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

0,1

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

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

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

Prog

(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

Crossrefs

Cf. A266586, A266578.

Sequence in context: A119589 A316915 A352575 * A268449 A289826 A309390

Adjacent sequences: A266795 A266796 A266797 * A266799 A266800 A266801

Keyword

nonn,base

Author

M. F. Hasler, Jan 01 2016

Status

approved