A260461 -id:A260461 - cp's OEIS frontend

A096091 Numbers n with property that largest number formed from digits of n (A004186(n)) is divisible by smallest number formed from digits of n (A004185(n)).

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 20, 22, 30, 33, 40, 44, 50, 55, 60, 66, 70, 77, 80, 88, 90, 99, 100, 101, 105, 108, 110, 111, 150, 180, 200, 202, 220, 222, 300, 303, 330, 333, 400, 404, 405, 440, 444, 450, 500, 501, 504, 505, 510, 540, 550, 555, 600, 606

Offset: 1

Amarnath Murthy, Jun 22 2004

The number N = d*10^m*(10^k-1)/9 is a member for all m, k where 1 < d < 10, since the quotient is 10^m. E.g., for d = 7, m = 4, k = 8 we get N = 777777770000.
Conjecture: There are infinitely many terms besides these.
From Jon E. Schoenfield, Jul 26 2015: (Start)
Every number whose nonzero digits are all identical (e.g., 70770070777) is a term in the sequence (so the sequence is infinite). Also, if k is a term, then so is k*10 (hence, so is k*10^m for m >= 1). Removal of all terms that satisfy either of the above criteria still leaves an infinite number of terms, beginning with 105, 108, 150, 180, 405, 450, 501, 504, 510, 540, 801, 810, ... (see A260461).
If any integer k is a term, then so is every integer obtained by permuting the digits of k, except for some (not necessarily all) permutations beginning with 0. E.g., since 12000 is a member, so are all the other permutations of its digits that begin with 1 (i.e., 10002, 10020, and 10200), and all those that begin with 2 (i.e., 20001, 20010, 20100, and 21000), as well as the permutations that begin with a single 0 (which, after leading zeros are removed, reduce to 1002, 1020, 1200, 2001, 2010, and 2100), but not those that begin with more than one 0 (i.e., the sequence does not include 12, 21, 102, 120, 201, or 210). Aside from those terms whose nonzero digits are all identical, it appears that only a small number of patterns result from sorting the digits in increasing order (and discarding the zeros, which, of course, are all leading zeros): these "primitives" begin with 12, 15, 16, 18, 24, 25, 36, 45, 48, 125, ... (see A260462). (End)

			110 is a member as 110/011=10.

Giovanni Resta, Table of n, a(n) for n = 1..10000

Cf. A004185, A004186, A096089, A096090, A260461, A260462.

isA096091 := proc(n)
    if modp( A004186(n),A004185(n))= 0 then
        true;
    else
        false;
    end if;
end proc:
for n from 1 to 1000 do
    if isA096091(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Jul 26 2015

Select[Range[999], (d = Sort@ IntegerDigits@ #; Divisible@@ FromDigits/@ {Reverse@ d, d})&] (* Giovanni Resta, Jul 26 2015 *)

Edited, corrected and extended by Robert G. Wilson v, Jun 23 2004

Entry revised by Editors of the OEIS, Jul 26 2015

A260462 Numbers k such that the digits of k are in increasing order and k divides (reverse(k) * 10^m) for some sufficiently-large integer m.

Original entry on oeis.org

12, 15, 16, 18, 24, 25, 36, 45, 48, 125, 128, 144, 168, 225, 256, 288, 1125, 1344, 2688, 12288, 111888

Offset: 1

Jon E. Schoenfield, Jul 26 2015

This sequence consists of the set of distinct numbers that result from taking the terms of A260461, sorting the digits of each term in ascending order, and discarding the leading zeros.
(Equivalently, this sequence consists of the set of distinct numbers that result from taking the terms of A096091 whose nonzero digits are not all the same, sorting the digits of each term in ascending order, and discarding the leading zeros.)
Through a(21) = 111888, the digits 7 and 9 do not appear.
After a(21) = 111888, there are no more terms through 10^27. Presumably, the sequence is full. Is there a proof?

Cf. A096091, A260461.

cp's OEIS Frontend

A096091 Numbers n with property that largest number formed from digits of n (A004186(n)) is divisible by smallest number formed from digits of n (A004185(n)).

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