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

Author

Kritsada Moomuang, Oct 14 2023

Keywords

• nonn
• base
• easy

Comments

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

Links

• 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).

Crossrefs

Intersection of A008589 and A010785.

Cf. A002281 (a subsequence).

Cf. A305322 (divisor 3), A002279 (divisor 5), A083118 (the impossible divisors).

Programs

PARI

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

Python

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

Formula

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)