A266586 -id:A266586 - cp's OEIS frontend

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

M. F. Hasler, Jan 01 2016

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

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

Cf. A266586, A266578.

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

a(n,d=digits(n),L=10^(1+#d))=for(k=1,L,Set(digits(k+n))==Set(concat(d,digits(k)))&&return(k))

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

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

A266578 Least number N such that the product n*N has the same digits as the concatenation (n,N) (counting repetitions), or 0 if no such number exists.

Original entry on oeis.org

0, 8714, 51, 0, 251, 21, 0, 86, 351, 0, 9209, 86073, 0, 926, 93, 0, 9635, 6012, 0, 8714, 6, 0, 9017, 651, 0, 401, 81, 0, 3701, 51, 0, 926, 40611, 0, 41, 936, 0, 3251, 6882, 0, 35, 678, 0, 9203, 3141, 0, 371, 2913, 0, 251, 3, 0, 635, 846, 0, 2171, 834, 0, 845, 21, 0, 1814, 585, 0, 281, 9843

Offset: 1

David W. Wilson and M. F. Hasler, Jan 01 2016

See A266586 for the variant where repeated digits are not counted.
One has a(3k-1) = 3*b(k)-1 with  b = (2905, 84, 29, 3070, 309, 3212, 2905, ...), and a(3k) = 3*c(k) with c = (17, 7, 117, 28691, 31, 2004, ...).
If n = 1 (mod 3), then the sum of digits of n*N = N (mod 3) is always different from the sum of digits of concat(n,N) which is 1+N (mod 3), therefore a(3k+1) = 0 for all k.

			For n = 1 there cannot be a number N such that n*N (= N) has the same digits as concat(n,N) (= "1N"), therefore a(1)=0.
For n = 2 and N = 8714 we have n*N = 17428 which has the same digits (1,2,4,7,8) as concat(n,N) = 28714. This N is the smallest such number, therefore a(2) = 8714.
Since 3*51 = 153 has the same digits than concat(3,51), and 51 is the smallest such number, a(3) = 153.
For n = 4 there is again no N with the desired property, thus a(4) = 0.
Since 5*251 = 1255 has the same digits (with repetition) than "5" and "251" together, a(5) = 1255.

David W. Wilson, Table of n, a(n) for n = 1..10000

Table[If[Mod[n, 3] == 1, 0, k = 1; While[Sort@ IntegerDigits[n k] != Sort@ Join[IntegerDigits@ n, IntegerDigits@ k], k++]; k], {n, 66}]

a(n,L=if(n%3!=1,9e9),d=digits(n))=for(k=2,L,vecsort(digits(k*n))==vecsort(concat(d,digits(k)))&&return(k))

from itertools import count
from collections import Counter as Ctr
def a(n):
    r = n%3
    if r == 1: return 0
    s = str(n)
    return next(N for N in count(r, 3) if Ctr(str(n*N)) == Ctr(s+str(N)))
print([a(n) for n in range(1, 67)]) # Michael S. Branicky, Nov 15 2022

a(3k+1) = 0 for all k >= 0; a(3k+2) = 2 (mod 3) for all k >= 0; a(3k) = 0 (mod 3) for all k >= 1.

cp's OEIS Frontend

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

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