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<DIV><FONT face=Arial size=2>More simply put...</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Let n be a positive integer coprime to 10 which
does not include every</FONT></DIV>
<DIV><FONT face=Arial size=2>positive digit. Then n divides some repunit
k(n). Multiply k(n) by each</FONT></DIV>
<DIV><FONT face=Arial size=2>positive digit not in n, concatenate the resulting
values to get m(n).</FONT></DIV>
<DIV><FONT face=Arial size=2>If n does not contain digit 0, append digit 0 to
m(n). m(n) is then a</FONT></DIV>
<DIV><FONT face=Arial size=2>multiple </FONT><FONT face=Arial size=2>of n that
includes all and only digits not in n. This shows that</FONT></DIV>
<DIV><FONT face=Arial size=2>A078249(n) <= m(n) exists.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV>----- Original Message ----- </DIV>
<DIV style="FONT: 10pt arial">
<DIV style="BACKGROUND: #e4e4e4; font-color: black"><B>From:</B> <A
title=jack@brennen.net href="mailto:jack@brennen.net">Jack Brennen</A> </DIV>
<DIV><B>To:</B> <A title=seqfan@ext.jussieu.fr
href="mailto:seqfan@ext.jussieu.fr">seqfan@ext.jussieu.fr</A> </DIV>
<DIV><B>Sent:</B> Monday, January 19, 2004 11:27 AM</DIV>
<DIV><B>Subject:</B> Re: [seqfan] Interesting Recreational Sequence</DIV></DIV>
<DIV><BR></DIV>> Will this sequence eventually become "all zeros" or can it
be shown<BR>> that non-zero entries exist ad infinitum?<BR><BR>I believe you
can show that any integer whose representation does<BR>not contain a '1' digit,
and which ends in '3', '7', or '9', has<BR>a multiple consisting of just the
digit '1' repeated some number<BR>of times. This is sufficient to prove an
infinite number of<BR>non-zero entries, since there are an infinite number of
integers<BR>which contain every digit but '1', and which end in '3', '7', or
'9'.<BR>For instance:<BR><BR> 203456789 divides evenly into
(10^101710278-1)/9.<BR><BR></BODY></HTML>