login
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.
1

%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