A266798 Least positive integer N such that n+N has the same digits as n and N together (without counting repetitions).

10, 100, 100, 100, 100, 100, 100, 100, 100, 89, 99, 1000, 1000, 818, 1000, 1000, 1000, 1000, 168, 90, 100, 1000, 1000, 1000, 1000, 727, 336, 247, 1000, 899, 100, 1000, 1000, 1000, 1000, 1000, 326, 636, 1000, 899, 100, 1000, 1000, 1000, 1000, 405, 1000, 227, 1000, 545, 100, 1000, 1000, 1000, 450, 494, 1000, 1000, 1000, 899

Offset: 0

Author

M. F. Hasler, Jan 01 2016

Keywords

• nonn
• base

Comments

Such an N always exists since 10^(1 + number of digits of n) satisfies the property.

a(n) = 1 for almost all n (in the sense of natural density). - Charles R Greathouse IV, Nov 15 2022

What is the largest number in this sequence? It is somewhere between a(9911111111) = 302345678 and 203456789111111111. - Charles R Greathouse IV, Dec 09 2022

Links

• Robert Israel, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A266586, A266578.

Programs

Maple

digs:= proc(n) option remember;
  local t;
  t:= n mod 10;
  if n < 10 then {t}
  else {t} union procname((n-t)/10)
  fi;
end proc:
f:= proc(n)
  local k,Ln;
  Ln:= digs(n);
  for k from 1 do
     if Ln union digs(k) = digs(n+k) then return k fi
  od
end proc:
seq(f(n),n=0..100); # Robert Israel, Jan 03 2016

PARI

a(n,d=digits(n),L=10^(1+#d))=for(k=1,L,Set(digits(k+n))==Set(concat(d,digits(k)))&&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(60)]) # Michael S. Branicky, Nov 15 2022

Formula

a(n) <= 10^(1 + number of digits of n).

a(n) <= 203456789111111111 < 2.04 * 10^17. (This can probably be improved by a few orders of magnitude.) - Charles R Greathouse IV, Nov 15 2022