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A323711 Numbers k such that k, 2*k, and 3*k are anagrams of each other.

%I A323711 #45 Feb 04 2019 14:26:51
%S A323711 142857,285714,1402857,1428570,1428597,1429857,2857014,2857140,
%T A323711 2859714,2985714,14002857,14028570,14028597,14029857,14285700,
%U A323711 14285970,14285997,14298570,14298597,14299857,15623784,15843762,17438256,17562438,18243756,21584376,23784156,24375618,24381756
%N A323711 Numbers k such that k, 2*k, and 3*k are anagrams of each other.
%C A323711 We assume entries have no leading zeros, so that n = 53617824 is not in the sequence, even though 2*n = 107235648 and 3*n = 160853472 are anagrams of 053617824.
%C A323711 From _Chai Wah Wu_, Feb 01 2019: (Start)
%C A323711 The first digit of terms is either 1, 2 or 3. Numbers of the form 140..028570..0 and 29..98570..0140..0 are terms where the number of 9's and 0's can be zero.
%C A323711 More generally, let a number n be written in decimal as xxxzzz where x and z are arbitrary digits and xxx, zzz are not empty strings. Let m be the number that is written as zzz in decimal and k be the least power of 10 that is strictly greater than m.
%C A323711 If 3*m < k, then n is a term if and only if xxx0..0zzz0..0 is a term. Note that this condition is satisfied if the first digit of m is 0, 1 or 2.
%C A323711 If 2*k <= 3*m, then n is a term if and only if xxx9..9zzz0..0 is a term. Note that this condition is satisfied if the first digit of m is 7, 8, or 9.
%C A323711 Not all terms with digits 0 and 9 are formed this way, see for instance the terms 137965842 and 157836042.
%C A323711 The first term where the first digit is 3 is a(1507) = 3051267489.
%C A323711 (End)
%C A323711 From _David A. Corneth_, Feb 02 2019: (Start)
%C A323711 Terms are multiples of 9.
%C A323711 Proof: as 3*k and k have the same digits, k is divisible by 3. If k isn't divisible by 9 then it has a different digital sum from 3*k. Therefore, k is divisible by 9. (End)
%H A323711 Chai Wah Wu, <a href="/A323711/b323711.txt">Table of n, a(n) for n = 1..10000</a>
%e A323711 The first entry, 142857, is well known for having n, 2*n, 3*n, 4*n, 5*n and 6*n all being anagrams. The next two numbers for which that happens are 1428570 and 1429857.
%o A323711 (Java) char[] digits1, digits2, digits3;
%o A323711 int val1, val2, val3;
%o A323711 for (int value=10; value<25000000; value++) {
%o A323711      digits1 = Integer.toString(value).toCharArray();
%o A323711      digits2 = Integer.toString(2*value).toCharArray();
%o A323711      digits3 = Integer.toString(3*value).toCharArray();
%o A323711      if (digits1.length == digits3.length) {
%o A323711           Arrays.sort(digits1);
%o A323711           Arrays.sort(digits2);
%o A323711           Arrays.sort(digits3);
%o A323711           val1 = Integer.parseInt(new String(digits1));
%o A323711           val2 = Integer.parseInt(new String(digits2));
%o A323711           val3 = Integer.parseInt(new String(digits3));
%o A323711           if ((val1 == val2) && (val1 == val3)) {
%o A323711                System.out.print(value + ",");
%o A323711           }
%o A323711      }
%o A323711 }
%o A323711 (Python)
%o A323711 A323711_list = [n for n in range(9,10**7,9) if sorted(str(n)) == sorted(str(2*n)) == sorted(str(3*n))] # _Chai Wah Wu_, Feb 02 2019
%Y A323711 Subsequence of A023086, numbers where n and 2*n are anagrams.
%K A323711 easy,nonn,base
%O A323711 1,1
%A A323711 _Darrah Chavey_, Jan 24 2019