A244859 Least positive multiple of n which when written in base 10 is either a repunit or of the form 111...000.

0, 1, 10, 111, 100, 10, 1110, 111111, 1000, 111111111, 10, 11, 11100, 111111, 1111110, 1110, 10000, 1111111111111111, 1111111110, 111111111111111111, 100, 111111, 110, 1111111111111111111111, 111000, 100, 1111110, 111111111111111111111111111, 11111100

Offset: 0

Author

Chai Wah Wu, Jul 07 2014

Keywords

• nonn
• base

Comments

a(1017) has 1008 digits. - Michael S. Branicky, Feb 22 2024

a(0) = 0 by convention: It can be considered as a repunit with zero digits, A002275(0) = (10^0-1)/9, and it is a positive multiple of n in the sense of k*n with k > 0. - M. F. Hasler, Mar 04 2025

Links

• Michael S. Branicky, Table of n, a(n) for n = 0..1016 (terms 101..936 from Robert Israel, terms 0..100 from Chai Wah Wu).
• Chai Wah Wu, Pigeonholes and repunits, Amer. Math. Monthly, 121 (2014), 529-533.

Crossrefs

Equal to A004290 for n = 1 .. 6.

Cf. A002275, A244927.

Programs

Maple

A244859:= proc(n) local m,d2,d5;
d2:= padic:-ordp(n,2);
d5:= padic:-ordp(n,5);
m:= n/2^d2/5^d5;
10^max(d2,d5)*(10^numtheory:-order(10,9*m)-1)/9
end proc:
A244859(0):= 0:
seq(A244859(n),n= 0..100); # Robert Israel, Jul 08 2014

PARI

apply( {A244859(n, m=Map(Mat([0,0])))=for(L=1,n, my(r=10^L\9); iferr(return(r-mapget(m,r%n)), E, mapput(m, r%n, r)))}, [0..33]) \\ M. F. Hasler, Mar 04 2025

Python

def a(n):
    if n == 0: return 0
    moddict = {0: 0}
    for e in range(1, n+2):
        repe = (10**e-1)//9
        r = repe%n
        if r in moddict:
            return repe - moddict[r]
        else:
            moddict[r] = repe
print([a(n) for n in range(29)]) # Michael S. Branicky, Feb 22 2024

Formula

a(n) = n*A244927(n). - M. F. Hasler, Mar 04 2025

a(3^k) = (10^(3^k)-1)/9. For n > 0, A055642(a(n)) <= n. If n > 2 is not a power of 3, then A055642(a(n)) < n. - Chai Wah Wu, Mar 04 2025