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 grobie 2005-09-24 16:46

Number of digits display

I was wondering how do I change the ini or what ever for my primes to display total digits. I have seen some posts that have it that way. Thanks in advance

 Pconfig 2005-09-24 19:05

You can calcutlate the number of digits yourself

log(base) * log(n) + log(k)
Round this number to above (ceil) and you have the number of digits

 TTn 2005-09-24 22:45

Do you want a windows application that does just this?
I can make a separate application accesory.

RMA.NET displays this on it's main window, along with the number of primes you have found in the prime.txt file.

 grobie 2005-09-25 05:48

where can I get this RMA.net

 TTn 2005-09-25 06:31

Right now it is available as a pre-release candidate here:

[url]http://groups.yahoo.com/group/primeform/files/RMA/[/url]

Join the primeform group in order to view, and download files from the menu, on the left.

You must have Microsoft.NET framework installed to use this software.
Let me know if you have any problems.

TTn

 grobie 2005-09-25 09:59

Hmm, tryed to d/l exe file and this pops up:

Application has generated an exception that could not be handled.

Process id=0x85c (2140), Thread id=0xae0 (2784).

Do you have a clue whats wrong?

 TTn 2005-09-25 18:33

grobie Hmm, tryed to d/l exe file and this pops up:
Application has generated an exception that could not be handled.
Process id=0x85c (2140), Thread id=0xae0 (2784).
Do you have a clue whats wrong?

You need Microsofts' .NET framework, which is available at the windows update site. I prefer it from CD though.
If you do already have the framework, it must be broken!
.NET applications always work on machines, with the proper framework installed. This application has been tested, and works.

I hope this helps!

 Dougy 2005-09-26 06:48

Just to clarify. Let $$N$$ be any natural number. It is clear that the number of digits of $$N$$, $$D(N)$$, satisfies $$D(N)= \left\lfloor log_{10}(N)+1\right\rfloor$$.

For Riesel-type numbers $$N=k \cdot 2^n-1$$ then $$D(N)=\left\lfloor log_{10}(N)+1\right\rfloor = \left\lfloor log_{10}(k \cdot 2^n-1)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ log_{10}(2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)+1\right\rfloor$$, using log laws, except when $$k=5^n$$ (i.e. $$N+1=k \cdot 2^n = 10^n$$) when we've overcalculated by one digit, and for this special case $$D(N)=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)\right\rfloor$$.

For Proth-type numbers, $$N=k \cdot 2^n+1$$ and $$D(N)= \left\lfloor log_{10}(N)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n+1)+1\right\rfloor=\left\lfloor log_{10}(k \cdot 2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ log_{10}(2^n)+1\right\rfloor=\left\lfloor log_{10}(k)+ n \cdot log_{10}(2)+1\right\rfloor$$. There are no special cases here since a special case would require $$N$$ to be written $$10000...000$$ in decimal notation, which is impossible since $$N$$ is odd.

 grobie 2005-09-26 08:40

Ok, show me with this prime how to calculate it: 290499495 61514 I know its small but I really want to learn how to figure out how many digits a given prime is.

 Cruelty 2005-09-26 10:01

[QUOTE=grobie]Ok, show me with this prime how to calculate it: 290499495 61514 I know its small but I really want to learn how to figure out how many digits a given prime is.[/QUOTE]
Paste this into Excell:
=LOG(290499495)+61514*LOG(2)+1 :smile:

 TTn 2005-09-26 10:54

Solution

Grobie,
I found that security settings will make this exact error come up.

The solution is to right click on RMA.exe, and choose "Save Target As..."
This will work for you.

Please let me know what you think.
TTn

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