cp's OEIS Frontend

Based on the first 2591 terms of A084687.

Trivial permutations are identified as (1) permutation = n, or (2) when n mod 10=0, permutations of n's digits which result in shifting only trailing zeros to the most significant side of n where they drop off, such that permutation = n/10^z, where z <= the number of trailing zeros of n. So if n were 1809000, the following permutations would be excluded as trivial: 1809000, 0180900, 0018090, 0001809.

A031877 (numbers which are multiples of their reversals) and both A084687 and A090053 (numbers divided by number formed by sorting their digits), are subsets of this sequence. This sequence differentiates itself by including terms such as 7425 which is divided by 2475 (a rearrangement of 7425's digits that is neither a reversal or an ascending sort.)

Trivial permutations are identified as those where the permutation = k itself. Digit loss occurs when a permutation has 0 in the most significant position, which drops off, leaving a number with fewer digits. For example, when k is 3105, the permutation 0315 is excluded because 315 has fewer digits than 3105. These exclusions make this sequence a subsequence of A090055. A084687 is a subsequence of this sequence.

Apparently each term of this sequence is divisible by 3. This has been confirmed for the first 100 terms.

From David A. Corneth, Jun 08 2025: (Start)

All terms are divisible by 3. Proof: Suppose a term t is not divisible by 3.

Then for some m < t that is an anagram of t with the same number of digits as t we have m * c = t where 2 <= c <= 9. If c > 9 then t has more digits than m and if c = 1 then m is a trivial anagram of t, excluded by definition.

Since m and t are anagrams, 9 | (t - m) = (c - 1)*m. If t is not divisible by 3 then m is not divisible by 3 and so 9 | c - 1. This is a contradiction since 2 <= c <= 9 for which no c is divisible by 9 which completes the proof.

In addition if t is not divisible by 9 then c = 4 or 7. (End)

Trivial cases are identified as (1) values of k where the digits are already in ascending order, like 123 or 2228, such that ASort(k)=k, or (2) values of k where k mod 10 = 0 and all digits other than trailing zeros are in ascending order, like 12000 or 333500, such that ASort(k)=k/10^z, where z = the number of trailing zeros of k. In case (1), k/ASort(k) is equivalent to k/k (as in 123/123). In case (2), k/ASort(k) is 10^z (as in 12000/12). Neither of these cases is very interesting.

Sequence A084687 is a subsequence of this sequence, but that sequence excludes any value of k with 1 or more zero digits.

Trivial cases are identified as those values of n where the digits are already in descending order, like 3210 or 8222, such that DSort(n)=n. In such cases DSort(n)/n is equivalent to n/n (as in 3210/3210).

a(1) and a(2) are primitive. Clearly if DSort(n) mod n = 0, then dsort(n x 10) mod (n x 10) = 0. Therefore since 1750842 is a member, so will be 17508420, 175084200, 1750842000 and so on. The nonprimitive member 19750842 sets up the implication that 1(9...)750842 is a member. A quick test of 199750842, 1999750842 and 19999750842 seems to confirm this.