%I
%S 1,4,6,14,45,370,588,3364,11115,168496,271458,2442138
%N Least k such that any sufficiently long repunit multiplied by k contains all nonzero digits in base n.
%C Trailing terms of rows of A277058.
%C Written in base n, the terms read: 1, 11, 12, 24, 113, 1036, 1114, 4547, 11115, 105659, 111116, 67676A, ...
%F Conjecture:
%F for n=2m, a(n) = (n^m1)/(n1) + m  1;
%F for n=4m+1, a(n) = (n^(2m)1)(n^2+1) / (2(n^21)) + m;
%F for n=4m1, a(n) = (n^(2m2)1)(n^2+1) / (2(n^21)) + m + n^(2m1).
%e Any binary repunit itself contains a 1, so a(2)=1.
%e kth decimal repunit for k>4 multiplied by 11115 contains all nonzero decimal digits (see A277057) with no number less than 11115 having the same property, so a(10)=11115.
%Y Cf. A002275, A277056, A277057, A277058.
%K nonn,base,more
%O 2,2
%A _Andrey Zabolotskiy_ and _Altug Alkan_, Sep 26 2016
