Author 
Message 
Bob Delaney science forum beginner
Joined: 05 Apr 2005
Posts: 16

Posted: Tue Jun 13, 2006 12:51 am Post subject:
Re: $100,000 and 2^p 1



In article <e6k80r$pm2$1@reader2.panix.com>, Paul Ciszek
<nospam@nospam.com> wrote:
Quote:  In article <1147869023.102499.325610@i39g2000cwa.googlegroups.com>,
bert <bert.hutchings@btinternet.com> wrote:
The web must offer dozens of good interactive calculator programs
for multipleprecision integers,
Could you recommend one? For that matter, I can think of some
fun things to do with a very extended precision floating point
calculator as well.

Try my freeware RPN calculator:
http://homepage.mac.com/delaneyrm/MPCalcRB.html
Bob 

Back to top 


Dan in NY science forum beginner
Joined: 26 Mar 2005
Posts: 20

Posted: Thu Jun 15, 2006 12:26 am Post subject:
Re: $100,000 and 2^p 1



"Jens Kruse Andersen" <jens.k.a@NOSPAMget2net.dk> wrote in message
news:jTYag.69$Eo2.27@news.get2net.dk...
Quote:  Dan in NY wrote:
"Jens Kruse Andersen" <jens.k.a@NOSPAMget2net.dk> wrote in message
news:sYqag.116$c47.85@news.get2net.dk...
Patrick Hamlyn wrote:
"Dan in NY" <Someone@NY.net> wrote:
"What is the smallest prime, p, such that 2^p 1 has ten million
decimal
digits?"
I would like to know the value of p for the "smallest" Mersenne
number
of at least a million decimal digits, written as 2^p 1. I want
to
know a the prime number, p, not any of the ten million digits of the
Mersenne number.
What's wrong with just taking (log 2) of 10^10 000 000+1?
It leads to a number with 10 000 001 digits.
(log 2) of 10^9 999 999 = 33219277.6269
The next prime is 33219281.
But 2^33219281 also has 10 000 001 digits.
So there is no answer to the first question which didn't say "at
least".
The second question asks for at least a million digits.
(log 2) of 10^999 999 = 3321924.7730.
The next prime is 3321937.
So the answer is 2^33219371 which has 1 000 003 digits.
Good thing he didn't want "the ten million digits".
I would have been 8 999 997 short.
Thank you for your reply. Because of it, I see an error in the my first
post of this thread, that I missed until now. I can't be sure, but you
probably figured that out. In the quote of my post above (from the
background information for the puzzle) where I said, "at least a million
decimal digits", I meant to write, "at least ten million decimal digits."
Yes, I figured out what you meant. On Usenet, people may tease you by
taking
badly formulated questions literally.
Also, you are correct that in my puzzle, I missed writing, "at least". I
didn't think it was an error because if a number has more than ten
million
digits, it has ten million, but I should have been more clear. [That is,
if
I have 15 eggs and I am asked whether I have a dozen eggs, I would likely
say, "Yes," even if the questioner didn't say, "at least."]
That's a nonmathematical context where the questioner probably just
wants to know whether you have enough eggs for something.
If you were asked "Do you have 9 fingers?", I'm guessing you would say
"No, 10" (assuming you actually do!).
Mathematics usually require precision to avoid such speculation.
Your question was: "What is the smallest prime, p, such that 2^p 1 has
ten million decimal digits?"
I think most mathematicians would interpret this as exactly ten million.
In your reply to Patrick's comment, "What's wrong with just taking (log
2)
of 10^10 000 000+1?", I was surprised at your reply, "It leads to a
number
with 10 000 001 digits." It took me a few moments to realize what you
meant. "10^10 000 000+1" is a bit ambiguous but that hadn't occurred to
me
when I read what Patrick wrote. I figured that the one really shouldn't
be
added at all.
To clear up the ambiguity I am referring to, add parentheses after the
"^".
But there are two ways to do this. Either "10^(10 000 000)+1 or "10^(10
000
000+1). If the "+1" is included inside as part of the exponent, I
thought
it should have been added to make it "10^10 000 001" so I didn't consider
that idea. However, doing it that way does do what you said, it, "It
leads
to a number with 10 000 001 digits."
No, it would give 10 000 002 digits.
How many digits in 10^2? In 10^3? See a pattern?
When you say, "But 2^33219281 also has 10 000 001 digits," I disagree
with
you but I really do want to know whether you are correct. I think it has
exactly ten million digits and I could be wrong. In fact I posted my
puzzle
partly in an attempt to verify that result of mine. Do you know of a
smaller integer (or prime) than 2^33219281 that has at least ten million
digits? I think the integer, 2^33219280, has 9 999 999 digits.
No, it has 10 000 000.
(log 10) (2^33219280) = 33219280*(log 10) 2 ~= 9 999 999.714
So 2^33219280 ~= 10^9 999 999.714
Remember the pattern?

Jens Kruse Andersen

&&&
Greetings Jens and other readers,
Thank you Jens and others who read and considered my questions. (In the
following where I say 10^7 digits, I mean at least 10,000,000 decimal
digits.) I asked for the integer, m, in the smallest number of the form
2^m 1, having 10^7 digits. I also asked for the prime number, m, of the
form 2^m 1 having 10^7 digits.
It seems so simple, now. I see that 10^3 = 1,000 has three zeros but four
decimal digits. The pattern is that 10^n has n zeros but n +1 decimal digits
as indicated by Jens Kruse Andersen.
The smallest power of 2 having 4 decimal digits = 1 + Int(Log(base2, 10^3) =
10. 2^10 = 1024. The smallest power of 2 with d decimal digits = 1 + Int(Log
(base2, 10^(d1)). The smallest power of 2 with 10^7 digits is 1 +Int(Log(
base2, 10^(9,999,999))). The smallest power of 2 with 10,000,001 decimal
digits is 1 +Int(Log( base2, 10^10,000,000)).
I have access to a computer with Mathematica. Using it, the smallest integer
with 10^7 digits is 10^(9,999,999) = 2^(33,219,277.63 ...). The smallest
integer with 10,000,001 decimal digits is 10^(10,000,000) = 2^(33,219,280.95
....).
These numbers are consecutive primes:
prime decimal digits
33,219,253 9,999,992
33,219,281 10,000,001
33,219,283 10,000,001
33,219,313 10,000,010
The smallest power of 2 with 10^7 digits = 33,219,278
The second power of 2 with 10^7 digits = 33,219,279
The third power of 2 with 10^7 digits = 33,219,280
The smallest power of 2 with 10^(10,000,001) decimal digits = 33,219,281
My conclusion from this is that the smallest number of the form 2^m 1
having 10^7 digits is 33,219,278 and the smallest number with m prime of the
form 2^m 1 having at least 10,000,000 digits has m = 33,219,281, having
10^7 +1 decimal digits. Thus I suppose that the prize winning first prime of
ten thousand or more digits  if it is a Mersenne prime  will have an
exponent larger than 33,219,281.
I apologize if I have mixed up the number and exponent or made some other
error. If you want more explanation, please ask me for it.

Dan in NY
(for email change t with g in
dKlinkenbert at hvc dot rr dot com) 

Back to top 


Dan in NY science forum beginner
Joined: 26 Mar 2005
Posts: 20

Posted: Sat Jun 17, 2006 10:55 pm Post subject:
Re: $100,000 and 2^p 1



"Dan in NY" <Someone@NY.net> wrote in message
news:XO1kg.9491$ci.6499@newswrt01.rdcnyc.rr.com...
[previous context snipped]
Quote:  Greetings Jens and other readers,
Thank you Jens and others who read and considered my questions. (In the
following where I say 10^7 digits, I mean at least 10,000,000 decimal
digits.) I asked for the integer, m, in the smallest number of the form
2^m 1, having 10^7 digits. I also asked for the prime number, m, of the
form 2^m 1 having 10^7 digits.

[most of my post snipped]
Quote:  My conclusion from this is that the smallest number of the form 2^m 1
having 10^7 digits is 33,219,278 and the smallest number with m prime of
the form 2^m 1 having at least 10,000,000 digits has m = 33,219,281,
having 10^7 +1 decimal digits. Thus I suppose that the prize winning first
prime of ten thousand or more digits  if it is a Mersenne prime  will
have an exponent larger than 33,219,281.
I apologize if I have mixed up the number and exponent or made some other
error. If you want more explanation, please ask me for it.

Dan in NY
(for email change t with g in
dKlinkenbert at hvc dot rr dot com)
&&& 
Greetings to you,
I did mix up a number and its proper exponent. Above, in my conclusion, I
have spelled out an error. I wrote, "ten thounand or more digits" where I
meant, "ten million decimal digits." This was pointed out to me by private
email.
The late Senator Everett Dirkson once remarked, "A billion dollars here, a
billion dollars there, pretty soon, we're talking real money." For anyone
who would consider $50,000 to be real money, I could now write "a thousand
digits here and a thousand digits there and soon we're talking real money.
I had even named the file with my analysis of this method, "Ten thousand
digit prime," so I had actually made that mistake before I wrote the post.
Maybe I mixedup the number of digits and the number of dollars in my head.

Dan in NY
(for email, exchange y with g in
dKlinkenbery at hvc dot rr dot com) 

Back to top 


AGDer2 science forum beginner
Joined: 18 Jun 2006
Posts: 1

Posted: Sun Jun 18, 2006 6:57 pm Post subject:
Re: $100,000 and 2^p 1



CONSTANTS OF DOUBLE DIGIT PRIME PROGRESSION
(P1, P2),(P3,P4)

 P1  P2 

 P3  P4 

(P1, P2)= Double DigitPrime Twins
(P3, P4)= Double DigitProgressive Prime Twins
(A,B/ C,D/ E,F)
A= (P3P1)
B= (P4P1)
C= (P3P2)
D= (P4P2)
E=A
F=D
BC=4
AC=2
BD=2
{[(P3P1)  (P3P2)] + [(P4P1)  (P4P2)]}= [(P4P1)  (P3P2)]
"Dan in NY" <Someone@NY.net> wrote in message
news:XO1kg.9491$ci.6499@newswrt01.rdcnyc.rr.com...
Quote:  "Jens Kruse Andersen" <jens.k.a@NOSPAMget2net.dk> wrote in message
news:jTYag.69$Eo2.27@news.get2net.dk...
Dan in NY wrote:
"Jens Kruse Andersen" <jens.k.a@NOSPAMget2net.dk> wrote in message
news:sYqag.116$c47.85@news.get2net.dk...
Patrick Hamlyn wrote:
"Dan in NY" <Someone@NY.net> wrote:
"What is the smallest prime, p, such that 2^p 1 has ten million
decimal
digits?"
I would like to know the value of p for the "smallest" Mersenne
number
of at least a million decimal digits, written as 2^p 1. I want
to
know a the prime number, p, not any of the ten million digits of
the
Mersenne number.
What's wrong with just taking (log 2) of 10^10 000 000+1?
It leads to a number with 10 000 001 digits.
(log 2) of 10^9 999 999 = 33219277.6269
The next prime is 33219281.
But 2^33219281 also has 10 000 001 digits.
So there is no answer to the first question which didn't say "at
least".
The second question asks for at least a million digits.
(log 2) of 10^999 999 = 3321924.7730.
The next prime is 3321937.
So the answer is 2^33219371 which has 1 000 003 digits.
Good thing he didn't want "the ten million digits".
I would have been 8 999 997 short.
Thank you for your reply. Because of it, I see an error in the my
first
post of this thread, that I missed until now. I can't be sure, but you
probably figured that out. In the quote of my post above (from the
background information for the puzzle) where I said, "at least a
million
decimal digits", I meant to write, "at least ten million decimal
digits."
Yes, I figured out what you meant. On Usenet, people may tease you by
taking
badly formulated questions literally.
Also, you are correct that in my puzzle, I missed writing, "at least".
I
didn't think it was an error because if a number has more than ten
million
digits, it has ten million, but I should have been more clear. [That
is,
if
I have 15 eggs and I am asked whether I have a dozen eggs, I would
likely
say, "Yes," even if the questioner didn't say, "at least."]
That's a nonmathematical context where the questioner probably just
wants to know whether you have enough eggs for something.
If you were asked "Do you have 9 fingers?", I'm guessing you would say
"No, 10" (assuming you actually do!).
Mathematics usually require precision to avoid such speculation.
Your question was: "What is the smallest prime, p, such that 2^p 1 has
ten million decimal digits?"
I think most mathematicians would interpret this as exactly ten million.
In your reply to Patrick's comment, "What's wrong with just taking (log
2)
of 10^10 000 000+1?", I was surprised at your reply, "It leads to a
number
with 10 000 001 digits." It took me a few moments to realize what you
meant. "10^10 000 000+1" is a bit ambiguous but that hadn't occurred
to
me
when I read what Patrick wrote. I figured that the one really
shouldn't
be
added at all.
To clear up the ambiguity I am referring to, add parentheses after the
"^".
But there are two ways to do this. Either "10^(10 000 000)+1 or
"10^(10
000
000+1). If the "+1" is included inside as part of the exponent, I
thought
it should have been added to make it "10^10 000 001" so I didn't
consider
that idea. However, doing it that way does do what you said, it, "It
leads
to a number with 10 000 001 digits."
No, it would give 10 000 002 digits.
How many digits in 10^2? In 10^3? See a pattern?
When you say, "But 2^33219281 also has 10 000 001 digits," I disagree
with
you but I really do want to know whether you are correct. I think it
has
exactly ten million digits and I could be wrong. In fact I posted my
puzzle
partly in an attempt to verify that result of mine. Do you know of a
smaller integer (or prime) than 2^33219281 that has at least ten
million
digits? I think the integer, 2^33219280, has 9 999 999 digits.
No, it has 10 000 000.
(log 10) (2^33219280) = 33219280*(log 10) 2 ~= 9 999 999.714
So 2^33219280 ~= 10^9 999 999.714
Remember the pattern?

Jens Kruse Andersen
&&&
Greetings Jens and other readers,
Thank you Jens and others who read and considered my questions. (In the
following where I say 10^7 digits, I mean at least 10,000,000 decimal
digits.) I asked for the integer, m, in the smallest number of the form
2^m 1, having 10^7 digits. I also asked for the prime number, m, of the
form 2^m 1 having 10^7 digits.
It seems so simple, now. I see that 10^3 = 1,000 has three zeros but four
decimal digits. The pattern is that 10^n has n zeros but n +1 decimal
digits
as indicated by Jens Kruse Andersen.
The smallest power of 2 having 4 decimal digits = 1 + Int(Log(base2, 10^3)
=
10. 2^10 = 1024. The smallest power of 2 with d decimal digits = 1 +
Int(Log
(base2, 10^(d1)). The smallest power of 2 with 10^7 digits is 1 +Int(Log(
base2, 10^(9,999,999))). The smallest power of 2 with 10,000,001 decimal
digits is 1 +Int(Log( base2, 10^10,000,000)).
I have access to a computer with Mathematica. Using it, the smallest
integer
with 10^7 digits is 10^(9,999,999) = 2^(33,219,277.63 ...). The smallest
integer with 10,000,001 decimal digits is 10^(10,000,000) =
2^(33,219,280.95
...).
These numbers are consecutive primes:
prime decimal digits
33,219,253 9,999,992
33,219,281 10,000,001
33,219,283 10,000,001
33,219,313 10,000,010
The smallest power of 2 with 10^7 digits = 33,219,278
The second power of 2 with 10^7 digits = 33,219,279
The third power of 2 with 10^7 digits = 33,219,280
The smallest power of 2 with 10^(10,000,001) decimal digits = 33,219,281
My conclusion from this is that the smallest number of the form 2^m 1
having 10^7 digits is 33,219,278 and the smallest number with m prime of
the
form 2^m 1 having at least 10,000,000 digits has m = 33,219,281, having
10^7 +1 decimal digits. Thus I suppose that the prize winning first prime
of
ten thousand or more digits  if it is a Mersenne prime  will have an
exponent larger than 33,219,281.
I apologize if I have mixed up the number and exponent or made some other
error. If you want more explanation, please ask me for it.

Dan in NY
(for email change t with g in
dKlinkenbert at hvc dot rr dot com)



Back to top 


Google


Back to top 



The time now is Tue Apr 24, 2018 6:01 pm  All times are GMT

