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a(n+1) = a(n) + sum of digits of a(n)^2.
0

%I #6 Nov 13 2010 13:17:08

%S 3,12,21,30,39,48,57,75,93,120,129,147,165,183,210,219,246,264,300,

%T 309,336,363,390,399,417,453,471,489,507,534,561,579,597,624,660,678,

%U 714,750,768,804,831,858,885,912,939,966,993,1029,1056,1074,1101,1110,1119

%N a(n+1) = a(n) + sum of digits of a(n)^2.

%Y Cf. A033298.

%K nonn,base,easy,less

%O 1,1

%A Miklos SZABO (mike(AT)ludens.elte.hu)