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%I #12 Jul 09 2020 16:49:45
%S 9513,81816,93513,94143,95193,816816,888216,933513,934143,935193,
%T 941493,951993,2491578,8166816,8868216,9333513,9334143,9335193,
%U 9341493,9351993,9414993,9519993,24915798,49827156,81666816,87127446,88668216,93333513
%N Nontrivial numbers k containing no zero digits which are divisible by the number formed by writing the digits of k in ascending order.
%C 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.
%C 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.
%C 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
%H Robert Israel, <a href="/A084687/b084687.txt">Table of n, a(n) for n = 1..208</a>
%e 9513/1359 = 7; 9876543192/1234567899 = 8; etc.
%p S:= [seq([i],i=1..9)]: R:= NULL: count:= 0:
%p for d from 2 to 8 do
%p S:= map(t -> seq([i,op(t)],i=1..t[1]), S);
%p for s in S do
%p x:= add(s[i]*10^(d-i),i=1..d);
%p if x mod 3 <> 0 then next fi;
%p for m from 2 to 10^(d+1)/x do
%p if sort(convert(m*x,base,10))=s then
%p count:= count+1; R:= R, m*x;
%p fi
%p od
%p od
%p od:
%p sort([R]); # _Robert Israel_, Jul 09 2020
%t Select[ Range[ 10^8], IntegerQ[ # /FromDigits[ Sort[ IntegerDigits[ # ]]]] && # != FromDigits[ Sort[ IntegerDigits[ # ]]] && Count[ IntegerDigits[ # ], 0] == 0 & ]
%Y Cf. A086083.
%K base,nonn
%O 1,1
%A Chuck Seggelin (barkeep(AT)plastereddragon.com), Jun 30 2003
%E Edited by _Robert G. Wilson v_, Jul 07 2003
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