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A365933 a(n) is the period of the remainders when repdigits are divided by n. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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.
LINKS
EXAMPLE
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]
PROG
(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
CROSSREFS
Cf. A305322 (divisor 3), A002279 (divisor 5), A366596 (divisor 7).
Cf. A083118 (the impossible divisors).
Sequence in context: A103952 A103955 A359531 * A109041 A227900 A010817
KEYWORD
nonn,base
AUTHOR
Karl-Heinz Hofmann, Nov 07 2023
STATUS
approved

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Last modified July 14 05:06 EDT 2024. Contains 374291 sequences. (Running on oeis4.)