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%I #6 Mar 30 2012 19:00:25
%S 1,2,3,4,5,6,7,8,9,11,22,33,44,55,66,77,88,99,1111111111111111111,
%T 2222222222222222222,3333333333333333333,4444444444444444444,
%U 5555555555555555555,6666666666666666666,7777777777777777777,8888888888888888888,9999999999999999999
%N Numbers all of whose divisors are repdigit numbers.
%C Subset of A010785, A190220 and A190221.
%e Number 99 is in sequence because all divisors of 99 (1, 3, 9, 11, 33, 99) are repdigit numbers.
%K nonn,base
%O 1,2
%A _Jaroslav Krizek_, May 06 2011
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