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%I #7 Mar 23 2014 03:19:06
%S 1,2,3,4,5,6,7,8,9,16,24,32,40,48,56,64,72,80,88,96,108,216,324,432,
%T 540,648,756,864,972,1216,2432,3648,4864,8000,9216,10000,18000,20000,
%U 28000,30000,38000,40000,48000,50000,58000,60000,68000,70000,78000,80000,88000,90000,98000,108000
%N Numbers n in which the last K digits of n form an integer divisible by K^3, for K = 1, 2, ..., M, where M is the number of digits in n.
%e a(33)=4864 because 4 is divisible by 1^3, 64 by 2^3, 864 by 3^3, 4864 by 4^3.
%t okQ[n_]:=Module[{idn=IntegerDigits[n]},And@@(Divisible[#[[2]],#[[1]]^3]&/@Table[{i,FromDigits[Take[idn,-i]]},{i,Length[idn]}])]
%t Select[Range[200000],okQ] (* _Harvey P. Dale_, Dec 10 2010 *)
%K base,nonn
%O 1,2
%A Sudipta Das (juitech(AT)vsnl.net), Feb 03 2003
%E Additional terms from _Harvey P. Dale_, Dec 10 2010
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