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%I #26 Jul 25 2019 01:10:44
%S 1,2,3,4,5,24,29,43,48,62,67,81,86,190,285,484,579,683,778,882,977,
%T 1981,2976,3980,4975,6974,7969,8973,9968,19972,29967,39971,49966,
%U 59970,69965,89964,99959,199963,299958,399962,499957,599961,699956,799960,899955,1999954,2999949
%N Numbers n such that A = n + digitsum(n) is divisible by the highest power of 10 <= A.
%e 48 + (4 + 8) = 60 is divisible by the highest power of 10 less than 60 (10^1). So 48 is a member of this sequence.
%o (PARI) DS(n)={t=0;for(i=1,#digits(n),t+=digits(n)[i]);return(t)}
%o for(n=1,10^7,if((n+DS(n))%(10^(#Str(n+DS(n))-1))==0,print1(n,", ")))
%Y Cf. A007953.
%K nonn,base
%O 1,2
%A _Derek Orr_, May 31 2014