[jasong]

First, welcome to the Forum. I'm going to try to answer your questions before someone only skims your post and decides to flame you.

Originally Posted by rong123

I am kinda new to Prime95 and all but am very interested in the hunt for new prime numbers and I have a question. I am wondering firstly, is there a simple program that can generate a Mersenne number for the bit depth specified?

[quote]I'm sure there are people here who can help you with that. If you just want to get in the general area of a digit level, than <digit level>/log(2) would work. For instance to find the first 1,000,000-digit Mersenne, 1,000,000/log(2) is about 3321929, or thereabouts.

I am kinda interested in starting factoring for the billionth mersenne prime search and I would like to know the lowest, next lowest, etc then decide which I may wanna start factoring as a start on my own little project..

Judging from the stuff you wrote before this, I believe you've misspoke. The billionth Mersenne NUMBER would be a bit easier than the billionth Mersenne prime. It would simply be a matter of identifying the billionth prime(regular prime) and plugging that as n into 2^n-1.

Also would it be like this when doing Mersenne trial factoring... I find the billionth number for instance, then take that number say its n then do a Factor=n^1, then n^2, ....n^72 in that order??...I see on the billion project where they start at like n^74 then have worked down to like n^71 or so.. wouldnt it be more logical to start at n^1 then work up to n^74, n^75, etc? when factoring?...Maybe I'm all wrong in my thinkiology here but just figuring all this stuff out..

Not totally sure what you're referring to, but that's okay.

Since Mersenne numbers are of a special form, we can make assumptions about their factors. Unfortunately, I'm not totally certain what that form is, so I'll let someone else tackle that. I will say that a TON of numbers are disqualified through number theory, making it literally millions of times faster to find a factor for a Mersenne number than if we simply tried all the primes in a given range.