%I #5 Mar 30 2012 17:27:33
%S 1,1,1,1,1,2,2,2,2,2,2,2,2,2,1,1,1,2,2,2,2,2,2,2,2,2,1,0,3,3,3,3,3,1,
%T 3,0,3,1,3,3,3,3,3,1,3,1,3,1,3,3,3,3,3,1,3,1,3,1,3,3,3,3,1,3,3,1,3,3,
%U 3,3,3,1,3,1,3,1,3,3,3,3,3,1,3,1,3,1,3,3,3,3,3,1,3,1,3,1,2,2,2,3,2,3,2,1,3
%N a(n) = number of 'Reverse and Add!' operations that have to be applied to the n-th term of A063055 in order to obtain a term in the trajectory of 1997.
%H <a href="/index/Res#RAA">Index entries for sequences related to Reverse and Add!</a>
%e 3995 is a term of A063055. One 'Reverse and Add!' operation applied to 3995 leads to a term (9988) in the trajectory of 1997, so the corresponding term of the present sequence is 1.
%Y A023108, A033865, A063054, A063055.
%K base,nonn
%O 0,6
%A _Klaus Brockhaus_, Jul 07 2001