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%I #14 Jun 13 2015 00:52:02
%S 1,4,16,42,87,156,254,386,557,772,1036,1354,1731,2172,2682,3266,3929,
%T 4676,5512,6442,7471,8604,9846,11202,12677,14276,16004,17866,19867,
%U 22012,24306,26754,29361,32132,35072,38186,41479
%N a(n) = (5*n^3+12*n^2+n+6)/6.
%C Column 3 of A114202. Third differences are 1,1,7,5,5,5,5,5,... with g.f. (1+6x^2-2x^3)/(1-x).
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: (1+6*x^2-2*x^3)/(1-x)^4 = (1+3*(x/(1-x))+9*(x/(1-x))^2+5*(x/(1-x))^3)/(1-x).
%F a(n) = sum{k=0..n, C(n, k)*C(3, k)*J(k+1)} where J(n)=A001045(n).
%F a(0)=1, a(n)=a(n-1)+(n-1)*(n+2)+A104249(n).
%e [1,3,9,5]=[1*1,3*1,3*3,1*5]=[C(3,0)*J(1),C(3,1)*J(2),C(3,2)*J(3),C(3,3)*J(4)].
%t CoefficientList[Series[(1+6x^2-2x^3)/(1-x)^4,{x,0,75}],x] (* _Harvey P. Dale_, Mar 06 2011 *)
%K easy,nonn
%O 0,2
%A _Paul Barry_, Nov 17 2005
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