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%I #3 Mar 30 2012 17:35:24
%S 1,2,2,3,2,4,2,4,3,5,2,6,3,5,4,7,3,8,4,7,5,4,4,8,4,6,6,7,5,9,3,7,4,6,
%T 5,9,4,7,5,9,4,9,5,6,8,7,5,10,7,7,5,7,4,10,4,11,6,7,5,13,5,6,8,9,7,8,
%U 6,9,7,9,3,13,4,6,7,9,4,11,5,11,8,6,6,13,7,8,8,8,7,13,6,8,5,7,6,13,5,10,6
%N a(n) = total number of distinct divisors of n and all of its substrings.
%C Note that we are counting 0 when it occurs as a digit of n, but are not counting any other integers as divisors of 0. (If we did, there would be infinitely many of them; every integer divides 0.) [From _Franklin T. Adams-Watters_, May 29 2010]
%e a(56) = 11 because divisors of 56 are d1= {1, 2, 4, 7, 8, 14, 28, 56}; 56 has two substrings 5,6; divisors of 5 are d2= {1, 5}, and divisors of 6 are d3= {1, 2, 3,6} ; union of d1,d2,d3 gives 11 distinct divisors of 56 and all of its substrings: {1, 2, 3, 4, 5, 6, 7, 8, 14, 28, 56}.
%t Table[id = IntegerDigits[n]; FLA = Flatten[Table[Partition[id, k, 1], {k, Length[id]}], 1]; fd = Union[FromDigits /@ FLA]; dv = Length[Union[Flatten[Divisors /@ fd]]], {n, 200}]
%Y Cf. A177834
%K base,nonn
%O 1,2
%A _Zak Seidov_, May 28 2010
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