[seqfan] Interesting Recreational Sequence
David Wilson
davidwwilson at comcast.net
Mon Jan 19 18:38:41 CET 2004
More simply put...
Let n be a positive integer coprime to 10 which does not include every
positive digit. Then n divides some repunit k(n). Multiply k(n) by each
positive digit not in n, concatenate the resulting values to get m(n).
If n does not contain digit 0, append digit 0 to m(n). m(n) is then a
multiple of n that includes all and only digits not in n. This shows that
A078249(n) <= m(n) exists.
----- Original Message -----
From: Jack Brennen
To: seqfan at ext.jussieu.fr
Sent: Monday, January 19, 2004 11:27 AM
Subject: Re: [seqfan] Interesting Recreational Sequence
> Will this sequence eventually become "all zeros" or can it be shown
> that non-zero entries exist ad infinitum?
I believe you can show that any integer whose representation does
not contain a '1' digit, and which ends in '3', '7', or '9', has
a multiple consisting of just the digit '1' repeated some number
of times. This is sufficient to prove an infinite number of
non-zero entries, since there are an infinite number of integers
which contain every digit but '1', and which end in '3', '7', or '9'.
For instance:
203456789 divides evenly into (10^101710278-1)/9.
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