A113599 Least n-digit multiple of n whose digit permutations yield at least n distinct multiples of n, or 0 if no such number exists.

1, 20, 102, 1004, 10005, 100002, 1000013, 10000008, 100000008, 1000000010, 10000000010, 100000000008, 1000000000012, 10000000000004, 100000000000005, 1000000000000016, 10000000000000016, 100000000000000008, 1000000000000000018

Offset

1,2

Comments

Conjecture: No term is zero. The proof should be simple.

There are no zeros for n<=500. - Sean A. Irvine, Apr 10 2013

The conjecture is true: There are 9*10^(n-1) n-digit numbers, meaning at least floor(9*10^(n-1)/n) n-digit multiples of n. There are binomial(n+9, 9) multisets of n digits. Thus, by the pigeonhole principle, one of these multisets contains at least ceiling(floor(9*10^(n-1)/n)/binomial(n+9, 9)) multiples of n, and this number is at least n whenever n > 3. - James Rayman, Feb 14 2023

Links

Table of n, a(n) for n=1..19.

Example

a(4) = 1004; the multiples of 4 arising as a digit permutation are 1004, 1040, 1400, 4100.

Crossrefs

Sequence in context: A039394 A043997 A297585 * A077757 A126859 A071334

Adjacent sequences: A113596 A113597 A113598 * A113600 A113601 A113602

Keyword

base,easy,nonn

Author

Amarnath Murthy, Nov 07 2005

Extensions

More terms from Sean A. Irvine, Apr 10 2013

Status

approved