a(1) = 1, then smallest n-digit square which leaves a square at every step if most significant digit and least significant digit are deleted until a one-or two-digit digit square is obtained. a(2n) = 0 if no such square exists. a(2n+1) = 10^2n only if no nontrivial candidate exists.
Beginning with term a(6) the following pattern applies: a(4k)=0; a(4k+1)=10^4k=(10^2k)^2; a(4k+2)=0; a(4k+3)=(10^(2k+1)+2)^2. - Ray Chandler, Aug 03 2003
