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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

<nospam@nospam.com> wrote:

 Quote: In article <1147869023.102499.325610@i39g2000cwa.googlegroups.com>, bert wrote: The web must offer dozens of good interactive calculator programs for multiple-precision 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
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...

&&&

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^(d-1)). 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
--
Dan in NY
(for email change t with g in
dKlinkenbert at hvc dot rr dot com)
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@news-wrt-01.rdc-nyc.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 mixed-up 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)
AGDer-2
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= (P3-P1)
B= (P4-P1)
C= (P3-P2)
D= (P4-P2)
E=A
F=D

B-C=4
A-C=2
B-D=2

{[(P3-P1) - (P3-P2)] + [(P4-P1) - (P4-P2)]}= [(P4-P1) - (P3-P2)]

"Dan in NY" <Someone@NY.net> wrote in message
news:XO1kg.9491\$ci.6499@news-wrt-01.rdc-nyc.rr.com...

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