A244927 -id:A244927 - cp's OEIS frontend

A079339 Least k such that the decimal representation of k*n contains only 1's and 0's.

1, 5, 37, 25, 2, 185, 143, 125, 12345679, 1, 1, 925, 77, 715, 74, 625, 653, 61728395, 579, 5, 481, 5, 4787, 4625, 4, 385, 40781893, 3575, 37969, 37, 3581, 3125, 3367, 3265, 286, 308641975, 3, 2895, 259, 25, 271, 2405, 25607, 25, 24691358, 23935, 213, 23125

Offset: 1

Benoit Cloitre, Feb 13 2003

From David Amar (dpamar(AT)gmail.com), Jul 12 2010: (Start)
This sequence is well defined.
In the n+1 first repunits (see A002275), there are at least 2 numbers that have the same value modulo n (pigeonhole principle).
The difference between those two numbers contains only 1's and 0's in decimal representation. (End)
This actually proves the stronger statement that there is always a multiple of the form 111...000 (Thm. 1 in Wu, 2014), cf. A244859 for these multiples and A244927 for the k-values. - M. F. Hasler, Mar 04 2025

			3*37 = 111 and no integer k < 37 has this property, hence a(3)=37.

Popular Computing (Calabasas, CA), Z-Sequences, Vol. 4 (No. 34, A pr 1976), pages PC36-4 to PC37-6, but there are many errors (cf. A257343, A257344).

Robert G. Wilson v, Table of n, a(n) for n = 1..10000 (terms 1..1999 from T. D. Noe, terms 2000..9998 from N. J. A. Sloane [based on A004290])
Chai Wah Wu, Pigeonholes and repunits, Amer. Math. Monthly, 121 (2014), 529-533.

Cf. A004290, A070189, A078241-A078248, A096681-A096688, A257343, A257344, A257345.

d(n,i)=floor(n/10^(i-1))-10*floor(n/10^i);
test(n)=sum(i=1,ceil(log(n)/log(10)),if(d(n,i)*(1-d(n,i)),1,0));
a(n)=if(n<0,0,s=1; while(test(n*s)>0,s++); s)

a(n) = A004290(n)/n.
a(n) < 10^(n+1) / (9n). - Charles R Greathouse IV, Jan 09 2012
a(n) <= A244927(n), with equality for n <= 6. - M. F. Hasler, Mar 04 2025

More terms from Vladeta Jovovic and Matthew Vandermast, Feb 14 2003

Definition simplified by Franklin T. Adams-Watters, Jan 09 2012

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

Original entry on oeis.org

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

Chai Wah Wu, Jul 07 2014

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

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.

Equal to A004290 for n = 1 .. 6.

Cf. A002275, A244927.

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

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

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

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

cp's OEIS Frontend

A079339 Least k such that the decimal representation of k*n contains only 1's and 0's.

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