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(1/12)*(n+5)*(n+1)*n^2.
3

%I #33 Oct 21 2022 21:09:53

%S 0,1,7,24,60,125,231,392,624,945,1375,1936,2652,3549,4655,6000,7616,

%T 9537,11799,14440,17500,21021,25047,29624,34800,40625,47151,54432,

%U 62524,71485,81375,92256,104192,117249,131495,147000,163836,182077,201799,223080,246000

%N (1/12)*(n+5)*(n+1)*n^2.

%C Partial sums of A084990. - _Arkadiusz Wesolowski_, Jan 25 2014

%H Vincenzo Librandi, <a href="/A014205/b014205.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 a(n) = 2*C(n+3,4) - C(n+1,2). - _Zerinvary Lajos_, May 02 2007

%F G.f.: x*(x^2-2*x-1)/(x-1)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009

%F a(0)=0, a(1)=1, a(2)=7, a(3)=24, a(4)=60, a(n)=5*a(n-1)-10*a(n-2)+ 10*a(n-3)- 5*a(n-4)+a(n-5). - _Harvey P. Dale_, Aug 10 2014

%p seq(2*binomial(n+3, 4)-binomial(n+1, 2), n=0..32); # _Zerinvary Lajos_, May 02 2007

%t Table[((n+5)(n+1)n^2)/12,{n,0,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,7,24,60},50] (* _Harvey P. Dale_, Aug 10 2014 *)

%t CoefficientList[Series[x (x^2 - 2 x - 1)/(x - 1)^5, {x, 0, 50}], x] (* _Vincenzo Librandi_, Aug 11 2014 *)

%o (Magma) [(1/12)*(n+5)*(n+1)*n^2: n in [0..50]]; // _Vincenzo Librandi_, Aug 11 2014

%o (PARI) a(n)=n^2*(n+1)*(n+5)/12 \\ _Charles R Greathouse IV_, Oct 21 2022

%Y Cf. A084990.

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_.