A305322 Repdigit numbers that are divisible by 3.

0, 3, 6, 9, 33, 66, 99, 111, 222, 333, 444, 555, 666, 777, 888, 999, 3333, 6666, 9999, 33333, 66666, 99999, 111111, 222222, 333333, 444444, 555555, 666666, 777777, 888888, 999999, 3333333, 6666666, 9999999, 33333333, 66666666, 99999999, 111111111, 222222222

Offset: 1

Author

Kritsada Moomuang, May 30 2018

Keywords

• nonn
• base
• easy

Comments

The terms > 0 are (10^d-1)*k/9 for k=1..9 if d is divisible by 3, and for k=3,6,9 otherwise. - Robert Israel, Jun 01 2018

Repdigit remainders A010785(k) mod 3 have period 27. - Karl-Heinz Hofmann, Nov 11 2023

Examples

			111 / 3 = 37;
222 / 3 = 74;
333 / 3 = 111;
444 / 3 = 148;
555 / 3 = 185.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..4996
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,1001,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1000).

Crossrefs

Cf. A008585, A010785.

Cf. A002279 (divisor 5), A366596 (divisor 7), A083118 (the impossible divisors).

Programs

Maple

L:= proc(d) if d mod 3 = 0 then [$1..9] else [3,6,9] fi end proc:
0,seq(seq((10^d-1)/9*k,k=L(d)),d=1..9); # Robert Israel, Jun 01 2018

Python

def A010785(n): return (n - 9*((n-1)//9))*(10**((n+8)//9) - 1)//9
def A305322(n):
    d0, d1 = divmod(n-1,15)
    if d1 < 7: return A010785(d0 * 27 + d1 * 3)
    return A010785(d0 * 27 + d1 + 12) # Karl-Heinz Hofmann, Nov 26 2023

Formula

From Alois P. Heinz, May 30 2018: (Start)

{ A008585 } intersect { A010785 }.

G.f.: 3*(300*x^20 + 200*x^19 + 100*x^18 + 330*x^17 + 220*x^16 + 110*x^15 + 333*x^14 + 296*x^13 + 259*x^12 + 222*x^11 + 185*x^10 + 148*x^9 + 111*x^8 + 74*x^7 + 37*x^6 + 33*x^5 + 22*x^4 + 11*x^3 + 3*x^2 + 2*x + 1)*x^2 / ((x-1) *(x^2 + x + 1) *(x^4 + x^3 + x^2 + x + 1) *(10*x^5-1) *(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1) *(100*x^10 + 10*x^5 + 1)).

a(n) = 1001*a(n-15) - 1000*a(n-30). (End)

From Karl-Heinz Hofmann, Nov 11 2023: (Start)

a(n) = A010785(floor((n-1)/15)*27 + ((n-1) mod 15)*3) iff (n-1 <= 6 (mod 15)).

a(n) = A010785(floor((n-1)/15)*27 + ((n-1) mod 15) + 12) iff (n-1 > 6 (mod 15)).

(End)

Extensions

Name clarified by Felix Fröhlich, Jun 01 2018