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a(1)=3; a(n)=floor((20+sum(a(1) to a(n-1)))/6).
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%I #4 Jul 15 2012 14:47:00

%S 3,3,4,5,5,6,7,8,10,11,13,15,18,21,24,28,33,39,45,53,61,72,84,98,114,

%T 133,155,181,211,246,287,335,391,456,532,621,724,845,986,1150,1342,

%U 1565,1826,2131,2486,2900,3383,3947,4605,5373,6268,7313,8532,9954,11613

%N a(1)=3; a(n)=floor((20+sum(a(1) to a(n-1)))/6).

%t Module[{lst={3}},Do[AppendTo[lst,Floor[(20+Total[lst])/6]],{100}];lst] (* _Harvey P. Dale_, Jul 15 2012 *)

%Y Cf. A073941, A112088, A072493.

%K nonn

%O 1,1

%A _Graeme McRae_, Jun 10 2006

%E More terms from _Harvey P. Dale_, Jul 15 2012