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%I #74 Jan 23 2024 13:25:27
%S 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,
%T 54,243,54,252,27,135,9,54,144,54,81,27,162,54,9,45,54,189,18,81,198,
%U 414,27,378,9,432,54,117,243,18,54,162,252,522,27,540,135,162,9,54
%N a(n) is the period of the remainders when repdigits are divided by n.
%C 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.
%H Karl-Heinz Hofmann, <a href="/A365933/a365933_3.txt">Table with additional information</a>.
%e For n = 6: Remainders of A010785(1..54) mod n.
%e A010785( 1...9) mod n: [1, 2, 3, 4, 5, 0, 1, 2, 3]
%e A010785(10..18) mod n: [5, 4, 3, 2, 1, 0, 5, 4, 3]
%e A010785(19..27) mod n: [3, 0, 3, 0, 3, 0, 3, 0, 3]
%e So the period is 3*9 = 27. Thus a(n) = 27. And the pattern seen above starts again:
%e A010785(28..36) mod n: [1, 2, 3, 4, 5, 0, 1, 2, 3]
%e A010785(37..45) mod n: [5, 4, 3, 2, 1, 0, 5, 4, 3]
%e A010785(46..54) mod n: [3, 0, 3, 0, 3, 0, 3, 0, 3]
%o (Python)
%o def A365933(n):
%o if n == 1: return 1
%o remainders, exponent = [], 1
%o while (rem:=(10**exponent // 9 % n)) not in remainders:
%o remainders.append(rem); exponent += 1
%o return (exponent - remainders.index(rem) - 1) * 9
%o (Python)
%o def A365933(n):
%o if n==1: return 1
%o a,b,x,y=1,1,1%n,11%n
%o while x!=y:
%o if a==b:
%o a<<=1
%o x,b=y,0
%o y = (10*y+1)%n
%o b+=1
%o return 9*b # _Chai Wah Wu_, Jan 23 2024
%Y Cf. A010785, A002275.
%Y Cf. A305322 (divisor 3), A002279 (divisor 5), A366596 (divisor 7).
%Y Cf. A083118 (the impossible divisors).
%K nonn,base
%O 1,2
%A _Karl-Heinz Hofmann_, Nov 07 2023
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