A266586 The least nonnegative integer N such that n*N has the same digits as n and N together, not counting repetitions.

1, 6163, 51, 416, 251, 21, 967, 86, 255, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1255, 1, 781, 973, 26, 265, 24, 81, 1139, 1135, 51, 1, 291, 186, 151, 41, 936, 3001, 886, 982, 416, 1, 341, 315, 1464, 181, 734, 371, 958, 1921, 251, 1, 2412, 635, 846, 221, 1801, 125, 948, 845, 21, 1, 251, 585, 2213, 281, 1076

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

a(n) = 1 whenever n has a digit '1', i.e., n in A011531.

a(n) <= A266578(n) unless A266578(n) = 0.

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

Cf. A266578, A266798.

Sequence in context: A281265 A007992 A033288 * A057880 A151967 A214556

Adjacent sequences: A266583 A266584 A266585 * A266587 A266588 A266589

Keyword

nonn,base

Author

M. F. Hasler, Jan 01 2016

Status

approved