OFFSET
1,2
COMMENTS
See A266578 for the variant where repeated digits are counted.
a(n) = 1 for 100 <= n <= 199 (and whenever n has a digit 1, cf. A011531), but then the sequence continues nontrivially with a(200,...) = (1255, 1, 751, 621, 251, 99, 511, 97, 101, 101, ...).
Record values are a(2) = 6163, a(2953) = 6521, a(3597) = 7209, a(5904) = 8047, a(23222) = 7681, a(39808) = 8011, a(39993) = 8231, a(44444) = 10151, ...
For small k=1,...,6, the graphs over the range 1 .. 10^(k+1) are roughly ("self"-)similar, because of the ranges 10^k .. 2*10^k-1 and (m+1/10)*10^k .. (m+2/10)*10^k-1 (with m=2,...,9) etc., on which a(n) = 1, while generically a(n) has values ranging quite uniformly between 1 and 10^4. For larger k, the picture changes, since pandigital numbers (and therefore also numbers having a digit '1') have asymptotic density one.
LINKS
M. F. Hasler, Table of n, a(n) for n = 1..10000
FORMULA
EXAMPLE
a(2) = 6163 since 2*6163 = 12326 has the same digits (1, 2, 3 and 6) as concat(2,6163) = 26163, and 6163 is the least N with this property.
MAPLE
f:= proc(n) local k, Ln, Lk, Lnk;
Ln:= convert(convert(n, base, 10), set);
if has(Ln, 1) then return 1 fi;
for k from 2 do
Lk:= convert(convert(k, base, 10), set);
Lnk:= convert(convert(n*k, base, 10), set);
if Lnk = Ln union Lk then return k fi
od
end proc:
map(f, [$1..100]); # Robert Israel, Jan 01 2016
PROG
(PARI) A266586(n, L=9e9, d=digits(n))=for(k=1, L, Set(digits(k*n))==Set(concat(digits(k), d))&&return(k))
(Python)
from itertools import count
def a(n):
digs = set(str(n))
return next(N for N in count(1) if digs | set(str(N)) == set(str(n*N)))
print([a(n) for n in range(1, 67)]) # Michael S. Branicky, Nov 15 2022
CROSSREFS
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Jan 01 2016
STATUS
approved