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A365933
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a(n) is the period of the remainders when repdigits are divided by n.
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1
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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]
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PROG
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(Python)
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)
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
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CROSSREFS
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Cf. A083118 (the impossible divisors).
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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