A365933 a(n) is the period of the remainders when repdigits are divided by n.

1, 9, 27, 9, 9, 27, 54, 9, 81, 9, 18, 27, 54, 54, 27, 9, 144, 81, 162, 9, 54, 18, 198, 27, 9, 54, 243, 54, 252, 27, 135, 9, 54, 144, 54, 81, 27, 162, 54, 9, 45, 54, 189, 18, 81, 198, 414, 27, 378, 9, 432, 54, 117, 243, 18, 54, 162, 252, 522, 27, 540, 135, 162, 9, 54

Offset: 1

Author

Karl-Heinz Hofmann, Nov 07 2023

Keywords

• nonn
• base

Comments

For n>1: Periods are divisible by 9 (= a full cycle in the sequence of repdigits). a(n)/9 is the period of the remainders when repunits are divided by n. So the digit part of the repdigits has no effect on periods generally. For most n the beginning of the periodic part is always A010785(1). If n is a term of A083118 the periodic part starts later after some initial remainders that do not repeat.

Examples

			For n = 6:                Remainders of A010785(1..54) mod n.
A010785( 1...9) mod n:      [1, 2, 3, 4, 5, 0, 1, 2, 3]
A010785(10..18) mod n:      [5, 4, 3, 2, 1, 0, 5, 4, 3]
A010785(19..27) mod n:      [3, 0, 3, 0, 3, 0, 3, 0, 3]
So the period is 3*9 = 27. Thus a(n) = 27. And the pattern seen above starts again:
A010785(28..36) mod n:      [1, 2, 3, 4, 5, 0, 1, 2, 3]
A010785(37..45) mod n:      [5, 4, 3, 2, 1, 0, 5, 4, 3]
A010785(46..54) mod n:      [3, 0, 3, 0, 3, 0, 3, 0, 3]
		

Links

• Karl-Heinz Hofmann, Table with additional information.

Crossrefs

Cf. A010785, A002275.

Cf. A305322 (divisor 3), A002279 (divisor 5), A366596 (divisor 7).

Cf. A083118 (the impossible divisors).

Programs

Python

def A365933(n):
    if n == 1: return 1
    remainders, exponent = [], 1
    while (rem:=(10**exponent // 9 % n)) not in remainders:
        remainders.append(rem); exponent += 1
    return (exponent - remainders.index(rem) - 1) * 9

Python

def A365933(n):
    if n==1: return 1
    a,b,x,y=1,1,1%n,11%n
    while x!=y:
        if a==b:
            a<<=1
            x,b=y,0
        y = (10*y+1)%n
        b+=1
    return 9*b # Chai Wah Wu, Jan 23 2024