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%I #14 Dec 28 2015 20:53:47
%S 1,11,11,1111,10,11,1001,10111,1001,110,11111,1111,101,1001,110,10111,
%T 101111,1001,110111,11110,1001,101011,100101,11111111,100,101,11011,
%U 11011,1000011,110,111,110101,1011101,111001,10010,11011,101011,111001,1111,101110,10111
%N Least positive multiple of n written in base 5 using only 0 and 1.
%H Robert Israel, <a href="/A004285/b004285.txt">Table of n, a(n) for n = 1..10000</a>
%p N:= 100: # to get a(1) .. a(N)
%p S:= [[1,1]]:
%p A[1]:= 1:
%p Todo:= {$2..N}:
%p count:= 1:
%p for d from 2 while count < N do
%p S:= map(t -> ([5*t[1],10*t[2]],[5*t[1]+1,10*t[2]+1]), S);
%p for p in S do
%p s:= p[1];
%p Ts:= select(t -> s mod t = 0, Todo);
%p Todo:= Todo minus Ts;
%p count:= count + nops(Ts);
%p for m in Ts do
%p A[m]:= p[2];
%p od
%p od;
%p od:
%p seq(A[n],n=1..N); # _Robert Israel_, Dec 28 2015
%o (PARI) a(n) = {k=1; while (vecmax(digits(k*n, 5)) != 1, k++); subst(Pol(digits(k*n, 5)), x, 10);} \\ _Michel Marcus_, Dec 27 2015
%K nonn,base
%O 1,2
%A _David W. Wilson_
%E a(33) corrected by _Sean A. Irvine_, Dec 26 2015
%E More terms from _Michel Marcus_, Dec 27 2015
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