%I #6 Oct 03 2014 11:09:48
%S 2,1,1,1,2,1,2,1,3,2,2,1,1,1,2,1,2,1,3,2,2,1,1,1,2,1,2,1,3,2,2,1,1,1,
%T 2,1,2,1,3,2,3,2,2,2,3,2,3,2,4,3,2,1,1,1,2,1,2,1,3,2,3,2,2,2,3,2,3,2,
%U 4,3,2,1,1,1,2,1,2,1,3,2,4,3,3,3,4,3,4,3,5,4,3,2,2,2,3,2,3,2,4,3,3,2,2,2,3
%N Number of connected components remaining when decimal expansion of the number n is cut from a piece of paper.
%e We assume 1,2,3,5 have no hole; 0,4,6,9 have 1 hole; 8 has two holes. So cutting 8 from a piece of paper creates three connected components: one for each hole and one for the remainder of the paper. Hence a(8)=3.
%Y Cf. A064531. Equals A064692 + 1.
%K base,easy,nonn
%O 0,1
%A _Matthew Conroy_, Oct 11 2001