[seqfan] Re: possible sequence

David Wilson davidwwilson at comcast.net
Sun Mar 9 03:23:07 CET 2014


A power of 2 can end in arbitrarily many nonzero digits.

Specifically, for any k, there is a unique sequence of k 1's and 2's that
can end a power of 2 in base 10 (see A023396).

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Charles
> Greathouse
> Sent: Friday, March 07, 2014 2:46 PM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: possible sequence
> 
> Ah, of course, sorry. In that case use 2^129 for which the last 36 digits
are
> nonzero. Or better, use 2^7879942137257 for which the last 308 digits are
> nonzero. So you'll need a cycle at least znorder(Mod(2,5^309)) > 10^215
long,
> which isn't going to fit in the universe.
> 
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
> 
> 
> On Fri, Mar 7, 2014 at 1:33 PM, <franktaw at netscape.net> wrote:
> 
> > That's what I started to say, but then I realized that it's the
> > opposite condition of what we are trying to prove.
> >
> >
> > Franklin T. Adams-Watters
> >
> > -----Original Message-----
> > From: Charles Greathouse <charles.greathouse at case.edu>
> > To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> > Sent: Fri, Mar 7, 2014 11:14 am
> > Subject: [seqfan] Re: possible sequence
> >
> >
> > I don't think it's workable, since the cycles mod 10^n must contain
> > 2^n which will have leading zeros for n > 1.
> >
> > Charles Greathouse
> > Analyst/Programmer
> > Case Western Reserve University
> >
> >
> > On Fri, Mar 7, 2014 at 10:48 AM, <franktaw at netscape.net> wrote:
> >
> >  If 2^86 is zero-free, with 26 digits, then the 26 digit cycle
> > contains
> >> zero-free numbers. So this would have to be at least the 27-digit
> >>
> > cycle.
> >
> >>
> >> Franklin T. Adams-Watters
> >>
> >>
> >> -----Original Message-----
> >> From: Allan Wechsler <acwacw at gmail.com>
> >> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> >> Sent: Fri, Mar 7, 2014 8:35 am
> >> Subject: [seqfan] Re: possible sequence
> >>
> >>
> >> I can vaguely imagine proving an upper bound on A007377, since any
> >>
> > fixed
> >
> >> number of low-order digits of the powers of two must eventually enter
> >>
> > a
> >
> >> cycle.  If one could actually step through, say, the 15-digit cycle,
> >>
> > and
> >
> >> find that every step had a zero ...
> >>
> >>
> >> On Thu, Mar 6, 2014 at 9:06 PM, Charles Greathouse <
> >> charles.greathouse at case.edu> wrote:
> >>
> >>  2^86 (26 digits) is conjectured to be the last 0-free power of two,
> >>
> >>> see A007377.
> >>>
> >>> 2^184 * 3^88 (98 digits) seems to be the last 0-free 3-smooth number.
> >>>
> >>> 0-free numbers cannot contain both 2s and 5s, and the best {3,
> >>>
> >>>  5}-smooth
> >>
> >>  number I can find is 3^28 * 5^90 (77 digits), so it looks like the
> >>>
> >>>  largest
> >>
> >>  0-free 5-smooth number is 2^184 * 3^88 again.
> >>>
> >>> I don't know what the largest 0-free 7-smooth number is, but it's at
> >>> least 2^298 * 3^69 * 7^7 (129 digits).
> >>>
> >>> One problem is that this sequence grows very quickly. Another is
> >>> that
> >>>
> >>>  no
> >>
> >>  terms are actually known...! But it is interesting to think about.
> >>>
> >>> Charles Greathouse
> >>> Analyst/Programmer
> >>> Case Western Reserve University
> >>>
> >>>
> >>> On Thu, Mar 6, 2014 at 7:57 PM, David Wilson
> >>>
> >> <davidwwilson at comcast.net
> >
> >> >wrote:
> >>>
> >>> > Largest zeroless p-smooth number for the first few primes p.
> >>> >
> >>> >
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