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%I #3 Mar 31 2012 12:38:28
%S 2,9,25,56,108,187,299,450,646,893,1197,1564,2000,2511,3103,3782,4554,
%T 5425,6401,7488,8692,10019,11475,13066,14798,16677,18709,20900,23256,
%U 25783,28487,31374,34450,37721,41193,44872,48764,52875,57211,61778
%N Averages of four consecutive cubes.
%C (0^3+1^3+2^3+3^3)/4=9,..
%F a(n)=(2*n-1)*(n^2-n+4)/2 = (2*n-1)*A089071(n+1) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4). G.f.: x*(1+x)*(2*x^2-x+2)/(x-1)^4. [From _R. J. Mathar_, Mar 31 2010]
%t f[n_]:=(n^3+(n+1)^3+(n+2)^3+(n+3)^3)/4;Table[f[n],{n,-1,5!}]
%Y Cf. A027575, A173960, A173961, A173962
%K nonn
%O 1,1
%A _Vladimir Joseph Stephan Orlovsky_, Mar 03 2010
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