[quote=science_man_88;213387]see testing each f won't help as that's like checking 2^129-1,2^131-1, etc. every gap of 2. so it's most important to find suitable f values to test.[/quote]
Do you realize your idea (besides to come up with an idea :smile:) is that you're looking for a certain number that's a common gap between Mersenne primes?
Not happenin'.

I never said it had to be common I just said i know k would be of form 4^f

if p must be of form p%2^n=2^n-1 then (2^n*x+2^n-1+1/2^n)*2^n-1 = (2^n*x+2^n/2^n)*2^n-1 = (x+1)*2^n-1 the hard part is connecting x with n.

and then I can see 2 possible solutions:

1) log2(x+1)+log2(2^n)+log2(1) -> n+log2(x+1)= 0
2) 1=log2(x+1)+log(2^n)

[QUOTE=science_man_88;213496]and then I can see 2 possible solutions:

1) log2(x+1)+log2(2^n)+log2(1) -> n+log2(x+1)= 0
2) 1=log2(x+1)+log(2^n)[/QUOTE]

I haven't followed the discussion, but since both solutions require x>=0, it follows that n gets negative in both cases. And solution (2) is even more restrictive: 0<=x<=1

Maybe I missed something crucial...

Thomas I'm probably [B][U]wrong[/U][/B] as I suck at math obviously but that's what i got actually if you go back a few steps these are disproved then because it came from k which if I'm right uses a positive x so I messed it up no surprise.

Whatever happened to [url=http://www.mersenneforum.org/showpost.php?p=186490&postcount=168]this post[/url]?

I thought this had more to do with something so I tried choosing somewhere else to put it akruppa's probably the only one not to look at it I think.

[QUOTE=science_man_88;213704]I thought this had more to do with something so I tried choosing somewhere else to put it akruppa's probably the only one not to look at it I think.[/QUOTE]

Wrong!