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A291288 a(n) = binomial(n+3, 3)*(1 + binomial(n+2, 3)/4).

%I A291288 #15 Mar 12 2024 19:18:39
%S A291288 1,5,20,70,210,546,1260,2640,5115,9295,16016,26390,41860,64260,95880,
%T A291288 139536,198645,277305,380380,513590,683606,898150,1166100,1497600,
%U A291288 1904175,2398851,2996280,3712870,4566920,5578760,6770896,8168160,9797865,11689965
%N A291288 a(n) = binomial(n+3, 3)*(1 + binomial(n+2, 3)/4).
%H A291288 Robert Israel, <a href="/A291288/b291288.txt">Table of n, a(n) for n = 0..10000</a>
%H A291288 Isaac Ahern, Sam Cook, <a href="http://www.ajuronline.org/uploads/Volume_13_3/AJUR%20Vol%2013%20Issue%203%2008.25.16%20pp.27-32.pdf">Affine Symmetry Tensors in Minkowski Space</a>, American Journal of Undergraduate Research, Volume 13 | Issue 3 | August 2016.
%H A291288 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7, -21, 35, -35, 21, -7, 1).
%F A291288 From _Robert Israel_, Aug 28 2017: (Start)
%F A291288 a(n) = (n+1)*(n+2)*(n+3)*(n+4)*(n^2-n+6)/144.
%F A291288 n*a(n) - (2+3*n)*a(n-1) + (8*n-16)*a(n-2) - (12+6*n)*a(n-3) = 0.
%F A291288 G.f.: (6*x^2-2*x+1)/(1-x)^7. (End)
%p A291288 f:=n->binomial(n+3,3)*(1+binomial(n+2,3)/4);
%p A291288 [seq(f(n),n=0..40)];
%t A291288 Table[Binomial[n+3,3](1+Binomial[n+2,3]/4),{n,0,40}] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{1,5,20,70,210,546,1260},40] (* _Harvey P. Dale_, Mar 12 2024 *)
%K A291288 nonn
%O A291288 0,2
%A A291288 _N. J. A. Sloane_, Aug 28 2017