%I #4 Dec 05 2013 19:56:14
%S 1,11,21,211,311,321,421,4211,5211,5311,6311,6321,7321,7421,8421,
%T 84211,94211,95211,105211,105311,115311,116311,126311,126321,136321,
%U 137321,147321,147421,157421,158421,168421,1684211,1784211,1794211,1894211
%N Let P(k) = floor(k/2). Start with n, apply P repeatedly until reach 1. a(n) = concatenation of numbers obtained.
%F Let P(k) = floor(k/2). Start with n, apply P repeatedly until reaching 0. a(n) = concatenation of the differences of the successive numbers obtained. - _David Wasserman_, Oct 25 2004
%e 11 -> 5 -> 2 -> 1, hence a(11) = 6311.
%e 11 -> 5 -> 2 -> 1 -> 0, hence a(11) = 6311.
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, Apr 26 2003
%E More terms from _David Wasserman_, Oct 25 2004