OFFSET
1,1
COMMENTS
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.
From Chai Wah Wu, Feb 01 2019: (Start)
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.
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.
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.
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.
Not all terms with digits 0 and 9 are formed this way, see for instance the terms 137965842 and 157836042.
The first term where the first digit is 3 is a(1507) = 3051267489.
(End)
From David A. Corneth, Feb 02 2019: (Start)
Terms are multiples of 9.
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)
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
EXAMPLE
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.
PROG
(Java) char[] digits1, digits2, digits3;
int val1, val2, val3;
for (int value=10; value<25000000; value++) {
digits1 = Integer.toString(value).toCharArray();
digits2 = Integer.toString(2*value).toCharArray();
digits3 = Integer.toString(3*value).toCharArray();
if (digits1.length == digits3.length) {
Arrays.sort(digits1);
Arrays.sort(digits2);
Arrays.sort(digits3);
val1 = Integer.parseInt(new String(digits1));
val2 = Integer.parseInt(new String(digits2));
val3 = Integer.parseInt(new String(digits3));
if ((val1 == val2) && (val1 == val3)) {
System.out.print(value + ", ");
}
}
}
(Python)
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
CROSSREFS
KEYWORD
easy,nonn,base
AUTHOR
Darrah Chavey, Jan 24 2019
STATUS
approved