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Number of connected components remaining when decimal expansion of the number n is cut from a piece of paper.
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%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