OFFSET
1,1
COMMENTS
For all n, a(n) exists (see proof in References).
REFERENCES
Peter Winkler, Mathematical Puzzles (revised edition), CRC Press, 2024, p. liii.
LINKS
Michael S. Branicky, Table of n, a(n) for n = 1..10000
FORMULA
a(10^e-1) <= 1^e 0 1^(8*e), where ^ denotes repeated concatenation of digits on the right-hand side. - Michael S. Branicky, Feb 22 2024
MATHEMATICA
a[n_]:=Min[Select[FromDigits/@Tuples[{0, 1}, n+1],
Divisible[#, n]&&Union[IntegerDigits[#]]=={0, 1}&]]; a/@Range[23]
PROG
(Python)
from itertools import count
def a(n): return next(d for k in count(1) if ("0" in (b:=bin(k)[2:])) and (d:=int(b))%n==0)
print([a(n) for n in range(1, 24)]) # Michael S. Branicky, Feb 22 2024
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Ivan N. Ianakiev, Feb 22 2024
EXTENSIONS
More terms from Michael S. Branicky, Feb 22 2024
STATUS
approved