A229029 - OEIS

%I #10 Sep 14 2013 02:23:19

%S 2,5,2,43,2,3,2,71,2,13,2,11,2,3,2,79,2,21,13,59,2,5,2,55,2,29,2,211,

%T 2,3,2,1055,2,13,2,11,5,11,2,967,2,5,2,3,2,5,2,111,2,29,2,19,12,3,20,

%U 7,2,61,2,19,2,3,2,1087,2,69,2,11,2,5,2,7,2,5,2,547,2,3,2,303,2,85,2,11,2,5,2,391,2,13,2,3,2,5,2,95,2,21

%N Minimal positive integer m such that, in base n, powers of m can end with an arbitrary number of consecutive 1's.

%e In the decimal number system (n=10), powers of numbers smaller than 71 cannot end with 1111. On the other hand, 71^13 == 1111 (mod 10^4), 71^1513 == 11111 (mod 10^5), etc. and it can be shown that 71^k can end in any number of ones. So a(10)=71.

%K nonn,base

%O 3,1

%A _Max Alekseyev_, Sep 11 2013