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%I #20 Sep 14 2017 04:15:32
%S 2,2,12,32,32,32,192,512,512,512,3072,8192,8192,8192,49152,131072,
%T 131072,131072,786432,2097152,2097152,2097152,12582912,33554432,
%U 33554432,33554432
%N Smallest k divisible by the n-th power of its last decimal digit > 1.
%C Conjecture: a(n) == 2 (mod 10).
%F a(n) = 3*2^n if n mod 4 = 2; 2^(n+2-((n+1) mod 4)) otherwise. - _Jon E. Schoenfield_, Sep 12 2017
%e a(0) = 2 because 2 is divisible by 2^0 = 1.
%e a(1) = 2 because 2 is divisible by 2^1 = 2.
%e a(2) = 12 because 12 is divisible by 2^2 = 4.
%t Do[k=1;While[!Total[Transpose[IntegerDigits[k][[-1]]>0&&Mod[k,IntegerDigits[k][[-1]]^n]==0&&!Mod[k,10]==1],k++]];Print[n," ",k-1],{n,0,25}]
%Y Cf. A132359.
%K nonn,base
%O 0,1
%A _Michel Lagneau_, Apr 22 2014
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