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 A277059 Least k such that any sufficiently long repunit multiplied by k contains all nonzero digits in base n. 3

%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^m-1)/(n-1) + m - 1;

%F for n=4m+1, a(n) = (n^(2m)-1)(n^2+1) / (2(n^2-1)) + m;

%F for n=4m-1, a(n) = (n^(2m-2)-1)(n^2+1) / (2(n^2-1)) + m + n^(2m-1).

%e Any binary repunit itself contains a 1, so a(2)=1.

%e k-th 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

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Last modified November 13 11:09 EST 2018. Contains 317133 sequences. (Running on oeis4.)