%I #5 Feb 11 2014 19:05:21
%S 1,12,111213,111212111213111214,
%T 111212111213111213111212111213111214111212111213111215
%N a(1) = 1; to form a(n+1), append a(n)-1, a(n)+1 at the beginning and the end of a(n), respectively. Ignore any leading 0's.
%e To form the third term from the second term 12, append 11, 13 to the beginning and end of 12, respectively, to get 111213.
%K base,easy,nonn
%O 1,2
%A _Joseph L. Pe_, Feb 09 2002