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a(n) = (n+2)*(n+1)*(n^2 + 7*n - 12)/24.
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%I #36 Oct 21 2022 21:14:16

%S -1,3,15,40,84,154,258,405,605,869,1209,1638,2170,2820,3604,4539,5643,

%T 6935,8435,10164,12144,14398,16950,19825,23049,26649,30653,35090,

%U 39990,45384,51304,57783,64855,72555

%N a(n) = (n+2)*(n+1)*(n^2 + 7*n - 12)/24.

%H Vincenzo Librandi, <a href="/A014309/b014309.txt">Table of n, a(n) for n = 1..725</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 + x^4 - 5*x^3 + 10*x^2 - 8*x)/(x-1)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009

%F E.g.f.: (24 - (24 - 48*x^2 - 16*x^3 - x^4)*exp(x))/24. - _G. C. Greubel_, Jun 12 2019

%F a(n) = binomial(n+2,4) + 2*binomial(n+2,3) - binomial(n+2,2). - _Étienne Tétreault_, Sep 02 2020

%t Table[(n+2)(n+1)(n^2+7n-12)/24,{n,40}] (* _Harvey P. Dale_, Feb 20 2011 *)

%o (Magma) [(n+2)*(n+1)*(n^2+7*n-12)/24: n in [1..40]]; // _Vincenzo Librandi_, Apr 25 2011

%o (PARI) {a(n) = (n+2)*(n+1)*(n^2+7*n-12)/24}; \\ _G. C. Greubel_, Jun 12 2019

%o (Sage) [(n+2)*(n+1)*(n^2+7*n-12)/24 for n in (1..40)] # _G. C. Greubel_, Jun 12 2019

%o (GAP) List([1..40], n-> (n+2)*(n+1)*(n^2+7*n-12)/24) # _G. C. Greubel_, Jun 12 2019

%K sign,easy

%O 1,2

%A _N. J. A. Sloane_