%I #26 Sep 27 2021 07:07:56
%S 1,2,3,4,5,6,7,8,9,12,24,36,48
%N Numbers n divisible by the sum of any k-subset of digits of n with k >= 1.
%C No additional terms less than 20000000. - _T. D. Noe_, Aug 14 2013
%C Terms > 9 must be even since any pair of digits has an even subset. Since terms must also be zeroless, they cannot be divisible by 5, which means no further terms could have 5 or more digits by the Pigeonhole Principle. Therefore, this sequence is complete. - _Charlie Neder_, May 31 2019
%e 48 is here because 48 is divisible by 4, 8, and 4+8.
%t okQ[n_] := Module[{s = Total /@ Rest[Subsets[IntegerDigits[n]]]}, ! MemberQ[s, 0] && And @@ IntegerQ /@ (n/s)]; Select[Range[10000], okQ] (* _T. D. Noe_, Aug 14 2013 *)
%Y Subset of A051004 and of A346535.
%K nonn,base,fini,full
%O 1,2
%A _Derek Orr_, Aug 02 2013
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