A335012 Number of different remainders when the first n terms of 1, 11, 111, 1111, ... are divided by n.

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, 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, 4, 60, 15, 18, 6, 6, 6, 33, 17, 66, 6, 35, 11, 8, 3, 4, 19, 6, 6, 13, 4, 81

Offset

1,3

Comments

a(n) = n if and only if n is a power of 3.

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.

If gcd(n,30) = 1 then a(n) = A084680(n). - Robert Israel, Jun 25 2020

Links

Table of n, a(n) for n=1..81.

Example

a(4) = 2 since when 1, 11, 111, 1111 are divided by 4 the remainders are 1, 3, 3, 3, two different numbers.
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.

Maple

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);
# Simpler:
f:= n -> nops({seq(((10^i-1)/9) mod n, i=1..n)}):
map(f, [$1..100]); # Robert Israel, Jun 25 2020

Mathematica

Table[Length@ Union@ Array[Mod[(10^# - 1)/9, n] &, n], {n, 81}] (* Michael De Vlieger, Jun 28 2020 *)

Prog

(PARI) a(n) = #Set(vector(n, k, (10^k-1)/9) % n); \\ Michel Marcus, Jun 15 2020

Crossrefs

Cf. A000042, A006883, A084680.

Sequence in context: A035572 A325531 A115215 * A158275 A147750 A089942

Adjacent sequences: A335009 A335010 A335011 * A335013 A335014 A335015

Keyword

nonn,base,look

Author

Sen-Peng Eu, May 19 2020

Status

approved