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%I #29 Jun 28 2020 18:49:12
%S 1,1,3,2,1,3,6,3,9,1,2,4,6,6,3,4,16,9,18,2,6,2,22,5,2,6,27,7,28,3,15,
%T 5,6,16,6,10,3,18,6,3,5,6,21,3,9,22,46,6,42,2,48,7,13,27,2,8,18,28,58,
%U 4,60,15,18,6,6,6,33,17,66,6,35,11,8,3,4,19,6,6,13,4,81
%N Number of different remainders when the first n terms of 1, 11, 111, 1111, ... are divided by n.
%C a(n) = n if and only if n is a power of 3.
%C Conjecture: a(n) = n-1 if and only if n is a long period prime (A006883), that is, n is a prime and the decimal expansion of 1/n has period n-1.
%C If gcd(n,30) = 1 then a(n) = A084680(n). - _Robert Israel_, Jun 25 2020
%e a(4) = 2 since when 1, 11, 111, 1111 are divided by 4 the remainders are 1, 3, 3, 3, two different numbers.
%e a(6) = 3 since when 1, 11, 111, 1111, 11111, 111111 are divided by 6 the remainders are 1, 5, 3, 1, 5, 3, three different numbers.
%p with(ListTools): a := proc (n) return add(10^i, i = 0 .. n-1) end proc: r := proc (n) return seq(`mod`(a(i), n), i = 1 .. n) end proc: seq(nops(MakeUnique([r(n)])), n = 1 .. 243);
%p # Simpler:
%p f:= n -> nops({seq(((10^i-1)/9) mod n,i=1..n)}):
%p map(f, [$1..100]); # _Robert Israel_, Jun 25 2020
%t Table[Length@ Union@ Array[Mod[(10^# - 1)/9, n] &, n], {n, 81}] (* _Michael De Vlieger_, Jun 28 2020 *)
%o (PARI) a(n) = #Set(vector(n, k, (10^k-1)/9) % n); \\ _Michel Marcus_, Jun 15 2020
%Y Cf. A000042, A006883, A084680.
%K nonn,base,look
%O 1,3
%A _Sen-Peng Eu_, May 19 2020
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