2^53,619,497
Clive Tooth
group of 14 consecutive zeros in its decimal representation. This 16,141,077
digit integer contains the sequence
...631770000000000000085024...
after about 1.6 million digits. Anybody wishing to check this ... probably
has a lot of time on their hands.
--
Clive Tooth
http://www.pisquaredoversix.force9.co.uk/
End of document
Virgil
<cl...@pisquaredoversix.force9.co.uk> wrote:
I can't help wondering how and why this curious factiod came to light.
scott...@my-deja.com
"Clive Tooth" <cl...@pisquaredoversix.force9.co.uk> wrote:
Is there a general theory here that can be proved/disproved regarding
the length of consecutive strings of zeros in arbitrarily high powers
of 2?
Is there an upper limit to the length?
If not can it be shown that there is not?
Just curious.
Scott
Sent via Deja.com
http://www.deja.com/
David Moews
|Is there a general theory here that can be proved/disproved regarding
|the length of consecutive strings of zeros in arbitrarily high powers
|of 2?
|
|Is there an upper limit to the length?
|
|If not can it be shown that there is not?
log 2/log 10 is irrational, so the set {(log 2^k/log 10) mod 1} is dense in
the unit interval. It follows that given any finite string of decimal digits,
there is a power of 2 whose decimal expansion starts with it, and in
particular, there are powers of 2 containing arbitrarily long strings of
zeroes.
--
David Moews dmo...@xraysgi.ims.uconn.edu
kfost...@my-deja.com
> Is there a general theory here that can be
proved/disproved regarding
> the length of consecutive strings of zeros in
arbitrarily high powers
> of 2?
> Is there an upper limit to the length?
> If not can it be shown that there is not?
> Just curious.
Any given finite string of digits appears.
This follows easily from the fact that
log_10(2) = log(2)/log(10)
is irrational, and that for any irrational real
number x, the fractional parts {nx}, as n runs
through the positive integers, are uniformly
distributed in the interval (0, 1).
Clive Tooth
Quite so. Arbitrarily long runs of zeros may also be found near the ends of
certain powers of two. In fact, for integer n>0, 2^(4*5^(n-1)+n) has a run
of n-floor(n*log_10(2)+1) zeros starting n digits from the end.
[2^(4*5^(n-1)+n) == 2^n (mod 10^n) may be proved by induction.]
Clive Tooth
>In article <94gvev$giq$1...@mail.pl.unisys.com>, "Clive Tooth"
><cl...@pisquaredoversix.force9.co.uk> wrote:
>
>> It appears to me that 2^53,619,497 is the smallest power of two to
>> contain a
>> group of 14 consecutive zeros in its decimal representation. This
>> 16,141,077
>> digit integer contains the sequence
>>
>> ...631770000000000000085024...
>>
>> after about 1.6 million digits. Anybody wishing to check this ...
>> probably
>> has a lot of time on their hands.
>
>I can't help wondering how and why this curious factiod came to light.
How: After computing for about six months I found it a few days ago.
Why: Several decades ago, while looking for something else, I came across
the title of a paper. The title was something like "The first power of two
to have 8 consecutive zeros in its decimal form." The paper itself was not
to hand and I wondered what that power of two was. Over the years, as
computers have become more powerful, I have intermittently continued the
search. I estimate that I will start looking for the run of 15 zeros in
about 5 years time. (So... plenty of time for somebody else to find it
before me.)
See also
http://www.deja.com/getdoc.xp?AN=661083482&fmt=text
The very first program which I wrote to do this searching was _much_
cleverer than the current one. (Computer time was _much_ more expensive
then.)
If 2^n contains a run of m zeros then 2^(n-1) must contain a run of m-1
zeros. That first program would examine the runs of zeros in the current
power of two and then usually multiply the decimal form by 16 [I think]
using various idiosyncrasies of the hardware in order to be as efficient as
possible.
The sequence is A006889 in Sloane's On-Line Encyclopedia of Integer
Sequences
http://www.research.att.com/~njas/sequences/Seis.html
I will send an update.