A084687 Nontrivial numbers k containing no zero digits which are divisible by the number formed by writing the digits of k in ascending order.

9513, 81816, 93513, 94143, 95193, 816816, 888216, 933513, 934143, 935193, 941493, 951993, 2491578, 8166816, 8868216, 9333513, 9334143, 9335193, 9341493, 9351993, 9414993, 9519993, 24915798, 49827156, 81666816, 87127446, 88668216, 93333513

Offset: 1

Author

Chuck Seggelin, Jun 30 2003

Keywords

• base
• nonn

Comments

Sequence excludes numbers which are already sorted, like 1234 or 133778, as sorting any such number yields the same number, which is of course divisible by itself, a trivial case.

All members of this sequence appear to be divisible by 3. Further, many of the terms of the sequence can be generated patternistically simply by inserting digits in certain places in earlier terms. The primitive terms of this sequence which cannot be patternistically generated are in sequence A086083.

All terms are divisible by 3. Proof: if the digits of x*y are a permutation of the digits of x, we must have x*y==x (mod 9), implying either x == 0 (mod 3) or y == 1 (mod 9). - Robert Israel, Jul 09 2020

Examples

			9513/1359 = 7; 9876543192/1234567899 = 8; etc.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..208

Crossrefs

Cf. A086083.

Programs

Maple

S:= [seq([i],i=1..9)]: R:= NULL: count:= 0:
for d from 2 to 8 do
  S:= map(t -> seq([i,op(t)],i=1..t[1]), S);
  for s in S do
    x:= add(s[i]*10^(d-i),i=1..d);
    if x mod 3 <> 0 then next fi;
    for m from 2 to 10^(d+1)/x do
      if sort(convert(m*x,base,10))=s then
        count:= count+1; R:= R, m*x;
      fi
    od
  od
od:
sort([R]); # Robert Israel, Jul 09 2020

Mathematica

Select[ Range[ 10^8], IntegerQ[ # /FromDigits[ Sort[ IntegerDigits[ # ]]]] && # != FromDigits[ Sort[ IntegerDigits[ # ]]] && Count[ IntegerDigits[ # ], 0] == 0 & ]

Extensions

Edited by Robert G. Wilson v, Jul 07 2003