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a(1) = 1000; for n>1, a(n) = a(n-1) + digitsum(a(n-1)).
1

%I #2 Mar 30 2012 16:51:44

%S 1000,1001,1003,1007,1015,1022,1027,1037,1048,1061,1069,1085,1099,

%T 1118,1129,1142,1150,1157,1171,1181,1192,1205,1213,1220,1225,1235,

%U 1246,1259,1276,1292,1306,1316,1327,1340,1348,1364,1378,1397,1417,1430,1438,1454

%N a(1) = 1000; for n>1, a(n) = a(n-1) + digitsum(a(n-1)).

%p f:=proc(n) global S; option remember; if n=1 then RETURN(S) else RETURN(f(n-1)+digsum(f(n-1))); fi; end; S:=1000; [seq(f(n),n=1..120)];

%Y First differences are A065075. Cf. A169732.

%K nonn,base

%O 1,1

%A _N. J. A. Sloane_, May 01 2010