A004290 Least positive multiple of n that when written in base 10 uses only 0's and 1's.

1, 10, 111, 100, 10, 1110, 1001, 1000, 111111111, 10, 11, 11100, 1001, 10010, 1110, 10000, 11101, 1111111110, 11001, 100, 10101, 110, 110101, 111000, 100, 10010, 1101111111, 100100, 1101101, 1110, 111011, 100000, 111111, 111010

Offset: 1

Author

David W. Wilson

Keywords

• nonn
• base
• nice

Comments

It is easy to show that a(n) always exists and in fact has at most n digits [Wu, 2014]. - N. J. A. Sloane, Jun 13 2014

a(n) = min{A007088(k): k > 0 and A007088(k) mod n = 0}. - Reinhard Zumkeller, Jan 10 2012

a(10^k) = 10^k and a(10^k - 1) = (10^(9k) - 1) / 9 for all k. Is a(n) < a(10^k - 1) for all n < 10^k - 1? - David Radcliffe, Aug 01 2025

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..9998 (first 2000 terms from T. D. Noe [and Ed Pegg Link])
• Ed Pegg Jr., 'Binary' Puzzle.
• Eric M. Schmidt, Sage code to compute this sequence.
• Chai Wah Wu, Pigeonholes and repunits, Amer. Math. Monthly, 121 (2014), 529-533.

Crossrefs

Cf. A004283-A004289, A078241-A078248, A079339, A096681-A096688, A257345.

Programs

Haskell

a004290 0 = 0
a004290 n = head [x | x <- tail a007088_list, mod x n == 0]
-- Reinhard Zumkeller, Jan 10 2012

Maple

f:= proc(n)
local L,x,m,r,k,j;
if n<2 then return n fi;
for x from 2 to n-1 do L[0,x]:= 0 od:
L[0,0]:= 1: L[0,1]:= 1;
for m from 1 do
   if L[m-1,(-10^m) mod n] = 1 then break fi;
   L[m,0]:= 1;
   for k from 1 to n-1 do
     L[m,k]:= max(L[m-1,k],L[m-1,k-10^m mod n])
   od;
od;
r:= 10^m; k:= -10^m mod n;
for j from m-1 by -1 to 1 do
    if L[j-1,k] = 0 then
      r:= r + 10^j; k:= k - 10^j mod n;
    fi
od;
if k = 1 then r:= r + 1 fi;
r
end proc:
seq(f(n),n=1..100); # Robert Israel, Feb 09 2016

Mathematica

a[n_] := For[k = 1, True, k++, b = FromDigits[ IntegerDigits[k, 2] ]; If[Mod[b, n] == 0, Return[b]]]; a[0] = 0; Table[a[n], {n, 0, 34}] (* Jean-François Alcover, Jun 14 2013, after Reinhard Zumkeller *)
With[{c=Rest[Union[FromDigits/@Flatten[Table[Tuples[{1,0},i],{i,10}], 1]]]}, Join[{0},Flatten[ Table[ Select[c,Divisible[#,n]&,1],{n,40}]]]] (* Harvey P. Dale, Dec 07 2013 *)

PARI

a(n) = {if( n==0, return (0)); my(m = n); while (vecmax(digits(m)) != 1, m+=n); m;} \\ Michel Marcus, Feb 09 2016, May 27 2020

PARI

apply( {A004290(n)=for(k=1,2^n,(t=fromdigits(binary(k)))%n||return(t))}, [1..44]) \\ M. F. Hasler, Mar 04 2025

Python

def A004290(n):
    if n > 0:
        for i in range(1,2**n):
            x = int(bin(i)[2:])
            if not x % n:
                return x
    return 0
# Chai Wah Wu, Dec 30 2014

Formula

a(n) = n*A079339(n). - Jonathan Sondow, Jun 15 2014

Extensions

Initial 0 deleted and offset corrected by N. J. A. Sloane, Jan 31 2024