A305322 -id:A305322 - cp's OEIS frontend

A366596 Repdigit numbers that are divisible by 7.

0, 7, 77, 777, 7777, 77777, 111111, 222222, 333333, 444444, 555555, 666666, 777777, 888888, 999999, 7777777, 77777777, 777777777, 7777777777, 77777777777, 111111111111, 222222222222, 333333333333, 444444444444, 555555555555, 666666666666, 777777777777

Offset: 1

Kritsada Moomuang, Oct 14 2023

7 divides a repdigit iff it consists of only digit 7, or has length 6*k (for any digit).

Repdigit remainders A010785(k) mod 7 have period 54. - Karl-Heinz Hofmann, Dec 04 2023

Karl-Heinz Hofmann, Table of n, a(n) for n = 1..2329
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,1000001,0,0,0,0,0,0,0,0,0,0,0,0,0,-1000000).

Intersection of A008589 and A010785.
Cf. A002281 (a subsequence).
Cf. A305322 (divisor 3), A002279 (divisor 5), A083118 (the impossible divisors).

r(n) = 10^((n+8)\9)\9*((n-1)%9+1); \\ A010785
lista(nn) = select(x->!(x%7), vector(nn, k, r(k-1))); \\ Michel Marcus, Oct 26 2023

def A366596(n):
    digitlen, digit = (n+12)//14*6, (n+12)%14-4
    if digit < 1: digitlen += digit - 1; digit = 7
    return 10**digitlen // 9 * digit # Karl-Heinz Hofmann, Dec 04 2023

From Karl-Heinz Hofmann, Dec 04 2023: (Start)
a(n) = A010785(floor((n-2)/14)*54 + ((n-2) mod 14) + 41),   for (n-2) mod 14 >  4.
a(n) = (10^(6*floor((n-2)/14) + 6)-1)/9*(((n-2) mod 14)-4), for (n-2) mod 14 >  4.
a(n) = A010785(floor((n-2)/14)*54 + ((n-2) mod 14)*9 + 7),  for (n-2) mod 14 <= 4.
a(n) = (10^(6*floor((n-2)/14) + 1 + ((n-2) mod 14))-1)/9*7, for (n-2) mod 14 <= 4.
(End)

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

Original entry on oeis.org

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

Karl-Heinz Hofmann, Nov 07 2023

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.

			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]

Karl-Heinz Hofmann, Table with additional information.

Cf. A010785, A002275.
Cf. A305322 (divisor 3), A002279 (divisor 5), A366596 (divisor 7).
Cf. A083118 (the impossible divisors).

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

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

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A366596 Repdigit numbers that are divisible by 7.

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

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