t Table[n+n(n-1)(n-2)(n-3),{n,0,40}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{0,1,2,3,28},40] (* _Harvey P. Dale_, Feb 02 2012 *)

%o (Magma) [n + n*(n-1)*(n-2)*(n-3): n in [0..35]]; // _Vincenzo Librandi_, Apr 30 2011

%o (PARI) vector(35, n, (n-1) + 4!*binomial(n-1,4)) \\ _G. C. Greubel_, Aug 26 2019

%o (Sage) [n + 24*binomial(n,4) for n in (0..35)] # _G. C. Greubel_, Aug 26 2019

%o (GAP) List([0..35], n-> n + 24*Binomial(n,4)); # _G. C. Greubel_, Aug 26 2019

%Y Cf. A001095, A001096, A052762.

%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